Production Note Cornell University Library produced this volume to replace the irreparably deteriorated original. It was scanned using Xerox software and equipment at 600 dots per inch resolution and compressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell’s replacement volume on paper that meets the ANSI Standard Z39.48-1984. The production of this volume was supported in part by the Commission on Preservation and Access and the Xerox Corporation. 1992.(punteli îlttiuEcaity Hihrary Stíjara, N*w Hork FROM Joh n tì.enry "Tã.imer. MTHmnœ /F?/'ELEMENTS OF QUATERNIONS A. S. HARDY, Ph.D., PROFESSOR OF MATHEMATICS, DARTMOUTH COLLEGE. BOSTON: PUBLISHED BY GINN & COMPANY. 1887.Entered according to Act of Congress, in tlie year 1881, by A. S. HARDY, in the office of the Librarian of Congress, at Washington. J. S. Cushing & Co., Printers, Boston.PREFACE. HE object of the following treatise is to exhibit the elementary principles and notation of the Quaternion Calculus, so as to meet the wants of beginners in the class-room. The jElements and Lectures of Sir William Rowan Hamilton, while they may be said to contain the suggestion of all that will be done in the way of Quater- nion research and application, are not, for this reason, as also on account of their diffuseness of style,, suitable for the purposes of elementary instruction. Tait’s work on Quaternions is also, in its originality and conciseness, beyond the time and needs of the beginner. In addition to the above, the following works have been consulted: Calcolo dei Quaternione. Bellavitis; Modena, 1858. Exposition de la Méthode des Équipollences. Traduit de l’Italien de Giusto Bellavitis, par C.-A. Laisant ; Paris, 1874. (Original memoir in the Memoirs of the Italian Society. 1854.) Théorie Elémentaire des Quantités Complexes. J. Hoüel ; Paris, 1874. Essai sur une Manière de Représenter les Quantités Imaginaires dans les Construction Gcéométriques. Par R. Argand; Paris, 1806. Second edition, with prefaceÏY PREFACE. by J. Hoüel ; Paris, 1874. Translated, with notes, from the French, by A. S. Hardy. Van Nostrand’s Science Series, No. 52; 1881. Kurze Anleitung zum Reclinen mit den (Hamilton*scherì) Quaternionen. J. Odstrcil; Halle, 1879. Applications Mécaniques du Calcul des Quaternions. Laisant; Paris, 1877. Introduction to Quaternions. Kelland and Tait; Lon- don, 1878. A free use has been made of the examples and exercises of the last work ; and, in Article 87, is given, by permis- sion, the substance of a paper from Volume I., page 379, American Journal of Mathematics, illustrating admirably the simplicity and brevity of the Quaternion method. If this presentation of the principles shall afford the undergraduate student a glimpse of this elegant and pow- erful instrument of analytical research, or lead him to follow their more extended application in the works above cited, the aim of this treatise will have been accomplished. The author expresses his obligation to Mr. T. W. D. Worthen for valuable assistance in the preparation of this work, and to Mr. J. S. Cushing for whatever of typographical excellence it possesses. A. S. HARDY. Hanover, N.H., June 21, 1881.CONTENTS CHAPTER I. Addition and Subtraction of Vectors ; or* Geometric Addition and Subtraction. Article. Page. 1. Definition of a vector. Effect of the minus sign before a vector .................................................. 1 2. Equal vectors............................................... 2 3. Unequal vectors. Vector addition............................ 2 4. Vector addition, commutative................................ 3 5. Vector addition, associative ............................... 3 6. Transposition of terms in a vector equation............. 4 7. Definition of a tensor...................................... 4 8. Definition of a scalar...................................... 5 9. Distributive law in the multiplication of vector by scalar quantities............................................ . 6 10. If 2a+ 2/3 = 0, then 2. 84. The parabola..............................................178 85. Tangent to the parabola.................................. 178 86. Examples on the parabola..................................180 87. Relations between three intersecting tangents to the para- bola ....................................................185 88. The ellipse...............................................191 89. Examples on the ellipse...................................192 90. The hyperbola.............................................195 91. Examples on the hyperbola .... -..........................195 92. Linear equation in quaternions. Conjugate and self-conju- gate functions............................................199 93. General equations of the conic sections....................201 94. The ellipse..............................................204 95. Examples...................................................206 96. The parabola.............................................214 97. Examples.................................................216 98. The cycloid................................................222 99. Elementary applications to mechanics.......................223 100. Miscellaneous Examples..................................231ELEMENTS OF QUATERNIONS.QUATERNIONS CHAPTER I. Addition and Subtraction of Vectors, or Geometric Addition and Subtraction* 1. A Vector is the representative of transference through a given distance in a given direction. Thus, if a, b are any two points, vector ab implies a trans- lation from a to B. A vector may be represented geometrically by a right line, whose length denotes the distance over which transference takes place, and whose direction denotes the direction of the trans- ference. In thus designating a vector, the direction is indicated by the order of the letters. Thus, ab (Fig. 1) denotes transference A Flg‘lm___________B from a to b, and ba from b to a. Retaining the algebraic signification of the signs + and —, if ab denotes motion from a to b, then —ab will denote motion from b to a, and AB = —BA, —AB = BA . . . . (1). Hence, the effect of a minus sign before a vector is to reverse its direction. The conception of a vector, therefore, implies that of its two elements, distance and direction; it was first defined as a directed right line. It is now applied more general^ to all quantities determined by magnitude and direction. Thus, force, the path2 QTTATEEOTONS. of a moving body, velocit}^, an electric current, etc., are vector quantities. Analytically, vectors are represented by the letters of the Greek alphabet, a, ¡3, y, etc. 2. It follows, from the definition of a vector, that all lines which are equal and parallel may he represented by the same vec- tor symbol with like or unlike signs. If equal and drawn in the same direction, they will have the same sign. Hence an equality between two vectors implies equality in dis- tance with the same direction. Thus, if AB (Fig. 2), CD, BE, EF and hg are equal and drawn in the same direction, they may be represented by the same vector symbol, and AB = CD = BE = EF = HG = a . . . . (2) . Fig. 2. c D H G Fig. 3. 3. It follows also from the definition of a vector that, if vec- tors are not parallel, they cannot be represented by the same vector symbol. Thus, if the point a (Fig. 3) move over the right line ab, from a to b, and then over the right line bc, from b to c, and ab = a, bc must be denoted by some other symbol, as ¡3. The result of these two succes- sive translations of the . point a is the same as that of the single and direct translation AC=y, from a to c ; in either case a is found at the extremity of the diagonal of the parallelogram of which ab and bc are the sides. This combina- tion of successive translations is called addition, and is written in the ordinary way, a + /? = y.........................(3). This expression would be absurd if the symbols denoted mag- nitudes only. It means that transference from a to b, followedGEOMETRIC ADDITION AND SUBTRACTION. 3 by transference from b to c, is equivalent to transference from A to c. The sign + does not therefore denote a numerical ad- dition, or the sign = an equality between magnitudes. It is, however, called an equation, and read, as usual, “a plus ¡3 is equal to y” This kind of addition is called geometric addition. 4. If the point a (Fig. 3), instead of moving over the sides ab, BC of the parallelogram abcd, had moved in succession over the other two sides, ad and dc, the result would still have been the same as that of the single translation over the diagonal ac. But since ab and bc are equal in length to dc and ad respect- ively, and are drawn in the same direction, we have (Art. 2) ab = dc and bc = ad, and if the first two translations are represented by ab and bc, the second two may be represented by bc and ab, or a + /? = /3 + a = y...............(4). Hence the operation of vector addition is commutative, or the sum of any number of given vectors is independent of their order. 5. If the point a (Fig. 4) move in succession over the three edges ab, bc, co of a parallelopiped, we have Eig. 4. and or AB + BC = AC, AC + CG = AG, (AB -f- BC) + CG == AG. or In like manner BC + CG = BG, AB + BG = AG, AB + (BC + CG) = AG. H G Hence (ab + bc) + CG = ab + (bc + cg) • (5), and the operation of vector addition is associative, or the sum of any number of given vectors is independent of the mode of grouping them.4 QTTATEKNIONS. 6. Since, if ac = y (Fig. 3), then ca = —y, we have a + /3 —y= 0, or, comparing with equation (3), « + ft = y, a term may be transposed from one member to another in a vector equation by changing its sign. Also, in every triangle, any side may be considered as the sum or difference of the other two, depending upon their direc- tions as vectors. Thus (Fig. 3) y —¡3 = a, y-a = p. It is to be observed that no one direction is assumed as posi- tive, as in Cartesian Geometry. The only assumption is that opposite directions shall have opposite signs. The results must, of course, be interpreted in accordance with the primitive as- sumptions. Thus, had we assumed ba = a (Fig. 3), y and ft being as before, then _ ft — a = y, a — P = — y- 7. If two vectors having the same direction be added together, the sum will be a vector in the same direction. If the vectors be also equal in length, the length of the vector sum will be twice the length of either. If n vectors, of equal length and drawn in the same direction, be added together, the sum will be the product of one of these vectors by n, or a vector having the same direction and whose length is n times the common length. If then (Fig. 2) V AF = XAB = XCD = Xa, where a, b and f are in the same straight line, cd = ab, and x is a positive whole number, x expresses the ratio of the lengths of af and a. From the case in which x is an integer we pass, > by the usual reasoning, to that in which it is fractional or in- ƒ commensurable. Vectors, then, in the same direction, have the same ratio as the corresponding lengths.I GEOMETRIC ADDITION AND SUBTRACTION. 5 If ab = a be assumed as the unit vector, then af = ma, in which m is a positive numerical quantity and is called the Tensor. It is the ratio of the length of the vector ma to that of the unit vector a, or the numerical factor by which the unit vector is multiplied to produce the given vector. Any vector, as /?, may be written in general notation p = TpTJp. In this notation, Tp (read ‘4 tensor of P”) is the numerical factor which stretches the unit vector so that it shall have the proper length ; hence its name, tensor. It is, strictly speaking, an abstract number without sign, but, to distinguish between it and the negative of algebra, it may be said to be always posi- tive. Tjp (read “ versor of P”) is the unit vector having the direction of ¡3 ; the reason for the name versor will appear later. T and U are also general symbols of operation. Written be- fore an expression, they denote the operations of taking the tensor and versor, respectively. Thus, if the length of p is n times that of the unit vector, T (0) = n, where T denotes the operation of taking the stretching factor, i.e. the tensor. While W=Uj8 indicates the operation of taking the unit vector, that is, of reducing a vector ¡3 to its unit of length without changing its direction. 8. If ne (Fig. 5) be any vector, and ba = yBC, then ba = ab = 2/BC ; Fig5> and, in general, if ba and bc be b__________c_______________a any two real vectors, parallel and of unequal length, we may always conceive of a coefficient y which shall satisfy the equation ba = 2/BC,6 QUATERNIONS. where y is plus or minus, according as the vectors have the same or opposite directions, y may be called the geometric quotient, and is a real number, plus or minus, expressing numerically the ratio of the vector lengths. This quotient of parallel vectors, which may be positive or negative, whole, fractional or incom- mensurable, but which is always real, is called a Scalar, because it may be always found by the actual comparison of the parallel vectors with a parallel right line as a scale. It is to be observed that tensors are pure numbers, or signless numbers, operating only metrically on the lengths of the vectors of which they are coefficients : while scalars are sign-bearing numbers, or the reals of Algebra, and are combined with each other by the ordinary rules of Algebra ; they may be regarded as the product of tensors and the signs of direction. Thus, let a— aUa. Then Ta = a. If we increase the length of a by the factor b, & is a tensor, but the tensor of the resulting vector is ba. If we operate with — b, — b is not a tensor, for a is not only stretched but also reversed ; the tensor of the resulting vector is as before ba ; in other words, direction does not enter into the conception of a tensor. As the product of a sign and a tensor, — b is a scalar. The operation of taking the scalar terms of an expres- sion is indicated by the symbol S. Thus, if c be any real alge- braic quantity, S(— &aUa + c) = c, for — ba Ua is a vector, and the only scalar term in the expres- sion is c. 9. It is evident from Art. 7 that if a, b, c are scalar coeffi- cients, and a any vector, we have (a -f* b -f- c) a = aa -f- ba -f- ca . . . . (6) • Furthermore, if (Fig. 6) OA = a, AB = p, BC = y, OAf = ma,GEOMETRIC ADDITION AND SUBTRACTION. 7 then, a'b' being drawn parallel to ab and b'c' to bc, a'b' = m/3, b'c' = my. Now oc = a + /? + y, and oc' = moc = m (a + /? -f* y). But we have also Hence Fig. 6. OC' = Oa' + a'b' + B'c' = ma + m/3 + my. m (a -j- /? -f- y) == ma -f- m/3 -f- my . . . (7), or the distributive law holds good for the multiplication of scalar and vector quantities. 10. It is clear that while a — a = 0, a± ¡3 cannot be zero, since no amount of transference in a direc- tion not parallel to a can aifect a. Hence, if na + m/3 = 0, since a and ¡3 are entirely independent of each other, we must have na = 0 and m/3 = 0, or n = 0 and m = 0. Or, if ma + n¡3 = m'a + ti'/?, then m = m' and n = And, in general, if 3a -f- 3/? = 0, 3a = 0 and 3/3 = 0 then8 QTTATEBNIONS. Three or more vectors may, however, neutralize each other. Thus (Fig. 7) Fig. 7. a + /3 + y + S = 0, e — ¡3 — a = 0, and this whether abcd be plane or { B gauche. In any closed figure, there- fore, we have a + /? + y + S +.... = 0, where a, /?, y, 8, .., are the vector sides in order. 11. Examples. 1. The right lines joining the extremities of equal and parallel right lines are equal and parallel. Let OA and bd (Fig. 8) be the given lines, and oa = a, bo = /?, da = y. Then, by condition, bd = a. Now, BA = BO + OA = P -{- a ; also, BA = BD + DA = a + y ; or, equating the values of ba, = y. Hence (Art. 2), y = /?, and bo is parallel and equal to da. Fig. 8. 2. The diagonals of a parallelogram bisect each other. In Fig. 8 we have bd = oa = op + pa ; so BD = BP -+- PD ; „ .*. OP + PA = BP + PD. But, op and pd being in the same right line, op = mPD. Similarly PA = WBP.GEOMETRIC ADDITION AND SUBTRACTION. 9 Hence and mPD + wbp = PD + bp, m == 1, n = 1, OP = PD, bp = PA. 3. If tivo triangles, having an angle in each equal and the including sides proportional, be joined at one angle so as to have their homologous sides parallel, the remaining sides will be in a straight line. Let (Fig. 9) ab = a, AE = f3. Th by condition, dc = æa, db = x¡3. Now cb = CD + db = X (/? — a). But BE — (3 — Hence (Art. 2), b being a common point, cb and be are one and the same right line. 4. If two right lines join the alternate extremities of tico parallels, the line joining their centers is half the difference of the parallels. We have (Fig. 10) AB = AD + DC + CB, and, also, AB = AE + EF + FB. Adding 2 AB = (ad + ae) + (dc + ef) + (cb + fb) = EF — CD ; Fig. 10. Fig. 9. C B JS or, as lines, ab = i (ef — cd) .10 QUATERNIONS. 5. The mediais of a triangle meet in a point and trisect each other. Fig. 11. Let (Fig. 11) bo = a, cd = /?. Then oc = a, da = /3. Now BA=2a+2/?=2 (a + jff), and, since od = (a + /3), ba and od are parallel. Again BP + PA = BA = 2 OD = 2 (op + Pd) . But bp and pd, as also op and pa, lie in the same direction, and therefore bp = 2 pd and pa — 2 op. Hence the mediais oa and db trisect each other. Draw cp and pe. Then bp = 2pd = |bd = | (2a + /2), and CP = CB + BP = f (2a-b/?) — 2a = § (/? — a), PE = PB + BE==a + yG — f (2a + 0) (jS — a). Hence pe and cp are in the same straight line, or the mediais meet in a point. 6. In any quadrilateral, plane or gauche, the bisectors of opposite sides bisect each other. We will first find a value for op (Fig. 12) under the supposi- tion that p is the middle point of ge. We shall then find a value for op, under the supposition that p is the middle point of fh. If these expressions prove to be identical, these middle points must coincide. In this, as in many other problems, the solution depends upon reaching the same point by different routes and comparing the results. Fig. 12.GEOMETRIC ADDITION AND SUBTRACTION. 11 Let OA = a, OB = /?, OC = y. 1st. oc + cg = oe + eg. (a) But CG = iCB = i(j3-y), which, in (a), gives y + i(P — j) = ia + BG- ep = |eg = |(y + j8 — a), OP = OE + EP = $a + \ (y + ¡3 — a) = i (a + P + y) • (P) 2d. PH — £ab = FO + OA, or FH — i (fi — a) = — by + a. FP = ^FH = i(a + ^ — y), OP = OF + FP = £y + i(a + /3 — y) = i(a + fi + y)i which is identical with (&). Hence, the middle points of fh and ge coincide. 7. If ABCD (Fig. 13) be any parallelogram, and op any line parallel to dc, and the indicated lines be drawn, then will mn be parallel to ad. Let AM = a, BM = /?. Then AO = ma, AD = na +J?/?5 OD = — ma + na +p/3. Fig. 13. We have in which NM = NO + OM = NP + FM, NO = X ( — ma + na +pp) 9 OM= (1 — m) a, NP = æ ( — m/3 + na + jpjS) ? pm = (1 — m) ¡3.12 QUATERNIONS. Substituting in the above equation, we obtain, by Art. 10, Fig. 13. 1 — m x =--------• c m Substituting this value in NM = no + OM, NM = --— ( — ma + na +pP) + (1 — m)a m 1 — m, . 1 —• m (na +p/3) =---AD. m ' m Hence ad and nm are parallel. 8. If through any point in a parallelogram, Zmes drawn parallel to the sides, the diagonals of the two non-adjacent parallelograms so formed will intersect on the diagonal of the Let (Fig. 14) OA = a, os = ¡3.. Then or = ma, oe = n/3. We have RD = RO + OE-hED = 7l^+(l—m) a, ES = EO + OR+RS =Wla+ (1—n) (3. Also FO = FR 4- RO = XRD + RO = X [n/3 + (1 — m) a] — ma, (a) and FO = fe + EO = 2/Es + EO = y [ma -f (1 — n) /?] — n/3. (b) From (a) and (b) nx = y (1 — n) — n and x (1 — m) — m = ym. Eliminating y « =----------. 1 — m — n original parallelogram. Fig. 14. FGEOMETRIC ADDITION AND SUBTRACTION. 13 Substituting this value of x in (a) FO = -——------[nfi + (1 — m) a] — ma 1— m — n 1 mn ■m — n (/3 + a), or, FO and oc = (fi + a) are in the same straight line. 9. If, in any triangle oab (Fig. 15), a line od be drawn to the middle point of ab, and be produced to any point, as f, and the sides of the triangle be produced to meet af and bf in h and R, then will hr be parallel to ab. Let OA = a, OB = fi. Then or = xa, oh — y fi, ab = fi — a. Now OD = OA + AB = \ (a + fi) . Fi g. 15. H Also, of = æ (a-{-fi), that is, some multiple of od. Then, 1st. BR=£>BF, — fi+ xa=p (—/3 + Of) =p [— p + z (“ + £)] ; x ~pz and — l = pz — p. (a) Eliminating z And, 2d. J) = CC +1. ah = gAF, a + yfi = q (—a + OF) = q [-— a + % (a + fi)] î . y = qz and — 1 = qz — q. (») Eliminating z From (a) and (b) q = y + l. z p q14 QUATERNIONS. and, since p = x + 1 and q = y + 1, Fig. 15. x —y and p~q* B —RH = no + oh = 2//? — xa = x(fi — a) = SCAB, F . or, RH and ab are parallel. 10. If any line pr (Fig. 16) be drawn, cutting the two sides of any triangle abc, and be produced to meet the third side in Q, then Fig. 16. PC . BQ • RA = CR . AQ . BP. Let bp = a, cr = ft. Then pc — i>a, ra = rjB and ba = bc + ca = (1 +p) a + (1 + 0 /?• We have AQ = ojba = x [(1 +p) a + (l +r) ÆL as also AQ == AR + RQ = — r/3 + y\PR = — r/3 + y {pa + ft) • x (1 +_p) = yp and x (1 + r) = — r + y. Eliminating y whence x = (1 + x) pr ; AQ__BQ PC RA BA~BA BP CR’ or PC . BQ . RA = CR . AQ . BP. 11. If triangles are equiangular, the sides about the equal angles are proportional. Let (Fig. 17) bc = a, ca = ¡B. Then be = ma, ed = n/3, BD = ma + nfi and ba = a + /?. Now BD = ^)BA, ma +nfi — p {a + p). m = p, n = p and m = n. be : bc : : ed : ca. WhenceGEOMETRIC ADDITION AND SUBTRACTION. 15 12. If through any point o (Fig. 17), within a triangle abc, lines he drawn parallel to the sides, then will ED GF , hi 0 -----------— CA CB AB Let ca = fi, CB = a. Then ab = a — ft ed = mft hi = p (a — fi) and GF = na. We have Now, as lines, co = CG + GO == CH + HO. (a) GF _GA___ CB CA ’ EB _ ED __ CB CA ’ DB__DE___ AB AC ’ CG = CA — GA = (1 — n) ft GO = CE = CB — EB = (1 — Tïl) a. HO = AD = AB — DB == (1 — m) (a — jÖ). Substituting in (a) (1 — n) p + (1 — m) a = pp + (1 — m) (a — fi), or (Art. 10) n + m+ p = 2. 12. Complanar vectors are those which lie in, or parallel to, the same plane. If a, fi, y are any vectors in space, they are complanar when equal vectors, drawn from a common origin, lie in the same plane. If a, fi, y are complanar, but not parallel, a triangle can al- ways be constructed, having its sides parallel to and some mul- tiple of a, fi, y, as aa, bfi, cy. If we go round the sides of the triangle in order, we have aa + bfi + Cy = 0. If a, ft y are not complanar, conceive a plane parallel to two of them, as a and fi. In this plane two lines may be drawn parallel to and some multiple of a and fi, as aa and bfi ; and these two vectors may be represented by pS (Art. 3).16 QUATERNIONS. Now pS, being in the same plane with aa and b/3, cannot therefore be equal to y, or to any multiple of it ; pS and y can- not therefore (Art. 10) neutralize each other. Hence p8 + cy = eta + bj3 + cy cannot be zero. If then, we have the relation aa + 5/3 + cy = 0 between non-parallel vectors, they are complanar; or, if a, ¡3, y be not complanar, and the above relation be true, then, also, a = 0, b = 0, c = 0. 13. Co-initial vectors are those which denote transference from the same point, (a). If three co-initial vectors are complanar, and give the relations, , . , ' (a) aa -f- b/3 -{- cy = 0 ^ ^ (b) a+6+c=0 ƒ........................ they will terminate in a straight line. For, let OA = a (Fig. 15), ob = ¡3, od = y. Then da = a — y, BA = a — ¡3. From Equation (9), (6) (a + b + c) a = 0, from which, subtracting (a) of Equation (9), b (a — ft) + C (a — y) = 0, 5ba + CDA = 0 ; and, since these two vectors neutralize each other, and have a common point, they are on the same straight line. Hence, a, d and b are in the same straight line. (b). Conversely, if a, /?, y are co-initial, complanar and ter- minate in the same straight line, and a, ò, c have such values as to render then will For aa + b/3 + cy = 0, a + 5 + c = 0. da = a — y and BA = a — /?.GEOMETRIC ADDITION AND SUBTRACTION. 17 But, by condition, a — ¡3 = X (a — y), or (1 — x) a — ¡3 + Xy = 0, in which (1 — x) — 1 + x = 0. 14. Examples. 1. The extremities of the adjacent sides of a parallelogram and the middle point of the diagonal between them lie in the same straiqht line. V Fig. 18. Let OA = a, OB = ¡3, oc = y. A D Then , OD = OB + BD, 2y — /3 — a = 0. o¿ But, also, 2 — 1 — 1 = 0, hence, b, c and a are in the same straight line (Art. 13). 2. If two triangles, abc and smn (Fig. 19), are so situated that lines joining corresponding angles meet in a point, as o, then the pairs of corresponding sides produced will meet in three points, p, Q, R, which lie in the same straight line. Let OA = a, OB = /?, OC = y. Then os = ma, om = n[3, ON = J)y, BA = a — /?, ms = ma — nf3, br = x (a — /3) and MR = y (ma — n¡3). 1st. BM = BR — MR, or n(3 ~ ¡3 ~ x (a — /5) — y (ma — n¡3), n — 1 = — x + 2/n, x — my — 0. ÍC == — m (n — 1) m — n Eliminating y18 QUATERNIONS. Also OR = OB + BR = P -J- X (a — ¡3) = ¡3 — m (n — 1) m — n (a - £)> whence or = n (m — 1) /3 — m (n — 1) a m — n (a) 2d. or CN = CP — NP, Py — y=v (P — y) — w (nfi — py). , p — 1 = — V + wp, V — wn = 0. Eliminating w Also n {p — 1) 71 — p /ft ( W) _ OP = OC + CP = y + 'l> (/3-y) = y--------m_n - (/? whence op = 71 — p p (n — 1) y — n (p — 1) P n — p y)> (6) 3d. In the same manner, we obtain 00 == m(p-l)a-p (m — 1) y ^ p — m From (a), (6) and (c) we observe that, clearing of fractions, and multiplying (a) by p — 1, (b) by m — 1, (c) by n — 1, and adding the three resulting equations, member by member, the collected coefficients of a, /?, y, in the second member of the final equation, are separately equal to zero. Hence the first member or (m—n) (p — 1) + op (ti—p) (m — 1) + oq (p—m) (n— 1) = 0. But (m — n) (p — 1) + (n — p) (m — 1) + (p — m) {n — 1) = 0. Hence, r, p and q are in the same straight line.GEOMETRIC ADDITION AND SUBTRACTION. 19 3. Given the relation aa + bp + Gy = 0. Then a, p, y are complanar ; but, if co-initial (as they may be made to be, since a vector is not changed by motion parallel to itself, i.e. by translation without rotation), and a + 'Plg* 20* b + c is not zero, they do not terminate in a straight line. Hence, if o is the ori- gin, and a, B, c, their ter- minal points, a, b and c are not collinear. Let these points be joined, forming the triangle abc (Fig. 20), and OA, OB, oc prolonged to meet the sides in a' b' c! To find the relation between the segments of the sides, let whence OA'=a'=#a, OB '=/3f=y/?, OC' = y' = Zy, y Substituting these in succession in the given relation, -a' + bp + Cy = 0, x aa + ~/^+ Gy = 0j aa -f- bp + -*/= 0, z whence, since a' c, b are to be collinear, ^-f-ö + c = 0, X20 QUATERNIONS, and, for a like reason, Whence X-. and b + c b + c a H------f— c = 0, y a + b + - = 0. z y = - a, £'= a + c & a + c ci + b a -f- b or, from the given relation, + cr /?/_ cr + ö + c’ P c+a’ (Xa + a + & Whence ò(a'-/?)=c(y-a'), c (ft— y) = a (a — ft), a (y'— a) = Ô Q3 — y'), and ba' __ c cb' __ a ac' b a'c ~ V b'a ~~ c’ c'b ~ a or, multiplying, ba' . cb' . ac' . = a'c . b'a . c'b. 4. If o (Fig. 20) be any point, awe? abc any triangle, the transversals through o cmc? the vertices divide the sides into seg- ments having the relation ba' • cb' . ac' . = a'c • b'a . c'b. Let a'c = a, Be = aa, cb'= /3, ca = bp. Then ba = aa + bp. \ Also let BO = ÆBbJ OA =s 2/a'a, BC'= mBA, CC' = ZQO.GEOMETRIC ADDITION AND SUBTRACTION. 21 Then bo = æbb' = X (bc + cb') = X (aa + ¡3), OA = y A1 A = y (a'c + ca) = y (a + b/3), bc'= mBA = m (aa + b/3), cc' = 2CO = z (cb + bo) = z [— aa -f- X (aa + P) ] • From the triangle boa we have bo + oa + ab = 0, X (aa + /3) + y(a + b/3) — b¡3 — aa = 0. xa + y — a = 0, x + yb — b = 0. Eliminating y „ _ & (1 — a) 1 -ba' From the triangle bcc' bc + cc'+ c'b = 0, aa -h z [— aa + X (aa + /?)] — m (aa + b/3) — 0, whence, as usual, and substituting the above value of æ, or 1 — m 1 — b m 1 —a Substituting for m, 6 and a, c'a__ ab' ca' bc' b'c # a'b’ Fig. 20. P which is the required relation. 5. If (Fig. 20) lines be drawn through a\ b' cj and produced to meet the opposite B sides of the triangle in p, q, r, then are p, q and r col- linear. 'R22 QUATERNIONS. With the notation of the last example, a — 1 BC = mB a = ■ a -f- b — 2 1st. From the triangle c'ba' (aa + b/3)• c'a'= c'b + ba' _ _ a —1 a + ò — 2 a -—1 (aa+ &/?) +(a — 1)<* Also a + & —2 [(&_ 2) «_&£]. a'r = æc'a'= a'c + CR = a'c — y/?, a —1 a + b — 2 [ (& — 2) a — bff\ = a — y Pi and br = BC + cr = aa-B. b-2h 2d. From the triangle c'ab' c'b'= c'a + ab' = (1—m) (aa + b^) + (l—b)¡3 b~1 :[«a-(a-2)/?]. Also a -f- b — 2 a + 6~2 )'b'= b'( [aa — (a — 2) yS] = — P + b'q = æc'b'= b'c + cq = b'c + ya> &-1 and 3d. ••• 2/ =--- a — 2 bq = BC + CQ = (a + 2/) a = -^a'~1-a- a — ¿ a'p = #a'b'= æ (a + ¡3), a'p = a'b + bp = (1 — a) a + y (&<* + &/0 5 a — 1GEOMETRIC ADDITION AND SUBTRACTION. 23 and bp = 2/BA = -—- (aa + b/3). cl — b («) Multiplying the second members of (a), (&), (c), by (a—1) (b — 2), — (a — 2) (& — 1), (a — b) respectively, their sum is zero. Hence (a — l)(b — 2) BR — (a — 2) (b — 1) bq + (a — b) bp = 0. But (a — 1) (& — 2) — (a — 2) (b —1) + (a — &) = 0. Hence r, q and p are collinear. 6. If pc (Fig. 20) and po be produced to meet aa' and bc, then T and s are collinear with c! A similar proposition would obtain for q and r. With the following notation, ba = a, ba'— /3, bb'= aa h/3. we have BO = BA + Ab'+ b'o = Ba'+ a'o, a + bp — (1 — a) a + x(aa + b/3) = p + y (a — /?). also Fig. 20. ,P BP = ba'+ a'p = ba + ap ; fi-\-z[aa-\- (b — 1)j8J =a-f~Wa< a — 1 -f- b .*. w =----------—, B, and bc = ba'+ a'c = ba + AC, R P + v/3 = a + u[(l-a)qL — 6/3], %24 QUATERNIONS. .. „_2±»=i, 1— a BC = bp l — a Now to find BS, bc' and bt, we have Fig. 20. 1st. BS = æ'ba' = BP + y' PO, x’= — BS = — - 1-2b-a bf3 1 -2b-a 2d. BC' = v’bA = BC + MCO, a v'= 2a-j-b — l’ BC — ■ 2a-\-b — 1 3d. BT = ba'+ a't = ba'+ z* a'o = bp + wf PC, */__ a + & a — — Hy+P), Fig. 24. c and that to the mean point of the tetraedron formed by joining them is i a + Æ + y 3 a + /3 a + y 3 ^ 3 (a + /3 + y), which is the vector to the mean point of oabc. The same is true of the tetraedron formed by joining the mean points of the edges ab, bc and ca with o, since a ra + ft i P+y i a + r 4 2 ^ 2 ^ 2 = ì (a-|-/? + y).GEOMETRIC ADDITION AND SUBTRACTION. 29 The above is, of course, independent of the origin, and would be true were o not taken at one of the vertices. 2. The inter section of the bisectors of the sides of a quadri- lateral is the mean point. Let (Fig. 25) OA = a, ob = fi, oc = y, od = 8, or = p. Then (Art. 15) p = i (op + oe) = i lì (a + + \ (y + fi)'] = ^(a + ^ + y + 8). If o is at a, then oa = a = 0, and p = 1 (P + y +g) • 3. If the sides (in order) of a quadrilateral be divided propor- tionately, and a new quadrilateral formed by joining the points of division, then will both quadrilaterals have the same mean point. Let a, fi, y, 8 be the vectors to the vertices of the given quadrilateral, from any initial point o. Then, for the vector to the mean point, we have i (a + fi + y + S). If m be the given ratio, and aj fi[ y' S' the vectors to the ver- tices of the second quadrilateral, then a'= a + m (fi — a) = (1— m) a + mfi, fi'=(l — m) fi + my, y' = (1 — m) y + mS, S' = a + (1 — m) (8 — a) = 8 — m (8 — a) ; whence h (/?'+ y'+ S'+ O = i (a + fi + y + S). Fife. 25. D30 QUATERNIONS. 4. In any quadrilateral, plane or gauche, the middle point of the bisector of the diagonals is the mean point. Let (Fig. 26) OA = a, ob == /?, oc = y, os = iy. Then (Art. 15) op = I (oq + os) = HK« + 0)+iy] = i(“ + ^ + y)- 5. If the two opposite sides of a quadrilateral be divided pro- portionately, and the points of division joined, the mean points of the three quadrilaterals will lie in the same straight line. Let cj a' (Fig. 27) be the points of division, and m the given ratio. Then, if oa = a, bc = y, oa'= ma, c'c = my, ab = ¡3 and o is the in- itial point, the vectors to the mean points p, p,' p" are op =:±(8a + 2j8 + y), op' = J[(m + 2)a + 2/? + (2 —m)y], op" = i[(m + 3)a + 2 /3+(l—m)y] ; ... pp' = __(r_a), p"p= _(y_a). Therefore, p¡ p{' p are in the same straight line. 20. Exercises. 1. The diagonals of a parallelopiped bisect each other. 2. In Fig. 58, show that bo and ch are parallel. 3. If the adjacent sides of a quadrilateral be divided propos tionately, the line joining the points of division is parallel to the diagonal joining their extremities. Fig. 27. CC' BGEOMETRIC ADDITION* AND SUBTRACTION. 31 4. The medial to the base of an isosceles triangle is an angle- bisector. 5. In any right-angled triangle abc (Fig. 58), the lines bk, Cf, al meet in a point. 6. Any angle-bisector of a triangle divides the opposite side into segments proportional to the other two sides. 7. The line joining the middle point of the side of any paral- lelogram with one of its opposite angles, and the diagonal which it intersects, trisect each other. 8. If the middle points of the sides of any quadrilateral be joined in succession, the resulting figure will be a parallelogram with the same mean point. 9. The intersections of the bisectors of the exterior angles of any triangle with the opposite sides are in the same straight line. 10. If ab be the common base of two triangles whose vertices are c and d, and lines be drawn from any point e of the base parallel to ad and ac intersecting bd and bc in f and a, then is FG parallel to dc.CHAPTER II. Multiplication and Division of Vectors, or Geometric Multipli- cation and Division. Fig. 28. 21. Elements of a Quaternion. The quotient of two vectors is called a Quaternion. We are now to see what is meant by the quotient of two vectors, and what are its elements. Let a and ft (Fig. 28) be two vec- tors drawn from o and o' respectively and not lying in the same plane ; and let their quotient be designated in the usual way by Whatever their relative positions, we of' ^ B' may always conceive that one of these vectors, as ¡3' may be moved parallel to itself so that the point o' shall move over the line o'o to o. The vectors will then lie in the same plane. Since neither the length or direction of ¡3’ has been changed during this parallel motion, we have j3 = f3[ and the quotient of any two vectors, a, ¡3', will be the same as that of two equal co-initial vectors, as a and B. We are then to determine the ratio —, in which a and B ft lie in the same plane and have a common origin o. Whatever the nature of this quotient, we are to regard it as some factor which operating on the divisor produces the dividend, i.e. causes ¡3 to coincide with a in direction and length, so that if this quotient be, g, we shall have, by definition, 32 q(3 = a when - = q , (14).GEOMETRIC MULTIPLICATION AND DIVISION. 33 If at the point o' we suppose a vector o'c = y to be drawn, not parallel to the plane aob, and that this vector be moved as before, so that o' falls at o, the plane which, after this motion, y will determine with a, will differ from the plane of a and (3, so that if the quotient q and q' will differ because their planes differ. Hence we con- clude that the quotients q and qf cannot be the same if a, [3 and y are not parallel to one plane, and therefore that the position of the plane of a and ¡3 must enter into our conception of the quotient q. Again, if y be a vector o'c, parallel to the plane aob, but differing as a vector from ¡3', then when moved, as before, into the plane aob, it will make with a an angle other than boa. Hence the angle between a and ¡3 must also enter into our con- ception of q. This is not only true as regards the magnitude of the angle, but also its direction. If, for example, y have such a direction that, when moved into the plane aob, it lies on the other side of a, so that aoc on the left of a is equal to aob, then the quotient q[ of -, in operating on y to produce a must turn y 7 in a direction opposite to that in which q = ~ turns ¡3 to produce a. Therefore q and q' will differ unless the angles between the vector dividend and divisor are in each the same, both as regards magnitude and direction of rotation. Of the two angles through which one vector may be turned so as to coincide with the other is meant the lesser, and it will therefore, generally, be < 180? Finally, if the lengths of ¡3 and y differ, then ^ = q will still differ from - = ql Therefore the ratio of the lengths of the vec- r tors must also enter into the conception of q. We have thus found the quotient g, regarded as an operator which changes ¡3 into a, to depend upon the plane of the vectors, the angle between them and the ratio of their lengths. Since34 QUATERNIONS. two angles are requisite to fix a plane, it is evident that q depends upon four elements, and performs two distinct opera- tions : 1st. A stretching (or shortening) of /?, so as to make it of the same length as a ; 2d. A turning of /?, so as to cause it to coincide with a in direction, the order of these two operations being a matter of indiffer- ence . Of the four elements, the turning operation depends upon three ; two angles to fix the plane of rotation, and one angle to fix the amount of rotation in that Fig. 28. plane. The stretching operation de- pends only upon the remaining one, ¿.e., upon the ratio of the vector lengths. As depending upon four elements we observe one reason for calling q a quaternion. The two ope- o-'----------—b' rations of which q is the symbol being entirely independent of each other, a quaternion is a complex quantity, decomposable, as will be seen, into two factors, one of which stretches or shortens the vector divisor so that its length shall equal that of the vector dividend, and is a signless number called the Tensor of the quaternion ; the other turns the vector divisor so that it shall coincide with the vector dividend, and is therefore called the Ver sor of the quaternion. These factors are symbolical^ repre- sented by Tg and Ug, read u tensor of g” and “versor of g,” and g may be written g = Tg . Ug. 22. An equality between two quaternions may be defined di- rectly from the foregoing considerations. If the plane of a and ¡3 be moved parallel to itself ; or if the angle aob (Fig. 28), remaining constant in magnitude and esti- mated in the same direction, be rotated about an axis through o perpendicular to the plane ; or the absolute lengths of a and ¡3GEOMETRIC MULTIPLICATION AND DIVISION. 35 vary so that their ratio remains constant, q will remain the same. Hence if , a -i cl' f — = q and —-= q, P /?' then will when 1st. The vector lengths are in the same ratio, and 2d. The vectors are in the same or parallel planes, and 3d. The vectors make with each other the same angle both as to magnitude and direction. The plane of the vectors and the angle between them are called, respectively, the plane and angle of the quaternion, and the expression -, a geometric fraction or quotient. It is to be observed that q has been regarded as the operator on ¡3, produc- ing a. This must be constantly borne in mind, for it will sub- sequently appear that if we write q/3 = a to express the operation by which q converts /3 into a, q/3 and /3q will not in general be equal. 23. Since g, in operating upon f3 to produce a, must not only turn ¡3 through a definite angle but also in a definite direction, gome convention defining positive and negative rotation with reference to an axis is necessary. By positive rotation with reference to an axis is meant left- handed rotation when the direction of the axis is from the plane of rotation towards the eye of a person who stands on the axis facing the plane of rotation. [If the direction of the axis is regarded as from the eye towards the plane of rotation, positive rotation is righthanded. Thus, in facing the dial of a watch, the motion of the hands is positive rotation relatively to an axis from the eye towards the dial. For an axis pointing from the dial to the eye, the motion of the hands is negative rotation. Or again, the rotation of the earth from west to east is negative relative to an axis from north to south, but positive relative to an axis from south to north.] On the above assumption, if a person stand on the axis, fac- ing the positive direction of rotation, the positive direction ofQUATERNIONS. Fig. 31 (Ms). the axis will always be from the place where he stands towards the left. If ij ft, j (Fig. 31) be three axes at right angles to each other, with directions as indicated in the figure, then positive rotation is from i to from j to ft, and from ft to rela- tively to the axes ft, i, j respectively. A precisely opposite assumption would be equally proper. The above is in accordance with the usual method of estimating positive angles in Trigo- nometry and Mechanics. 24. Let OA and on (Fig. 29) be any two co-initial vectors whose lengths are a and 5, a and (3 being unit vectors along oa and ob, so that OA :—■ Ctot, Fig. 29. a OB = b(3. Let the angle boa between the vectors be represented by ; also draw ad perpendicular to ob, and let the unit vector along da be 8. The tensor of od is evidently a cos <£ and that of da a sin <£. If we assume that, as in Algebra, geometrical quotients which have a common divisor are added and subtracted by adding and subtracting the numerators over the common denominator, so that a ± y then, since we have « + y_ P OA OB OA = OD + DA, OD -f“ OA OB _ a cos . /? “ &T/T~ __ a f cos 0.13 ~b\~r~ OD DA OB OB a sin . 3 ~ b.p sin . 8\ + PGEOMETRIC MULTIPLICATION AND DIVISION. 37 We have already defined (Art. 8) the quotient of two parallel vectors as a scalar, and in the first term of the parenthesis, ¡3 right angles to each other. This quotient is to be regarded, as before, as a factor which, operating on the divisor ¡3, produces 8, ¿.e., turns ¡3 left-handed through an angle of 90° ; and this quotient must designate the plane of rotation and the direction of rotation. If we define the effect of any unit vector, operating as a multiplier upon another at right angles to it, to be the turn- ing of the latter in a positive direction through an angle of 90° in a plane perpendicular to the operator, then the unit vector €, drawn from o perpendicular to the plane of 8 and /3, and in the direction indicated in the figure, will be the factor which oper- ating on ¡3 produces 8, and The unit vector <•, as an axis, determines the plane of rotation ; its direction determines the direction of rotation, and by defini- tion its rotating effect extends through an angle of 90° ; as a quotient, therefore, it completely determines the operator which changes ¡3 into 8. Equation (a) thus becomes or, if OA and ob be themselves denoted by a and (3, and the ten- sors of a and ¡3 by Ta and T/?, being a unit vector, - = 1, and (a) — = - (cos <£ -f e sin <£), OB o q = —^ (cos <£ + e sin <£) .... (15),38 QUATERNIONS. Ta in which — is the tensor of g, being the ratio of the vector lengths, and cos <¡> + e sin <£ is the versor of g, its plane, deter- mined by the axis c, and angle being the plane and angle of the quaternion. Ta When a and ¡3 are of the same length, or Ta=T/3, Tg=—- = 1, i/j and the effect of g as a factor, or operator, is simply one of version. Like T, the symbol U is one of operation, indicating the oper- ation of taking the versor, so that Vq = cos <£ + e sin cj>. Fig. 29. This operation takes into account but one of the two distinct acts which we have seen the quotient q must perform, as an agent converting ¡3 into a, namely, the act of version ; it thus eliminates the quantitative element of length. In this respect it is similar to the reduction of a vec- tor to its unit of length, an opera- tion which also eliminates this same element of length, and has been designated by the same symbol U. When a and ¡3 are at right angles to each other, «ƒ> — 90° and the ver- sor cos H- € sin cj> reduces to the unit vector c, which has been de- fined, as an operator, to be a versor turning a line at right angles to it through an angle 90? Any vector, therefore, as a, contains, in its unit vector in the same direction, a versor element or factor of which Ua is the symbol, U indicating the reduction of a to its unit of length or the taking of its versor factor. Hence the appellation versor of a (Art. 7). If in Equation (15) the vectors be reduced to the unit of length,GEOMETRIC MTTLTIPLICATIOK AND DIVISION. 39 25. We may now express the relation | = f|(cos^ + £sin^) = ? in the symbolic notation or - = T-.U- p p p > q = Tq .Vq . (Eq. 15) (16), and say that the quotient of two vectors is the product of a tensor and a ver sor ; and that 1st. The tensor of the quotient, tensors; 2d. The versor of the quotient, (cos + € sin <£), is the cosine of the contained angle plus the product of its sine and a unit vector, at right angles to their plane and such that the rotation which causes the divisor to coincide in direction with the dividend shall be positive. 26. If, for - = g, we write ^ = q\ it is evident that q' differs P a from q both in the act of tension and ver- sion ; the tensor of g' being the reciprocal of the tensor of g, and the unit vector c, while still parallel to its former position, is reversed in direction (Art. 23) since the direction of rotation is reversed (Fig. 30). Hence — = (cos — € sin <£) .... (17). a Ta Fig. 30. is the ratio of their P is called the reciprocal of -. As already remarked, the a fi positive direction of € is a matter of choice. It is only neces- sary that if we have + c in U^, we must have — e in U or p a conversely.40 QUATERNIONS. Fig. 31. 27. Let J, ft (Fig. 31) represent unit vectors at right angles to each other. The effect of any unit vector acting as a multi- plier upon another at right angles to it, has been defined (Art. 24) to be the turning of the latter in a positive direc- tion in a plane perpendicular to the ope- rator or multiplier through an angle of 90? Thus, i operating on j produces ft. This operation is called multiplica- tion, and the result the product, and is expressed as usual ij=h (18). The quotient of two vectors being a factor which converts the divisor into the dividend, we have also ft_ Í (19), either the product or quotient of two unit vectors at right angles to each other being a unit vector perpendicular to their plane. This multiplication is evidently not that of algebra; it is a revolution, which for rectangular vectors extends through 90? Nor is ft in Equation (18) a numerical product, nor i in Equa- tion (19) a numerical quotient. This kind of multiplication and division is called geometric. In accordance with the above definition we may write the fol- lowing equations : (20). II h-i j jk = i II II •. 1 II ji = — Jc ~ft__ . i kj = — i ^=ft JGEOMETRIC MULTIPLICATION AND DIVISION. 41 Is5 II 1 Vs. ft ^ i(-j) =-k -ft . -J Ì = ft — i — = ft -i J (—&) = — * — i II ft . . . (20). Since the effect of ft, j as operators is to turn a line from one direction into another which differs from it by 90? they are called quadrantal ver sors. 28. Since we have or iXj = 7c and i x ft = — j = — 1X i, i X i X j = — 1 X j, i x i = — 1. We may denote the continued use of i as an operator by an exponent which indicates the number of times it is so used. This is consistent with the meaning of an exponent in algebraic notation. In both cases it denotes the number of times the operator is used, in one instance as a numerical factor, in the other as a versor. Thus i/w* = i2*3, in jj y 3 -, etc. r In conformity to this notation the above equation becomes i2=-l (21),42 QUATERNIONS. and in a similar manner, . . (22). Hence the square of a unit vector is — 1. The meaning of the word 44 square ” is more general than that which it possesses in Algebra, as was that of the word 44 product” in Art. 27. The propriety of this ex- tension of meaning lies in the fact that for certain special cases, the processes above defined reduce to the usual alge- braic processes to which these terms were originally restricted. The conclu- sion i2 — — 1 is seen to follow directly from the definition, since if i operates twice in succession on either ± j or ± 7c, it turns the vector, in either case suc- cessively through two right angles, so that after the operation it points in the opposite direction. A similar reversal would have resulted if the minus sign had been written before the vector. Thus — (± j) = T j. Hence i x i, or ¿2, as an operator, has the effect of the minus sign in revers- ing the direction of a line. 29. It is to be observed that so long as the cyclical order i, J, ft, i, f ft, i, .... is maintained, the product of any two of these three vectors gives the third ; thus ij=lc, jk = i, M = j\ and therefore (ij) ft = ftft = ft2=—1, (jk) i = ii = i2 = — 1, (kï)j = jj =ƒ = -i; as also i(jk)= ii = i2=— 1, j (In) = .Ü = ƒ = — 1) = ** = *2=—1,GEOMETRIC MULTIPLICATION AND DIVISION. 43 hence i(jlc) = (ij)k, j (Jci) = (jk) i, h (ij) = (ki)j, which involves the Associative law. We may therefore omit the parentheses and write . ifk = jld = Idj = — 1 .... ; (23), or, the continued product of three rectangular unit vectors is the same so long as the cyclical order is maintained. But 7c(ji)=]c(-lc) = -7(-*) (-J)(- 0= 10. J-=-k. 12. — i i -Jc_ 14. -j _ j Jc 16. Jc2 . ~=J- — A; 18. Jà__ i J iJc_ ^ 20. Jc __ j V li- 22. 1 Jc_ ft j * ¿ ijJc __ 24. i j Jc __ kji k' j’ i 25. Is it correct to write, in general, the product of any frac- tions, as - . in the form — ? j j m ft ^ ~4^ is correct or not, and why. 26. State whether ■ Jc Já 27. i2j2Jc2 = -(ijJc)2.GEOMETRIC MULTIPLICATION AND DIVISION. 49 38. Resuming Equation (15), ç=!=ï?(cos*+£sin*); the quaternion q was shown (Art. 25) to be the product of a tensor and a versor. It may also be regarded as the sum of two parts, the first of which ieh \p0 lT/3 COS (f) is a scalar, whose sign is second that of the cosine of the angle (<ƒ>) between the vectors, while the ~Ta 1 — sin • e is a vector at right angles to their plane, whose sign depends upon the direction of rotation of the fraction This may be expressed symbolically in the notation* so that we have both and P P q = IqXJq q=Sq + Yq. P (35), The second member of this last equation is read “ scalar of q plus vector of q” S q and Yq being respectively symbols for the scalar and vector parts of the quaternion. As already explained in the case of the symbol $, Y is a symbol of operation, denoting the operation of taking the vector terms of the expression before which it is written. The quotient of two vectors is, therefore, the sum of a scalar and a vector. _ T Ta The scalar of the quotient I Sg = — cos <£ is the ratio of the tensors times the cosine of the contained angle. The tensor of Ta ‘T fi the sine of the contained angle. The versor of the vector part [UYg = e] is a unit vector perpendicular to their plane, having a [■ the vector part | TYg = ^ sin J is the ratio of the tensors times50 QUATERNIONS. direction such that the direction of rotation of the divisor is posi- tive or left-handed. Letting a and b be the tensors of a and /?, and collecting the preceding expressions for facility of reference, we have Vq = cos<£ + €sin<£ Sg=2 cos Yg=^ sin cj> . € o TYg=- sin b UY g=€ SUg=cos cf> YUg=sin cf> . e TYUg=sin<£ These expressions require no further explanation than that derived from a simple inspection of Equation (15) in connection with the meaning already assigned to T, U, S and Y as symbols of operation. 39. De Moivre’s Formula. The following considerations will explain why the parenthesis (cos + e sinc/>) as a versor turns fS left-handed through an angle . They also contain the quaternion interpretation of imaginary quantities. Let V = sin cf> and z = cos <£. Differentiating, dv = cos d, dz = — sin <¡> d, or dv = zd, dz = — vdcf). (a) (à)GEOMETRIC MULTIPLICATION AND DIVISION. 51 Multiplying (a) by V—1, and adding the result to (b), and therefore, from (e) and (ƒ), (cos<£ + sin<£ . V—3)m= cos m + sinm<£ . V^ï (37), which is the well-known formula of De Moivre. This formula may be made the basis of a system of analytical trigonometry. Thus, for example, to deduce the formulae for the sine and cosine of the sum of two angles, we have from (d) or dz + dv . V—1 = (— V + z V—1) d, dz + dv • V — 1 = (v V— 1 + z) V—1 d<£, whence (<0 which may be written whence or z + v V— 1= cos + sin • V — 1 = cosmcj) + sinmc£ . V— 1= (d) w But we have from (d) (cos <£ + sin<£ • V— l)m = (/) cos<£ + sin<£ V — 1 = e0s/~S cos 0 + sin 0 V— 1 = eöv31* Multiplying member by member, fo) But from De Moivre’s formula cos (<£ + 0) + sin ( + 0) V— 1 = +®(Ji)52 QUATERNIONS. Equating the first members of (g) and (h), since in any equa- tion between real and imaginary quantities these are separately equal in the two members, we have cos (9 + <£) = cos 0 cos <£ — sin 0 sin . sin {9 + <£) = sin 9 cos + cos9 sin <£. These formulae, while they may be of course demonstrated independently of De Moivre’s formula, are here deduced from imaginary expressions. It would therefore appear that these expressions admit of a logical interpretation. If any positive quantity m be multiplied by (V—l)2 the re- sult is — m. That is, in accordance with the geometrical inter- pretation of the minus sign, we may regard the above factor (V=I)2 as having turned the linear representative of m about the origin through an angle of 180? If, instead of multiplying m by (V —1 )2, we multiply it by V — 1, we may infer from analogy that the line m has been turned through an angle of 90° about the origin. If, too, we ob- serve that each of the four expres- sions m, mV—1, —m, —mV—1 is obtained from the preceding by multiplying by the factor V —1, they ma}- be regarded as denoting in order a distance m on the co-ordi- nate axes OX, OF, OX/ OF' (Fig. 33), V—1 being, as a factor, a versor turning a line left-handed through a quadrant. These expressions therefore locate a point on the axes, both as to dis- tance and direction from the origin. Since every imaginary expression can be reduced to the form ± a ±b V—1, we may, in accordance with the above interpre- tation of V—1, regard such an expression as defining the posi- tion of a point out of the axes. Thus oa = a (Fig. 34) and Fig . 33. Y /—r\ mV—1 \ X,l\ —m 0 m \X \ \ \ \ i —mV—L / \ N / X y Y*GEOMETRIC MULTIPLICATION AND DIVISION. 53 AP = &, laid off at a at right angles to oa since b is multiplied by V — 1 ; so that in passing over oa and ap in succession we reach the point p. It is also evident that such an expression implicitly fixes the position of p by polar co-ordinates, since Va2 + b2 = op and tanpoA = -. In like manner ___ a -b + a V—1 would locate a point p¡ oa' having a length = a, but laid off perpendicular to oa, since V—I is a factor, and a'p'= — b. As before, we have implicitly op'= Va2 + b2 and , , a tan P OA = — b Furthermore, if we operate on the first expression, a + b V—1, which fixes the point p, with V—1, we obtain the second, -ò-f-aV—1, or V-l as a factor turns op through 90° so as to make it coincide with op! As an operator, therefore, we may regard V—1, like i, J, ft, as a quadrantal versor, turning a line through a quadrant in a positive direction. Algebraically it denotes an impossible operation. (In Algebra quantities are laid off on the same line in two opposite directions, + and —. It was because quan- tities are so estimated only in Algebra that Sir W. Hamilton called it the Science of Pure Time, since time can be estimated only into the future or the past.) But it is unreal or imaginary only in an algebraic sense. If the restrictions imposed by Al- gebra are removed, by enlarging our idea of quantity and at the same time modifying the operations to which it is subjected, this imaginary character disappears. In applying the old nomen- clature to these new modifications, it will be seen that the prin- ciple of permanence is observed, ¿.e., the new meaning of terms is an extension of the old ; and when the new complex quantities reduce to those of Algebra, the new operations become identical with the old. If now we operate upon Pig. 34.54 QUATERNIONS. which, if we regard a = OA (Fig. 35) and W—1=ap as vectors, is equivalent to op, with the Fig. 35. expression cos -F sin ; also pa" and al perpendicular and as par- allel to OX l Then a cos — b sin = ol — a"l = oa" a sin cf> + b cos cf> = la + sp = a"p. Make oa'= oa" and lay oif a'p'= a"p perpendicular to OX, since it has V — 1 as a factor ; then (a cos — b sin cf>) +V — 1 (a sin -f b cos ) = oa'+ a'p'= op,f and p'op = . But the formulae for passing from a set of rectangular axes OX, 0 Y, to another rectangular set OX' 0 Yf, are X = cc'cos + y'sin <£, y == y1 cos <ƒ> — æ'sin. <£, in which XOX'=<£, æ = OA, y = AP, æf= oa" yf=PA¡' or OA = OK + KA, AP = NP — a"k, a"k being perpendicular and a"n parallel to OX. Hence the effect of the operator has been to turn op left- handed through an angle <£, which is equivalent to turning the axes right-handed through the same angle.GEOMETEIC MULTIPLICATION AND DIVISION. 55 + 1,-1 and V—1 are particular cases of the general versor cos + sin • V—1, namely, when is 0° 180° and 90° respectively, +1 preserv- ing, —1 reversing and V—1 semi-inverting the line operated upon. We may now see the meaning of De Moivre’s formula (cos<£ + sincj> . V—l)m = cosm<£ + sinm<£ . V — 1. As operators, the first member turns a line through an angle <£ successively m times, while the second member turns it through m times this angle once, producing the same result. The expressions cos + sin . V—1 and cos <£ + sin <£ . c are identical, except that in the latter the plane of rotation is not indeterminate, being perpendicular to e, V—1 being any unit vector with in- determinate direction in space. Equation (37) may be put under the form cosm (271-ti + ) + sinm (Zirn + ) • V—1 = [cos (2?m -f- <£) + sin (27rn + cj>) • V—l]m. In the second member if <£ = 0 and m = we have Vl for all integral values of n, while the first member for n — 0, n = 1, n = 2 becomes 1, — i + ^V—T, — \ — ^ V— 1, the three roots of unit}7. In the same way for m = i,_________ 1+^V-i, -i, _ the six roots of unity. The real roots lie on the axis, along which direction is assumed plus and minus, while the imaginary56 QUATERNIONS. roots are vectors in a direction not that of the axis, and are the sum of two vectors, one of which is in the direction of the axis and the other perpendicular to it. 40. Let a and ¡3 be unit vectors along oa and ob (Fig. 36). Resolve oa = a into the two vectors Fis-36- od, da. Then OA = a = 0D + DA. But OD = COS <£./?, da = € (sin . /?) = sin <£ . c/3, € being a unit vector perpendicular to the plane aob, as in the figure. Hence a = cos cj> . ¡3 + sin<£ . e/L (a) Now when a and ¡3 are unit vectors, we have by definition ^ • ¡3 = (cos + €sin<£)/3 = a ; or, comparing with (a), (cos<£ + csinc£)/?:=cos<£ . /?-f sin<£ . e/3. The distributive law, therefore, applies to the multiplication of a vector by the scalar and vector parts of a quaternion; for if a and ¡3 are not unit vectors, the tensors, as merely numerical factors, can be introduced without affecting the versor conclu- sion. Resolve ¡3 into the vectors oc, cb, cb being perpendicular to oa. Then ob = ¡3 = oc + CB. But Hence oc = cos cf> • a, CB = — € (sin (j) • a) . COS <ƒ> . a — sin 0 • ea = ¡3, or, by the distributive principle, (C0S<£ — sin<£ . e) a = (3.GEOMETEIC MULTIPLICATION AND DIVISION. 57 Using the two members of this equation as multipliers on the corresponding members of (a) (cos — sin • e) aa = fi (cos . fi + sin (f) • efi), or, since a2 = — 1, — cos + * sin = fia.........(38). If a and fi are not unit vectors, fia = TfiTa ( — cos + € sin ) . . . (39). Operating with each member of (a) on fi, afi= (cos. fi + sin(j> . efi)fi = cos <£ • fi2-\- sin c/> . efi2 = — cos (f> — e sin ..................(40), or, if a and fi are not unit vectors, afi = TaTfi ( — cos <ƒ> — e sin <£) . . . (41). The product of any two vectors zs, therefore, a quaternion, which, as before, may be regarded either as the sum of a scalar and a vector or the product of a tensor and a versor. In gen- eral notation a/2 = Sa/?+Ya/?=Sg+Yg .... (42), afi = Tq • Ug.........................(43). The scalar of the product [Safi = — TaT/3 cos <£] is i/ze product of the tensors and the cosine of the supplement of the contained angle. The vector of the product [ff afi = — TaT/? sin . e] 7zas /or #s tensor [TYafi= TaTfi sin <ƒ>] the product of the tensors and the sine of the contained angle, and for a versor [UYa/? = — e] a unit vector at right angles to their plane such that rotation about it as an axis is positive or left-handed.58 QTTATEBNIONS. Representing the tensors of a and ¡3 by a and ò, we have, as in Art. 38, from Equation (41), Fig. 36. T q = ah Ug = — cos <£ — csin Sg = — ab cos Yg = — aòsin<£ . e TTg = ab sin <ƒ> UYg = - € (44). SUg = — cos c/> YUg = — sin • c TYUg = sin (TY : $) g = — tan <£ 41. Resuming the expressions for the products and quotients of a and /?, /3a = T/3Ta (— cose£ + €sin<£), (a) a/3 = TaT/3 ( — cos — c sin ^>), Õ) - = (cos — * sia ), a la (c) - = ^ (cos + e sin ), ft TftK r (d) we observe 1st. That if a and ¡3 be interchanged the sign of the vector part is changed. It is equivalent to a reversal of the angle <£, and consequently a change in the direction of rotation. Hence UY/?a = c = —UYa/? V UV- =£ = —UY^ [...................O5)- (3 a J Vector multiplication is not therefore in general commutative. 2d. If the vectors are unit vectors, fta = — @, aft = ^ . . . a ¡3 . . (46),GEOMETRIC MULTIPLICATION AND DIVISION. 59 the product being expressed also by a quotient. This is of course always possible, as appears from (a), (5), (c) and (d), and the transformation may be effected thus : or - = — 7¡r- = sr (cos<£ — esin<£), [Eq. (81)] a Ta Ta — fia = T/3Ta (cos — e sin <£) ; /?a = T/3Ta (— COS<£ + esincjy) . 3d. If = 0, then in either (a) and (5) or (c) and (cl) the vector part of q becomes zero, and the quaternion de- grades to a scalar. When = 0 the vectors are parallel, and aj3 = — TaT3 = — a&, as in Art. 35 ; also - = — = as in H H ’ p T/3 5 Art. 8. If at the same time a and B are unit vectors - = - =1 /? a [or = aa~1 = — a2 = 1] and a/? = a2 = — 1, as in Arts. 33 and 28. If then q he any quaternion and Yg = 0, ¿/¿e vectors of ivhich q is the quotient or product are parallel, 4th. If <£ = 90° then in either (a) and (h) or (c) and (d) the scalar part of q becomes zero, and the quaternion degrades to a vector ; and either the product or quotient of two rectangu- lar vectors is therefore a vector at right angles to their plane, aj3 reducing to — ape and ~ to ^e, as in Art. 34. If at the same time a and j3 are unit vectors, a/3 = — e and - = e, as in Art. 27. ^ If then q he any quaternion and Sg = 0, the vectors of ivhich q is the quotient or product are perpendicular to each other. 5th. If an equation involves scalars and vectors, the vector terms having been so reduced as to contain no scalar parts, then since the scalar terms are purely numerical and independent of the others, the sums of the scalars and vectors in each member are separately equal. Thus if X -f- au + h/3 = d + y + afa + (h'~ hn)/3 1 f. (47), x = d + y and aa + h/3 = a'a + (hf— h")jBj then60 QUATERNIONS. which might also be written (Art. 38) S (x -f- act -j- bf3^) = S -f~ y oJcl -j- ( bf— Y (x + aa + &/?) = Y\_d + y -f a'a + (bf— b") /?]. 6th. Ë being the quotient which operates on a to produce /?, a we have by definition 7th. TYa/?, or ab sin , is the area of a parallelogram whose sides are equal in length to a and b and parallel to a and ¡3. Sa/?, or — a5cos, is numerically the area of a parallelogram whose sides are a and 5, and angle ab is the complement of . 8th. Since the scalar symbol S indicates the operation of taking the scalar terms, Sa = 0.............. (49), and, for a similar reason, Ya = a ....... (50). Again, since a2 is a scalar, V(a2)=0 . ...............(51), s (a2) = — a2.............(52). Y (a2) may be written Y. a2, as also S(a2) = S • a2, but these forms must be distinguished from (Ya)2 and (Sa)2, which latter are also sometimes written Y2a and S2a. 9th. Comparing (a) and (6), Sa/? = S/?a............. and Ya/? = —Y/?a............ Adding and subtracting (a) and (b), we have also a¡3 + Pa = 2 Sa/? a/?-£a=2Ya¡3 . . . (53) , (54) . . . (55), • . (56).GEOMETKIC MULTIPLICATION AND DIVISION*. 61 10th. a/3 . ¡3a = (Sa/3 +Ta/3) (Sa/3 - Va/3) [Eqs. (53) and (54)] = (Sa/3)2— Sa/3Va/3 + Sa^Va/3 - (Ya/?)2. Hence a/3 . /?a = (Sa/?)2-(Ya/?)2 . . . . (57), or, from Equation (44), a/3 • /3a = (Ta/3s)2.........(58). 42. Powers of Vectors. The symbol ¿m, m being a positive whole number, has been seen (Art. 28) to represent a quadrantal versor used m times as an operator ; the exponent denoting the number of times i is used as a quadrantal versor. By an extension of this meaning of the exponent, i™ would naturally represent a versor which, as a factor, produces the —th part of a quadrantal rotation. Thus ft produces a rotation through one-third, and ft through three-fifths of a quadrant, respectively. With the additional meaning attached to the negative exponent (Art. 32), as indi- cating a reversal in the direction of rotation, we may in general define where i is any vector-unit and t any scalar exponent, as the representative of a versor ivliich would cause any right line in a plane perpendicular to i to revolve in that plane through an angle t X 90° the direction of rotation depending upon the sign of t. Hence ever}^ such power of a unit vector is a versor, and, conversely, every versor may be represented as such a power. 24> Since the angle (<£) of the versor is t x -, we have £ = —, and any versor cos + e sin <£ may be expressed and 20 cos 4> + e sin = e71 . 20 COS — €sin<£ = €“ 7T . . (59), • . (60), the vector base being the unit vector about which rotation takes place, and the exponent the fractional part of a quadrant through which rotation occurs.62 QUATERNIONS. The operation of which fi is the agent is one-half that of which i is the agent, and therefore two operations with the former is equivalent to one with the latter ; or, as in Algebra, fi fi=i = fi+1*...................(61), or, employing the other versor form, if a, /?, y are complanar unit vectors so that a . 20 — = cos cjy + e sm <£ = e * ? /3 20 “ = COS 0 + €sin 0 = €7r> then since a ¡3 a * y ~ y we have (cos + e sin 4>) (cos 0 + e sin 0) = cos <£ cos 0 + €2 sin cjy sin 0 + € (sin h etc., cosm7T = 0 and the versor degrades to the vector ±e. If the vector is not a unit vector, as xi = p, to interpret the exponent, say p*, so as to satisfy the formula phpkssp....................(63), which is analogous to Equation (61), we must combine with the conception of rotation through half a quadrant an act of tension represented by the square root of the tensor of p. Thus, if X = 16, and we write pi = (16¿)i= 164£4, then pV = (16M) (16èíè)=16i = p, or, if x= V8, pJ = ^8 . = . i\ pW = (V2 . iJ) ( V2 . i¡) ( V2 . i1) = V8 . i - p. And, in general, pt=z(xi)t = ^. il..............(64), or the tensor of the power is the power of the tensor, and the versor of the power is the power of the versor. Symbolically T. ^ = (TP)«...................(65), U. p? = (Up)‘..................(66). Any such power (p*), as the representative of the agent of both an act of tension and version, is therefore a quaternion, whose tensor and versor can be assigned by the above rules, and, conversely, every quaternion can be expressed as the potoer of a vector, which quaternion may degrade to either a scalar or a vector as seen in the preceding versor conclusions. Hence it follows that the index-law of Algebra is applicable to the powers of a quaternion.64 QUATERNIONS. 43. Relation between the Vector and Cartesian deter- mination of a point. If £, j, k are three unit vectors perpendicular to each other at a common point, then the vector from this point to any point p may be written p = xi + yj+zk.................(67), in which o?, y, z are the Cartesian co-ordinates of p. If the vec- tors are not mutually perpendicular and are represented by a, ¡3, y, then p = xa + y¡3 4- zy.............(68), in which 05, y, z are the Cartesian co-ordinates of p referred to the oblique axes. So long as the vectors a, ¡3, y are not com- planar, p refers to any point in space. Since any quaternion q ma}r be expressed as the sum of a sca- lar and a vector, if w be any scalar, then q = w + xa + y¡3 + zy.............(69). As composed of four terms, we observe an additional reason for calling this complex expression a quaternion. Any vector equation p = o- = aa + b¡3 + cy, involves three numerical equations, as X = a, y=b, z = c, unless the vectors are complanar ; in which case we may write y = na + m/3, and p = (x + zn) a + (y + zm)P, <7= (a + cn) a -f (b + cm)¡3, which, for p = o-, involves but two equations X -f- zn = a + cn, y-{- zm = bgeometric multiplication and division. 65 Resuming the quadrinomial form of g, when the component vectors are at right angles, we have q = w + tá + yj+ zk 8q = w Yq = xi + yj+ zk (70). Since (TYq)2 = - (Vq)2 = *2 + y2 + z2, we have T Yq = '/x2 + y2 + z2 j UYQ ^Yq = Xi + W+ Zk [ * # - (71)‘ TVg y/x2 + y2 -j- z2 j Also, since (Art. 41, 10th.) (Tg)2 = (Sg)2 — (Vq)2 == tv2 + a?2 + ƒ + z2, Tg = Vîtr + af + F + z g w + + yj + zk U g “-------------:------ " Tg vV + a2 + y2 + z2 SUg = TYUg Sg =____________________ Tg Vw2 + æ2 + 3/2-M2 ít2-}- ƒ + z2 w¿ + x2 + y2 + z2 _TYg_ ï ~ Tq~% • (72). 44. The plane of a quaternion has been already defined as the plane of the vectors or a plane parallel to them. The axis of á quaternion is the vector perpendicular to its plane, and its angle is that included between two co-initial vectors parallel to those of the quaternion. If this angle is 90° the quaternion is called a Right Quaternion. Any two quaternions having a common plane, or parallel planes, are said to be Complanar. If their planes intersect, they are Diplanar. If the planes of several quaternions intersect in, or are parallel to, a common line, they are said to be Collinear. It follows that the axes of collinear quaternions are complanar, being perpendicular to the common line. Complanar quaternions are always collinear, and66 QUATERNIONS. complanar axes correspond to collinear quaternions, but the lat- ter may of course be diplanar. o Let —and —7- be any two quaternions. If complanar, they o'b od may be made to have a common plane ; and, if diplanar, then" planes will intersect. In the former case let oe be any line of their common plane, or, in the latter, the line of intersection of their planes. Now, without changing the ratios of their vec- tor lengths, the planes, or the angles of the given quaternions, two lines, of and 00, ma}" alwa}Ts be found, one in each plane, or in their common plane, such that with oe we shall have o'a of -, o"c OG — = — and — = — ; OB OE OD OE and, therefore, any two quaternions, considered as geometric fractions, can be reduced to a common denominator ; or, in the above case OA o"c_OF OG______OF -f- OG o'b o"d oe + oe“ OE Moreover, a line oh, in the plane ao'b, may always be found such that o'a _ OE o'b“ oh’ and therefore o''c o'a __ og oe __ OG 0"d o'b OE * OH ~ oh’ and G'a ' °"c OF ^ OG OF OE OF O'B * 0"d OE # OE ~ OE OG ~ OG 45. Reciprocal of a Quaternion. The reciprocal of a scalar is another scalar with the same sign, so that, as in Algebra, if x be any scalar, its reciprocal isGEOMETRIC MULTIPLICATION AND DIVISION. 67 The reciprocal of a vector has been defined (Art. 83), so that, if a be any vector, - = a-1 = — ~ Ua. J a Ta The reciprocal of a quaternion has also been defined (Art. 26) ; thus a p~q being any quaternion, Ê- =— = g_1 a q Fig. 37. is its reciprocal. The only difference between the quotients - and Ê. (Fig. 37) is that, as opera- ft a tors, one causes ft to coincide with a, while the other causes a to coincide with ft. A quaternion and its recipro- cal have, therefore, a common plane and equal angles as to magnitude, but opposite in direction ; that is, their axes are opposite. Or Z - = Zq and axis - = — axis q. q q Since !:*=£:£«=£.£ = £, and = ft ft ft ft a a a ft ft the product of two reciprocal quaternions is equal to positive unity, and each is equal to the quotient of unity by the other; we have, therefore, as in Algebra, -q=l and g = T, and no q 1 Ì new symbol is necessary for the reciprocal. - is, however, sometimes written Eg, R being a general symbol of operation, namely, that of taking the reciprocal. It follows from the above that68 QUATERNIONS. or, the tensors of reciprocal quaternions are reciprocals of each other ; while the ver sors differ only in the reversal of the angle. If then 1 Ta q = - = — (cos á + c sin é>) H p T/r . r J we shall have Kg = i = g_1 = - = — (cos — e sin ) q a Ta (74). Fig. 37. 46. Conjugate of a Quaternion. If ft (Fig. 37) be taken complanar with p and a, and making with a the same angle that p does, T/3' being also equal to T/?, then, if ^ = g, is called the conjugate of q, and is written Kg. The symbol K indicates the operation of taking the conjugate. A quaternion and its con- jugate have, therefore, a common plane and tensor, as also, in the ordi- nary sense, equal angles ; but their axes are opposite ; or ZJLq = Zq = Z- TKg = Tg = and If then we shall have TÌ . 1 axis Kg = — axis g = axis- Q q = - = — (cos ò + € sin d>) P T/3k r Tr» Kg = T^(C0S^“€ SÍn^) (75). (76), or, the tensors of conjugate quaternions are equal, and the versors differ only in the reversed of the angle. Regarding a scalar and a vector as the limits of a quaternionGEOMETRIC MULTIPLICATION AND DIVISION. 69 (Art. 41, 3d and 4th), we see from Equation (76) that the con- jugate of a scalar is the scalar itself, and that Ka = — a = — TaTJa............(77) , or, the conjugate of a vector is the vector reversed. In general notation we may write q = sg + Vq, whence it follows from the above that Kg = Sg —Yg ) ’ or (Art. 43) V .... (78), K q = iu — xi — yj —zk) that is, the scalar of the conjugate of a quaternion is the scalar of the quaternion, and the vector of the conjugate of a quaternion is the vector of the .quaternion reversed ; a result which may be expressed symbolically «** = ** I ..... . (79). VK q = -lq) v ’ These are Equations (53) and (54). If we add and subtract the two conjugate quaternions we have q = $q +Y q, Kg = Sg — Yg, g -f- Kg = 2 Sg } g-Kg=2Yg}‘ (80). The sum of tico conjugate quaternions is, therefore, always a scalar, positive or negative as the Zq is acute or obtuse. If Z q = -, this sum is evidently zero. 2 Since, if g is a scalar, Kg = g, then, converse^, ifKq = q, q is a scalar.70 QUATERNIONS. 47. Opposite Quaternions. If, for Opposite of g, and is evidently — g, for As appears from the figure, opposite quaternions have a com- mon plane and tensor, supplementary angles and opposite axes ; or T ( — g) = Tg, Z —g = 7r —Z g and axis ( —g) = — axis g. Since f* f* the sum of two opposite quaternions is zero, or Fig. 37. 3+ (-?) = <>. Also, since — a # a_ — a /? ‘y8, y3 'a ___ — a yS - =-l a or, quotient is negative unity. If then we shall have . . (81). If Z q = Kg = — g ; and, conversely, if Kg = — g, g is a vector. 'GEOMETRIC MULTIPLICATION AND DIVISION. 71 48. Since Ug is independent of the vector lengths, and only dependent upon relative directionversors are equal whose axes and angles are the same. Hence But (Art. 24) and, Equation (82), UKg = uI. 9 a Ua •. uì= — q U q UKg = Ug l.Ua 'U/8 (82). (83), Again, since the conjugate of a versor is the same as the re- ciprocal of that versor, we have, from Equations (82) and (88), UKg = KUg (84). 49. Representation of Versors by spherical arcs. If a, /?, y,.. are co-initial unit vectors, their extremities will p being any Fig. 38. all lie on the surface of a unit sphere (Fig. 38). a quaternion, turns ft from the position on to OA, and this versor may be repre- sented by the arc ba joining the vector extremities ; for this arc determines the plane of the versor as also the magnitude and direction of its angle, the direction of rotation being indicated by the order of the letters as in the case of vectors. This representation of versors by vector arcs is of importance in the theorems re- lating to the multiplication and division of quaternions, and may be made upon a unit sphere ; for, if a, /3, y,.. are not unit vectors, the quaternions will differ from the versors by a nu- merical factor only, the introduction of which cannot affect the72 QUATERNIONS. Fig. 38. versor conclusions. Disregarding, then, the tensors, since ver- sors are equal whose planes are parallel and angles equal (in- cluding direction), equal arcs on the same great circle and estimated in the same direction represent equal versors, for any arc may be slid over the great circle on which it lies without change of length or reversal of direction. On this plan b'a = ab will represent the recipro- cal or conjugate of ba, and a quadrantal versor would have for its representative BC, an arc of 90? Also, the versors of all complanar quaternions will be repre- sented by arcs of the same great circle, while arcs of different great circles will represent the versors of diplanar quaternions, which are always unequal. If M, N and p are the vertices of a spherical triangle, the vector arcs MN, np and pm will represent versors, and it will be seen that by taking the geometric sum of two of these arcs in a cer- tain order, the remaining arc will represent the versor of their product ; so that if q1 be represented by pm and q by np, q'q may be constructed by a process of spherical addition represented by pm + np = NM, nm representing the versor q'q ; but that because q'q and qq' are not generally equal, this process of spherical ad- dition, as representing versor multiplication, is not commutative as was that of vector addition, pm + np and np + pm representing diplanar versors. 50. Addition and Subtraction of Quaternions. Since a quaternion is the sum of a scalar and a vector, in finding the sum or difference of several quaternions the sum or difference of their scalar and vector parts may be taken sepa- rately. The former will be a scalar and the latter a vector; consequently, the sum or difference of several quaternions is a quaternion.GEOMETRIC MULTIPLICATION AND DIVISION. 73 1. Both the associative and commutative principles being applicable to the summation of scalars, as also to that of vectors (Arts. 4, 5), they also hold good for the addition and subtrac- tion of quaternions ; or q+r=r+q and 2 + (r+s) = (g + r) + s If then g = Sg+Yg r = Sr +Yr s = g+r +........ Ss-f-Ys; in which Ss = S (g + r +...) = Sg + Sr +..., Ys = Y(g + r +....) =Yq + Yr +...., and, in general, S2g= 2Sq I Y%q = SYg Í (86), or, in quaternion addition and subtraction, S and Y are distribu- tive symbols. 2. If g + r -\-p +.= s, then, Equation (78), Kg+ Kr + Kp + *...= Sg+ Sr + Sp +.....-Yq-Yr-Yp-......... = Ss-Ys = Ks. .*. 2Kg = K2g.................(87), K, like S and Y, being a distributive s}unbol. 3. Again, since the conjugate of a scalar is the scalar itself, KSg = Sg. But Sg = SKg. Hence KSg = Sg = SKg...............(88). Also, since the conjugate of a vector is the vector reversed, KYg = -Yg.74 QUATERNIONS. But — Yg = YK g. Hence KYg = — Yg = YKg.............(89); hence K ¿s commutative with S and Y. Fig. 38. 4. Since any two quaternions may be reduced to a common denominator (Art. 44), so that a y __ qf+ yf 0 + 8' ’ and since Ta'+ Tyf > T (af+ y') unless af = xy’ andæ>0, it follows that Tg + Tg'>T(g-}-g') unless g = æg'and x>0. Hence, in general, T2g is not equal to STg. Moreover, since USg is a function of the tensors under the 2 sign, while SUg is independent of the tensors, USg is not equal to 2Ug. This also appears from the representation of ver- sors by spherical arcs (Fig. 38). Hence, in the addition and subtraction of quaternions, T and U are noi, in general, dis- tributive symbols. 51. Multiplication of Quaternions. 1- Let g = Sg +Vg, r = Sr +Vr be any two quaternions. Then p = qr = SgSr + SgYr + SrVg +YgYr. The last member, being the sum of a scalar and a vector, is a quaternion. Hence, the product of two quaternions is a quater- nion, and p = $p +Yp = Sgr +Ygr, in which Sgr = SgSr + S . YgYr .... (90), and Yqr = SgYr + SrYg + Y. YqYr . . . (91 ).GEOMETEIC MULTIPLICATION AND DIVISION. 75 If we multiply g by r, we obtain S rq =SrSg + S. YrYg, T rq = Sr T g + SgYr +Y. YrYg. But, Equation (53), S . YrYg = S . YgYr. Srg = Sgr..................(92). But, Equation (54), Y. YgYr = — Y. YrYg, and therefore the products gr and rq are not equal. Hence, quaternion multiplication is not in general commutative. If, however, g and r are complanar, Y g and Yr are parallel, and Y . YgYr = 0 ; in which case gr = rg. Conversely, if gr = rg, g and r are complanar. Since Reciprocal, Conjugate and Opposite quaternions are complanar, they are commutative, or gKg = Kg . g 11 ii qq = qq = qq =q q 9 (-?) = - m (93). 2. It has been shown (Art. 44) that any two quaternions g, g' can be reduced to the forms - and Z having a common a a denominator, or to the forms - and Z. Hence 8 a We have then a a a p p Ti' = TV=Ty = Ty< Ta = Ty. T^_ ' ■ g p Tp Ta T/î Ta’ Ta ~ q ' q q ¡i U/J Ua U/i Ua Ua 1 *. (94).76 QUATERNIONS. In a similar manner T(g'g) = T a sj S TS Ta U(g'ç) = ïj|=Ug'U? O Ta __ , ¡¡j§ “ T« • T? (95). Hence the tensor of the product (or quotient) o/ any two qua- ternions is the product (or quotient) of their tensors, and the ver- sor o/ the product (or quotient) is the product (or quotient) of their versoi's. In fact, tensors being commutative, we have, in general, Tllg = IITg....................(96), Ilg = Tllg • Uiig = IITg • IlUg, Ung = nUg......................(97). 3. The multiplication and division of tensors being purely arithmetical operations, we proceed to the corresponding opera- tions on the versors. It has been shown (Art. 44) that any two versors g, q'} may be reduced to tlie forms £ = — g'=f=—' a OA* /? Ob’ (Fig. 39), A, B, C( sphere. being the vertices of a spherical triangle on a unit Then q pa a the versors qr and rq being represented by the arcs ca' and ac' respective^. These arcs, though equal in length, are not in the same plane, and therefore the versors rq and qr are not equal. Constructing these versors, by spherical addition we should have bc' -f- ab = ac', AB + bc' = ba' + CB = ca', a change in the order giving unequal results.78 QUATERNIONS. Hence, unless ac' and ca' lie on the same great circle, in which case q and r are complanar, quaternion multiplication is not commutative. 5. Other results, hereafter to he obtained symbolically, may be readily proved by means of spherical arcs, as follows : If ab (Fig. 39) represents the versor of q = -, a'b = ba repre- 1 a sents the versor of Kg or — The spherical sum of ab + ba q 1 being zero, the effect of the versors in the products gKg and g- Q is to annul each other. Hence, if the vectors are not unit vectors, Again, from we have gKg = Kg . g = (Tg)2 1 1 g- = -g = 1. g g AB + BC' = CAf, (98), qr = —, y and the versor of K (qr) will therefore be represented by a'c. But a'c = BC + AfB, whence K(gr) = KrKg..................(99), or, the conjugate of the product of two quaternions is the product of their conjugates in inverted order. 6. The product or quotient of complanar quaternions is readily derived from the foregoing explanation of versor products and quotients as dependent upon a geometric composition of rota- tions. For, disregarding the tensors, the vector arcs which represent the versors, since the latter are complanar, will lie on the same great circle, and the processes which for diplanar ver- sors were geometric now become algebraic. P Thus for g —ÿjGEOMETRIC MULTIPLICATION AND DIVISION. 79 and, Fig. 39, ba' -f- AB = AB + ba' = AAf ; also for qn= - and q'= a a Qn a1 # ¡3 af a a/ q1 a * a a B B* and ba + aa' = ba'. The product or quotient of any two complanar quaternions is therefore obtained by multiplying or dividing their tensors and adding or Fig. 39. subtracting their angles. Thus pq = Tp . Tq [cos (<£ + (9) + esin (<£ + $)]. If P = 0, ç2 = (Tg)2 (cos 2 <£ + e sin 2 <£), or, generally, gn=(Tg)n(cosn<£ + €sinw<£) . . . (100), whence result the following general formulae, T(qn) = (Tq)n u(V) = (Ug)n $TJ(gn) = cos nZq TYU (qn) = sin nZ.q , . . (101), which are all involved in Art. 42. 52. 1. Distributive and Associative Laws in Vector and Quaternion Multiplication. Having assumed (Art. 24) P ,y P + y - + -=------5 a a a80 QUATERNIONS. whence fia~l + ycT1 = (¡3 + y) cT1, since a is any vector, we have /3a + ya = (/3 + y) a. (a) Taking the conjugate of (/3 + y)a, K[(/? + y)a] = KaK(/? + y) [Eq. 99] = Ka(K/? + Ky). [Eq. 87] Taking the conjugate of (/3a + ya), K(/3a + ya) = K/3a + Kya = KaK¡3 + KaKy. Hence Ka (K/3 + Ky) = KaK£ + KaKy, or *'(£'+y0 = *7*'+“V. (fi) Hence, from (a) and (b), the multiplication of vectors is a doubly distributive operation, and (fi + y) (a + 8) = /?a + ya + /38 + yS (102). ff 2. Let g = y be any quaternion and a any vector ; also /3 a vector along the line of intersection of a plane perpendicular to a with the plane of g. Then another vector, S, may be found in the latter plane, such that g=-^, ^ having the same angle, plane 6’ ° ° and axis as y. Also let y be a vector in the intersecting plane, such that ï = a. If now a be any scalar, (a + a)g = (a+^)! = (^+^)§. _a/3 + y /3_a/? + y /3 * 8 8 “ 8+8~ 8+/3 8 = ag + ag.GEOMETRIC MULTIPLICATION AND DIVISION. 81 Taking the conjugates as above, g'(a'-f* a') = gfaf+ g'a'. Hence, in general, (a + a) (af+ af) = aa' + aa' + a'a + aa'; (c) or regarding a, a' and a, a' each as the sum of two scalars and two vectors respectively, ($!+ Ct2 + al + a2) 1 + Cl 2 "t" a i d- a 2) ^ (% + a2) (a1! + a 2) + (% + a2) (a'x + o/2) + (a'x + a'2) (ai+ a2) + (al + a2) (a*l + a2) = (&i + ax) (a 1 + a 1) + (cq + aT) (a 2 + a 2) + ( + 6) + e sin (<£ + 0) = (cos <£ + e sin <£) (cos 6 + e sin0) = cos cos 0 + c(sin <£ cos 0 + cos sin6) + e2 sin 0 sin sin0, sin(<£ + 0) = sin<£ cos0 + cos sin0. y a yGEOMETRIC MULTIPLICATION AND DIVISION. 87 in which ~ = cos (\j/ — 0) — €sin(i/r — Ö), P - = cos# + esind, a 7 ~ = cos ij/ — c sin t^, 7 and, as in the preceding example, cos(ijf — 6) = cosd cos if/ -f" sind sin \fr, sin (ij/ — 0) = cos# sin i/r — sin# cosi/r. 11. If a straight line intersect tico other straight lines so as to make the alternate angles equal, the tico lines are parallel. Let a and y (Fig. 44) be unit vectors along ab and cd, and fi a unit vector along ac. Then aj3 = — cos# + csind, fiy — — cosd — €sin0 ; whence afi — fiy = 2 Ya/?, and therefore, Equation (56), y = a. If a = ab' then a/? = cos d — c sin 0, /?y = — cos# — csin0, afi — fiy = 2 Sa/3 ; y = -a. [Eq. (55)] 12. If a parallelogram he described on the diagonals of any parallelogram, ¿/¿e area o/ the former is twice that of the latter. Let a and /? represent the sides as vectors ; then the diagonals are a + fi and a — /?, and V(a + 0) (a - 0) =V(j8a - a/3) = 2 Y fia, Fig. 44. since Ya2 =Yfi2 = 0 and —Yafi = Y/?a.88 QUATERNIONS. But, from the order of the factors, UV(a + /?) (a — ¡3) = UV/?a, hence TV (a + P) (a - P) = 2 TV/?a, which is the proposition (Art. 41, 7). 13. Parallelograms on the same base and between the same parallels are equal. Fig. 45. Hence We have (Fig. 45) BE = BA -f- AE = BA + ÆBC. Operating with V • bc X V(bc . be)=V(bc . ba), since Vícbc2 = 0. BC . BE Sin EBC = BC . BA Sin ABC, which is also true when the bases are equal, but not co-incident. 14. Fig. 46. If, from any point in the plane of a parallelogram, per- pendiculars are let fall on the diag- onal and the two sides that contain it, the product of the diagonal and its perpendicular is equcd to the sum, or difference, of the products of the sides and their respective per- pendiculars, as the point lies with- out or within the parallelogram. Let OA = a, Then But Hence OB = /?, op = p (Fig. 46). Vap+V/?p=V(a + /3)p. UVap = Ü Y/3p = UY (a + P)p. TVap + TYfip = TV (a + P)p.GEOMETRIC MULTIPLICATION AND DIVISION. 89 For p' =¿= op', we have UYapf= - UY/V= ± UY(a + j8)p'; .-. TVap'~TY/V= TY(a+ /?)¿ 15. If on any two sides of a triangle, as ac, ab (Fig. 47), ¿iüo exterior parallelograms, as acfg, abde, 5e constructed, cmcZ ¿/¿e sides ed, gf, produced to meet in h, then to ill the sum of the areas of the parallelograms he equal to that tohose sides are equal and parallel to cb and ah. Let AE = a, ab = /?, ac = y and AG = 8. Then AH = AE + EH = a — X/3, Operating with x Y . ¡3 Y (ah • ¡3 ) = Ya/3. (a) We have also AH = AG + GH =8 — yy- Operating with X Y. y Y(AH.y)=Y8y. (b) Fig. 47. D Hence, from (a) and (5), Yah(/3 — y) = Ya/3 —Y8y, ‘ Y (ah . cb) =Yap - Y8y = Ya/? +Yy8. These vectors have a common versor ; whence the proposition. If one of the parallelograms, as ad', he interior, then ae'= — a and ah' = — a — x'¡3 = 8 + y' y, and Y(ah'. £) = —Ya/?, Y (ah', y) =Y8y ; Yah.'(¡3 - y) = - Ya/3 - Y8y =Y/3a - Y8y.90 QUATERNIONS. But in this case IJY (ah' . cb) = - UY/3a = - TJYSy, and the area of the parallelogram on ah', cb, is the area of af minus the area of at>. 16. To find the angle betiueen the diagonals of a parallelogram. Let AD = BC = a (Fig. 48), and ba = cd = /?, d and d' being the tensors of the diagonals. Then AC • db = — (a — ¡3) (a + ¡3) = -a2-(a/3-/3a) + f? = — a2 — 2 Ya/3 + /32. dd’= a2 - b2. dd'—2ab sin0, /. tanDoc = -tanrf, = M^. v a2 — b2 17. The sum of the squares on the diagonals of a parallelo- gram equals the sum of the squares on the sides. In Fig. 48 BD2 = (a + £)2 = a2 + 2Sa£+/32, CA2 = (/? — a)2 = /?2 — 2 Sa/2 + a2 ; /. CA2 + BD2 = 2 a2 + 2 /52, or BD2 -f CA2 = BA2 + AD2 + DC2 + CB2. 18. The sum of the squares of the diagonals of any quadri- lateral is twice the sum of the squares of the lines joining the middle points of the opposite sides. Fig. 48. Taking the scalars cos doc • Taking the vectors sin doc . since UY(ac . db) = — UYa/LGEOMETRIC MULTIPLICATION AND DIVISION. 91 Let ab =: a, ad =/?, DC = y (Fig. 49). For the squares of the diagonals, we have Whence the proposition readily follows. 19. The sum of the squares of the sides of any quadrilateral exceeds the sum of the squares on the diagonals by four times the square of the line joining the middle points of the diagonals. Let AB = a, AC = /?, AD = y (Fig. 50). The squares of the Fig. 50. sides as vectors are d (/? + y)2 + 03-a)2, Fig. 49. and for the bisecting lines HP+y—i(fi+y—'a)]2 + [/?+á-y—ì*]2- A F a or 2 (a2 + /?2 + y2) - 2 s/?a - 2 Sy/?. The squares of the diagonals are or /32 + (y -«)*, )02 + y2 + a2 — 2 Sya. The former sum exceeds the latter by 92 QUATEKNIONS. But 7. = ao, and — 2 5 2 we obtain sa. Substituting these values, 4(ao + sa)2, or 4so2, which is also true of the vector lengths. 20. In any quadrilateral, if the lines joining the middle points of opposite sides are at right angles, the diagonals are equal. With the notation of Fig. 49, we have FE . GH = [£(y — Now * (Æ + y) = cy — y (y + <*)• Operating with Y • (y + a) x Hence x cYay VyP+Vap+Yay AO = æ (/? 4~ y) = ______cYay Yy/3 + Yafi +Yay (fi+y) 9 or, since a, /?, y are unit vectors, AO = - c smB sin a + sinB + sine 08+ y). Squaring, to find the length of ao, we have, since (P + y)2=s 2 (1 -j- cos a) , — ao2= AO = [i esms sin a + sinB 4-sinc c sinB sin a -f sin b + sin c ______c sinB______ sin a + sinB -f sine “ 2 2 (1-f cosa), V2 (14- cosa), 2 cos-Ja. 25. If tangents be drawn at the vertices of a triangle inscribed in a circle, their intersections ivith the opposite sides of the triangle will lie in a straight line.96 QUATERNIONS. Let o be the center of the circle (Fig. 54) whose radins is r, and OA = a, on = /?, oc = y. Since oa and ap are at right angles, S(oa . ap) = 0. But AP = AB + BP = AB + yBC = /3 — a + y(y — /3) ; Fig- 54. hence, substituting this value above, Sa[/3 — a + y(y — /3)] = 0, Sa2 = Sa/3 + ï/S(ay-a/3), and r» + gqff . Say — Sa¡3 ’ Therefore OP = OB + BP = B + 1JBC = B-? (y — R) H H Say-Sa/T7 P> _ (r2 + Say)^-(r2 + Sa^)y Say — Sa/3 Similarly, or, by a cyclic change of vectors, 00 _ (f + Saô)y — (r2 + S/?y)a °Q- g ’ -r _ 0* + Sgy)q ~(r2 + Say)B Sfiy — Say Whence (Say — Sa/3) OP + (Sa/3 — S/?y)oQ + (S/?y — Say) OR = 0. But also (Say — Sa/5) + (Sa£ — S/3y) + (S/3y — Say) = 0. Hence p, q and r are collinear. 26. The sum of the angles of a triangle is two right angles.GEOMETRIC MULTIPLICATION AND DIVISION. 97 Let a, ¡3, y be unit vectors along bc, ca and ab (Fig. 55). Then (Art. 42) ' ' UMov SS 2 Hence -(<£ + 6 + *A) = an even multiple of 2 (Art. 42), as 2 n, 7r as we go round the triangle n times. In taking the arithmetical sum, or passing once round, we take the first even multiple of 2, or ?(<¿,+(9 + ^) = 4; 7T (fi 0 ifi *— 2 7T, and the sum of the interior angles is 3 7r —• 2 7r = 7r, or two right angles. 27. The angles at the hase of an isosceles triangle are equal to each other. Let a and (3 (Fig. 56) be the vector sides Fis* 56- of the triangle, and Ta = T/L Then, if the proposition be true, = K P or a — /3 ¡3 — cl a(a - ¡3) “x = K0O3 - a) 08 - a) a(/3~a) = (a — f3)(3\ .-. a2 = /32, which is true, since Ta = T/L98 QUATERNIONS. 28. To find a point on the base of a triangle such that, if lines be drawn through it parallel to and limited by the sides, they will be equal. Draw de (Fig. 57) and df parallel to the sides. From similar triangles, if AE = ÆAC, X ■■ Fig. 57. AE _ FB AC AB AB — AF whence Now 1 — æ = AF AB AD = AF + FD, or, since fd = ae, : (1 — fl?) AB + XAC. But, since fd is to be equal to ed, and therefore (1 — #)Tab = xTac = y ; (1 — æ)Tabüab = 2/Uab, æTacüac = 2/Uac, ad = y( Uab + Uac) , and d is on the angle-bisector. 29. If any line be drawn through the middle point of a line joining two parallels, it is bisected at that Fig. 58. point. G H 30. If the diagonal of a parallelogram is an angle-bisector, the parallelogram is a rhombus. 31. In any triangle the sum of the squares of the lines gh, ke, df (Fig. 58) is three times the sum of the squares of the sides of the triangle. 32. The sum of the angles about two right lines which intersect is four right angles.GEOMETRIC MULTIPLICATION AND DIVISION. 99 33. If the sides of any polygon he produced so as to form one angle at each vertex, the sum of the angles is four right angles, 34. Find the eight roots of unity (Art. 39). 35. The square of the medial to any side of a triangle is one- half the sum of the squares of the sides which contain it, minus one-fourth the square of the third side, 55. Product of two or more Vectors. 1. Let q = a/3, r = y, Then, since Sgr = Srg, Sa/3y = Sya/L Let q = ya, r = j3, Then 8qr = $?V = Sya/3 = S/3ya ; .-. Sa/3y = S/5ya = Sya/3..............(108), or, the scalar of the product of three vectors is the same if the cyclical order is not changed. This may also he shown by means of the associative law of vector multiplication as follows : afiy = (a/?) y = (Sa/3 +Ya/3)y. Taking the scalars Sa/?y=S(Sa/3+Vo¡S)y = S(Yafi . y), since S(Sa/2 . y) = 0, = S . yYaj3 ; introducing the term S • ySa/3 = 0, = S • yYa/3 + 'S . yS a/3 = S • y (Sa/? +Vaj8) = Sy(a/?) = Sya¡3,100 QUATEKmONS. In a similar manner Sa/3y = S . a(S/?y+Y/?y) = S . aYySy = S(V0y .a) = s(v/3y+ S/?y)« = S/3ya, and, as before, Sa/?y = S/?ya = Sya/?. 2. Again Sa/?y = S . a(S/3y + Y/3y) = S . aY/?y = - S . aYy/3 = — Sa(Yy/3 -f- Sy/3) 5 Sa/3y = — Say/3 ..•••• (109), or, a change in the cyclical order of three vectors changes the sign of the scalar of their product. 3. Resuming a/3y = a(/?y) and taking the vectors, Ya(3y = Y . a (S/?y +Y/?y) = aSf3y +Y • aV/3y- Also Yy/3a — Y (8y/3 + Yy¡3) a = Y. aSy/3 —Y • aYy/3 = Y . aSy/?+Y. aY/2y = Y • a(Sy/3 + Y/3y) = aS/3y + Y.aY/5y; Yafîy = Yyf3a......................(110), or, the vector of the product of three vectors is the same as the vector of their product in inverted order. 4. Geometrical interpretation of Saj3y. Let a, /?, y be unit vectors along the three adjacent edges oa, ob, oc (Fig. 59) of any parallelopiped, 0 being the angle be-GEOMETRIC MULTIPLICATION AND DIVISION. 101 tween a and ¡3, and 0' the angle made by y with the plane aob. Then a¡3 — — cos0 + * sin#, e being a vector perpendicular to the plane aob. Operating with X S . y Sa¡3y = S( — COS 0 + esill0)y = S(sin 0 • ey). But Sey = — cos of the angle between e and y = — sin 0' ; .\ Sof3y = — sin 0 sin 6'. Fig. 59. Now, if a, /?, y represent as vectors the edges oa, ob, oc, whose lengths are a, 6, c, Sa/?y = — TaTyOTy sinO sin 0' = — abc sin 0 sin 0[ But ab sin 0 = area of the parallelogram whose sides are a and &, and csinfl' = perpendicular from c on the plane aob. Hence — Sa/2y = volume of a parallelopiped ivhose edges are a, b and c, drawn parallel to a, ¡3 and y. Cor. 1. Whatever the order of the vectors, the volume is the same ; hence, as already shown, ± Sa/?y = ± S/?ya = ± Syaf3 = ^ SayyS, etc. Cor 2. If Sa/3y = 0, neither a, y8, nor y being zero, then either 0 = 0, or 6f = 0, and the vectors are complanar. Cor. 3. Conversely, if a, /?, y are complanar, Sa/?y = 0. Cor. 4. The volume of the triangular pjTamid of which the edges are oc, ob, oa, is — £ Sa/3y. 5. We have seen that when a, f3 and y are complanar, Sa/2y=0, and therefore a¡3y is a vector. To find this vector, suppose a102 QUATERNIONS. triangle constructed whose sides ab, bc, ca have the directions of a, fi and y respectively, a vector not being changed by motion parallel to itself. Since the tensor of the vector sought is the prod- uct of the tensors of a, ¡3 and y, we have to find U (ab . bc • ca) , ¿.e., its direction. Circumscribe on the triangle abc a circle and draw a tangent at a, represented by t'at. Since the angles tab and bca are equal, we have whence U(bc • ca) = U(ab • at') [= U(ba . at)]. Introducing Uab x U(aB . BC • ca) = U(aB . AB . ATf) [ = U(AB . BA • Aï)], or, since U(ab • ba) = — (U. ab)2= 1, U (ab . BC . ca) = — U . ATf = U . AT. Hence, if a, b, c are any three non-collinear points in a piane, or if a, (3) y are the sides of a triangle joining them, in order (in either direction, since Ya(3y = Yy/3a), a/?y, /?ya, ya/3 are the vector tangents to the circumscribing circle at the angles of the triangle. Again, if a, B, c are any three points in a plane, not in a straight line, and a and ¡3 are two vectors along the two succes- sive sides ab, bc of the triangle which they determine, and cd a vector drawn from c parallel to y, intersecting the circumscribed circle at d, then is da parallel to Ya¡3y = 8. For S = a/?y = aj?/3y = a/32/3_1y = — (T/?)2a/?_1y = - (T£)2|y, whence U . which turns ¡3 parallel to — a, turns y into a direction 8 = da, the opposite angles of an inscribed quadrilateral being supplementary.GEOMETRIC MULTIPLICATION AND DIVISION. 103 If y have a direction such that cd crosses ab, or the quadri- lateral is a crossed one, it is evident on construction of the figure that _ y 6 U8 = Ua/2y = U (ad) = — US. Hence the continued product of the three successive vector sides of a quadrilateral inscribed in a circle is parallel to the fourth side, its direction being towards or from the initial point as the quadrilateral is uncrossed or crossed ; and, conversely, no plane quadrilateral can satisfy the above formula ± U8 = Ua/3y, unless a, B, c and i) are con-circular. The continued product of the four successive sides of an inscribed quadrilateral is a scalar, for a/3y8 = (a/?y)8 = ± S2 = d2. Since the product of two vectors is a quaternion whose axis is perpendicular to their plane, while the product of a quaternion by a vector perpendicular to its axis is another vector perpen- dicular to its axis, and so on, it follows that the continued product of any even number of complanar vectors is generally a quaternion whose axis is perpendicular to their plane, while the product of any odd number of complanar vectors is a vector in the same plane. Hence the formulae Sd=0, Sa/3y=0, Sa/?ySo-=0, etc., for complanar vectors. If, however, the given vectors are parallel to the sides of a polygon abc....MN inscribed in a circle, then U(ab • BC . CD MN . NA) = U(ab*. BC . CA) U(AC . CD . DA) X U (am . MN . na) . But each of the products U (ab • bc . ca) is equal to U • at, at being the tangent to the circle at a. Hence U(AB . BC • CD....MN . NA) == (U . AT)n, which reduces, according as n is even or odd, to ±1 or ±U . at. Hence the product of the vectors will be a scalar or a vector104 QUATERNIONS. according as their number is even or odd, and in the latter case this vector is parallel to the tangent at a. If the vectors are not complanar, but parallel to the successive sides of a gauche polygon inscribed in a sphere, the polygon may be divided as above into triangles, for each of which the product of the three successive sides is a vector tangent to the circumscribing circle, all these vectors lying in the tangent plane to the sphere at the initial point. If the number of sides is even, their product will be a quaternion whose axis is perpendicular to the tangent plane, i.e., lies in the direction of the radius of the sphere to the initial point ; if odd, the product is a vector in the tangent plane. Hence, if a, b, c and d are four given points, not in a plane, ab = a, BC = /?, cd = y being given vectors, and p any other point such that dp = o-, pa = p, if p lies on the surface of a sphere through the four given points, we have the necessary and sufficient condition a/3y(Tp = pcry/3a, for each member is equal to minus the conjugate of the other, and must therefore (Art. 46) be a vector. 6. From Equation (56), Py — y/3 = 2 Y£y. Operating with T.oX 2Y. aV/3y = Y . a(fiy-yP). Introducing in the second member pay — Pay, =V (aPy — ayP + /3ay — fiay) = Y(a/Î + Po.) y — Y (ay/3 + ya/?) = V. 2(Sa£)y-Y(ay+ya)/î = 2 ySa/3 — 2 fîSay. Y . aV/?y = ySayS — /8Say .... (111). HenceGEOMETRIC MULTIPLICATION AND DIVISION. 105 This formula may be extended. Thus, for a write YaS, and we have Y . YaSY/fy = yS(YaS)/? - £S(YaS)y, Y.YaSY/?y = ySa8/?--/3Sa8y . . . . (112). An inspection of this formula shows that it gives a vector complanar with y and ¡3. Moreover, since Y . YaSY/?y = Y . Yy/3Ya8 = 8Sy/?a - aSy/?8, it is also complanar with a and 8, and is, therefore, parallel to the line of intersection of the planes of a, 8, and /?, y. Similarly Y . Y/3yYaS = 8S£ya - aS£y8 = - Y . Ya8Y/3y . (113). Adding Equations (112) and (113) 8S/?ya — aS/?y8 + ySaS/3 — /2SaSy = 0 . . (114), or 8Sa/5y = aS/?y8 — /3 Say8 + ySa/?8 . . . (115), a formula expressing a vector 8 in terms of any three given di- planar vectors, a, /?, y ; so that, if S/3y8 = 5, — Say8 = Sya8 = c, Sa/58 = a, Sa/?y = m, 8 = (ba -j- c/3 -j- , or, since Tp = T/?, cosò = cos. Similarly Yap = Y/?a, and sinò = sin . Hence 0 = & and a bisects the angle between /? and p. 19. Show that p = a/?a_1 == a-1 (Sa/? — Ya/?). 20. p being any vector, show that Y . YapYp/? = xp. 21. If Sa/? = — a2, show that a is perpendicular to /? — a. /? 0 22. What are the relative directions of a and /?, if K- =---? no a a108 QUATERNIONS. 57. Examples. 1. The altitudes of a triangle intersect in a point. Let (Fig. 60) ac = /?, CB = a, ab = y. Fig. 60. Then vectors along c'c, b'b and a'a are Operating with x S . /?, we have, since y$e/32 = 0, M. S- CM = AM — AC = i/3 — £y + i J^|ey> Sey/3 ma'= AA'- AM = 1^8 --Jy + |^ ey. »erp .*. ma' = 2cm, and, since, as vectors, they are multiples of each other, and have a common point, they form one and the same straight line. 4. To find the condition that the perpendiculars from the angles of a tetraedron to the opposite faces shall intersect. With the notation of Fig. 52, the perpendiculars from a and b on the opposite faces are Y/?y and Yya. If the}7 intersect, at p say, then must a, b, p lie in one plane. Hence, Art. 55, 4, Cor. 3, S [(/? — a) Y^yVya] = 0,GEOMETRIC MULTIPLICATION AND DIVISION. Ill or S (¡3 — a) [S • YfiyYya + Y • YpyYya] = 0, S (¡3 — a)V . YfiyYya = 0. But, Equation (117), Y • YfîyYya = — yS/?ya ; . *. — (S/?y — Say) S ¡3ya = 0, or S/3y = Say. (a) From the figure, we have BC2 + OA2 = (y — P)2 + a2 = y2-2Sy/3 + /? + a2 or, from (a), = y2 — 2 Say + /32 + a2 = (y-a)2 + i32 = AC2 + OB2. Hence the condition is that the sums of the squares of each pair of opposite edges shall he the same. 5. Interpret Equation (118), SSa/3y = Y/?ySaS + YyaS/?S + Ya/3Sy8, under the condition that a, p, y be complanar with 8. If a, /?, y are complanar, Sa/3y = 0, and therefore, 8 being in or out of the plane, SaSY/?y + S/2SYya + SySYa/2 = 0. (a) If 8 be in the plane, we have for any four co-initial lines OA, OB, oc, OD, sin BOC COS AOD + sin COA COS BOD + sillAOB COS COD = 0, and, for a line perpendicular to od, sin boc sin AOD + sin coa sin bod + sinAOB sin cod = 0. If 8 is perpendicular to the plane, the terms in (a) vanish separately. Fig. 52 (Ms). C112 QUATERNIONS. 6. If X) Y, Z be the angles made by any line op loith three rectangular axes, then cos2 X + cos2 Y + cos2 Z=l. From Equation (67) ip == xi2 + yij + zik = -x + ylc-- zj, whence (Sip)2 = X?. Operating in a similar manner with S *j X and S . Jc x we obtain -^=(sw*+ ($/,)*+(s**)*. If Tp = r, then p2 = — r2, $¿p = — r cos X, etc. Hence op2 = op2 (cos2 X + cos2 Y + cos2 Z), or cos2X+cos2F+cos2^= 1. Applications to Spherical Trigonometry. Let abc (Fig. 63) be any spherical triangle on the surface of a unit sphere whose center is o ; a, /?, y afta ft a or cos & = cosa cose + sin a sine Sa'y', in which Sa.y = — cos ö' = cos b. cosb = cosa cose -f- sina sine cosb. Similarly, or directly by cyclic permutation in (c), (p) (c) cos c = cos b cos a -f- sin b sin a cos c. From the relation ft1 ft1 a' 7 “ 7 7 may be deduced in like manner — cosa = cose cosb — sine sine cosa.114 QUATERNIONS. 8. Resuming the equation 7 « r of the last example, and taking the vectors, we have [Equa- tion (91)], But V- = S-V- + S-Y- -f V . Y-V-* y ay y a r ay TP (a) Y- = — a/sin a, 7 B a S-Y- = cosc(^fsin&)=coscsin& • /?{ a B S-Y - = cosò(y,sinc) = cos&sinc . yj y a Æ a Y • Y-Y-=s Y(y'sine) (/3'sin&) = sinc sin 6 Yy'/?' a y = sin c sin 6 (—a sin af ) = — sin c sin 6 sin a . a. Substituting in (a), —sin a • a/= cose sinò • /?'+cosö sine « yf—sine sin & sin a • a. (&) Operating with X $ . yf_1, af B’ y' a — sina . S-= cose sinôS^+cosô sine SA—sine sinö sinAS-ij 7 7 7 7 in which S —, = cosò' = — COSB, 7 $■—, = — COSA, 7 Hence S —, = 0, since a and y1 are at right angles, sin a cos b = cos b sin c — cos c sin b cos a,GEOMETEIC MULTIPLICATION AND DIVISION. 115 and in the same manner, or by a cyclic permutation of the letters, sinò cose = cose sina — cosa sine cos b, sin c cos a = cos a sin ò — cos ò sin a cos c. 9. Operating on Equation (ò) of the last example with X Y . y'-1 instead of X S . yf_1, —sinaY^ = cose sinòY—, + cosò sine Y^ — sine sinò sin a Y—,- But V—i = fisinb' = /?sinB, S' Y — a sin a ' = — a sin a, r v^=o. y Fig. 63. C' Substituting these values — sin a sin b . ft = — cos c sin ò sin a . a a — sin c sin ò sin A . Y~; yi Operating with x a"1, and substituting for £ - = cos c + yf sm c, we obtain or — sina sins cose — sina sinB sine • yf = — cose sinò sinA — sine sinò sin a • y. Equating the scalar or vector parts, we have in either case sina sinB = sin a sinò, sina : sinò : : sin a : sinB. The formulae of the preceding examples have all been deduced from the equation ~ ^ -• The product as well as the quotient may also be employed, as follows :116 QUATERNIONS» 10. Assuming the vector product Ya/3Y/?y, and taking the vector part, we have [Equation (117)], Y . YapYPy = - pSafiy. (a) But Y . Ya/3Y/fy = Y(y'sinc) (a'sina) = sine sina sinB . ¡3, and, Art. 55, 4, Sa/?y = — sine sin0j 0' being the angle made by oc with the plane of c. Substituting in (a), sine sin a sins • ft = sine sin0'. /?, Fig. 63. & or sin#' = sin a sìub. By permutation, from (a), Y . YyaYa/3 = - aSya/3 = - aSa/3y, or sinb sine sinA . a = sine sin#' • a, sin 0' = sin b sin a. Equating these values of sin#' we have, as in Example 9, sin a : sin b : : sin a : siiiB. 11. Let pa, _p6, pc represent the arcs drawn from the vertices of abc perpendicular to the opposite sides. Resuming Equation (a) of the preceding example, and taking the tensors, TY • YafiYfiy = Sa/?y = sine sinpc, = S/?ya = sin a siny>a, = $ya/? = sin 5 sin_p6,GEOMETRIC MULTIPLICATION AND DIVISION. 117 and, taking the tensor of Y • YafiYfiy from the last example, sine sin a sins = sin a sinjpa = sin& sinp6 = sine sinp0, or smpa = sine sinB, sin c sin a . smp6 =-----------— sinB, sino sinpc = sin a sinB. 12. Show that if abc, a'b'c' be two tri-rectangular triangles on the surface of a sphere, cosaa' = cosbb' cosce' — cosb'c cosbcJ the triangles being lettered in the same order. Let a, /?, y, a¡ y' be the vectors to the vertices. These being at right angles, in each triangle, we have COSAA' = - Saa' = - S . YySyY/5'y! or, Equation (122), cosaa' = S/2/3'Syy' - S/3'yS/ty = cos bb' cos cc' — cos b'c cosbc'. [The vectors of Equation (122) are arbitrary, but we may divide both members by the tensor of the product of the vectors, so that S(YUa/3YUy8) = SUaSSU/?y - SUaySU£8, for the unit sphere.] 13. Let ABCD be a spherical quadrilateral whose sides are ab = a, bc = 6, cd = c, da = d, the vectors to the poles of these arcs being a, y\ 8' respectively. Then Ya/3 = a'sin a, YyS = y'sine.118 QUATEBNIONS. From Equation (122), S • Ya/3YyS = SaSS^y — SayS^S, or sin a sin c Sa'y11 = ( — cos da) ( — cos bc) — (—cos db) (—cos ac) . But Safyf = — cosL, L being the angle formed by the arcs ab and cd where they meet, the arcs being estimated in the directions indicated by the order of their terminal letters. Hence sin ab sin CD cosl = cosac cosbd — cos ad cos BC, a formula due to Gauss. 14. Retaining the above notation, abcd being still a spherical quadrilateral, denote the angles at the intersections of the arcs ab and cd, ac and db, ad and bc, by l, m and n respectively. Then, from Equation (125), S[Ya£YyS + YayY8/3 + YaSY/?y] = 0, we have identically sin ab sin cd cosl + sinAC sinBD cosm + sin ad sinBC cosn = 0. Were the points a, b, c, d on the same great circle, the angles L, m and n would be zero, and the above reduces to sin ab sin cd + sinAC sinBD + sin ad sinBC = 0, and for a line oaJ perpendicular to oa and in the same plane, dropping the accent, we have COS AB sin CD + COSAC shlBD + COS AD SHIBC = 0, which are the results of Example 5 of this article.GEOMETRIC MULTIPLICATION AND DIVISION. 119 58. General Formulae. 1. We have seen, Equation (86), that SS = SS and YS = ST ; but (Art. 50, 4) that ST is not equal to TS, nor SU to US. We have also seen, Equations (96) and (97), that TII = IIT and Uil = nU ; but SII is not equal to IIS, nor YII to IIY : for, 1st, SII is independent of the factors under the II sign, provided the product remains the same, while IIS is dependent upon them ; and, 2d, because (Art. 55, 5) IIY is not necessarily a vector. 2. Resuming Equation (92), S rq = Sgr, and, since r is arbitrary, writing rs for we have, by the asso- ciative law (Art. 52), S(rs)q = $q(rs), $r(sq) = S (sq)r, Srsg = Ssgr = Sgrs .... (126), a formula which may evidently be extended. Hence, the scalar of the product of any number of quaternions is the same, so long as the cyclical order is maintained. 3. Let p, g, r, s be four quaternions, such that qr=ps. (a) Operating with Kg x , Kg . gr = (Kg . q)r = (gKg)r = Kg . ps, since conjugate quaternions are commutative. Hence (Tg)2r = Kg .ps, or r = = = ■ • • (127). Operating on (a) with xKr, we have qr . Kr = ps . Kr,120 QUATERNIONS. or q(Tr)2 = psJLr ¡isKr, 1 psRr =ps- . . . (128). ••• 2 (IV)2 Hence, in any equation of the products of tivo quaternions, the first factor of one member may be removed by writing its con- jugate as the first factor of the second member, and dividing the latter by the square of the terisor, or simply by introducing the reciprocal as the first factor in the second member. By substi- tuting the word last for first, the above rule will apply to the second transformation. 4. Resuming, for facility of reference, the equations 5. It has been already shown (Art. 54, Fig. 40) that (Ta)2 + (Tp)2= (Ty)2, and (Art. 54, Fig. 42) that Ta=Ty. cos<¿>, T/? = Ty. sin ; and therefore (Ty)2 COS2 + (Ty)2 sin2 4* = (Ty)2, or », sin2<£ -f- cosz<£ =1- Hence, from Equations (44), = — (cos<¿> + esin<¿>) = Ty. Uy = Sy + Vy, (J.) (CO (B) we observe directly that Sy = S(Ty.Uy) = Ty. SUy . Vq = TV?. UVy = Ty. VUy . » TYy = Ty . TVUy = TVKy . . (129) , (130) , (131) . (SUy)2 + (TVUy)2 = 1 .... (132).GEOMETRIC MULTIPLICATION AND DIVISION. 121 This important formula might have been written at once by assuming the above well-known relation of Plane Trigonometry. 6. From Equations (129) and (131), we may write Equa- tion (132) under the form (Sg)2 + (TYg)2 = (Tg)2...................(133), or, from Equation (107), (Sg)2 - (Yg)2 = (Tg)2 = (S g)2 + (TYg)2 . (134), since c2 = —1. 7. Comparing (^1), (B) and ((7), SUg = SU^ = SUKg . . . (135), TYUg = TVU- = TYUKg . . (136), and from Equations (129) and (135), 1 Sg = Tg. SUg = Tg. SU- = Tg. SUKg. . (137). 8. Since Tg = TKg, we have Tg • TKg == (Tg)2.............(138), and Tg being a positive scalar, KTg = TKg................(139). As exercises in the transformation of these and the following symbolical equations, some of the results already obtained will be deduced anew. Thus, to prove that T(gg') == TgTg', whence T . g2 = (Tg)2, we have (Tgg')2= (gg')K(gg') Equation (107) = gg'Kg'Kg Equation (99) = g(g'Kg')Kg = (Tg')2gKg = (Tg')2(Tg)2, .*. Tgg' = TgTgi122 QUATEKNIOHS. 9. Substituting for Sg and TYg their values from Equations (79) and (131) (SKg)2 + (TVKg)2 = (Sg)2 + (TVg)2 . 10. Resuming from Art. 51, 1, the expressions Yrq = SrVg + SgVr + Y. VrVg, Vqr = SgYr + SrYg + V. YgYr, Sgr=SgSr+ S . VgVr, we have, by adding and subtracting, Y gr + Y rq = 2 SgYr + 2 SrYg "| Vgr - Yrg = 2 V. YgYr J And, if g = r, from (a) and (c), whence V. g2 = 2SgVg S . g2 = (Sg)2 + (Yg)2 (140). (a) (b) (c) (141) . (142) , (143) . (144) , (145) . g2 = (Sg)2 + 2 SgYg + (Yg)2. . Dividing Equations (142) by (Tg)2 SU . g2 = (SUg)2 + (VU?)2| YU . g2 = 2 SUg . VUg ) since, evidently, S . g2 = (Tg)2SU . g2| V. g2 = (Tg)2 VU. Again, substituting in the second of Equations (142) the value of (Yg)2 from Equation (134), we have S . g2 = 2(Sg)2-(Tg)2.............(146), and dividing by (Tg)2 SU . g2=2(SUg)2-l..................(147). Substituting (Sg)2 from the same equation S. g2=2(Vg)2 + (Tï)2 • • • • • (148).GEOMETRIC MULTIPLICATION AND DIVISION. 123 Equations (146) and (148) may be written (g-j_T)g2= 2(Sg)2 and (S — T)g2 = 2(Yg)2. Introducing in (a), or (5), the condition that q and r are complanar, we have, after substituting versors, YUgr = YUgSUr + YUrSUg, since, under the condition, Y(YUgYUr) = 0. Taking the tensors, since q and r are complanar, TYUgr = TYUgSUr + SUgTYUr . . . (149), and, interpreting, Art. 51, 6, sin (0 + <£) = sin 0 cos <£ + cos 0 sin . Introducing the same condition of complanarity in (c) Sgr=SgSr-TYgTYr, or, substituting versors as above, SUgr = SUgSUr — TYUgTYUr . . . (150), or, interpreting, cos (6 + ) = cos 0 cos — sin <£ sin 0. 11. Putting Equation (146) under the form í S . g2 + T • g2 S<7 = \-----2----’ and writing Vq for q, we have SV5 = ViW+W) • • • • (151). 12. Taking the tensors of the first of Equations (142), we have TV . or M'*=y, pP = yi and therefore Kp • y = Tip . p/3 = (Kp . p)/3 = (Tp)2/?. Now (KgKp)y = Kg(Tp)2/? = (Tp)2Kg . ¡3 = (Tp)2Kg . ga = (Tp)2(Tg)2a = (Tpg)2a = Kpg . pq . a = Kpg . y .*. Kpg = KgKp, which, by the Associative law, gives Kn = n'K.................(158). 14. Show that K(— g) = — Kg. 15. Show that T(p + q)2 = (p + g) (Kp + K g) = (Tp)2 + (Tg)2 -f- 2 S . pKg = (Tp)2 + (Tg)2 + 2TpTgSU . pKg = (Tp + Tg)2 — 2TpTg(l—SU . pKg), and therefore that T(p + g) cannot be greater than the sum or less than the difference of Tp and Tg. 16. Show that gUYg"1 = TVg - SgUYg. 59. Applications to Plan© Trigonometry. 1. For formulae involving 20, let Then g = Tg(cos20 + €sin20). Vg = g' = VTg(cos0 + €sin0).126 QUATERNIONS. From Equation (142), S . #2 = (S#)2 + (V#)2, we then have sg = (Sg')2+(W> or, dividing out Tg, and, interpreting, SUg = (SUg')2 + (vu?')2; cos 20 = cos2 6 — sin2 0. Again, from Equation (147), SU . — esin<£), q and r being complanar, we have TYUgr"1 = TYUgSUr — SUgTYUr . . (159), SUgr'1 = SUgSUr + TYUgTYUr . . (160), or, interpreting, sin (0 — <£) = sin 0 cos c/> — sin cos0, cos (0 — <£) = cos 0 cos <£ -f- sin0 sin <¡>.GEOMETRIC MULTIPLICATION AND DIVISION. 127 3. Adding Equations (149) and (159), TYUgr + TYUgr-1 = 2 SUrTYUg, in which, if gr = jp, qr~1 = t, g = Vp£, r = Vp£_1(Art. 58, 3), TVUp + TVUí = 2SU(V¿F1)T^u(V^) . (161), or sino; + sin y = 2 cos^(æ — y) sin^(æ + y). Similarly, by subtracting the same equations, TYUgr - TYUgr-1 = 2 SUgTYUr, TVUp-l'VUi = 2SU(Vpi)TVU(VpF) . (162), or sinæ — siny = 2cos|-(a; + y) sin|-(æ — y). 4. From Equations (150) and (160), by addition and sub- traction, we obtain, in a similar manner, SUp + SUi = 2SU(V#)!SIJ(Virri) . . . (163), and ___ SUp — SUi = — 2 TVU ( VpO TVU ( VpF1), whence cos# + cosy = 2cos^-(# + y) cos|-(# — y), cosy — cosa? = 2 sin £(x + y) sin i(# — y). 5. Resuming Equation (152), TVVg = V|(Tg — Sg), it may be put under the form 2(TVUVg)2 = l — SUg, or 2sin2i0=l —cos0. and, in a similar manner, from Equation (151), sVg = V i(Sg + Tg), 2(SUVg)2 = SUg + 1, or 2cos2^0 =1+ cos0.128 QUATERNIONS. 6. From Equation (142) or (tv: S)g2 = 2 SgTVg (8q)*+(jQ)* 2T Vg (Sg)2 ~w 2(TV : S)g 1— [(TV : S)gJ2’ tan 2 # 2 tan# 1 — tan2# And, in a similar manner, cot 20 = cot2# —1 2 cot# 7. From Equations (90) and (91), g and r being complanar, Sqr = SgSr + S • YqYr = SgSr - TYgTYr, TY qr = SgTYr + SrTY g, we have, by division, or Also or (TY: S)qr = SgTYr + SrTYg SgSr — TYgTYr (TY: S)r + (TY: S)q 1-(TV: S)g(TV : S)r tan(# -}-) = tan cj> + tan # l-~tan<£ tan# (TY : S)gr"1 (TY: S)g-(TY: S)r 1 + (TV : $)g(TV : S)r’ tan(# —-<£) = tan # — tan eft 1 + tan# tan ± (TV: S)t _ TYpSt ± TYtSp _ TVUpSUi ± TVÜíSUp SpSi SUpSUi Hence, from Equations (149) and (159), (TV : S),» ± (TV : S) i = . sin (x ± y) tana? ± tan y =------*---— • coso? cos y or By a similar process, coto? ± cot y _ sin (y ± x) sino? sin y 9. From Equations (161) and (163) TYUp + TYUi TVUVpi = ^ ■» SUVpi = whence 2 suVpr1 SUp + su< 2SUVpF1’ or or (TVU : SU) Vpt = (TV : S) Vpi= ^ w . \ sino? -b sin y tan|(æ + y) =------------ v coso? + cosy And, in a similar manner, from Equations (162) and (163), /f_T m ri=i TYUp-TYUi (TV : S) Vpi = SUp + SUt > tan i(x — y) =-----:----- coso? + cos y130 QUATERNIONS. 10. Similar formulae may be deduced for functions of other ratios of an angle. Thus, from Equation (90), writing rs for and making q = r = s all complanar, we have, by Equation (142), S . (f — (Sg)3 — 3Sg(TTg)2, or cos30 = cos30 — 3cos0 sin20, or, under the more familiar form, cos 30 = 4cos30 — 3cos0.CHAPTER III. Applications to Loci* 60. Any vector, as p, may be resolved into three component vectors parallel to any three given vectors, as a, /?, y, no two of which are parallel, and which are not parallel to any one plane. Thus p = xa + y/3 + zy...............(164) refers to any point in space. If the variable scalars a?, y, 2 are functions of two independ- ent variable scalars, as t and u, p is the vector to a surface, which, if the functions are linear, will be a plane. We may, therefore, write p = $(t,u)...................(165) as the general equation of a surface. If a?, y and % are functions of one independent variable scalar, as p is the vector to a curve, which, if the functions are linear, becomes a right line. We may, therefore, write P = Ht)...................(166) as the general equation of a curve in space. If a, /?, y are complanar, we may replace either two of the vectors in Equation (164) by a single vector, in which case p = contains but two variable scalars, functions of £, and is the equation of a plane curve, or of a straight line if the func- tions are linear. The essential characteristic of the various equations of a straight line is that they are linear, and involve, explicitly or implicitly, one indeterminate scalar. 131132 QUATERNIONS. 61. Assuming 9 = ssa + y fa (a) in which X and y are variable scalars, functions of a single vari- able and independent scalar, as £, as the general form of the equation of a plane curve, by substituting in any particular case the known functions x = f(t), y = f’(t), or x = f"(y), we may avail ourselves of the Cartesian forms and apply to the resulting function in p the reasoning of the Quaternion method. For example, suppose a and /3 are unit vectors along the axis and directrix of a parabola, the origin being taken at the focus. In this case we have the Cartesian relation f = 2 px +p*, (b) or, substituting in (a), P = ^ (Z - f ) a + y/3, as the vector equation of the parabola. Or, again, a and ft being any given vectors parallel to a diam- eter and tangent at its vertex, p—|a(c) is the vector equation of a parabola, in terms of a single inde- pendent scalar t. 62. Let f(x) be any scalar function as, for example, f(x) = x2. Then d [ ƒ (x) ] = 2 xdx = [ ƒ ' (íc) ] dæ. If, however, f{q) be a function of a quaternion g, as, for example, in the above case, f{q) = q\ then f(q + dg) = (g + dg)2 = g2 + gdg + dg . g + (dq)2, d [ƒ(; j limit dP = n = aon136 QUATERNIONS. then (y + &y) = (x + Ax)* -, whence, as usual, Ay —2x Ax + (A æ)2, or, n being a positive whole number, nAy=:2xnAx-\- n~l(n A x)2. If, now, the differences A y and A x tend together to zero, while n increases and tends to infinity in such a manner that nAx tends to some finite limit, as a, we have, for the other equimultiple n A y, n A y = 2 xa + n~la2. But, since a, and therefore a2, is finite, n~la2 tends to zero, and, at the limit, nAy = 2xa. Hence the limits of the equi- multiples n Ax and n Ay are respectively a and 2 æa, and clx = a, cly = 2xa by definition; from which dy = 2 xdx. For a vector function we should write dp = ^Íníf(P + n~ldP)-f(p)l • • (169>> and for a scalar function, p = Cip = cl |> (0] = (t + f) - (Í)] • (170), in which latter t and dt are independent and arbitrary scalars. 64. As a further illustration of the definition, let P—W)APPLICATIONS TO LOCI. 137 be the equation of any plane curve in space, and op = p (Fig. 64) a vector from the origin to a point p of the curve ; t being any arbitrary sca- lar representing time, for example ; so that its value, for any other point p' of the curve, represents the interval elapsed from any definite epoch to the time when the point generating the curve has reached p If p' be the vector to pj then p'— p = ppr= Ayo is strictly the finite difference between p and p, and, if the corre- sponding change in t be Aí, Fig. 64. pp'= (p + A p) — p = A p= <£(i + A i) — <£(i) = A ; where op'= (i + A i), and A t is the interval from p to p! In -|A i, p would have reached some point as p", for which op"= cf>(t -f|-A i), on the supposition that pp" is described in i A i. On the basis of this closer approximation to the velocity at p, p would have been found at pn, had this velocity remained unchanged, such that ppn= 2pp"= 2(op,f—op)= 2[<£(i-j-JA £)_<£(£)]. For a closer approximation to the vector described in A t with the velocity at p, suppose at the end of -J A t the point is at p'", for which op,f'= (t + ^At). Under this supposition, the vec- tor described in A t would have been Tp'"= 3pp'"= 3(opm— op) = 3[<£(í-f-¿A i) — <£(i)], and, at the limit, representing the multiple of the diminishing chord by dp, 138 QUATERNIONS. 65. Eesuming Equation (167), dp = df(q) = lf(q + n~ldq) -ƒ(?)], (a), the second member may be written /(g, dq), but not, as ordi- narily, f(q)dq. In/(g, dg), dq may be composed of parts, as g' gf', g'",., with reference to which/(g, dq) = /(g, gf+g"+.) is distrib- utive. To prove this, let dq = q'+q"; we are to prove that f(q, q'+ q") =f(q, q') +f(q, q") • Since before passing to the limit, the second member of (a) is a function of n, q and dg, we may express this function by the symbol /n(g, dg), and write f(q, dq) = «[ƒ(? + -/(?)]=ƒ»(?> *7), or f(q + rrldq) =f(q) + rr'fn(q, dq). Replacing dg by gf and g" in succession, we have f(q + n-'q') =f(q) + n^f^q, q'), f(q + n-'q") =f(q) + n^f^q, q"), and, following the same law of derivation, f(q + n~Jq” + ft-1 q') =f(q + q") + n~lfn(q + m"1 q", q'), f(q + n-'q'+n-1 q") =f(q) + w_1/„(g, q’+ q"), from which fn{q, q'+ q") =f«(q> q") +L(q+q'), the limiting form of which, for n = co, is f(q,q'+q")=f(q,q")+f(q,q') . . (171),APPLICATIONS TO LOCI. 139 which may, in like manner, be extended to the case of dg = g'+g”+g'"-i-........................... It follows from the above that, if p =f(q-> xdq), f(q,xdq) = xf(qidq) .... (172). If Q = F(q, r,...), whence, Equation (168), ãQ = ã[F(q, r, .....)] = ÍÍÍ « TO + n-1 dÿ, r + n-1 dr, )-F(q,r, ) ], the last member will be a linear and homogeneous function of dg, dr, ..., and distributive with reference to each of them. Hence, to differentiate such a function, we do so with reference to each factor, and take the sum of thé results obtained, as usual ; taking care, however, not to make use of the commutative prop- erty. Thus d(gr) = dq . r + gdr, but not rdg + qdr. 66. When g is a function of any variable scalar £, represent- ing time, for example, then, if t be given a finite increment A for which the corresponding one of g is A g, we have A g = A w + A xi + A yj -J- A zk ; and, if the several parts of the quaternion vary continuously with the independent variable i, at the limit we may form, as usual, the differential coefficient dq dw .dx. , dy . . dz, dt dt dt dt dt The successive differential coefficients, as also the partial ones, when g = <£(£, v,..), are derived from the quadrinomial form in the same manner.140 QUATERNIONS. 67. Examples. 1. To find dTg. dTq __ dVw2 -f-x*+y2 + & dt dt dq or dq dTq dt ~W~~3 * SVq' 2. (TPy=-P2. The first member being a scalar, we have 2TpdTp. From the second member 3. To find dUg. We have TgUg = q ; dTg . Ug -+- dUg . Tg = dg, Equating TpdTp = — Spdp. From this we may obtain dTp = - S • Updp = S^' dTn dn orAPPLICATIONS TO LOCI. 141 whence dTg . Uq dUg . Tg __ dq TqVq TqVq q ’ or dUg __ dq dTq U q q Tq5 and, substituting from Ex. 2, dUg _ dq ^ dq . Vq ~ q q dUg__vdg • *• TT — * i Uq q or cl\]q = V^2 . uq. q 4. From the above expressions for dTq and dUg, we have dq = dTg . Ug + TgdUg = /s^2 + Y^W V U q U q) — is— + Y— \ q qj as the form under which the differential of a quaternion may always be written. 5. To find dUp. We have, from p = TpUp, dp = dTp • Up -f" TpdUp, dp __ dTp . dUp p “ Tp + Up = gÆp dUp from Ex. 2, P V dUp__dp dp___Trdp__ vdp • p Vpdp W_7_s7 7 ’ whence, also, dUp pT . dpp (Tp)3 ’142 QUATERNIONS. 6. From the above expressions for dTp and düp, dp = dTp • Up + p Y . dpp (Tp)2 ’ 7. That S, Y and K are commutative with d is seen from the following : g = Sg + yg, whence dq = dSg + dyg, and, since dq is a quaternion, dq = Sdg + ydg, hence dSg = S dq and Again whence Kg = Sg — yg, dYq = ydg. (а) (б) («0 dKg = dSg — dyg, and, taking the conjugate of dq in either (b) or (a), we have, with or without (c), dKg = Kdg. 8. (Tg)2 = gKg. 2 TgdTg = n [(g + rr\lq) (Kg + n^dKq) - gKg] = limit [dg(Kg + n-1Kdg) + gKdg] = dq . Kg + gKdg = K • gKdg + gKdg = 2 S . gKdg =28. Kgdg, [Equation (80)] or, since Tg = TKg and UKg = U- = —, g Ug dTg = S . U^dg = S . Ug^dg. If g = a vector, as p, then, since Kp = —■ p5 this becomes dTp = — S • Updp, as in Ex. 2.APPLICATIONS TO LOCI. 143 9. r = + dr • rKr = rdr . Kr + (Tr)2dr, or, adding, rdq -f dq » Kr = [r2 -f- (Tr)2]dr + rdr(r + Kr) = [r2 + (Tr)2 + 2 Sr . r]dr, which gives dr = dVg in terms of dq. 11. qq~1 = 1. We have qd(q~l) + dq . g_1 = 0. Operating with g-1 x q~1qd(q~l) + g-1dg . g"1 = 0, ,1 1 7 1 d- =----dq . — g g g144 QUATEEMONS. If g = a vector, as p, 71 1 j 1 cl- =--ap- P P P = — -dp- + \dp — i icfy P P P P P = ^--(-dp + dP.l) P p\p PJ P2 P P = (*£-2S^- = -K- P/P l 12. Differentiate SUg. cZg d$Ug = SdUg = S . Yy Ug dq = S.-|TUg = s . ^UYgTYUg <1 = -S . dq gUYg 13. Differentiate YUg. TYUg. cïVUg = Y. dVq =V.Yy Ug = T.TJg-1T(dg . g-1). 14. Differentiate TYUg. ÆTVUg = S^*- H UVg clq ,*rg /TUg “ ^UYg UYg = S . dg gUYg SUg. 1 • —• P [Exs. 7 and 3.] [Exs. 7 and 3.] [Ex. 2.]APPLICATIONS TO LOCI. 145 The Right Line. As in Cartesian coordinates, the form of the equations of a right line, as of other loci, will depend upon the assumed con- stants, and in any given problem one form may be more con- veniently used than another. 68. Right line through the origin. If o be the initial point, or origin, and p = or a variable vec- tor in the prolongation of a = oa, then p=zXa.....................(173) is the equation of a right line through the origin in the direction of the constant vector a. The equations ...............<™> obviously refer to the same right line. Since any line, represented as a vector by a, is parallel to p=xa, we may say that the above equations are those of a right line through the origin parallel to a given line ; or, a being a point given by a = OA, they are the equations of a right line through the origin and a given point. 69. Parallel lines. If p = on be a constant vector to a given point b, then p = /3-{-Xa.........................(1^5) is the equation of a right line through a given point, and parallel to a given line, as p'= xa through the origin. Or, a being a given vector, it is the equation of a right line through a given point and having a given direction. If a is an undetermined vector, it becomes the general equation of any one of the infinite num- ber of right lines which may be drawn through a given point. If o and b coincide, ¡3 = 0, and, as before, p = xa.146 QUATERNIONS. a remaining the same, and ft = ob' being a vector to any other point b' for the equations of two parallels, we have p — P +xa ) p = ft'+x’a ) (176), or, since a and p —¡3 are parallel, Ya(p-/?)=0j Ya(p-/3')=0j (177). 70. Right line through two given points. If OA = a (Fig. 65), ob = p are the vectors to the given points, and p the variable vector to any Fig. 65. point R of the line whose equation is re- R quired, we have and ar = æab = x(p — a), OR = OA -f- AR, or, for the required equation, p = a-{-x(p — a) (178), which, if one of the points, as a, coincides with the origin, becomes p = xp, as before. We have seen, Art. 55, that if $a/?y = 0, a, p and y are com- planar. Replacing y by the variable vector p, $a.Pp = 0....................(179) is the equation of a plane, since it expresses the condition that p is complanar with a and p. If we have also Sayp = 0, the two equations, taken together, represent the line of intersection of these two planes. * These equations may be obtained from the line p = xa by ope- rating with S(YaP) X and S(Yay)x ; or, conversely, to find the equation of the line in terms of known quantities, having given Sa/fy) = 0, Sayp = 0,APPLICATIONS TO LOCI. 147 write these latter under the form S • pYa/3 = 0, S . pYay = 0, whence it appears that p is perpendicular to both Yafi and Yay, and is consequently parallel to the axis of their product ; therefore P = yY • YafiYay = y (y Sa/3a — a&afty) [Eq. (112)] = — 2/aSa^y, or, putting — ySafty = æ, p = xa. 71. Right line perpendicular to a given line. 1. Let 8 = od (Fig. 66) be a vector through the origin. To find the equation of dc through its extremity Fig> ^ and perpendicular to it. Now p — 8 is a p r c vector along dr, and therefore by condition S8(p-8) = 0. Whence SSp = -(T8)2, or o SSp = Cy a constant ..... (180). In order that p, p — 8 and 8 be complanar, we must have S • 8p(p —• 8) = 0, or S • (Y8p) (p — 8) = 0. 2. p — 8, being perpendicular to both 8 and Y8p, will be parallel to the axis of their product, or to Y . 8Y8p. Hence, if y = oc be a vector to any point c, in the plane of od and dr, the equation of a right line through a given point c, perpendicular to a given line od, will be p y -f xY • SY8y...............(181). /148 QUATERNIONS. 3. If the perpendicular is to pass through the origin, then, from Equation (180), SSP = 0.....................(182), or, in another form, from Equation (181), y being parallel to Y. SY8y, p = yY.my.....................(183). 4. The student will find it useful to translate the Quaternion into the Cartesian forms. Thus, from Equation (180), if rod= 0, SSp = — TST/o cos 6, whence, if r and d represent the tensors, rd cos 0 = d2, or r = - — , cos# the polar equation of a right line. 5. Equation (181), of a line through a given point and per- pendicular to a given line through the origin, may be otherwise obtained, as follows : Let y and 8, as before, be vectors to the point and along the given line, respective^, and fi a vector along the required per- pendicular, whose equation will then be p = y + xfi. (a) To eliminate fi we have the conditions &8fi = 0, since 8 and fi are perpendicular to each other, and SySfi = 0, since y, 8 and fi are complanar. But Y8y is perpendicular to this plane, and therefore Y • 8Y8y is parallel to fi ; hence, substitut- ing in (a), p = y + xY. SY8y, or simply p = y + a?8YSy.APPLICATIONS TO LOCI. 149 If SyS/3 ^ 0, y, 8 and ft are not complanar, and the problem is indeterminate ; which also appears from (a), by operating with X S . 8, whence, since S/38 = 0, Sp8 = Sy8, a result which is independent of ft, and an infinite number of lines satisfy the condition. 6. If the line to which the perpendicular is drawn does not pass through the origin, let p = ft + xa (a) be its equation. Then, if p be the vector to the foot of the per- pendicular, we have Sa(p — y) = 0, or Sa(#a + /? — y) = 0, (Ò) because the line is perpendicular to (a), or its parallel a. Hence, from (&), “ - —a ^«(y — ft), xa: or, fbr the perpendicular p — y, p — y = Xa-\- ft — y = a”1 Sa(y — /5) — a-1a(y — /?) ==_a-1Ya(y^), Its length is evidently TY[Ua . (y-0)]...............(184). 7. This perpendicular is the shortest distance from the point to the line. The problem may, therefore, be stated thus : to find the shortest distance from c to the line p = xa + ft. p being the vector from c to any point of the given line, this vector is P + Xa — y, and, in order that its length be a minimum, c?T (/3 + sea — y) = 0 = T(ft 4- xa — y)dT(ft + Xa — y) = — $\_(ft + Xa — y)a\dx = 0,150 QUATERNIONS. or S(/2 + Xa — y)a = 0, that is, the line must be perpendicular to p = xa + ¡3. 8. If the perpendicular distance from the origin to p = ¡3 + xa is required, p, being as before the vector to the foot of the per- pendicular, coincides with it ; hence, y being zero, and 8 repre- senting this value of p, § = xa + /?. Operating with X S . 8, since Sa8 = 0, — (T8)2 = S/33. Tlpnnp TS'__ Sg8_S,/3T8U8 T8 T 8 ’ or T8 = S . /2U8 . . . . . . (185). 72. We are to observe that the foregoing equations of a right line are, as remarked in Art. 60, all linear functions involving, explicitly or implicitly, a single real and independent variable scalar. Such is evidently the case for such equations as p = xa, [Eq. (173)] P = p + æa, [Eq. (175)] p = a + æ(/? — a). [Eq. (178)] So also for the implicit forms, as Tap = 0 [Eq. (174)] ; em- ploying the trinomial forms a = ai -\~bj eh) p = xi + yj + zk, we have ap = (bz — cy) i + (cx — az)j + (ay — bx) 7c — (ax + by + cz). Whence Yap = (bz — cy) i + (cx — az)j + (ay — bx) 7c = 0 ; .\bz = cy, cx — az, ay — bx, in which X and y are functions of z.APPLICATIONS TO LOCI. 151 The Plane. 73. Equation of a plane. 1. If, in the equation $ . 8fi == 0, which denotes that fi is per- pendicular to S, we replace j3 by the variable vector p, S . Sp = 0.................(186) is the equation of a plane through the origin perpendicular to 8. 2. Or, let 8 = OB (Fig. 66) be the vector- rig. 66 (Ms). perpendicular on the plane, and dr any line d_________r c of the plane. Then SS(p —8) = 0, S8p = 82 = -(TS)2, or o SSp = c, a constant . . (187) is the general equation of a plane perpendicular to 8. Here dr is any line of the plane ; and, if YSp = e, Sep = an indeterminate quantity . . . (188). If the plane pass through the origin, we have, as before, SSp = 0. Conversely, if SSp = c is the equation of a plane, 8 is a vector perpendicular to the plane. 3. The equation of a plane through the origin perpendicular to 8 may also be written in terms of any two of its vectors, as y and fi ; p = xfi + yy. Both of these indeterminate vectors may be eliminated by operating with S • 8 X, whence SSp = 0 as before ; or one may be eliminated by operating with Y • fi X, whence Yfip = zS,152 QUATERNIONS. from which we may again derive SSp = 0 by operating with V . 8 X ; for 1 Y . 8V/fy = YzS2 = 0 = [Eq. (Ill)] whence, since S8/2 = 0, SSp = 0. 4. The equation of a plane through a point b, for which ob = /?, and perpendicular to 8, is S8(p-/?) = 0.................(189). 5. Having the equation of a plane, SSp = c, to find its dis- tance from the origin, or the length of p when it coincides with 8, we have p = xS ; hence S 8p = c = SæS2 = x82, or 82’ which, in p = xS, gives c p = 8’ or tp=¿......................... (190>- 74. To jÆnd the equation of a plane through the origin, making equal angles with three given lines. Let a, /?, y be unit vectors along the lines. The equation of the plane will be of the form SSp = 0. By condition, Sa8 = S£8 = Sy8 = T8 sin cf> = æ, being the common angle made by the lines with the plane. HenceAPPLICATIONS TO LOCI. 153 To eliminate 8, we have, from Equation (118), SSaj3y = Ya/3SyS + YjSySaS + YyaS ¡88, and, by condition, 8Saj8y = x(Yaj8 + Yj8y + Yya). The vector represented by the parenthesis is, then, the per- pendicular on the plane, whose equation, therefore, is Sp(Ya/? + Y/?y + Yya) = 0 . . . . (191), and the sine of the angle is Safty T(Ya/?+Y/?y + Yya)’ 75. Equation of a plane through three given points. Let a, /?, y be vectors to the given points ; then are the lines joining these points, as (a — /?), (/3 — y), lines of the plane. If p is the variable vector to any point of the plane, p — a is also a line of the plane. Hence S(p - a) (a - £) (/3 - y) = 0, or $(pa/3 — pay — p/32 + p/3y — a2/3 + a2y + a/32 — a/?y) = 0. But S(-p/32) = 0, S(-a^) = 0, etc., S( — pay) = Spya = S • pYya, Spa/2 = S • pYa/3, etc., hence S . p(Ya/3 + Y/3y + Yya.) - Sa/3y == 0 . . (192), which, by making the vector-parenthesis = 8, may be written under the form Sp8-Sa/?y = 0,154 QUATERNIONS. in which 8 is along the perpendicular from the origin on the plane. When p coincides with this perpendicular, p = #8, and, from the above equation, x82 = Sa/3y, or, for the vector-perpendicular, p = ccS = 8 1 Safty = Sa/3y Yaft + Yfty + Vya 76. We observe again, from inspection of the equations of a plane, that, as remarked in Art. 60, they are linear and func- tions of two indeterminate scalars. Thus, for a plane through the origin S8p = 0, [Eq. (186)] employing the trinomial forms 8=ai-\-bj-\-cJc and p=xi+yj+z7cy we obtain Sp == (bz — cy)i -f (cx — az)j -f- (ay — bx)h — {ax + by + cz), the last term of which is the scalar part ; hence ax + by -\-cz = 0, the equation of a plane through the origin o, perpendicular to a line from o to (a, ö, c), which may be written f(x, y, z) = 0 ; or as a function of two indeterminates. In the same way, from an inspection of the other forms, p = xa + y/?, [Art. 73, 3] p = 8 + xa + y/3, S8p — c' = ax + by + cz — c' = 0, [Eq. (187)] we observe they are linear functions of two indeterminate scalars. 77. Exercises and Problems on the Right Line and Plane. 1. ft and y being vectors along two given lines which intersect at the point a, to which the vector is oa = a, to write the equation of a line perpendicular to each of the two given lines at their intersection.APPLICATIONS TO LOCI. 155 Y/3y is a vector in the direction of the required line, whose equation, therefore, is p = a-\-x\[3y . . . . . . (193). If a = oa' be a vector to any other point, as a' then is p = a1 -f- ocY/3y the equation of a line through a given point perpendicular to a given plane ; the latter being given by two of its Unes. 2. a and ¡3 being vectors to two given points, a and b, and SSp = c the equation of a given plane, to find the equation of a plane through a and b perpendicular to the given plane. 8, p — a and a — ¡3 are lines of the required plane, hence S(p — a) (a — /3)S = 0, or Sp(a-/?)S + Sa/?3=0...............(194) is the required equation. 3. oc = y being a vector to a given point c, and p = a + a?/?, p = a'+ x'(3' the equations of two given lines, to write the equation of a plane through c parallel to the two given lines. If lines be drawn through the given point parallel to the given lines, they will lie in the required plane. As vectors, ¡3 and ¡31 are such lines, and p — y is also a line of the plane. Hence m'(p~y) = o..................(195) is the required equation. If y = a, or a', it is the equation of a plane through one line parallel to the other. Or, if y is inde- terminate, it is the general equation of a plane parallel to two given lines. Otherwise : the equation of a plane through the extremity of y parallel to two given lines, whose directions are given by a and ¡3, is evidently p = y + xa + y¡3. 4. To find the distance betioeen two points. a and ¡3 being vectors to the points, y = /3-a. Squaring c2 = b2 + a2 — 2 ab cos c.156 QTTATEKNIONS. 5. A plane being given by two of its lines, /? and y, to write the equation of a right line through & perpendicular to the plane. Let OA = a. Draw two lines through a parallel to ¡3 and y. Then p = a + æV/3y.................(196). If the plane is given by the equation S8p = c, then p = a + x8................ . (197). 6. Find the length of the perpendicular from a to the plane, in the preceding example. Operating on Equation (197) with S . 8 x S8p = S8a -f- x82 = c, or x82 = c — S8a ; #8 = S"1^ — S8a)...............(198). 7. SS(p — /3) = 0, Equation (189), being the equation of a plane through b perpendicular to 8, to find the distance from a point c to the plane. Let y = oc. The perpendicular on the plane from c, being parallel to 8, will have for its equation p = y -f- x8. To find x, operate with S • 8 x , whence SSp = S8y + æ82, or, from the equation of the plane, $8y + x82 = 88p; .’. x8 =— 8~1S8(y — /?), and xT8 = T8“1S8(y - p) = S [U8 . (y - £)]. 8. TEnie the equation of a plane through the parallels p = a+æ/3, /o = a'+ æ/LAPPLICATIONS TO LOCI. 157 9. Write the equation of aplane through the line p = a -f- xft perpendicular to the plane SSp = 0. 10. Given the direction of a vector-perpendicular to a plane, to find its length so that the plane may meet three given planes in a point. Let 8 be the given vector-perpendicular, and Sap = a, S ftp = b, Syp = c the equations of the given planes. If the equation of the plane be written ' S8p = X, then a; must have such a value that one value of p shall satisfy the equations of all four of the planes. From Equation (118) we have pSa/?y = Ya/3Syp + Y/3ySap + YyaSftp = cYa/3 + aYfty + 5Yya. Operating with S . 8 X, to introduce x, xSafty = cS8aft + uSS^y + &S8ya. 11. To find the shortest distance between two given right lines. Let the lines be given by the equations p = a + xfi, (a) p = a’+x’/3l (b) The equation of a plane through either line, as (b), parallel to the other (a), is [Equation (195)] (c) Y/3ft' is a vector-perpendicular to this plane. Hence, if yYft ft' be the shortest vector distance between the lines, we have, since a — a'— yY ft ft' is a vector complanar with ft and ft[ S/?/na~a'-yY/?/3')==0,158 QUATERNIONS. or S(S/3/3' + Y/3/3') (a-a'- y\m = 0, whence - */(W)2 + S[W(“ — «')] = 0; or, dividing by T(Y/3/2')* TW')=TS[(a-a')UW')]. . . (199), the symbol T denoting that only the positive numerical value of the scalar is taken. Otherwise : since the distance is to be a minimum, tf*V-p) = o, whence S(p'-p) (P'dx'-I3dx) = 0, or S(p'-p)/? = 0 and S(p'-p)j8'=Q, or the shortest distance is their common perpendicular, whose length may be found as above. 12. Given SS1p = d1 and S82p = d2, the equations of tivo planes, to find the equation of their line of intersection. This equation will be of the form p = niSi -f- 7182 “h £rV8i82. (ci) To find m and n, we have, from (a), S8lp = m81* + n&8182, S82 p = wS22 + mSSA, from which we obtain _ S8lP- nSSi S2 _ S82p - nS22. m“ 8,* ” S8A ’ _S8182SS1p-S12S82p n~ (SSA)2-^2 ‘ But (S8182)2-S12822 = (VS182)2; n~ (YSAYAPPLICATIONS TO LOCI. 159 And similarly d2SS,S2- rZ,S22 m~ (VW ' Substituting these values in (a) d1S8182-d2Si% P~ (HS,)2 òí d2S8j Sg-djS,* (VSA)2 8j + i»ySj82, which is the equation of the required line, a less useful form than those of the two simple conditions of Art. 70. If the two planes pass through the origin, then also does their line of intersection ; and since every line in one plane is perpen- dicular to Si, and every line in the other to S2, Y81 S2 is a line along the intersection, as in (a), and the equation becomes p = x\S1S2 (200). 13. The planes being given as in Equation (189), S8(p-0) = O, («) SS'(p-/?') = 0, (ft) to find the line of intersection. The vector p to any point of the line must satisfy both (a) and (b). This vector may be decomposed into three vectors parallel to S, S' and YSS/ which are given, and not complanar, by Equation (118) ; whence pS . SS'YSS' = SpSY(S' • YSS') + SpS'Y(YSS'. S) + S(pYSS') YSS', or, from (a) and (6), - p(TYSS')2 = SS/3Y(S'. YSS') + SS'/3'Y(YSS'. S) + SSS'pYSS', or, since SSS'p is the only indeterminate scalar, putting it equal to 05, we have - p(TYSS')2 = SS/3Y(S'. YSS') + SS'£'Y(YSS'. S) + æYSS! If the planes pass through the origin, in which case ¡3 and (3[ . are zero, we have, as before, p = ojyss:160 QUATERNIONS. 14. To write the equation of a plane through the origin and the line of intersection of S8(p-j8) = 0, (a) S8,(p-j8,) = 0. (6) If p is such that $Sp = $8/3, and also SS'p = SS'yS% then both the above equations will be satisfied. Hence, from (a) and (p) S8pS8'0r- SS^SS'p = 0, which is also a plane through the origin. This equation may also be written S[(8SSy??--8'S8/?)p]==0, which shows that is a vector-perpendicular to the plane, and therefore to the line of intersection of (a) and (&). 15. To find the equation of condition that four points lie in a plane. If the vectors to the four points be a, /?, y, 8, then, to meet the condition, 8 — a, 8 — /?, 8 — y must be complanar, and therefore S(8-a)(8-0)(8-y) = O, whence SS/?y + Sa8y + Sa/?8 = Sa/?y . . . (201), which is the equation of condition. Or, X and y being indeterminate, we have also 3 = a + x(Ji-a) + y{y-p), or 8 + (x — l)a + (y — x)fi — yy = 0, 1 + (* — !) + {y — x) — y = 0. andAPPLICATIONS TO LOCI. 161 Or, in general, act -J- bf3 ~f* Cy -J- clS — 0 Ì —|— & —j— c —f— cZ = 0 3 . . (202), are the sufficient conditions of complanarity. These conditions are analogous to Equation (9). 16. Given the three planes of a triedral, to find the equations of planes through the edges perpendicular to the opposite /aces, and to shoio that they intersect in a right line. Taking the vertex as the initial point, let Sap = 0, (a) $/3p = 0, (b) s yp = o (C) be the equations of the plane faces. Then Ya/3 is a vector par^ allei to the intersection of (a) and (b), and Y • y Ya/3 is a vector perpendicular to the required plane through their common edge. Hence the equation of this plane is S(pY. yVa/3)=0. («') Similarly, or by a cyclic change of vectors, S(pV. aV/3y)=0, (b>) S(pV . fiYya) = 0 (c-) are the equations of the other two planes. If from their common point of intersection normals are drawn to the planes, then are Y • y Ya/3, Y • aV/3y and V • /3Yya vector lines parallel to them ; but, Equation (123), Y (yYa/3 + aV¡3y + jffVya) = 0. Hence these vectors are complanar, and the planes therefore intersect in a right line. Otherwise: from Equation (111) Y(aY/?y) = ySa/8-/?Sayi162 QUATERNIONS. hence, from (&f), S(pySa/2 — p/2 Say) = Sa/?Spy — SaySpfi = 0. Similarly, or by cyclic permutation, S/?ySpa - S/?aSpy = 0, SyaSp/? — Sy/?Spa = 0. But the sum of these three equations is identically zero, either two giving the third by subtraction or addition. 17. To find the locus of a point which divides all right lines terminating in two given lines into segments which have a com- mon ratio. Fig. 67. A Let da and d'b (Fig. 67) be the two given lines, a and ¡3 unit vectors parallel to them, ba any line terminating in the given lines, and r a point such that ra = WBR. Assume dd', a perpendicular to both the given lines, o, its middle point, as the origin, and od = 8, od' = — 8, or = p. Then Adding But OA — p -{- RA = 8 -}- Xa. OB = p-f-RB = — 8 -b y/3. 2p + RA + RB == Xa + y(3. (®) . m — 1 m — 1 / .cv. \ RA + RB = ----RA =-------( — p + 8 + Xa) , m m which substituted in (a) gives p — 8 — xa = m(y/3 — p — 8), (b) whence, since S8/3 = SSa =0, S8p(m + l) = 82(ì-m) = c, or the locus is a plane perpendicular to dd'.APPLICATIONS TO LOCI. 163 If the given ratio is unity, or br = ra, then m = 1 and SSp = 0, and the locus is a plane through o perpendicular to dd\ If a and /? are parallel, then (b) becomes p — 8 = m(xf a — p — S), whence S8p(m + 1) = (1 — ra)S2, a right line perpendicular to dd! If at the same time m = 1, SSp = 0 and p = xna, a right line through the origin parallel to the given lines. 18. If the sums of the perpendiculars from two given points on tico given planes are equal, the sum of the perpendiculars from any point of the line joining them is the same. Let a and b be the given points, oa = a, ob = /?, and SSp=d, S8'p = d' be the equations of the planes ; 8 and 8' being unit vectors, so that x8 and ySf are the vector-perpendiculars from a on the planes. Then X = SaS — d, y = SaS'- d\ and X + y = Sa (8 + S') — (d -f- d'). Similarly æ'+y=S/?(8+S')-(^ + cO- But, by condition, Sa(8 + S') = S/5(8 -f- 8'), or SOS-aXS + SO^O. (a) The vector from o to any other point of the line ab is a + z (/? — a) ; whence, for this point, 05" + 2/" = s [a + 2 08 ~ a) ] (S + S') - (d + d') , for which point, since (a) remains true, the sum therefore is unchanged.164 QUATERNIONS. 19. To find the locus of the middle points of the elements of an hyperbolic paraboloid. Let the equations of the plane director and rectilinear direc- trices be SSp = 0, p = a + æ/3 and p = a' + of/?! Also, let om = /X be the vector to the middle point of an ele- ment so chosen that the vectors to the extremities are a + xfi and a1 + x’fi'. Then, since m is the middle point, 2/¿ = a -f- xf3 -f- oJ -f- x'ff, (a) The vector element is — Xr¡3f — a! -{- a -f- xj3, and, being parallel to the plane director, SS(— af -f- a + xfi — X!/3') = 0. This is a scalar equation between known quantities from which we may find xf in terms of x ; substituting this value in (a), we have an equation of the form P = a± + æft, or the locus is a right line. 20. If from any three points on the line of intersection of two planes, lines be drawn, one in each plane, the triangles formed by their intersections are sections of the same pyramid. The Circle and Sphere. 78. Equations of the circle. The equation of the circle may be written under various forms. If a and ¡3 are vector-radii at right angles to each other, and Ta = T/?, we may write p = cos0 • a + sin0 • ¡3 .... (203), in terms of a single variable scalar 0.APPLICATIONS TO LOCI. 165 If a and ¡3 are unit vectors along the radii, P = xa + y/3, or, since æ2 + y2= r2, p = -y2)ha + yfi .... (204). The initial point being at the center, Tp = Ta -i T- = 1 — r2 - (205) are evidently all equations of the circle. If o (Fig. 68) be any initial point, c the center, to which the vector oc = y, p the variable vector to any point p, cp = a, then P p —y==a, whence (p-y)2 = -r1 • -(206), the vector equation of the circle whose radius is r. If Ty = c, it may be put under the form p2 — 2 Spy = c2 — r2 ...........(207). If the initial point is on the circumference, we still have (p — y)2 = — r2 ; but y2 = — r2, hence p2 — 2 Spy = 0 ................(208), or, since in this case Spy = Spa, p2 — 2 Spa = 0 ................(209). 79. Equations of the sphere. This surface may be conveniently treated of in connection with the circle ; for, since nothing in the previous article restricts the lines to one plane, the equations there deduced for the circle are also applicable to the sphere.166 QUATERNIONS. Another convenient form of the equation of a sphere is (Fig. 68) T (p — y) — Ta................(210), I the center being at the extremity of y and Ta the radius. 80. Tangent liyie and plane. A vector along the tangent being dp, we have, from Equation (208), dp = — sin# • a + cos9 • /?, and for the tangent line ir = p + xdp, 7T= cos# • a + sin# . ¡3 + X [—sin# . a + cos 0./?] (211), where tt- is any vector to the tangent line at the point corre- sponding to 0. From the above we have directly Spdp = 0, or the tangent is perpendicular to the radius vector drawn to the point of tangency. 69- By means of this property we may whence or follows : let tt be the vector to any point of the tangent, as b (Fig. 69), c being the initial point and p the vector to p, the point of tangency. Then Sp(w — p) = 0, Sp7T = — r2 ^ 7T S- = 1 P (212), are the equations of a tangent. Since nothing restricts the line to one plane, they are also the equations of the tangent plane to a sphere.APPLICATIONS TO LOCI. 167 The above well-known property may also be obtained by differentiating Tp = Ta ; whence, Art. 67, 2, Spdp = 0, and therefore p is perpendicular to the tangent line or plane. 81. Chords of contact. In Fig. 69 let cb = ¡3 be the vector to a given point. The equation of the tangent bp must be satisfied for this point ; hence, from Equation 212, or S/fcr = -r*..................(213), which is equally true of the other point of tangency p,' and being the equation of a right line, it is that of the chord of contact pp! And for the reason previously given, it is also the equation of the plane of the circle of contact of the tangent cone to the sphere, the vertex of the cone being at b. 82. Exercises and Problems on the Circle and the Sphere. In the following problems the various equations of the plane, line, circle and sphere are empk>3Ted to familiarize the student with their use. Other equations than those selected in any special problem might have been used, leading sometimes more directly to the desired result. It will be found a useful exercise to assume forms other than those chosen, as also to transform the equations themselves and interpret the results. Thus, for example, the equation of the circle (209), p2 — 2 Spa = 0 may be transformed into Hp{p — 2a) = 0,168 QUATERNIONS. Fig. 70. P which gives immediately (Fig. 70) the property of the circle, that the angle inscribed in a semi-circle is a right angle. Obviously, this includes the case of chords drawn from any point in a sphere to the extremities of a diameter, and the above equation is a statement of the prop- osition that, p being a variable vector, the locus of the vertex of a right angle, whose sides pass through the extremities of a and — a, is a sphere. Again, with the origin at the center, we have (Fig. 71), Fig. 71. D (p + a) + (a — p) = 2 a, and, operating with x S • (p — a), &(p + a) (p ~ a) = 0 ; .*. p is a right angle. This also follows from Tp=Ta, whence p2=a2 and S(p+a)(p— a) = 0. Again, from Tp = Ta, T(p + a) (p — a) = 2 TYap, The first member is the rectangle of the chords pd, pd' (Fig. 71), and the second member is 2 on • op sinDOP. Hence the rectangle on the chords drawn from any point of a circle to the extremities of a diameter is four times the area of a triangle whose sides are p and a. Also, from T p = Ta, and for any other point p'2 = —r2 ; ••• s(p'+pW-p)=o. But /of— p is a vector along the secant, and p'-f- p is a vector along the angle-bisector ; now when the secant becomes a tan-APPLICATIONS TO LOCI. 169 gent, the angle-bisector becomes the radius ; therefore the radius to the point of contact is perpendicular to the tangent. 1. The angle at the center of a circle is double that at the cir- cumference standing on the same arc. We have Tp = Ta, and therefore, Art. 56, 18, P = (p + a) xa(p+ a), whence the proposition. 2. In any circle, the square of the tangent equals the product of the secant and its external segment. 3. The right line joining the points of intersection of two circles is perpendicular to the line joining their centers. Let (Fig. 72) cc' = a, cp = p, cpf = p', and r, r' be the radii of the circles. Then also Hence or (p-a)2 = -r'2; (p'-a )2 = -r'2. Spa = Spfa, Sa(p-p') = 0; Fig. 72. hence ppf and ccf are at right angles.170 QUATERNIONS. 4. A chord is drawn parallel to the diameter of a circle; the radii to the extremities of the chord make equal angles with the diameter. If p and p be the vector-radii, 2 a the vector-diameter, then xa = the vector-chord, and (p'— a**)2 = - A (p+xa.y = — r!, whence the proposition. 5. If abc is a triangle inscribed in a circle, show that the vector of the product of the three sides in order is parallel to the tangent at the initial point. [Compare Art. 55.] If ab = /?, ca = y, and o is the center of the circle, then —Y(ab . Be • ca) = Y. /?(/? +y)y ,=n/32y+/V) = !?y + ŸP- c and b being points of the circumference satisfying p2 — 2 Spa = 0 [Eq. (209)], substituting and operating with S • a X S . aY (ab . BC • ca) = 2 Sa/3Say — 2 Sa/?Say = 0. Hence Y (ab • bc • ca) is perpendicular to a, or parallel to the tangent at a. 6. The sum of the squares of the lines from any point on a diameter of a circle to the extremities of a parallel chord is equal to the sum of the squares of the segments of the diameter. Let pp' (Fig. 73) be the chord parallel to the diameter dd¡ rig. 73. o the given point, and c the center of the circle. Let cp = p, cpf = p' oc = a, op = j3 and op' = /?! Then OP2 = — fi2 = — (a2 + 2 Sap + p2), op'2 = - J8,2= - (a2 + 2 Sap'+ p'2) ; . *. OP2 + OPf2 = 2 OC2 + 2 DC2 — 2 (Sap+Sap').APPLICATIONS TO LOCI. 171 But S(f> — p') (p + p') = S(p + p')xa = 0. Therefore Sap + Sap' = 0, and op2 + op'2 = do2 -f- od'2. 7. To find the intersection of aplane and a sphere. Let p2 = — r2 be the equation of the sphere, 8 a vector-perpen- dicular from its center on the plane and T8 = d. Then, if p be a vector of the plane, p = S + /3. Substituting in the equation of the sphere, since S¡38 = 0, we have /32 = — (r2 — d2), the equation of a circle whose radius is Vr2 — d2, and which is real so long as d < r. 8. To find the intersection of two spheres. Let the equations of the given spheres be (Eq. 207) p2 — 2 Spy = c2 — r2, p2 —- 2 Spy'= c'2 — r'2. Subtracting, we have 2 Sp(y — y') = a constant. The intersection is therefore a circle whose plane is perpen- dicular to y— yi the vector-line joining the centers of the spheres. Assuming (Eq. 21u) T (p — y) = Ta and T (p — y') = Ta' show that 2 Sp(y — y') = a constant, as above.172 QUATERNIONS. 9. Tlie planes of intersection of three spheres intersect in a right line. Let y', y", yw be the vector-lines to the centers of the spheres, and their equations p2-2Spy' = c', p2-2Spy" = c", p2_2Spy'"=c"'. The equations of the planes of intersection are, from the pre- ceding problem, 2S/o(y'-y") =c" -c', (a) 2sf»(y' — y"') —c"'—c', (b) 2$p(y"-y"')=c"'-c". (c) Now, if p be so taken as to satisfy (a) and (b), we shall obtain their line of intersection. But if p satisfies (a) and (&), it will also satisfy their difference, which is (c) ; the planes there- fore intersect in a right line. 10. To find the locus of the intersections of perpendiculars from a fixed point upon lines through another fixed point. Let p and pr be the points, ppf = a, and 8 a vector-perpen- dicular on any line through pj as p = a + x¡3. Then 8 = a + y£, and operating with S . 8 x 82 = SSa, which is the equation of a circle (Eq. 209) whose diameter is pp! 11. From a fixed point p, lines are drawn to points, as p', p",..of a given right line. Required the locus of a point o on these lines, such that pp' . po = m2. Let the variable vector po=p ; then pp !=xp. By the condition T(pp'*po)= m2, or T (xp . p)= m2; a?p2 = T-m2.APPLICATIONS TO LOCI. 173 If 8 be the vector-perpendicular from p on the given line, and T8 = ci, S&(xp — 8)= 0, or æSSp = — d2 ; .■.P=-S8P, hence the locus is a circle through p. 12. If through any point chords he drawn to a circle, to find the locus of the intersection of the pairs of tangents drawn at the points of section of the chords and circle. Let the point a he given by the vector oa = a, o being the initial point taken at the center of the circle. Let p' = or be the vector to one point of intersection r. The locus of r is required. The equation of the chord of contact is (Eq. 213) Sfa = — r2, which, since the chord passes through a, may be written Sp'a = - r2, where a is a constant vector. The locus is therefore a straight line perpendicular to oa (Eq. 180). 13. To find the locus of the feet of perpendiculars drawn through a given point to planes passing through another given point. Let the initial point be taken at the origin of perpendiculars, a the vector to the point through which the planes are passed, and 8 a perpendicular. Then SS (8 — a) =0, or S2-SaS=0 is true for any perpendicular. Hence the locus is a sphere whose diameter is the line joining the given points.174 QUATERNIONS. Otherwise : if the origin be taken at the point common to the planes, and the equation of one of the planes is SSp = 0, then the vector-perpendicular is (Eq. 198) 8-1SSa, and, if p be the vector to its foot, p = a — S-1SSa, or p — a = — S_1 SSa, whence (p — a)2 = 8“2(SSa)2, and Sap — a2 = — 8-2(S8a)2. Adding the last two equations p2 — Sap = 0, or T(p_ ia) = Tia, which is the equation of a sphere whose radius is and center is at the extremity of -, or whose diameter is the line joining the points. 14. To find the locus of a point p which divides any line os drawn from a given point to a given plane, so that op . os = m, a constant. Let op = p and os = cr ; also let SSo- = c be the equation of the plane. We have, by condition, and TpTcr == m, Up = Ucr ; and Ter = - Up cr mUp ~w mpAPPLICATIONS TO LOCI. “vmrofnuTK ^umnsnl’0* 175 Substituting in the equation of the plane mSSp + cp2 = 0, which is the equation of a sphere passing through o and having Tib — for a diameter. OD 15. To find the locus of a point the ratio of whose distances from two given points is constant. Let o and a be the two given points, oa = a, or = p, r being a point of the locus. Then, by condition, if m be the given ratio, T(p — a) = mTp, whence or p2— 2 Spa + a2 = rn2p2, (1 — m2) p2 = 2 Sap — a2 = 2 Sap — Î-—^-a" 1 — m2 2 __ 2 Sap 1 — m2 (1 — m2)2 (1 —m2)2’ ••• \ 1— m2J 1— m2 which is the equation of a sphere whose radius is T—~ I 1—m* whose center c is on the line oa, so that oc =--a. (Ea. 21 O') 1—m2 v ^ 7 ;a, and 16. Given two points a and b, to find the locus of p ivhen AP2 + bp2 = OP2. o being the origin, let OA = a, ob = /?, op =p. Then, by condition, p2 = (p —a)2 + (p — Z3)2* whence p2_2Sp(a + /S) = -(a2 + /82), [p — (a + /?)]2 = 2Saff, T[p —(a + /3)] = V— 2 Sa/?,176 QUATERNIONS. which is the equation of a sphere whose center is at the extremity of (a + /?), if Saft is negative, or the angle aob acute. If this angle is obtuse, there is no point satisfying the condition. If aob = 90°, the locus is a point. 83. Exercises in the transformation and interpretation of elementary symbolic forms. 1. From the equation (p + a)2 = (p — a)2 T^=l, p — a derive in succession the equations T(p + a) = T(p — a) and and state what locus the}7 represent. 2. From the equation KP- + P- = 0 a a derive symbolically the equations ap + pa = 0, s£=0, Su£=0, iu£|=-l, and TYU¿ = 1, and interpret them as the equations of the same locus. 3. Transform to the forms S- = 1 and SU-=T-’ and interpret. a a p 4. Transform S^^ = 0 to S- = S-> ancl interpret. a a a 5. Transform (p — /3)2 = (p — a)2 to T (p — /3) = T (p — a), and interpret. 6. What locus is represented by K^ — ~ = 0?APPLICATIONS TO LOCI. 17T 7. What by Q2=-l? By (£j = - a2? 8. What by U ^ = U ^ ? Up = U/3? lj£ = l? 9. — lí—? a a 10. fv- )W? \ a> 1 a 11. YP~ •^0? V- 0 L a 12. Y- = a = 0? 13. -K- = a2? a a 14. su- = su-? su- = -su-? (sU-Y==(sU-') ? a a a a \ ay \ ay 15. Tp~l? 16. Transform (p —- a)2 = a2 to T(p — a) = Ta, and interpret. 17. Under what other form may we write (p — a)2= (¡3— a)2? Of what locus is it the equation ? 18. p2 + a2 = 0 ? p2 —f-1 = 0 ? Translate the latter into Car- tesian coordinates, by means of the trinomial form, and so deter- mine the locus anew. 19. T(p-/J) = T(0-a)? 20. Compare SU - = T- and S- = 1 with the forms of Ex. 3. p a p 21. What locus is represented by Sftp + p2 = 0 when T/3 = 1 ? 22- (sCHt0’=i? 23. 24. Show that Y . Ya/SYap = 0 is the equation of a plane. What plane? [Eq. (112)].178 QUATERNIONS. The Conic Sections. Cartesian Forms. 84. The Parabola. Resuming tlie general form of the equation of a plane curve p = xa + yfi, from the relation y2 = 2px, we obtain ..............(214> for the vector equation of the parabola when the vertex is the initial point. If the latter is taken anywhere on the curve, from the relation y2 = 2p'x, we obtain ............(215); and if the initial point is at the focus, then y2 = 2px -\-p2 gives P = ^{y2-tf)a- + yP .... (216); or again, in terms of a single scalar £, P = |2a + w2, t is imaginary, and no tangent can be drawn ; in this case (a) becomes P=(j+aja + np, the point being within the curve.180 QUATERNIONS. 86. Examples on the parabola. 1. The intercept of the tangent on the diameter is equal to the Fig 74 abscissa of the point of contact. Since the tangent is parallel to the vector ta -f- /3, or to any multiple of it, it is parallel to t2 a ft/3 or to *-a + tfi + - a, that is, to (Fig. 74) Z 2 OP + ox. But TP = to + op ; to — ox. 2. If from any point on a di- ameter produced, tangents be drawn, the chord of contact is parallel to the tangent at the vertex of the diameter. If tf and t" correspond to the points of tangency, we have for the vector-chord of contact = Ça+ «'/?-Ça-«"/?, which is parallel to P- t'+t" or, from Equation (&), Art. 85, to ¡3 + na, which is independent of m. 3. To find the locus of the extremity of the diagonal of a rect- angle whose sides are two chords drawn from the vertex. Let op and op' be the chords. Then or = p = ^-a + yP, Op'=p'=|-a — y% 2p (a) (P)APPLICATIONS TO LOCI. 181 The vector-diagonal w' is p + p', or which may be put under the form of the equation of the parabola But, by condition, Spp* == 0. Hence, from (a) and (ò), Sa/? being zero, yy'~yy'=Vpy-> («*) which in (c) gives s'= a + (y - y)/? + 4pa. Changing the origin to the extremity of 4pa, Hence the locus is a similar parabola whose vertex is at a distance of twice the parameter of the given parabola from its vertex. Moreover, from (d), xx, = (2p)2. Hence the parameter is a mean proportional between the ordinates and the abscissas of the extremities of chords at right angles. 4. If tangents be drawn at the vertices of an inscribed triangle, the sides of the triangle produced will intersect the tangents in three points of a right line. Let opp' (Fig. 74) be the inscribed triangle, and one of the vertices, as o, the initial point. Then, for the points p and pf respectively, we have (0) 182 QUATERNIONS. Let 7r15 7t2, 7r3 be the vectors to the points of intersection ; then Also TTi = OP + PSi = ~ a + Í/5 + x(ta + fi) . A 7T1 = X,OP,= X,^a + t,fij ; ƒ2 /y>l fl2 • '• — + atf = 2~’ ¿ t2 2tt'-tn Hence 2tt' — tl2\ 2 In a similar manner 71-2 i'2 /* But Also -a +/? 2t'-t\2 tt3 = op + 7/pp' = op + 2/0'— p) = |a + ij8 + y r2— ^ 7T3 = zfi; f2 ft 2 /2 2+ y~2~ = °’ t+y(t' tt! Hence Now t + V iï o 2 t-t' 2t'—t f-tn .ft' ,a\ Jt ,, rt ——--------^*2- -^*3=tU*+/3j-t (-a+/3)-(t-t)P=0. Also 2 t-t’ 2 t'-t t2-tn t tt' : 0. Hence 7rx, 7r2 and 7r3 terminate in a straight line.APPLICATIONS TO LOCI. 183 5. The principal tangent is tangent to all circles described on the radii vectores as diameters. Fig. 75. Let AP = p (Fig. 75), a and ¡B being unit vectors along the axis n and principal tangent. Then, if the circle cut the tangent in t, and TC be drawn to the center, T(tc) = T(fc) = T(-^-fp) ; TC2 = ¿(p — ma)2. o Also TC = TA -f- AF + FC D = — ma + i(p— ma), TC2 = [— z/3 + ma+tK/o — ma)]2. Equating these values of tc2, we have, since S/?a = 0, z2fi2 — z$/3p + mSap = 0, z2-zy + |=0, which gives but one value for z. 6. To find the length of the curve. It has been seen (Art. 62) that, if p = (j>(t) be the equation of a plane curve, the differential coefficient is the tangent to the curve. Hence, if this be denoted by // = <£'(£), Tp'dt is an element of the curve whose length will be found hv integrating Tp' with reference to the scalar variable involved between proper limits ; or s — s0 = Ptp'. For the parabola184 QUATERNIONS. we have 7. To find the area of the curve. With the notation of the previous example, twice the area swept over by the radius vector will be measured by (Art. 41, 7) TYpp'dt. The area will then be found b}r integrating TYpp' with reference to the involved scalar between proper limits and taking one-lialf the rèsult ; or From the origin, where y0 = 0, to any point whose ordinate is the area \xy of the triangle, which, with the sector, makes up the total area of the half curve, we have f xy, or two-tñirds that of the circumscribing rectangle. The origin may be changed to any point in the plane of the curve, to which the vector is y, by substituting the value p = y + Pi in the equation of the curve, Pi being the new radius vector ; we may thus find any sector area limited b}r two positions of px, the vertex of the sector being at the new origin. Thus, transferring to an origin on the principal tangent, distant b from the vertex, p = b/3 + px ; which, in the equation of the parabola, gives For the parabola A-. A or, since a ¡3 = 90', APPLICATIONS TO LOCI. 185 integrating, as before, between the limits y = b and y = 0, Fig. 76. S3/ 87. Relations between three intersecting tangents to the Parabo- la. [“Am. Journal of Math.,” yol. i. p. 379. M. L. Holman and E. A. Engler.] Let px, P21 ps be the vectors to the three points of tangenc}r, Pi, p3 [Fig. 76], and 7Ti, 7t2, 7t3 the vectors to sl9 s25 s3i the points of intersection of the tan gents. Resuming Equa- tion (216), where the focus is the initial point, and a and ¡3 are unit vectors along the axis and the directrix, p yjfc Pi s2 2) (yi +p2) (Ttt2)2 = A (y82 +p2) (ÿ2 +JP2) (t^s)2 = A (p2 +-P2) (2/22+P2) and from (d) and (e) (Tx3)2 = TPlTp2 (T^2)2=TpITp3 (T-71-j)2 — Tp21pS (c). (d), 00, (ƒ)•APPLICATIONS TO LOCI. 187 - From (c), it appears that the distance of the point of intersec- tion of two tangents from the axis is the arithmetical mean of the ordinates to their points of contact. From ( ƒ ), that the distance from the focus to the point of intersection of tico tangents is a mean proportional to the radii vectores to the points of contact. 1st. If p2 becomes a multiple of /?, Pi = -^ <*’ -p2)a + 2/2/3 = «/3; ••• z = y2 = ±p. Or, the parameter is the double ordinate ttirough the focus, or twice the distance from the focus to the directrix. 2d. If p1 is the multiple of p2 (Fig. 77), then p2 — is a focal chord, and Xp2 = Pii or, from (a), -f)°-+2/2/3J=-^(y? -p2)a + yiP ; ___yï-p2 yi yi -p2 y2 whence188 QUATERNIONS. or 2/i(2/i2/2 +P2) = 2/2(2/12/2 +j>2), and yiÿ2+152=0. (g) From (a) and (c) S"3P1 = — ^(ï/22/i — -lñ-Í («/I + 2/2)2/l = -^i(2/i2+132) (2/12/2+f0 = ° ; (à) or, a line from the focus to the intersection of the tangents at the extremities of a focal chord is perpendicular to the focal chord. The vectors along the tangents are Pi — ttz and p2 — 7r3, and, from (7¿), S(pi — 7r3) (p2 — 7r3) = Spip2 + ^32 = 0, or, the tangents at the extremities of the focal chord are perpen- dicular to each other. Since, from (g), 2/i2/2 = -152, we have *3 = -^(2/12/2 -i>2)« + i(2/i +2/2)/3 = - + i(2/i + 2/*)fr or, the tangents at the extremities of a focal chord intersect on the directrix. 3d. If p2 becomes a multiple of a (Fig. 78), y2 = 0, and from (C) n = -^(2/22/1 -/)“ + ¿(2/1 + 2/2>/3 +f/*> (*•) or, ¿7¿e subtangent is bisected at the vertex.APPLICATIONS TO LOCI. 189 Also - y’\ y*R -~pa-2P' Operating with S . 7r3 x S^a(^3 — Pi) = = 0, 4 4 or, a perpendicular from the focus on the tangent intersects it on the tangent at the vertex. Again, since 7r3 is parallel to the normal at pl5 the latter may be written, from (i), ^^j = za + yip-, whence P Vi z = -æ|, 2/1 = æ|, X7TS = X[ —^a-f- or x = 2, « = —i>5190 QUATERNIONS. hence, the subnormal is constant; and the normal is twice the perpendicular on the tangent from the focus. The normal at px may be written xirs^—z'a+pu or # *(-■|«+=-«'«+- p2)a+yiP ; whence, from (b), x = 2, and z' = J- (yi14- f) = TPl ; ¿P or, the distance from the foot of the normal to the focus equals the radius vector to the point of contact, or the distance from the point of contact to the directrix, or the distance from the focus to the foot of the tangent. The portion of the tangent from its foot to the point of con- tact may be written za + pl5 in which z has just been found. Hence 2a+pl = ip +I^a + Yp ~^a+ or Vi Za- + P\=L^a- + yiP> O') the portion of the tangent from the foot of the focal perpendicu- lar to the point of contact is — *3 + Pi = ^ a — 2^ + 2^ (y1 P2) a + ViPi or -*3 + Pi = f^ + f/?, (k) or, comparing (j) and (k), the tangent is bisected by the focal perpendicular, and hence the angles between the tangent and the axis and the tangent and the radius vector are equal, and the tangent bisects the angle between the diameter and radius vector to the point of contact.APPLICATIONS TO LOCI. 191 (k) is also tlie perpendicular from the focus on the normal, and shows that the locus of the foot of the perpendicular from the focus on the normal is a parabola, tvhose vertex is at the focus of the given parabola and whose parameter is one-fourth that of the given parabola. 88. The Ellipse. 1. Substituting in the general equation p = xa-\-y¡3 the value of y from the equation of the ellipse referred to center and axes aV + òV = a262, we have p = £ca + m*(a2— x2)*j3 .... (220), b2 in which m = —0 and a and (3 are unit vectors along the axes. • ¡3 .... (222). 2. From Eq. (220) we have, for the vector along the tangent, a — m* (a2 — or) *x/3 = a----— —— - ¡3 = a-----(3 Vm Va2 — æ2 y = X {ya — mx/3) ; hence, for the equation of the tangent line, irz=:Xa + y(3-\-X(ya — mx/3) . . . (223); or, more simply, from Eq. (222), the vector-tangent is — sin (¡> • a + cos . /?,192 QUATERNIONS. and the equation of the tangent is 7T = cos • a + sin • /? + x( — sin <£ • a + coscj> • fi)> (224). Since — sin <¡> • a + cos . ft is along the tangent, cos <£ . a + sin<£ . ¡3 and — sin <£ . a-f-cos<£ . /? are vectors along conjugate diameters. 89. Examples on the Ellipse. 1. The area of the parallelogram formed by tangents drawn through the vertices of any pair of conjugate diameters is constant. We have directly TY [2 (cos cj> . a + sin • /?) 2( — sin . a + cos cf> • /?)] = 4 TYa/2 = a constant; namety, the rectangle on the axes. 2. The sum of the squares of conjugate diameters is constant, and equal to the sum of the squares on the axes. For, since Sa/2 = 0, (cos<£ . a-f sin<£ . (3)2 + ( — sin<£ . a + cos • /3)2 = a2 + /32. 3. The eccentric angles of the vertices of conjugate diameters differ by 90? The vector tangent at the extremity of is p = cos <ƒ> . a + sin <ƒ> . ¡3 (a) — sin cp • a + cos (j> • ¡3. This is also a vector along the diameter conjugate to p, and is seen to be the value of p when in (a) we write <£ + 90° for <£. 4. The eccentric angle of the extremity of equal conjugate diam- eters is 45° and the diameters fall upon the diagonals of the rectangle on the axes.APPLICATIONS TO LOCI. 198 5. The line joining the points of contact of tangents is parallel to the line joining the extremities of parallel diam- eters. 6. Tangents at right angles to each other intersect in the cir- cumference of a circle. 7. If an ordinate pd to the major axis he produced to meet the circumscribed circle in q, then QD : pd : : a : b. 8. If an ordinate pd to the minor axis meets the inscribed circle in Q, then QD : pd : : b : a. 9. Any semi-diameter is a mean proportional between the dis- tances from the center to the points lohere it meets the ordinate of any point and the tangent at that point. For the point p (Fig. 82) we have * p = cos cj> . a + sin $ . /?. Also OT = £COP = OQ + QT = æ(cos <£ • a + sin• ¡3) = cos<£r. a + sin<£' •/? + £(“ sin<£'. a + cos <£'. /3). Eliminating £, X = or But OT = XOV = 1 COS(cj> — <£') ______Í_______OP. COS(cfi — cj)') ON = £C(OP = OQ + QN = æ'(coscj> . a + sin. fi) = cos c/>\ a + sin <£' ./? + £'( — sin . a + COS cj> • /3) Eliminating t\ or x! = cos(<£ — <£'), on = cos(<£ — <£')op ; .*. ON . OT = 0P¿.194 QUATERNIONS. 10. To find the length of the curve. With the notation of Ex. 6, Art. 86, we obtain, from Eq. (222), p'= — sin . acos . /3, Tpf= V ( a2 — h2) sin2 <ƒ> + b2, _____________________ ; — s0 =J V (a2 — b2) sin2 + b2, which involves elliptic functions. If a = b, wè have, for the /»0 circle, s — s0 = I r = r( — 0). ^00 From Eq. (220), we obtain p'= a — 7)Ú(a2 — X2) “5£C/?, Tp'=Jl+^-X2=, ^ a2 — X2 *s/a2 — X2* a2 Jrx a L e2x2 l which may be expanded and integrated ; giving for the entire curve 2 Trail Se4 2.2 2.2.4.4 2 3.3.5e6 , \ --------------etc. 1, .2.4.4.6.6 J a converging series. If e = 0, we have, for the circle, 27rr. 11. To find the area of the ellipse. With the notation of Ex. 7, Art. 86, TY/opf= TV (cos • a + sin ./?)( — sin • a 4- cos • /3) = TY (cos2 . a/3 — sin2 • /3a) = TYa/3 ; A or, since a/3 = 90° ^ i Ç rHVpp'= \irab. The whole area is therefore wab.APPLICATIONS TO LOCI. 195 90. The Hyperbola. 1. Let a and ft be unit vectors parallel to the asymptotes. Then, from the equation, =m, . m a p — xa -I-p X we have, for the equation of the hyperbola, .... (225); or, if a and /3 are given vectors parallel to the asymptotes, p = ta+i....................(226); t or, again, in terms of the eccentric angle, p = sec<£ . a + tan<£ . j3 . . . . (227). 2. The equation of the tangent, obtained as usual, is from Eq. (226), -§)• . . . (228), p = ta + — + X^ta where ta — is a vector along the tangent. t 91.' Examples on the Hyperbola. 1. If, lühen the hyperbola is referred to its asymptotes, one diagonal of a parallelogram whose sides are the coordinates is the radius vector, the other diagonal is parallel to the tangent. If (Fig. 79) cx = ta, xp =^, then î CP = ta -f QX = ta ; t t196 QUATERNIONS. 2. A diameter bisects all chords parallel to the tangent at its vertex. Let (Fig. 79) cp be the diameter, t corresponding to the point p. The tangent at p is parallel to ta — ë. and cp = ta -j- p'p" being the parallel chord, ^ CP = CO -b 0Pf= X (\,a Also, if t' correspond to pJ CP'—t'a+^-, ••• (x + y)t = t', !Lzl = l, orAPPLICATIONS TO LOCI. 197 Hence, for every point, as o, determined by a?, there are two points p' and p", determined by the two corresponding values of y, which are equal with opposite signs. 3. The tangent at pA to the conjugate hyperbola is parallel to cp (Fig. 79). 4. The portion of the tangent limited by the asymptotes is bisected at the point of contact. 5. If from the point d (Fig. 79), inhere the tangent at p meets the asymptote, dn be drawn parallel to the other asymptote, then the portion of pn produced, which is limited by the asymptotes, is trisected at p and n. We have 6. The intercepts of the secant between the hyperbola and its asymj)totes are equal. and the equation of ss' is whence, for the points s, s' æ = —-1, X = 2. The vector along the tangent parallel to the secant is ta — I Hence (Fig. 79) t c r"=2'/2= X but op" s= op' (Ex. 2), and therefore pf,Rff = p'r!198 QUATERNIONS. 7. If through any point p" (Fig. 79) a line r'^p'r' he drawn in any direction, meeting the asymptotes in r" and rJ then Pf R,f • PfRf = PDf2. 8. If through p¡ p" (Fig. 79) lines he dr aim parallel to the asymptotes, forming a parallelogram of which p'p" is one diagonal, the other diagonal will pass through the center. The vector from c to the farther extremity of the required diagonal is +£+(* - +(?>“■?) 0 - *'a+? ■- b (fa+?)* But t"a + - is the vector from c to the other extremity of the required diagonal. 9. If the tangent at any point p meet the transverse axis in t, and pn he the ordinate of the point p ; then CT . cn = a2, c being the center and a the semi-transverse axis. From Eq. (227), substituting in ct = cp 4- pt, X sec <£ . a = sec . a + tan • /? + y (tan sec . a + sec2 • /?) ; and or ^ —-----77’ sec^<£ CT • CN = (xseecf} . a) (sec<£ . a) = a2, ct . cn = a2. 10. If the tangent at any point p meet the conjugate axis at t,' and pn' he the ordinate to the conjugate axis, then ct' . CNf = h\ c being the center and h the semi-conjugate axis.APPLICATIONS TO LOCI. 199 92. The preceding examples on the conic sections involve directly the Cartesian forms. A method will now be briefly indicated peculiar to Quaternion analysis and independent of these forms. 1. The general form of an equation of the first degree, or as it may be called from analogy, a linear equation in quaternions, is aqb + a'qb'+ a"qbn +...= c, or %aqb = c, (a) in which q is an unknown quaternion, entering once, as a factor only, in each term, and a, b, a'f b\.., c are given quaternions. It may evidently be written whence But and SSagö -f STaqb = Sc + Yc, 2S aqb = Sc, 2Yaqb = Yc. Saqb = Sqba = SgS&a + S . YgYöa, O) (<0 Vaqb = Y (Sa + Va) (Sç + Yq) (Sb + Yb) = V . S?(Sa + Va) (Sb + Yb) + V (SaYqSb + SaYqYb + YaYqSb + YaYqYb) = SqYab + V(SaS5 - SaYb + SbYa)Yq + Y . YaYqYb + V . YaYbYq ~ Y . YaYbYq [Eq. (116)] = SqYab + V (SaSô - SaYb + SbYa - YaYb) Yq + 2 VaS .YqYb = SqYab + V . a (KÒ)Vg + 2 VaS . YqYb. YYe have therefore, from (b) and (c), Sc = SqXSba + S . YqXVba, Yc = SqVfab + 5V. a(Kb)Yq + 2SVaS . YqYb, or, writing 2Sa5 = cl, Uab = 8, lYba = S| Sq-w, Yq = p,200 QUATERNIONS. we obtain Sc = wã + Spïï, Tc == wS + SY . a (Kb)p + 2 SYaS . pYb. We may now eliminate w between these equations, obtaining Yc . cl - Sc . 8 = dEYa(K&)p - SS/>8'+ d2SYaS . PYb which involves only the vector of the unknown quaternion g, and which, since Y and S are commutative, may be written under the general form B y = Yrp + 2/3Sap, in which y, a, a, ..., ¡3, /?J..are known vectors, r a known quaternion, but p an unknown vector. This equation is the general form of a linear vector equation. The second member, being a linear function of p, may be written Yrp + S/3Sap = p = y .... (229), where <£p designates any linear function of p. If we define the inverse function -1 by the equation \4>p) = p> the determination of p is made to depend upon that of c¡r1. 2. Without entering upon the solution of linear equations, it is evident on inspection that the function is distributive as regards addition, so that + #'+ . . . (230). Also that, a being ahy scalar, <£ap = ap................(231), and dp = dp...............(232). 3. Furthermore, if we operate upon the form <¡>p = S/?Sap + YrpAPPLICATIONS TO LOCI. 201 with S • cr X, cr being any vector whatever, Scrc/>p = 2S (cr/?Sap) + Scr(Yrp). But S(cr/3Sap) = Scr/3Sap = SpaSjScr = S(paSj3cr), and S(oYVp) = S[o-V(Sr + Yr)p] = SrScrp + Scr(Yr)p = SrSpo- - Sp(Vr) by the interchange of the letters a and /?, and Kr for r, we shall have, whatever the vectors p and cr, S(cr<¿p) = S(p<¿'cr). Functions, which, like and , enjoy this property, are called l conjugate functions. The function <ƒ> is said to be self-conjugate, that is, equal to its conjugate <£, when for any vectors p, cr, Scrc^p = Sp<£cr. 93. In accordance with Boscovich’s definition, a conic sec- tion is the locus of a point so moving that the ratio of its dis- tances from a fixed point and a fixed right line is constant. 1. Let F (Fig. 80) be the fixed point or Fig. so. focus, no the fixed line or directrix, and p FP D any point such that — = e, the constant ratio DP or eccentricity. Draw fo perpendicular to the directrix, and let fo=a, OD=yy, pd=xa and FP = p. By definition, -^ = e T(pd) ’ or p2 = eW. (a) Also p -J- Xa = a -f- yy.202 QUATERNIONS. Operating with S . a X , we have, since Say = 0 Sap + #a2 * = a2, or CC2a4 = (a2 — Sap)2. Substituting in (a) a2p2=e2(a2— Sap)2 .... (233), in which e may be less, greater than, or equal to unity, corre- sponding to the ellipse, hyperbola and parabola. 2. For the ellipse, Fig. 81, putting p — xa for the points a and aJ we have Fig. 81. D O A P e and X e X = 1+e 1—e or, since p = xa = æfo e FA =-----FO, 1+e FA'= —-—FO, 1—e whence and therefore FO a,APPLICATIONS TO LOCI. 203 which furnish the well-known properties of the ellipse, fa = a(l — e), fa'= a( 1 -f-e), of = ae, e 3. Changing the origin to the center of the curve, let cf = a' ; a2p2 = e2(a2 — Sap)2, remembering that a'2 = — a2e\ we obtain a2p,2 + (Sa'p')2 = - a4(l - e2), or, dropping the accents, c being the initial point, a2p2 + (Sap)2 = — a4(l — e2) . . . (234), the equation of the ellipse in terms of the major axis with the origin at the center. If p coincides with the axes, Tp = a or &, as it should. 4. Equation (234) may be deduced directly from Newton’s definition, thus : let cf = a (Fig. 81) as before, f and f' being the foci, and cp = p. Then - ] a ; whence V and, by definition, FP = p — a, f'p = p + a, FP + f'p = 2 a as lines ; or T (p — a) + T (p + a) = 2 a,204 QUATERNIONS. a being the semi-major axis. Whence V— (p — a)2 = 2 a — V — (p + a)2. Squaring — p2 + 2 Spa — a2 = 4 a2 — 4 a V — (p + a) 2 — p2 — 2 Spa — a2, Spa — a2 = — a V— (p + a)2. Squaring again ( Spa)2 - 2 a2 Spa + a4 = - a2 (p2 + 2 Spa + a2), a2p2 + (Spa)2 = —a4 — a2 a2, or, as before, ci2 p2 + (Spa)2 = — q4(l — e2). 94. 1. The equation of the ellipse a2p2 + (Sap)2= — a4(l — e2) may be put under the form S . p a2 p -j- aSap a4(f— e2) or, in the notation of Art. 92, writing CL2 p + aSap a4(l — e2) Pj the equation of the ellipse becomes Sp<£p = 1 (235). 2. By inspection of the value of p it is seen that, when p coincides with either axis, p and <£p coincide. Operating on p with S . p = a2 Scrp 4~ ScraSap . a4(l-e2) 5APPLICATIONS TO LOCI. 205 operating on écr = — —^ with S . p X , we have a4(l — e¿) hence c j a2Spo- + SpaSacr ap*(r = Sotp ...... (236), and is self-conjugate. 3. Differentiating Equation (235), we have Sclptjyp -j- Spdcfrp = 0, Sdpfo + Spc/>clp = 0, [Eq. (232) ] Spdp + Sp4>dp = 2 Sp<¿dp = 0. [Eq. (236) ] If 7T be a vector to any point of the tangent line, 7T = p -f- xdp, whence Sp^)(7r —p) = S(?r —p)^)p = 0, (q) or S7T^)p = Sp^)p = Sp<^7T = 1 . . . . (237) is the equation of the tangent line. From (a) we see that p is a vector parallel to the normal at the point of contact, being parallel to p only when p coincides with the axes, as already remarked. 4. To transform the preceding equations into the usual Car- tesian forms, let i be a unit vector along ca (Fig. 81), and j a unit vector perpendicular to it. If the coordinates of p are x and 2/> then, since a = cf, P = xi + yj, and __ a2p + aSap_______a2(xi + yj) + qeiS * aei(xi + yj) ^ ~ a4(l — e2) ~~ a4(l —e2) _ a2æi (1 — e2) + a2yj ““ a4(l — 62) '206 QUATERNIONS. -, 2 & or, since 1 — r = — cr ~ [ d2 + V)' and Spp = 1 = — S . (xi + yj)(^ + yl^, a2 if + b2x2 = n2b2. Again, if X' and y' are the coordinates of a point in the tangent, ir = x’i + y'j ; and a2yy'-\- b2xx' = a2b2. The above applies to the hyperbola when e > 1, that is, when b2 1 — e2 = — —, giving the corresponding equations a2y2 — b2x2 = — a2 ò2, a2yy'— = — a2ò2. 95. Examples. 1. To find the locus of the middle points of parallel chords. Let /? be a vector along one of the chords, as rq (Fig. 82), the length of the chord being 2 y, and let y be the vector to its middle point ; then P = y + VP and p = y — y¡3 are vectors to points of the ellipse, and s (y+yft) $ (y + yfi)=l, S(y-yj8)^(y-y/8)=i; whence, expanding, subtracting, and applying Equation (236), M>/? = 0,APPLICATIONS TO LOCI. 207 the equation of a straight line through the origin. Since /3 is parallel to the normal at the extremity of a diameter parallel to /?, the locus is the diameter parallel to the tangent at that point. Fig. 82. 2. Equation of condition for conjugate diameters. Denote the diameter op (Fig. 82) of the preceding problem, bisecting all chords parallel to /?, by a. Then Sa<£/J = 0, or S/fya = 0. In the latter, ¡3 is perpendicular to the normal at the ex- tremity of a, and is therefore parallel to the tangent at that point; hence this is the equation of the diameter bisecting all chords parallel to a. Therefore, diameters which satisfy the equation Sa<£/3 = 0 are conjugate diameters. 3. Supplementary chords. Let pp' (Fig. 82) and dd' be conjugate diameters, and the chords PD, pdf be drawn. Then, with the above notation, dp = a — a — acj>/3 + (3cf>a — .208 QUATERNIONS. But Saa = 1, Sficjyfi = 1, Sa<£/? = S/3$a ; S(a + /3)^>(a —/3) = 0. Hence, if dp is parallel to a diameter, PDf is parallel to its conjugate. 4. If two tangents be drawn to the ellipse, the diameter parallel to the chord of contact and the diameter through the intersection of the tangents are conjugate. Fig. 82. Let TQ (Fig. 82) and tr be the tangents at the extremities of the chord parallel to /?, and ot = 7r. Then OQ = Xa -f- yP, OR = Xa -j- y’p. From the equation of the tangent S?rp = 1, we have S7T^> (xa + yP)== 1, S7T ^(xa-^-y'P) = 1. Expanding and subtracting StTp = 0. Hence, Ex. 2, ir and /?, or op and od, are conjugate. The locus of T for parallel chords is the diameter conjugate to the chord through the center.APPLICATIONS TO LOCI. 209 5. If qoq' (Fig. 82) be a diameter and qr a chord of contact, then is q'r parallel to ot. rq being parallel to /?, and oq' = — oq, we have rq = 2 y¡3, rq' = yft — rxa — xa — y/3 ; whence, directly rq' = —- 2&a ; as also Srq<£rq' = 0, rq and rq' being supplementary chords. 6. The points in which any two parallel tangents as q't$ qt (Fig. 82) are intersected by a third tangent, as ttJ lie on conju- gate diameters. The equation of rt' is = 1, and that of q't' is S7r'p'=l. For the point t' tt = 7r' ; whence, by subtraction, S7T^)(p — p') = 0. 7. Chord of contact. The equation of the tangent, Spcf>ir = 1, is linear, and satisfied for both q and r. Hence, writing 7r = 1 is the equation of the chord of contact. 8. To find the locus of t for all chords through a fixed point (Fig. 82). Let s be a fixed point of the chord rq, so that os = o- = a constant. Then Scr7r = S7T<£cr = 1, a right line perpendicular to <£p = 1, we obtain Sa?fa (xa + y/3) = 1 ; whence, since Sacf>/3 = 0, or Xx'SacfHL— 1, xxr = 1 ; or •. Xa • Xra = a2, ON • OT = OP2. 10. Ifjyo1 (Fig. 82) and ppr are conjugate diameters, then are PD and pd' proportional to the diameters parallel to them. With the same notation whence DP = a — ¡3, DfP = a + /?, OE = m(a — ¡3), of = n(a + /3).APPLICATIONS TO LOCI. 211 From the equation of the ellipse Sm(a — (3)4>m(a — /8) = 1, and $?ì(a -f- ƒ8) cj>n (a -}- ƒ8) = 1. (а) (б) Now, from (a), since S¡3^/3 = Sa<£a = 1 and S/3<£a = Sa<£/? = 0, 2m2 = 1. Similarly, from (6), 2w2=l; Also m = w. DP : d'p : : T(a - p) : T(a + /?) : : Tm(a — /?) : T?i(a + /?) : : OE : of. 11. The diameters along the diagonals of the parallelogram on the axes are conjugate; and the same is true of diameters along the diagonals of any parallelogram whose sides are the tangents at the extremities of conjugate diameters. 12. Diameters parallel to the sides of an inscribed parallelo- gram are conjugate. Fig. 83. Let the sides of the parallelogram (Fig. 83) be and let PP' = a, PQ = j8, OP = p, OQ = pi OPr — p + a, OQ' = p' + a, p' — p = ¡3. Then212 QUATERNIONS. From the equation of the ellipse, Sp<£p=l, we have for qf and p' S(p'+a)<^)(p,+ a)= 1, S(p +a)p'+Sa<£a = 0, 2 Sa <£p -f- Sa<£a = 0. Spfa — Spc/>a = 0, S(pf — p)<£a = S/?c/>a = 0. 13. The rectangle of the 'perpendiculars from the foci on the tangent is constant, and equal to the square of the semi-conjugate axis. Let the tangent be drawn at r (Fig. 83) and or = p. Then €¡)p is parallel to the normal at r, that is, to the perpendiculars FD, f'd! Hence, of being a, od'= æ'<£p — a, OD = a -f- #<£p, which, since d and Df are on the tangent, in S7r<£p = 1 give S (x'^p — a)p)p = 1, æ'(<£p)2 = 1 + Sa p, X (<£p)2 = 1 — Sa<£p ; orAPPLICATIONS TO LOCI. 213 whence and But T x'p = f'd'= T 1 + Sa p = FD = T--------- 9P fBy_T1~~(Sa^)2 “ (#)2 ' FD X F '2 a2(a2p2) + 2 ft2 (Sap) 2 + a2 (Sap) 2 a8( 1 — e2)3 ’ /, N2 / a2p + aSapV_ ^ = V“ a4(l — e2) / “ or, substituting a2p2 from Equation (234) and a2 = — a2e2, Also _ (Sap)2— a4 a6(l — è2) ' 1 — (Sa<¿p)2 = 1 — |j a2 Sap + a2 Sap 2 a4 — (Sap)2 a4(l — e2) J a4 t t m^4 —(Sap)2 a6(l — e2) 2/ 2X » fdxf'd' = T------~t o—~4 == a2(l — e2) = &2. a4 (Sap)2 —a4 v 7 14. The foot of the perpendicular from the focus on the tangent is in the circumference of the circle described on the major axis. To prove this we have to show that the line od (Fig. 83) is equal to a. Now od = a + Xc¡>p — n I ÏPO—S^P) “ (#)2 from the preceding example. Hence /0t>\2 a i 2 Sap)2 - ae+ a4 (Sap)2-a4 == — a2e2 — a2(l — e2) = —a2 ; od = a.214 QUATERNIONS. The Parabola. 96. 1. Resuming Equation (233) and making e — 1, the equation of the parabola is a2p2 = (a2 — Sap)2 . which may be written p2 + 2 Sap — a-2 (Sap)2 1 : ~2 — I? (238), or Sp in which, if we put P H- 2 a — a — a 2aSap”|_____i 2 ^ I > a2 J . p — a 1 Sap w = —12—’ (239), we have for the equation of the parabola Sp(<£p + 2a“1)= 1 . . . . and, as in the case of the ellipse, Scr<£p = Speyer..............(240). Operating on p by S . a x, we obtain Sa<£p = 0..................(241); hence, 4>P ^ a perpendicular to the axis. Operating on p by S . p X Sp<£p = p2 — a 2(Sap)* = a2()2. . . (242). 2. Differentiating Equation (239), we have 2 Spcf>dp + 2 Sdpa-1 = 0. For any point of the tangent line to which the vector is 71-, 7r = p + ^p,APPLICATIONS TO LOCI. 215 from which, substituting dp in the above, S/op + Tra-1—pa"1) = 0 ; (a) or, since Spj>p =1 — 2Spa-1 [Eq. (239)], S7rp — 1+2 Spa"1 + Sira"1 — Spa-1 = 0, whence S7r(<£p + a--1) + Spa"1 = 1 . . . . (243), the equation of the tangent line. 3. From (a) we obtain S(?r — p) (p + a"1) = 0 ; 4 or, since 7r — p is a vector along the tangent, 4>p + a_1 is in the direction of the normal. 4. If a be a vector to any point of the normal, the equation of the normal will be o- = p + x(cf>p + a"1)...............(244). 5. The Cartesian form of Equation (239) is obtained by making p = xi + yj, a = Fo (Fig. 80) = — pi ; p=- xi + yj — -jr xpi ~¥ = yj. p29 whence, Equation (239) becomes y2 7" 1 ’ p p .-. y2=2pæ+p2, the equation of the parabola referred to the focus.216 QUATERNIONS. 97. Examples. 1. The subtangent is bisected at the vertex. Tig. Sé. We have (Fig. 84) ft = rea, which in the equation of the tangent Str( + a"1) + Spa”1 = 1 gives $xa(c¡)p + a-1) + Sa-1p = 1. But Sap = 0 ; hence X + Sa_1p = 1 ; (a) multiplying by a Xa + aSa-1p = a, (x — ^)a = a — -Ja — aSa_1p = \ a — aSa-1p, AT = — AF — aSa-1p. But the value of p gives a2 p = p — a-1 Sap ;APPLICATIONS TO LOCI. 217 and, since <£p is a vector along mp and a-1 Sap a vector along fm, from p = fm + mp we liave FM = a-1 Sap = aSa-1p, (Ò) MP = a2 cj>p ; (c) AT = — AF — FM = — AM, or, as lines, at = AM. 2. The distances from the focus to the point of contact and the intersection of the tangent with the axis are equal. or (Fig. 84), [Eq. (288)] Xa = a — aSa *p, (ft)2 = (a — aSa _1p)2 = (a — a 1Sap)2 __(a2 — Sap)2 .’. FP = FT. 3. The subnormal is constant and equal to half the parameter. The vector-normal being <¡>p + a-1 (Art. 96, 3), we have (Fig. 84) PN == 2(<£p + a-1) ; but PN = PM + MN = — a2cf>p + Xa ; [Ex. 1, (c)] or %(4>p + a l) — — a2p + a = FO + OD = FD. lFD = FQ = !a2<£/> + £a = i.a?p + FA ; *. ia2p, then, since Sap = 0 [Eq. (241)], z(4>py=^p(l>p = a-2((f>py‘i [Eq. (242)] Z = a2, and cr = + a2(<£p + a"1) = § a + a2<£p, or p + a-1) + Spa-1 = 1, or 2 S7T<£p + 2 Sa_17T + 2 Sa _1p = 1. Operating on (a) with x S . <£p, whence S^p = æ(c£p)2, the preceding equation becomes 2®(^»)*-^ + 2Sa-V=l. (6)220 QUATERNIONS. Also [Eq. (242)] Sp<£p = a2(<£p)2, which, in the equation of the parabola Sp(<£p + 2 a"1) = 1, gives a2(^>p)2 + 2Sa-1p=l. (c) Whence, from (b) and (c), by subtraction, c*pr But, from (a), ()2 = 2x 2a2x + a* it2—2 æSttcT1 -f- X2a~2 7T2 1 -------------------- _ + a? ÍC2 Equating these values of (p)2, and substituting the value of a?, 2 tt2 Sour — a27T2 + (Sa7r) 2 = 0, which is the equation of the locus required. To transform to Cartesian coordinates, make whence and 7r2 = 7T = xi + yj, and a = aí, — (x2 + y2) 5 Sa.7T = — ax, a2 æ3 2T a ----a; 2 aa the equation of the cissoid to the circle whose diameter is the distance from the vertex to the directrix. 7. If pp' (Fig. 75) be a focal chord, and pa, pa' produced meet the directrix in d¡ d, pd and p'd' 6e parallel to af. or ADf= — XAP = AO + OD,' x(i~p)=i+yy- Operating with S . a X — 2 Sap) = a?. (a)APPLICATIONS TO LOCI. 221 Now FP = p and fp' = — x'p are vectors to points on the curve, and hence satisfy its equa- tion. Whence [Eq. (238)] a2p2 = (a2 — Sap)2, X,2a2p2 = (a2 + of Sap) 2 ; . •. x'2 (a2 — Sap) 2 = (a2 -f- x''Sap) 2 ; or of (a2 — Sap) = a2 + âfSap, x\o? — 2 Sap) = a2. Hence, comparing with (a), X = x\ or, the sides produced of the triangle apf are cut propor- tionately, and therefore d'p' is parallel to af. Fig- 75 (bis). 8. If with a diameter equal to three times the focal distance, a circle he described with its center at the vertex, the common chord bisects the line joining the focus and vertex. The equation of the curve being a2p2 = (a2 — Sap) 2, (a) that of the circle whose center is a (Fig. 75), referred to f, is of the form [Eq. (210)] • T (p y) =T£, or, by condition, T P- Tfa; 2 =T?a' which, in (a), gives p2 = Sap + A Sap = 4’ which is the proposition.222 QUATERNIONS. 98. The Cycloid. 1. Let a and ¡3 be vectors along the base and axis of the cycloid and Tfi = Ta = r, the radius of the generating circle. Then, for any point p of the curve, x — r6 — rsm6 = r(6 — sin 0), y = r — rcos0 = r(l — cos0), and the equation of the cycloid is P = (0 — sin0)a + (1 — cos 0)(3. 2. The vector along the tangent is (1 — cos 0) a + sin 6 . /?, and the equation of the tangent is 7T = (0 — sin0)a + (1 — cos#)/? + ¿[(1— cos0)a + sin# . /?]. 3. The vector from p to the lower extremity of the vertical diameter of the generating circle through p is pc = — (1 — cos0)/3 + sin9 . a, and, from the above expression, for the vector-tangent pt, S(pc . pt) = 0 ; hence pc is perpendicular to the tangent, or the normal passes through the foot of the vertical diameter of the generating cir- cle for the point to which the normal is drawn, and the tangent passes through the other extremity. 4. If, through p, a line be drawn parallel to the base, intersecting the central generating circle in q, show that PQ = v(ir — 0) = arc QA, a being the upper extremity of the axis. /APPLICATIONS TO LOCI. 223 5. With the notation of Ex. 6, Art. 86, p' = (1 — COS#)a + sin# . /?, p12 = — [(1 — cos #)2 + sin2#]?*2, Tp1 = r Vl — 2 cos# + cos2# + sin2# = r V2 — 2 cos# the length of the entire curve. 6. With the notation of Ex. 7, Art. 86, TYppf= TV[(# — sin#)sin# . afi + (1 — cos#)2/?a] = TY[(# sin# — sin2# —(1— cos#)2]a/2 = 7^(0 sin# + 2 cos# — 2). the whole area of the curve. 99. Elementary Applications to Mechanics. 1. If b he the magnitude of any force acting in a known di- rection, the force, as having magnitude and direction, may be represented by the vector symbol /?, which is independent of the point of application of the force. In order, completely, to define the force with reference to any origin o, the vector OA=a, to its point of application a, must also be given. For concur- ring forces, whose magnitudes are b[ ......, we have, for the resultant, ¡3 = 2$ which is true, whether the forces are compla- nar or not, and is the theorem of the polygon of forces extended. For two forces, ¡3 = (3*3- ¡3" ; whence ¡32 = ¡3,2 + /3"2 + 2 Sor —(3 sin# — #cos# — 2#)~j = 3 nr2. ^ 197r ==3*rr*, 1° Í27r224 QUATERNIONS. b2 = bf2 + bm + 2b'b” cos #, which is the theorem of the parallelo- gram of forces. For any number of concurring forces, the con- dition of equilibrium will be 2/?'= 0. For a particle constrained to move on a plane curve whose equation is p = <£(£), dp being in the direction of the tangent, since the resultant of the ex- traneous forces must be normal to the curve for equilibrium, we have Sdp2¿8'=Sdpj3 = 0. (a) 2. If OAf=a,' and¡3' is a force acting at a¡ then TYa'/?'=a'5' sin# is the numerical value of the moment of the couple f3f at a1 and — ¡3' at o. Representing, as usual, the couple by its axis, its vector symbol will be YoJfil If — ¡3' act at some point other than the origin, as c' and ocf= y', the couple will be denoted by Y(af— y')/3i From this vector representation of couples, it fol- lows that their composition is a process of vector addition ; hence the resultant couple is 2Y(ay')/?,' and, for equilibrium, 2Y(a/— yf)y$f= 0. If the couples are in the same or parallel planes, their axes are parallel and T2 = 2T. Since a'—y' is independent of the origin, the moment of the couple is the same for all points. Since Y(af— y')/?'= Ya'/?'—Yyf$ the moment of a couple is the algebraic sum of the moments of its component forces. If the forces are concurring, and a1 is the vector to their common point of application, 2Yaf/?,== Y2a,/?,= Ya'2/?f= Ya'/?, or the moment of the resultant about any point is the sum of the moments of the component forces. When the origin is on the resultant, af coincides with ¡31 in direction, and Y a'(3 = 0 ; or the algebraic sum of the moments about any point of the resultant is zero. If a single force ¡3f acts at a' we may, as usual, intro- duce two equal and opposite forces at the origin, or at any other point c' and thus replace /?'A, by /3'0 and Yaf/3', or by /3!G> and Y (a' — y !)/3f. If f be a unit vector along any axis oz through the origin, then the moment of f3* acting at a', with reference £o the axis oz, will be — S/3'a/£, or — S . ÍY¡3'a’. If ¡3’ and £ are in the same plane, in which case they either intersect or are parallel ; or, if the axis passes through a' there will be no moment : in these cases, a' (31 and £ are complanar, and — S/3V£ = 0.APPLICATIONS TO LOCI. 225 3. If the forces are parallel, their resultant ft = 2/3r= 2&'U/?f = Up%V ; and, therefore, for equilibrium, 2T/3f== 0. The moment of a force with reference to any axis oz through the origin being — S/3V£, and the moment of the resultant being equal to the sum of the moments of the components, we have S/?a£==2S/?'a/£, which, for parallel forces, becomes S(2&'. Uft . a£) = S(U/32öV • £), which, being true for any axis, is satisfied for 26'. a = 2&V; 25' a! 26' ’ (0 which is independent of U/3, and hence is the vector to the cen- ter of parallel forces. When 2&r= 0, the above equations give /3 = 0 and a = ao, the system reducing to a couple. For a sys- tem of particles whose weights are w\ w"., we have the vec- 2wV tor to the center of gravity a = 2wf From this equation, 2wf(a — a') = 0 ; whence, if the particles are equal, the sum of the vectors from the center of gravity to each particle is zero; and, if unequal, and the length of each vector is increased propor- tionately to the weight of each particle, their sum is zero. For equal particles, a = Wr2 af :ü¡7 , or the center of gravity of a system of equal particles is the mean point (Art. 18) of the polyedron of which the particles are the vertices. For a continuous body whose weight is w, volume v, and density d at the extremity of 2d dva* 2d dv , in which 2 may be replaced by the integral sign if the density is a known function of the volume. For a homo- geneous body, a = 2 dvar 2 dv , which is applicable to lines, surfaces or solids, V representing a line, area or volume. Thus, for a plane curve p = <¡>(t) = a\ dv = ds — Tdp = T$f(£)di and Ç4>(t)T'(t)dt fü'(t)dt («)226 QUATERNIONS. 4. General conditions of equilibrium of a solid body. Let the forces /?' /?,"., act at the points aJ a¡'.of a solid body, and oa' = a' oa" = a" ...... Replacing each force by an equal one at the origin and a couple, the given system will be equiva- lent to a s}'stem of concurring forces at the origin and a system of couples. Hence, for equilibrium, 2/3'= 0, (d) 2Ya'/3'= 0. (e) Let £ be the vector to any point x. Then, from (d), Y • £2/3'= 0, and therefore, from (e), Y . £2/3'= 2Ya'/3' ; whence 2Y/3V- 2Y£'£ = 2Y/3'(a'- £) = 0. ( ƒ) Conversely, £ being a vector to any point, the resultant couple, for equilibrium, is 2Y(a'— £)/3'= 0 ; .*. 2Ya'/3'= 0 and 2/3'= 0. Therefore ( ƒ ) is the necessary and sufficient condition of equi- librium. This condition ma}r be otherwise expressed by the principle of virtual moments. Let 85 8','...be the displacements. Then the virtual moment of /3' is — S/3'8' ; and, for equilibrium, 2$/3'8'=0. This equation involves (d) and (e). Thus, if the displacement corresponds to a simple translation, 8'= 8" =8"' = etc. = a constant, and we may write 2S/3'8' = S32/3' = 0 ; whence, since 8 is real, 2/3'= 0. Again, if the displacement corresponds to a rotation about an axis £, £ being a unit vector along the axis, a'= Y£a') = - £S£a'- £Y£a' the last term being a vector perpendicular to the axis. For a rotation about this axis through an angle 0, this term becomes 20 — £?r £Y£af= — £ COS0 Y£a'-f- sin0 Y£a¡ and a' becomes a 1= - £S£a'— £ COS0 Y£a'+ sin<9 Y£a' which, for an infinitely small displacement, = — £S£a'— £Y£a'+ 0Y£a!APPLICATIONS TO LOCI. 227 Placing the scalar factor under the vector sign and writing £ simply for to denote the indefinitely short vector along oz, a/+S'=a'+Y£a'; or, 8'= Y£a! Hence SS/3'8'= SS/3'Y£a'= S£SYa'0' ; or, since £ is not zero, SYa’¡3'= 0. 5. Illustrations. (1) Three concurrent forces, represented in magnitude and direction by the mediais of any triangle, are in equilibrium. (See Ex. 2, Art. 17.) (2) If three concurring forces are in equilibrium, they are complanar. By condition, ¡3’+ /3"+ /3nl= 0. Operating with S . p(pnX , we have 0. (3) In the preceding case, operating with Y. /3fx, we have Vf}t/3,,+Yp,pw= 0 ; whence, since the forces are complanar, TYƒ?'£"= TYp’p'," or b'b" sin(# p") = W" sin(p’ pln). A sim- ilar relation may be found for any two of the forces ; whence V : bn : bw : : sin(/?J' p,n) : sin (P\ fi,n) : sin(/?i fin). (4) If two forces are represented in magnitude and position by two chords of a semicircle drawn from a point on the circum- férence, the diameter through the point represents the resultant. (5) A weight, w[ rests on the arc of a vertical plane curve, and is connected, by a cord passing over a pulley, with another weight, w!f Find the relation between the weights for equili- brium. (a) Let the curve be a parabola, and the pulley at the focus. Then, from Eq. (a) of this article, the equation of the curve be- ing P = (y2—p2)a-+yfi, we have s (I « + is) (- w'a + = 0,228 QUATEBNIONS. in which r = radius vector. Hence y i, yx „, y „ A 1 pr r ’ or, since r = x + p, wf=wi' Hence, if the weights are equal, equilibrium will exist at all points of the curve. ( b) Let the curve be a circle and the pulley at a distance m from the curve on the vertical diameter produced. With the origin at the highest point of the circle, p = xa + V2 nx — a?ft. Hence, r being the distance of the pulley from w[ s \ y ¡à + a W'a ■ y ft■'+ (m + x)ani n ) = 0; w" no' K + m (c) Let w be placed on the concave arc of a vertical circle, and acted upon by a repulsive'force varying inversely as the square of the distance from the lowest point of the circle. To find the position of equilibrium. The origin being at the lowest point of the circle, and r the distance required, let p be the intensity of the force at a unit’s distance ; then ~ will be its intensity for any distance r, and whence R-x \fxa + y/3p y P)\ r r2 r = (d) Let w1 rest on a right line inclined at an angle 0 to the horizontal, and connected with wn by a cord passing over a pul- ley at the upper end of the line. Find the relation between the weights. With the origin at the lower end of the line, its equa- tion is p = xa. If ft is in the direction of w[ then 8a(w'ft+w"a) = 0; w"= w1 sin0. (6) To find the center of gravity of three equal particles at the vertices of a triangle, a, b, c being the vertices, the vectorAPPLICATIONS TO LOCI. 229 from A to the center of gravity of the weights at A and b is Jab = ad. The vector to the center of gravity of the three weights is J(ab + ac) = Jab H-ædc = Jab+ æ(—Jab + ac) ; .*. æ=J, and the required point is the center of gravity of the triangle. (7) Find the center of gravity of the perimeter of a triangle. (8) Find the center of gravity of four equal particles at the vertices of a tetraedron. (9) Show that the center of gravity of four equal particles at the angular points of any quadrilateral is at the middle point of the line joining the middle points of a pair of opposite sides. (10) The center of gravity of the triangle formed by joining the extremities of perpendiculars, erected outwards, at the mid- dle points of any triangle, and proportional to the corresponding sides, coincides with that of the original triangle. Let abc be the triangle, bc = 2 a, ca = 2/3 and € a vector perpendicular to the plane of the triangle. Then, if m is the given ratio, b the initial point, and r15 r2, r3 the extremities of the perpendiculars to bc, ca, ab, respectively, BRX = a + me a, br2 = 2 a + /? + me/3, BR3 = a + ft — me(a +/?) ; • J-(BRi + br2 + BRs) == + 2/Î) = J[2a + 2 (a + /2)]. (11) To find the center of gravity of a circular arc. The equation of the circle p = r(cos0 . a + sin0 . /3), gives dp = r(— sin0 • a + cos0 . j3)dO ; For an arc of 90° integrating between the limits ~ and 0, u 2 a* 2r /“ a2 = —(a + /?), the distance from the center being — V2 ; which 7T 7r230 QUATERNIONS. may be obtained directly also by integrating between the limits 7 and — 7. For a semicircumference or arc of 60? we have, in 4 4 7 2 r 3 r like manner, — and_____ 7T 7r (12) If a, /?, y are the vector edges of any tetraedron, the origin being at the vertex, then p — a, ¡3 — y, a — /? are lines of the base, p being any vector to its plane. Hence this plane is represented by S (p - a) (0 — y) (a - £) = 0 ; .-. Sp (Ya¡3 + Yya + Y¡3y) — Sa¡3y = 0. If 8 be the vector perpendicular on the base, 8 = x(Ya/3 + Yya + Y/3y) = and, taking the tensors, Sa/?y Ya/3 + Y/3y + Yya’ T(Ya/5 + Y(3y + Yya) = —— = 2 area base. But Ya/3+ Y£y +Yya+ Y/?a + Yy/3 + Yay = 0, in which the last terms are twice the vector areas of the plane faces. The sum of the vector areas of all the faces is therefore zero. Since any polyedron may be divided into tetraedra by plane sections, whose vector areas will have the same numerical coefficient, but have opposite signs two and two, the sum of the vector areas of any polyedron is zero. These vector areas represent the pres- sures on the faces of a polyedron immersed in a perfect fluid subjected to no external forces. For rotation, since the points of application of these pressures are the centers of gravity of the faces, to which the vectors are , i(a + /3 + y), |.(^ + a), i(y + /?), i(a + y)5 we have the couples - i V K* + 0 + y) (Vaj8 +Y/3y+Yya) + (a + (3)Yf3a + (fi + y) Yy/? + (y+ a)Yayj = “iY(aY/?y + /JYya + yYa¡3), since aYa/3 + aY/?a = 0, etc. But, Equation (123), this sum is zero. Hence there is no rotation.MISCELLANEOUS EXAMPLES. 231 100. Miscellaneous Examples. 1. In Fig. 58, F, a and k are collinear. 2. In Fig. 58, AD2—AE2=AB2—AC2. 3. ín Fig. 13, if the lines from the vertices of the parallelo- gram through o and p are angle-bisectors, omitp is a rectangle. 4. If the corresponding sides of two triangles are in the same ratio, the triangles are similar. 5. /?, a, y being the vector sides of a plane triangle, if /?=a-f-y, show that &2=c2—cacosB+&5cose. 6. The sides bc, ca, ab of a triangle are produced to d, e, f, so that cd = mBC, ae = tica, bf == jxajb. Find the inter- sections Ql5 Q2, Q3 Of EB, FC ; FC, DA ; DA, EB. 7. In any right-angled triangle, four times the sum of the squares of the mediais to the sides about the right angle is equal to five times the square of the hypothenuse. 8. If abc be any triangle, m its mean point, and o any point in space, then ab2+ bc2+ ca2 = 3 (oa2+ ob2+ oc2) — (3 om)2. 9. If abcd be any quadrilateral, m its mean point, and o any point in space, then ab2 + bc2 + CD2 + DA2 = 4(oa2+ ob2+ oc2+ od2) — (4om)2— ac2— bd2. 10. If abc be any triangle, and c', Bf, a' the middle points of ab, ac, CB, then, o being any point in space, ab2+bc2+ca2=4(oa2+ob2+oc2) — 4(oB,2-f-oc'2-f oa'2). 11. If abc be any triangle and m its mean point, then ab2+ bc2+ ca2 = 3 (am2+ bm2+ cm2) . 12. Points p, Q, r, s are taken in the sides ab, bc, cd, da of a parallelogram, so that ap = mAB, bq — mBC, etc. Show that PQRS is a parallelogram whose mean point coincides with that of ABCD.232 QUATEKKLONS. 13. The sides of any quadrilateral are divided equably at p, q, r, s, and the points of division joined in succession. If PQRS is a parallelogram, the original quadrilateral is a i parallelogram, 14. The middle points of the three diagonals of a complete quadrilateral are collinear. 15. If any quadrilateral be divided into two quadrilaterals by any cutting line, the centers of the three are collinear. 16. If a circle be described about the mean point of a paral- lelogram as a center, the sum of the squares of the lines drawn from any point in its circumference to the four angular points of the parallelogram is constant. 17. A quadrilateral possesses the following property : any point being taken, and four triangles formed by joining this point with the angular points of the figure, the centers of gravity of these triangles lie in the circumference of a circle. Prove that the diagonals of this quadrilateral are at right angles to each other. 18. The sum of the vector perpendiculars from a, b, c,.... on any line through their mean point is zero. 19. a, 5, c are the three adjacent edges of a rectangular paral- lelopiped. Show that the area of the triangle formed by joining their extremities is -J-Vö2c2+ a2c2+a?b2. 20. Given the co-ordinates of a, b, c, d referred to rectangular axes. Find the volume of the pyramid o—abcd, o being the origin. 21. Any plane through the middle points of two opposite edges of a tetraedron bisects the latter. 22. The chord of contact of two tangents to a circle drawn from the same point is perpendicular to the line joining that point with the center. 23. If two circles cut each other and from one point of section a diameter be drawn to each circle, the line joining their extremities is parallel to the line joining their centers, and passes through the other point of section.MISCELLANEOUS EXAMPLES. 233 24. The square of the sum of the diameters of two circles, tan- gent at a common point, is equal to the sum of the squares of any two common chords through the point of tangency, at right angles to each other. 25. T is any point without a circle whose centre is c ; from t draw two tangents tp, tq, also any line cutting the circle in V, and pq in r ; draw cs perpendicular to tv. Then SR . ST = sv2. 26. If a series of circles, tangent at a common point, are cut by a fixed circle, the lines of section meet in a point. 27. In Ex. 26, the intersections of the pairs of tangents to the fixed circle, at the points of section, lie in a straight line. 28. If three given circles are cut by any circle, the lines of section form a triangle, the loci of whose angular points are right lines perpendicular to the lines joining the centers of the given circles. 29. The three loci of Ex. 28 meet in a point. 30. Given the base of an isosceles triangle, to find the locus of the vertex. 31. Find the locus of the center of a circle which passes through two given points. 32. Find the locus of the center of a sphere of given radius, tangent to a given sphere. 33. The locus of the point from which two circles subtend equal angles is a circle, or a right line. 34. Given the base of a triangle, and m times the square of one side plus n times the square of the other, to find the locus of the vertex. 35. Given the base and the sum of the squares of the sides of a triangle, to find the locus of the vertex. In Ex. 35, given the difference of the squares, to find the locus. 36.234 QUATERNIONS. 37. OB and oa are any two lines, and mp is a line parallel to os. Find the locus of the intersection of oq and bq drawn parallel to ap and op, respectively. 38. From a fixed point p, on the surface of a sphere, chords pp', pp", .... are drawn. Find the locus of a point o on these chords, such that pp', po = m2. 39. A line of constant length moves with its extremities on two straight lines at right angles to each other. Find the locus of its middle point. 40. Find the locus of a point such that if straight lines be drawn to it from the four corners of a square, the sum of their squares is constant. 41. Find the locus of a point the square of whose distance from a given point is proportional to its distance from a given line. 42. Find the locus of the feet of perpendiculars from the origin on planes cutting off pyramids of equal volume from three rectangular co-ordinate axes. 43. Given the base of a triangle and the ratio of the sides, to find the locus of the vertex. 44. Show that YapYp/3 = (Ya/3)2 is the equation of a hyperbola whose asymptotes are parallel to a and ¡3. 45. Find the point on an ellipse the tangent to which cuts off equal distances on the axes. 46. a and b are two similar, similarly situated, and concentric ellipses ; c is a third ellipse similar to a and b, its center being on the circumference of b, and its axes parallel to those of a and b : show that the chord of intersection of a and b is parallel to the tangent to b at the center of c. * Presswork by Ginn & Co., Boston.MATHEMATICS. 165 Wentworth & Hill's Exercises in Algebra. I. Exercise Manual. i2mo. Boards. 232 pages. Mailing price» 40 cts. ; Introduction price, 35 cts. — II. Examination Manual. i2mo. Boards. 159 pages. Mailing price, 40 cts.; Introduction price, 35 cts. Both in one volu?ne, 70 cts. Answers to both parts together, 25 cts. The first part (Exercise Manual) contains about 4500 problems classified and arranged according to the usual order of text-books in Algebra ; and the second part (Examination Manual) contains nearly 300 examination-papers, progressive in character, and well adapted to cultivate skill and rapidity in solving problems. Wentworth & Hill’s Exercises in Arithmetic. I. Exercise Manual. II. Examination Manual. i2mo. Boards. 148 pages. Mailing price, 40 cts.; Introduction price, 35 cts. Both in one volwne, 70 cts. Answers to both parts together, 25 cts. The first part (Exercise Manual) contains problems for daily practice, classified and arranged in the common order; and the second part (Examination Manual) contains 300 examination-papers, progressive in character. The second part has already been issuedf and the first part will be ready in August, 1886. Analytic Geometry. By G. A. Wentworth. i2mo. Half morocco. 000 pp. Mailing price, $0.00; for Introduction, The aim of this work is to present the elementary parts of the subject in the best form for class-room use. The connection between a locus and its equation is made perfectly clear in the opening chapter. The exercises are well graded and designed to secure the best mental training. By adding a supplement to each chapter provision is made for a shorter or more extended course, as the time given to the subject will permit. The book is divided into chapters as follows : — Chapter I. Loci and their Equations. “ II. The Straight Line. “ III. The Circle. “ IV. Different Systems of Co-ordinates. “ V. The Parabola. « VI. The Ellipse. “ VII. The Hyperbola. “ VIII. The General Equation of the Second Degree.188 MATHEMATICS. Peirce's Three and Four Place Tables of Loga- ritìntile and Trigonometric Functions. By James Mills Peirce, University Professor of Mathematics in Harvard University. Quarto. Cloth. Mailing Price, 45 cts. ; Introduction, 40 cts. Four-place tables require, in the long run, only half as much time ',s five-place tables, one-third as much time as six-place tables, and one-fourth as much as those of seven places. They are sufficient for the ordinary calculations of Surveying, Civil, Mechanical, and Mining Engineering, and Navigation ; for the work of the Physical or Chemical Laboratory, and even for many computations of Astron- omy. They are also especially suited to be used in teaching, as they illustrate principles as well as the larger tables, and with far less expenditure of time. The present compilation has been prepared with care, and is handsomely and clearly printed. Elements of the Differential Calculus. With Numerous Examples and Applications. Designed for Use ás a College Text-Book. By W. E. Byerly, Professor of Mathematics, Harvard University. 8vo. 273 pages. Mailing Price, $2.15 ; Intro- duction, $2.00. This book embodies the results of the author’s experience in teaching the Calculus at Cornell and Harvard Universities, and is intended for a text-book, and not for an exhaustive treatise. Its peculiarities are the rigorous use of the Doctrine of Limits, as a foundation of the subject, and as preliminary to the adoption of the more direct and practically convenient infinitesimal notation and nomenclature ; the early introduction of a few simple formulas and methods for integrating ; a rather elaborate treatment of the use of infinitesimals in pure geometry ; and the attempt to excite and keep up the interest of the student by bringing in throughout the whole book, and not merely at the end, numerous applications to practical problems in geometry and mechanics. James Mills Peirce, Prof, of Math., Harvard Univ. (From the Har- vard Register) : In mathematics, as in other branches of study, the need is now very much felt of teaching which is general without being superficial; limited to leading topics, and yet with- in its limits; thorough, accurate, and practical ; adapted to the communica- tion of some degree of power, as wellMATHEMATICS. 189 as knowledge, but free from details which are important only to the spe- cialist. Professor Byerly’s Calculus appears to be designed to meet this want. . . . Such a plan leaves much room for the exercise of individual judgment; and differences of opinion will undoubtedly exist in regard to one and another point of this book. But all teachers will agree that in selection, arrangement, and treatment, it is, on the whole, in a very high degree, wise, able, marked by a true scientific spirit, and calculated to develop the same spirit in the learner. . . . The book contains, perhaps, all of the integral calculus, as well as of the differential, that is necessary to the ordinary stu- dent. And with so much of this great scientific method, every thorough stu- dent of physics, and every general scholar who feels any interest in the relations of abstract thought, and is capable of grasping a mathematical idea, ought to be familiar. One who aspires to technical learning must sup- plement his mastery of the elements by the study of the comprehensive theoretical treatises.... But he who is thoroughly acquainted with the book before us has made a long stride into a sound and practical knowledge of the subject of the calculus. He has begun to be a real analyst. H. A. Newton, Prof, of Math. in Yale Coif New Haven : I have looked it through with care, and find the sub- ject very clearly and logically devel- oped. I am strongly inclined to use it in my class next year. S. Hart, recent Prof of Math, in Trinity Colf Conn.: The student can hardly fail, I think, to get from the book an exact, and, at the same time, a satis- factory explanation of the principles on which the Calculus is based ; and the introduction of the simpler methods of integration, as they are needed, enables applications of those principles to be introduced in such a way as to be both interesting and instructive. Charles Kraus, Techniker, Pard- ubitz, Bohemia, Austria ; Indem ich den Empfang Ihres Buches dankend bestaetige muss ich Ihnen, hoch geehr- ter Herr gestehen, dass mich dasselbe sehr erfreut hat, da es sich durch grosse Reichhaltigkeit, besonders klare Schreibweise und vorzuegliche Behan d- lung des Stoffes auszeichnet, und er- weist sich dieses Werk als eine bedeut- ende Bereicherung der mathematischen Wissenschaft. De Volson Wood, Prof, of Math., Stevens' Inst., Hoboken, N.J. : To say, as I do, that it is a first-class work, is probably repeating what many have already said for it. I admire the rigid logical character of the work, and am gratified to see that so able a writer has shown explicitly the relation between Derivatives, Infinitesimals, and Differentials. The method of Limits is the true one on which to found the science of the calculus. The work is not only comprehensive, but no vague- ness is allowed in regard to definitions or fundamental principles. Del Kemper, Prof, of Math., Hampden Sidney Coll., Va. : My high estimate of it has been amply vindi- cated by its use in the class-room. R. H. Graves, Prof of Math., Univ. of North Carolina : I have al- ready decided to use it with my next class ; it suits my purpose better than any other book on the same subject with which I am acquainted. Edw. Brooks, Author of a Series of Math. : Its statements are clear and scholarly, and its methods thoroughly analytic and in the spirit of the latest mathematical thought.190 MATHEMATICS. Syllabus of a Course in Plane Trigonometry. By W. E. Byerly. 8vo. 8 pages. Mailing Price, io cts. Syllabus of a Course in Plane Analytical Geom- etry. By W. E. Byerly. 8vo. 12 pages. Mailing Price, 10 cts. Syllabus of a Course in Plane Analytic Geom- etry (.Advanced Course.') By W. E. Byerly, Professor of Mathe- matics, Harvard University. 8vo. 12 pages. Mailing Price, 10 cts. Syllabus of a Course in Analytical Geometry of Three Dimensions. By W. E. Byerly. 8vo. io pages. Mailing Price, 10 cts. Syllabus of a Course on Modern Methods in Analytic Geometry. By W. E. Byerly. 8vo. 8 pages. Mailing Price, 10 cts. Syllabus of a Course In the Theory of Equations. By W. E. Byerly. 8vo. 8 pages. Mailing Price, 10 cts. Elements of the Integral Calculus. By W. E. Byerly, Professor of Mathematics in Harvard University. 8vo. 204 pages. Mailing Price, #2.15; Introduction, $2.00. This volume is a sequel to the author’s treatise on the Differential Calculus (see page 134), and, like that, is written as a text-book. The last chapter, however, — a Key to the Solution of Differential Equations, — may prove of service to working mathematicians. H. A. Newton, Prof, of Math., Yale Coll. : We shall use- it in my optional class next term. Mathematical Visitor : The subject is presented very clearly. It is the first American treatise on the Cal- culus that we have seen which devotes any space to average and probability. Schoolmaster, London : The merits of this work are as marked as those of the Differential Calculus by the same author. Zion’s Herald : A text-book every way worthy of the venerable University in which the author is an honored teacher. Cambridge in Massachusetts, like Cambridge in England, preserves its reputation for the breadth and strict- ness of its mathematical requisitions, and these form the spinal column of a liberal education.192 MATHEMATICS. Elements of the Differential and Integral Calculus. With Examples and Applications. By J. M. Taylor, Professor of Mathematics in Madison University. 8vo. Cloth. 249 pp. Mailing price, $i«95> Introduction price, $1.80. The aim of this treatise is to present simply and concisely the fundamental problems of the Calculus, their solution, and more common applications. Its axiomatic datum is that the change of a variable, when not uniform, may be conceived as becoming uniform at any value of the variable. It employs the conception of rates, which affords finite differen- tials, and also the simplest and most natural view of the problem of the Differential Calculus. This problem of finding the relative rates of change of related variables is afterwards reduced to that of finding the limit of the ratio of their simultaneous increments ; and, in a final chapter, the latter problem is solved by the principles of infinitesimals. Many theorems are proved both by the method of rates and that of limits, and thus each is made to throw light upon the other. The chapter on differentiation is followed by one on direct integra- tion and its more important applications. Throughout the work there are numerous practical problems in Geometry and Mechanics, which serve to exhibit the power and use of the science, and to excite and keep alive the interest of the student. Judging from the author’s experience in teaching the subject, it is believed that this elementary treatise so sets forth and illustrates the highly practical nature of the Calculus, as to awaken a lively interest in many readers to whom a more abstract method of treat- ment would be distasteful. Oren Root, Jr., Prof of Math., Hamilton Coll., N. Y. : In reading the manuscript I was impressed by the clearness of definition and demonstra- tion, the pertinence of illustration, and the happy union of exclusion and con- densation. It seems to me most admir- ably suited for use in college classes. I prove my regard by adopting this as our text-book on the calculus. C. M. Charrappin, S.J., St. Louis Univ. : I have given the book a thorough examination, and I am satis- fied that it is the best work on the sub- ject I have seen. I mean the best work for what it was intended,—a text- book. I would like very much to in- troduce it in the University. {Jan. 12, 1885.)194 MATHEMATICS. Metrica! Geometry: An Elementary Treatise on Mensuration. By George Bruce Halsted, Ph.D., Prof. Mathema- tics, University of Texas, Austin. i2mo. Cloth. 246 pages. Mailing price, $1.10; Introduction, $1.00. This work applies new principles and methods to simplify the measurement of lengths, angles, areas, and volumes. It is strictly demonstrative, but uses no Trigonometry, and is adapted to be taken up in connection with, or following any elementary Geometry. It treats of accessible and inaccessible straight lines, and of their inter- dependence when in triangles, circles, etc. ; also gives a more rigid rectification of the circumference, etc. It introduces the natural unit of angle, and deduces the ordinary and circular measure. Enlisting the auxiliary powers which modern geometers have recog- nized in notation, it binds up each theorem also in a self-explanatory formula, and this throughout the whole book on a system which renders confusion impossible, and surprisingly facilitates acquire- ment, as has been tested with very large classes in Princeton College. In addition to all the common propositions about areas, a new method, applicable to any polygon, is introduced, which so simplifies and shortens all calculations, that it is destined to be universally adopted in surveying, etc. In addition to the circle, sector, segment, zone, annulus, etc., the parabola and ellipse are measured ; and be- sides the common broken and curved surfaces, the theorems of Pappus are demonstrated. Especial mention should be made of the treatment of solid angles, which is original, introducing for the first time, we think, the natural unit of solid angle, and making spherics easy. For solids, a single informing idea is fixed upon of such fecundity as to place within the reach of children results heretofore only given by Integral Calculus. Throughout, a hundred illustrative examples are worked out, and at the end are five hundred carefully arranged and indexed exercises, using the metric system. OPINIONS. Simon Newcomb, Nautical Al- manac Office, Washington, D. C. : I am much interested in your Mensuration, and wish I had seen it in time to have some exercises suggested by it put into my Geometry. {Sept. 8, 1881.) Alexander MacFarlane, Exam- iner in Mathematics to the University of Edinburgh, Scotland : The method, figures, and examples appear excellent, and I anticipate much benefit from its minute perusalMATHEMATICS. 195 Elementary Co-ordinate Geometry. By W. B. Smith, Professor of Physics, Missouri State University. i2mo. Cloth. 312 pp. Mailing price, $2.15; for Introduction, $2.00. While in the study of Analytic Geometry either gain of knowledge or culture of mind may be sought, the latter object alone can justify placing it in a college curriculum. Yet the subject may be so pur- sued as to be of no great educational value. Mere calculation, or the solution of problems by algebraic processes, is a very inferior dis- cipline of reason. Even geometry is not the best discipline. In all thinking the real difficulty lies in forming clear notions of things. In doing this all the higher faculties are brought into play. It is this formation of concepts, therefore, that is the essential part of mental training. He who forms them clearly and accurately may be safely trusted to put them together correctly. Nearly every seeming mis- take in reasoning is really a mistake in conception. Such considerations have guided the composition of this book. Concepts have been introduced in abundance, and the proofs made to hinge directly upon them. Treated in this way the subject seems adapted, as hardly any other, to develop the power of thought. Some of the special features of the work are : — 1. Its size is such it can be mastered in the time generally allowed. 2. The scope is far wider than in any other American work. 3. The combination of small size and large scope has been secured through superior methods, — modern, direct, and rapid. 4. Conspicuous among such methods is that of determinants, here presented, by the union of theory and practice, in its real power and beauty. 5. Confusion is shut out by a consistent and self-explaining notation. 6. The order of development is natural\ and leads without break or turn from the simplest to the most complex. The method is heuristic. 7. The student’s grasp is strengthened by numerous exercises. 8. The work has been tested at every point in the class- room.196 MATHEMATICS. Determinants. The Theory of Determinants : an Elementary Treatise. By Paul H. Hanus, B.S., Professor of Mathematics in the University of Colorado. 8vo. Cloth, ooo pages. Mailing price, $0.00; for Introduction, $0.00. This is a text-book for the use of students in colleges and tech- nical schools. The need of an American work on determinants has long been felt by all teachers and students who have extended their reading beyond the elements of mathematics. The importance of the subject is no longer overlooked. The shortness and elegance imparted to many otherwise tedious processes, by the introduction of determinants, recommend their use even in the more elementary branches, while the advanced student cannot dispense with a knowl- edge of these valuable instruments of research. Moreover, deter- minants are employed by all modern writers. This book is written especially for those who have had no previous knowledge of the subject, and is therefore adapted to self-instruction as well as to the needs of the class-room. To this end the subject is at first presented in a very simple manner. As the reader ad- vances, less and less attention is given to details. Throughout the entire work it is the constant aim to arouse and enliven the reader’s interest by first showing how the various concepts have arisen naturally, and by giving such applications as can be presented with- out exceeding the limits of the treatise. The work is sufficiently comprehensive to enable the student that has mastered the volume to use the determinant notation with ease, and to pursue his further reading in the modern higher algebra with pleasure and profit. In Chapter I. the evolution of a theory of determinants is touched upon, and it is shown how determinants are produced in the process of eliminating the variables from systems of simple equations with some further preliminary notions and definitions. In Chapter II. the most'important properties of determinants are discussed. Numerous examples serve to fix and exemplify the prin- ciples deduced. Chapter III. comprises half the entire volume. It is the design of this chapter to familiarize the reader with the most important special forms that occur in application, and to enable him to realize the practical usefulness of determinants as instruments of research. [Ready June 1.MATHEMATICS. 197 - Examples of Differential Equations. By George A. Osborne, Professor of Mathematics in the Massachusetts Institute of Technology, Boston. i2mo. Cloth, viii + 50 pp. Mail- ing price, 60 cts.; for Introduction, 50 cts. Notwithstanding the importance of the study of Differential Equa- tions, either as a branch of pure mathematics, or as applied to Geometry or Physics, no American work on this subject has been published containing a classified series of examples. This book is intended to supply this want, and provides a series of nearly three hundred examples with answers systematically arranged and grouped under the different cases, and accompanied by concise rules for the solution of each case. It is hoped that the work will be found useful, not only in the study of this important subject, but also by way of reference to mathematical students generally whenever the solution of a differen- tial equation is required. Elements of the Theory of the Newtonian Poten- tlül Function. By B. O. Peirce, Assistant Professor of Mathematics and Physics, Harvard University. i2mo. Cloth. 154 pages. Mailing price, $1.60; for Introduction, $1.50. A knowledge of the properties of this function is essential for electrical engineers, for students of mathematical physics, and for all who desire more than an elementary knowledge of experimental physics. This book, based upon notes made for class-room use, was written because no book in English gave in simple form, for the use of students who know something of the calculus, so much of the theory of the potential function as is needed in the study of physics. Both matter and arrangement have been practically adapted to the end in view. Chapter I. The Attraction of Gravitation. II. The Newtonian Potential Function in the Case of Gravitation. III. The Newtonian Potential Function in the Case of Repulsive Forces. IV. Surface Distributions. Green’s Theorem. V. Application of the Results of the Preceding Chapters to Electrostatics.Mathematics.2 Introd. Prices. Byerly...........Differential Calculus........................$2.00 Integral Calculus.......................... 2.00 Ginn.............Addition Manual.....................^ . .15 Halsted..........Mensuration.................................. 1.00 Hardy ...........Quaternions.................................. 2.00 Hill.............Geometry for Beginners ...................... 1.00 Sprague..........Rapid Addition..................................10 Taylor ..........Elements of the Calculus..................... 1.80 Wentworth .......Grammar School Arithmetic ...............75 Shorter Course in Algebra.................. 1.00 Elements of Algebra........................ 1.12 Complete Algebra .......................... 1.40 Plane Geometry ............................. .75 Plane and Solid Geometry .................. 1.25 Plane and Solid Geometry, and Trigonometry 1.40 Plane Trigonometry and Tables. Paper.. .60 PI. and Sph. Trig., Surv., and Navigation . 1.12 PI. and Sph. 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(separate), each .30 Wheeler..........Plane and Spherical Trig, and Tables..... 1.00 Copies sent to Teachers for examination, with a view to Introduction, on receipt of Introduction Price. GINN & COMPANY. Publishers. BOSTON. NEW YORK. CHICAGO.