Digitized by the Internet Archive in 2022 with funding from University of Illinois Uroana-Champaign httos://archive.org/details/linearalgebra00pach LUBRARY. ion: SS Sn t+ a =e cs ba ot 1 " - oF. vea i. a Hel 5 _—" tel et cre 5 tl en ee “= “ nacciine if 4 ame ; _ LE LOE AA A LALA ALLEL A | AA A Re Ne i en LINEAR ALGEBRA. Ey we" WZ HUSSEIN TEVFIK PACHA. oe sat on SECOND EDITION, . \ REVISED AND ENLARGED. CONSTANTINOPLE: i ee pons ey re Orn LS: CHAPTER - ight Lines in Algebraic Expressions | wee ; a Beep De Vos os ew propositions on the employment of Lines in oe operations. . . 10 CHAPTER IL. CHAPTER TIL Merendiular:2 4... =... et Pk! cause eae sce och 7 ° ° ° ° ° 49 = e ° e ° ° ° 56 CHAPTER IV. | of a Straight Line 2 Plane. t Circle... ° ° . . e ° ° ° e . ° ° e Poteas ConicrSection ./- o's “ omurTeR as evolution e e ° e e e e ° faving a Circular Section ... . CHAPTER VL | ! Surfaces ‘of*the Second’ Order 2% at: sO ee ee ee General: Equation of the Second Order. . 2% J{2% 2 ean. 5 2 ee eee Examples ° ° e ° e e e e e e ° ©. e e e e e e e « e e e CHAPTER VII. On Some Additional, Applications . . . i°+°. 0°.) 2.0 325) Se Be FUXAMIpPles; ; wei ele . ° e e e e ° e e e e ¢ e CHAPTER VIL Complex Quantities. ~ .° a5. Gael pe? ae ee AUS PEPHIONS =. cho}. ie ave’ othe A Se ae ae ee FIXSIMDICS ge I ce, ee ett d aM iy - ~ reve BAAR Ae TRE a f a PLey & r 73 a hie BGhPeePea 8 i dab , MATHEMATICS Pi 12 BRARY. PREFACE. Linear Algebra, as treated in this pamphlet, grew out of an effort to ex- tend Argand’s system concerning ordinary complex or imaginary quantities to space of three dimensions. The system of Quaternions, ascience for which the world is indebted to the extraordinary genius of Sir William Rowan Hamilton, also grew out of a similar effort; notwithstanding which the two systems have very little in common. Argand’s system, only applicable to Plane Geometry, is not a special case of Hamilton’s; therefore, even after the invention by Hamilton of his most powerful instrument of calculus of directed lines, Argand’s Algebra remained incomplete. Cauchy, one of the great masters of Mathematics, has used Argand’s sys- tem insome of his most brilliant investigations. The beautiful ‘‘Méthode des Equipollences,” avery general system of plane analytical Geometry established by M. Bellavitis, is only a development of Argand’s Algebra. De Morgan’s Double Algebra is nothing else than Argand’s conception. The most simple and elegant demonstration of the fundamental proposition of the theory of algebraic equations is furnished by Argand’s system. And, this method of Argand on imaginary quantities gives the geometrical interpretation of such an expression as A+By = which is tently met in ordinary Algebra; thus Without Argand’s system even the ordinary Algebra cannot be considered plete. ‘his is my apology for attempting to han notwithstanding Hamilton’s | Calculus, a new Algebra, or to complete Argand’s system. Persons who _ to know something ee the history of the system of imaginary quan- may satisfy themselves by reading the preface by J. Hotiel and noles rof. A. S. Hardy to Argand’s book translated by Prof. Hardy, or the ndix to the elements of Co-ordinate Geometry by De Volson Wood. fhe addition and subtraction of lines in this Linear Algebra, as well as lamilton’s Calculus, are the same as Argand had conceived. The multi- tion of Linear Algebra being entirely different from that of Quaternions, prises, as a special case, Argand’s multiplication. Linear Algebra is ap- e to the Geometry of three dimensions as well as to Plane Geometry, hen applied to Plane Geometry is nothing but Argand’s Algebra with ome new notations. Multiplication of Linear Algebra is based on a geometrical conception and has nothing which may be regarded as quasi-metaphysical; and not only in its applications it bears a very remarkable analogy to the Calculus of Qua- ‘ternions but it enables us to re-establish all principles of Quaternions by 842630 othe way than that followed in the great book of Hamilton or that of Tait. We may safely quote in favor of this Linear Algebra what has been — said in favor of Quaternions: : ‘‘And what is of the utmost importance, in an educational standpoint of view, the reader of this subject does not require to encumber his memory _ with a host of conclusions already arrived at in order to advance. Every problem is more or less self-contained.” We read in the appendix to the elements of Co-ordinate Geometry by De. Volson Wood: ‘‘It is proper to note that M. Servois seems to have made the nearest approach to an anticipation of quaternions (Lectures, Preface, p. 57). He inferred from analogy, that if «, 8, y, bethe angles between aright line and the three rectangular axes, the following expression ought to be true: (p cos «-+q cos $-+-r cos y) (p’ cos «-+q' cos 8-+r’ cos 7) = cos? «-+cos* b-+-cos? y= 4; but he could not determine the values of p,q,r, p’,q,1, and asked ‘Will they be imaginaries, reducible to the general form ALBY =? Itis now known that they are identical with the +i, +j, +z, —i, —j, —k, of quaternions.” Yes, putting Hamilton’s i,j,k for p,q,r aan —i, ae —k for p,q,k, quaternion multiplication will give (i cos «-+j cos 8-+k cos y) (—i cos «—j cos B—k cos y) = 14. But Servois asked this question I think to see if Argand’s system can be extended to space of three dimensions. The answer given to his demand is equivalent to saying that the at- tempted extension is impossible. We read in Tait’s Quaternions: ‘‘Beyond this (Servois), few attempts were made, or at least recorded, in earlier times, to extend the principle to space of three dimensions; and though many such have been made within the last forty years, none, with |the single exception of Hamilton’s, have resulted in simple, practical methods; all, however in- — genious, seeming to lead at once to processes and results of fearful com=- | plexity.” Now in answer to the inquiry of Servois, we will add this: For that ex- | tension we do not need the reduction of p,q,r, p’, q’, andr’ to the i | form A+BY = (of ordinary Algebra); and to Hee (p cos «-+q cos B-+r cos ) (p’ cos «-+q/ cos B-+r’ cos y) = cos* «+cos? b+-cos? y= 4 . we have only to assume that p, q, and r are mutually perpendicular unit- — lines, and p’, q’, and r’ are, respectively, conjugates of p, q, and r. We — are unable to detect here any ‘‘fearful complexity.” ele To conclude, Ihave one remark to make: the Multiplication of Quaternions — obliges us ‘‘to work in space of three dimensions even when nominally — treating plane geometry;” while in Linear Algebra we do not need to consider necessarily a third dimension when treating only two dimensions. LINEAR ALGEBRA. CHAPTER I. Straight Lines in Algebraic Expressions. 4. The author here desires to say to the reader, that though the present chapter itains nothing new it is yet of unquestioned importance for a clear understanding of e chapters following. 2. If the lines AB and NO, for example, of a geometric figure are in different di- rections, and if not only their absolute lengths are considered, but their respective directions : well, itis evident that, though the lengths of these lines are equal, it cannot be said go IN O; 3. By the expression AB=NO, in Linear Algebra and in the science of Quaternions o, it is understood that the length of AB is equal to that of NO, and also that the ection of the line AB is the same as that of NO, that is to say they are either on » same straight line or parallel to each other in the same direction. In Numerical Al- bra it is the absolute equality of the lengths only of these lines which is understood. 4. ln describing a line AB for example, if we say the line AB or simply AB we ‘an the special line AB which has a determined direction and length. If we write the e AB, or simply AB, we mean that the length alone is considered. Sometimes N(AB) will be written for AB, which is read, number of AB, and N(a), 'a, which is similarly read, number of «. N2(AB) and N2(«) will also be written for A B)xN(AB) and N(«)xN(2), or for ABXAB and «xa, Thus the symbol formed by prefixing the letter N to the symbol of a line, or by tting a dash under that symbol, will represent the absolute length of the line, by pressing the ratio which the length bears to some assumed standard, or unit of length. 5. To represent the different lines of a figure, with regard to their directions as well lengths, Greek letters are often employed. For example, if p is put for the line AB, long as the problem is not changed, by this ep is understood the line AB which by yposition has a determined length and direction. 6. It is obvious that the lines AB, BD, having a determined direction and length, the line AD will also, necessarily have a determined direction and length; and if in departing from the point A after having traced the line AB in giving to itits direction and length, we trace BD commencing at B, giving to it also its length and direction, the distance from A to D will represent A the line AD with its special length and direction, ee A ine having a determined direction and length may be considered as representing the displacement of a point from its original to its final position. As for example, the line AB denotes transference from A to B, and BA from B to A. 7. The operation of tracing the line AB from the point A, and the line BD from the point B, and giving to them their respective directions, il be represented by the expression AB+BD; and to show that by this operation AD is found, the expressior AbB+BD=AD will be employed. 8. If in departing from the point A the lines AB, BD, DH, HN, NO, are succes sively traced in their respective directions, the line joining A to O, or AO, will b represented by A O=A B+B D+D H-+N 0. It isreadily seen that after having traced AB, if in place of tracing the other lines in th order given, we trace successively a line parallel and equal in length to each one of thes lines, in their respective directions, in whatever order, we shall still find the same line A O. It is needless to say that this manner of representation of straight lines is general. 9, Jt is now apparent what in Linear Algebra is meant by AB+BD=AD. If the lines AB, BD are found equal in length, it is evident the length of AD will diminish with the angle ABD; and finally AD will become zero whenever this angle does; in ths case the point D coincides with A, and the line BD with BA; for this reason AB+BD=0 or AB+BA=0. Thus in the expressions AB+BA or BA+AB AB and BA neutralize each other; therefore when a line measured in one direction is represented by a positive symbol, the same line measured in the opposite direction may be represented by the same symbol taken negatively, that is AB=—BA or BA=—AB, hence if the line AB is represented by pe, the line BA will be —p. 10. If AB andDE are on the same right line, and in the same direction, we F admit, as in Numerical Algebra, that AB is to DE as AB to DE, that is Now if AB=DE, then AB=DE and consequently AB+ED=0. 41. If AB and DE are parallel in the same direction, and AB=DE, we must admit M AB=DE. For if we take AN and DM onthe same right line AD, and AN=DM, we admit AN=DM (Art, 10), but DE compared toDM is "s situated ? *3 293 ae) ee exactly as AB compared to AN, and this similarity of position is so complete that. if we know AB from its relation to AN it will be exactly as if we know DE from its relation to DM. Therefore as DM is admitted to-be equal to AN we have a right to assume that AB equals DE. Thus AB=DE and AB+ED=0. And if AF>=a« DG we shall have ' AF=a. DG.. _ 42. It follows that, if a line AB is represented by a« (a being an abstract number, % aunit line in the direction AB), any line which is parallel to AB or placed on the same line, and in the same direction and.has the same length, can be designated also by a«. In the case that the second line is in an opposite direction it will be designated by —a«, 13. Wehave seen that AB, BD, DH drawn successively in their respec- tive directions, the line AH which closes the polygon ABDH can be 8 represented by AH=AB+BD+DH, or by designating the units of AB, BD, DH respectively by «, 8, y, and their lengths by x, y, z, and the line AH, by pe, then p= aatyb+zy. 44. It is obvious that, if the lines AB, BD, DH are not in the same plane we can consider the numbers x, y and z as Cartesian co-ordinates of the point H. 4 415. If we take the lines AB, AC and AD for example, and trace from the point B a line equal to AC, and from the end of this a line equal to AD and designate by AH the side which will close the polygon thus formed, the line AH will be called the swm of the lines AB, AG, AD, or Cc AH=AB+AC+AD. This operation we define as addition. A It will also be readily seen that the following operations AB+AD+AC, AC+AB+AD, AD+AC+AB ete. will give the same result. In ‘the case in which the lines to be added are in the same direction, this operation is reduced to the addition of Numerical Algebra. 2 Sie IAL) Se 16. Subtraction is the operation of finding one of two lines, when the other and their sum are given. To subtract AD from AB, or to find a line which [0 | added to AD will produce the line AB, it is evident that if we trace from the point B a line equal to DA, we shall have the line AN which added to AD will produce AB: N AN+AD=AB and AN=AB—AD. A few propositions on the employment of lines in Algebraic operations. 17. In case that we have AB+BD+DH=AH we shall have also nx AB+nxBD+nx DH=nxX AH, designating by n an abstract number. And if we have ax AB+nxXBD+nxDH=nxAH we shall have | AB+BD+DH=AH. In tracing the expressions AB+BD+DH, and nx AB+nxBD+%x DH, the truth of the proposition will be manifest. Thus if we designate the lines AB, BD, DH, HL, and AL by the Greek letters «, 8, y, 5, and »; and if we have, for example, 28 «% + 218+-77+A98= lho, we shall have also as in Numerical Algebra T(4e+38+y7+78)=7, 20 or hatdb+y+78—=2o, 18. If the lines « and ® have not the same direction, and we designate by a and b two abstract numbers, the lines a«, D8 cannot neutralize each other in Algebraic ex- pressions. Therefore if as a result of some operation we have, ax%x+bsi—0 we shall conclude that a=0, b=0, tH cc And again if « and 8 being in different directions, we have ax+bp—ha+lp we must also have (a—k)«+(b—1)b=0; da a—k=0 and b—l=0. 19. If, 8, and y are non-parallel lines in the same plane, it is always possible to find the numerical values of a, b, c, so that, ax+b8+cy shall=0. For as these «, B, and ¥ are in the same plane, a triangle can be constructed the sides of which shall be parallel respectively to «, 8, y. Now if the sides of this triangle taken in order be a, 68, cy respectively, we shall have, by going round the triangle, a%+bB+cy=0. 20. If«, 8, and y are three lines neither parallel, nor in the same plane, it is im- possible to find numerical values of a, b, c, not equal to zero, which shall render a«+b$+cy=0, for ax+bf can be represented by a line in the plane parallel to « and 8. Now cy is not in that plane, therefore the sum of a«+b8 and cy cannot equal 0. It follows that, if ax+b8+cy=0. and «, 6, and y are not parallel to each other, they are in the same plane. 21. There is but one way of making the sum. of the multiples of «, 8, and y equal to 0. Let : a%+bB+cy=0 and also aa+b8-+-cyvy=0., By eliminating y we get (ac—ca,)%«+(bo—cb)8=0; but as «, 6 are in different directions, ac—ca—0 and be—cb=0; e° — a¢6=ca 4 é ‘and be=cb, or Greet ae Os Uae so that the second equation is simply a multiple of the first. If we observe that the tri- angles which give the different values of a, b, e, are similar the last proposition will be accepted a priort. 22. If «,$, andy are coinitial coplanar lines, terminating in a straight line, then the =12— ey ween (SF But AC is a multiple of AB, or (ye a(h—e)=ah—ae; Higa a Rg amy or (o—1)«a—ah-+y=0; and as in this equation the coefficients of «, 8, y are *—1, —#, +1. which. correspond to a, b, ¢ in the first equation, and as (of) =, then. a+b+c==0. 2 Conversely, if a, 8, and y are coinitial, coplanar lines, and if both oat: heey ae and a+b+c=0, then do «, 8, and y terminate in a straight line. For by supposition, | Gtb+ce=), didrerire! feof | at ay+by+cx==0, and by subtraction es pore | Ai ioe a(y—«)+b(y—8)=0 or (2) +2 —B)=0. This shows that y—« is a multiple of y—B8 and therefore it is in the same straight line with it; «, 6, and y terminate in that straight line or in a line parallel to it.. 23. Examples. ; | Ex. 1. In a plane triangle are given one angle, an adjacent side, and the sum of the lengths of the other sides, to determine the triangle. Let ABD be the given angle, AB=b ” y 99 side, S + 3, 8um of the lengths of the other two sides. If in designating by « and B-two unit lines; we represent by «« the unknown side adjacent to the angle B, and by yf the opposite side.to this angle, we shall. have ‘dy Ph eRe AS AR and ¥3 aaa8 1y ‘ bi b £oh.' (on i SS pa+y, . ‘a ee AF tee by eliminating « ibis yhb+Sa—ya; yP+yeb+Sa, The last equation furnishes us a method of solution for this problem. For S« is known, and it represents the line BD which has the length S, the unit of this line being «; therefore b+-S« or AD is known also; and y B+ y « being equal to b+S« is equal to AD. But as « and8 are units, AD, y$ and y« in the expression AD=y+y« evidently form an isosceles triangle of which AD is the base, and yf, y« the equal sides (in lengths). Besides as « is a unit inthe direction BD, the side y« must necessarily be in this direc- tion. Thus evidently the angle BDA is one of the equal angles of that triangle. Then to find the other angle we have only to make an angle DAC equal to the angle BDA and thus we shall have CD=y« and AC=y8 which is the side opposite to the angle ABD in the required triangle. Ex. 2. The difference between the diagonal of a square and one of its sides being given, to determine the square. Let the defference Benveen the ee of the side AB an of the diagonal AC be d. ¥ «,8, ard y being three units, we will designate the side AB by w«, the side BC by «8 and the diagonal AC by yy, we shall now have, yY—uatoab, | 7 | yard, and from these two equations | a (%+B—y)=dy. The units «, 6, andy in this equation are known, for if we put the unit « on the line AK and the unit 8, perpendicularly upon «, the unit y will be found on the «+f. Therefore if we take AK=« and KI=8 and ln==—y, we shall have An=«+8—y, and in taking the length A m equal to the difference d, Am will be=dy, and consequently te AN Lem, But as An and Am are in the same direction we can say Therefore it is evident that in joining the points » and K, and tracing the line mB pa- rallel to nK we shall have psidi ts sigry Ane c= AB which is the length of the side of the required square. Ex. 3. The bisectors of the sides of a triangle meet in a point which trisects each of them. Let the sides of the triangle ABC be bisected in D, E, and F; and let AE and CD meet in G. Therefore, AG+GC=AC=AB+BC=2(DB+BE)=2(0G+GE) =2DG+2GE. But as AG—2GE is on the line AE, GOC—2DG on the line CD, their sum cannot be zero unless each one of them equals zero. Consequently, AG—2GE=0, or AG=2GE, and GC—2DG=—0, or GC=2DG. These equations show that EG is a third of EA and DG a third of DC. If now we suppose that the point G is the point where CD and BF meet, in the same manner it will be seen that DG is a third of DC, and FG a third of FB, and consequently the three bisectors must necessarily meet in the same point which separates one third of each. Ex. 4. The middle points of the lines which join the points of bisection of the op- posite sides of a quadrilateral, coincide, whether the four sides of the quadrilateral be in the same plan or not. Let AB=chA CH-8 eA D—7; X the middle point of EG. - We have AE+EG=AD+DG, ame $¢+EG=y+i(b—y); he a | EG=y+i(@—yv)—$4=26+y—2), and as AX=}¢+4EG, AX=}a+} (6-+y—a) =} (a+ 8-+1) ee he es which being symmetrical in «, 8, and y is the same for the line from A to the middle of HF, hence the middle points of the lines EG, FH must coincide. Being naturally desirous to publish this little work as economically as may prove compatible with clearness of statement, I have contented myself with putting into the present chapter some readily solved examples only. Should however any one wish further illustration, he can find very beautiful and sufficiently interesting solutions of a similar kind in the second chapter of the Introduction to Quaternions by Kelland and Tait. —— FP PPP bt Hee CHAPTER Il. MULTIPLICATION. 24. One of the various directions will be considered as the principal direction. In the following Figure OX is assumed to be such a direction. The multiplication of one by the other of any two coinitial lines not in the same plane with the principal direction, is shown in the following operation. Suppose we wish to find the result of the multiplication of the line OA into OB. Let OX, OY and OZ be at right angles to one another, OY being in the plane XOA, then OZ will be perpendicular to this plane. 4*. Increase or diminish the length of @B, till it becomes equal to that of OD, which is the product of the line OB and the ratio of the line OA to the unit of length. Suppose this product is OD. Let down a perpendicular from the point D on the plane which passes through the principal direction OX and the line OA. Let D’ be the foot of this perpendicular in the same plane. We thus have a right- angled triangle whose plane is perpendicular to the indefinite plane KOA, of which D D’ is the hight; OD’ the base, OD the hypothenuse, | wre AG 2*, Move this triangle around the axis OZ, keeping it always perpendicular to the plane XOA until its base OD‘, coincides with OD" which is in the plane XOA, and which makes with OX anangle equal to the sum ofthe angles that OD’ and OA, make with the principal direction OX. Let D,D" be the hight of this triangle after this ope- ration. For definiteness we will admit that the angles which the lines OA and OD’ make with the principal direction are formed and measured from OX on this side. That is, OZ being considered the positive axis of rotation, these angles are formed positively with reference to the principal direction, by the lines OA and OD’ making on the indefinite plane XOA a rotation similar to that of the hands of a watch. 3°, Let us imagine that the line D,D’, while remaining parallel to the plane ZO X, turns from left to right, like the hands of a watch, around the point D’ through the angle D,D’D, (the positive axis of this rotation being parallel to O Y) equal to the angle AOX which the line OA makes with the principal direction. Now suppose that the line OD,, which joins the point O to the point D, is the result of the action of OA on OB. We will call this operation, mulitplication, and the result OD, will be considered as the required product of OA multiplied into OB. 25. We shall see that the resulis of this multiplication have a great analogy with the results of ordinary multiplication, which in fact is but a particular case of it. Con- sequently we shall use the same signs that are used in Numerical Algebra. Thus in de- signating OA, OB, and OD, respectively by «, 8, and y, we shall write as in Nume- rical Algebra xB or oo. “Boor we po—y, 26. Had we wished to multiply OB into OA we should have had the same operation to make which we have just written, with this single change, that the plane XOB would have had to be taken instead of the plane XOA, and the line OA instead of the line O B. 27. It is readily seen that the product OAXOB is not generally the same as the product OB xX OA. Thus we cannot ordinarily have «.8—=8.«; that is the commutative law does not ordinarly apply to the factors of a product of two lines*. 28. Let D,F be parallel to XO (Fig., Art. 24), consequently per pendicular to D,D", It is useful to ‘know the value of this line D,F. Let us suppose that ¢ indicates the angle AO X; », the angle DOD’ which is between “In De Morgan’s Triple Algebra, the commutative law, or the property « 6= 6a, is maintained, (See Cambridge Phil. Trans., viii, 241.) afta OD and its projection in the plane XO A. Since the angle D,D” D, is equal to the angle AOX (Art. 24), we have Der ved, 818 but p,D'=D,D’=DD, therefore D,F=DD". sin ¢; but DD'=OD.sin o=O0A.OB.sin », consequently Oe Oe A) By Siting: Si ites or BD SAT, OB: Sit, 9. SiN, O.2; t being the unit-line in the principal direction OX, 29. If the question was of the product OB xX OA (Art. 26), FD, would have been directed in the opposite direction to OX and we would have FD,=—VB.OA.sin®. 4.1, ® being the angle BOX, ¥ the angle which is between OA and its projection in the plane XOB. It is easy to see that if 9 indicates the angle which is between the planes XOA, XOB, we shall have sin o—=sin ®., Sin 9, sin J=sin 9. sin 9. Rule of Signs. 30. Algebraists have laboriously attempted to demonstrate that ax—b=—ab, —axb=—ab and Hai) 2a ES Ne Nevertheless the demonstrations given in books of Numerical- Algebra on this matter are not rigorously logical. This need not appear strange. The definitions given for multi- cation are much less general than is the idea of a negative quantity. If therefore in employing only such definitions as are applicable to abstract numbers, algebraists have not succeeded in satisfactorily demonstrating the rules of sigus as above stated, it is not to be wondered at. dl. To perceive that, aX —B==—af, —a«xfPom—e«B, and —«x—f—=«8 we have merely to apply our definition, bee te Special cases of Multiplication. 32. If OB (Fig., Art. 24) is perpendicular to th plane which passes through OX and OA, the foot D’ of the hight DD’ and the foot D’ of the hight D’D" will coincide with the origin O; consequently the lines D’D,, D’D, and OD, will be found upon the plane which passes through OX and OB, and at the same time the lines OD, and D’D, will coincide with each other. . In such a case the multiplication of OA into OB consists in turning upon the plane XOB from left to right the line OD, which is the product of OB and the abstract number of the length of OA, until the angle “ D,OB shall be equal to the angle AOX=¢. OD, is the required —_ % product. It is readily seen that in this case, to find the product OB. OA, —--—~ it is only necessary to turn OD to the opposite side; the required aan Te ‘< product will be OH, if the angle HOB=g¢., 33. If the line OA (Fig., Art. 24) is perpendicular to the plane XOB at the same time that OB is perpendicular, as in the preceding case to the plane XOA, or in other terms, if the lines OA, OB and the principal direction OX form the three contiguous sides of a rectangular parallelopiped, the angle DOD, will be a right angle, and con- sequently the line OD, which is the product OA.OB will be found in the principal direction OX. If in this case the required product is OB.OA, the line which represents it will evidently fall on the direction opposite to OX, that is to say on the negative side of the principal direction. 34. Thus we have this important result, that when OA, OB and the principal di- rection are perpendicular to each other, in designating OA by « and OB by § we shall have aB—=—Ba or «8+ha=0. 30. Let OA and OB be perpendicular to the principal direction OX, without having OA perpendicular to OB; let us indicate by 9 the angle BOA. In this case the point D’ (Fig., Art. 24) is on the direction OA; D"’, on the opposite direction of OX; D,, on the direction D’ X; and we shall have D'O=D.O=0A . OB. cos 8, D'D,=D'D'=DD'=04. OB. sin 9; consequently OAXOB=O0D,=D'D,—D’0O =OA.OB.sin®.+—OB.OA. cos®.% —=OA.OB. (sin §— cos §) .¢. We shall have also OBxOA=—OB. OB. (sin 4+ cos 4). 7 (Art, 26), wes YO nt From these two relations we shall have OAXOB—OBXxXOA>=20A . OB sin 9.2; therefore in supposing that OA and OB, be two unities of length carried in the direc- tions OA and OB; and in indicating them by 8, 8, we shall have 68 —6 B—=2sin 0.7%. If in this case we have 6=4, we must have Onn OB == (Vy a product which is naught when neither of its two factors are naught. 36. If the lines OA, OB and the principal direction OX (Arts. 24, 26) are in the same plane, the hight DD’ will be reduced to zero, and at the same time the point D,, will coincide with the point D’, and the line OD, with the line OD’. We thus see that, when OA and OB and the principal direction are in the same plane, the product OA .OB will also be in the same plane and it will make with the principal direction an angle equal to the sum of the angles which OA and OB make with the principal direction. It is almost unnecessary to add that the product OB.OA is the same as OA.OB. Thus in this case the commutative law holds, that is, we have as in Nume- rical Algebra «8—=B«. This case consists of the multiplication of Argand’s Algebra, or of that of De Mor- gan’s Double Algebra. Let OA and OB both being in the plane XOY,, be represented by A and B, respectively. Conceive that the positive axis of the rotation from OX towards OY, is directed downwards, or below the plane XOY', that is the rotation from OX towards OY, will be positive; then ‘‘the length of AXB is the arithmetical product of the lengths of A and B, expressed in units;* and the angle of AXB with the unit line** is the swm of the angles of A and B.***” 0 X 37. Let us designate by 9 the angle which a line OA makes with the principal di- rection OX, and suppose that a line OB which is found in the same plane as OA and OX make with this OX an angle equal to 2*—y9. The product of these OA and OB will be found on the principal direction, that is to say, Oe leon Ol OA OA. O Be 7; this is a result of great importance in Linear Algebra. 38. If the lines OA and OB are on the same straight line, and in the same direc- tion, they will necessarily be in the same plane asthe principal direction, and the angle which their product makes with this direction (Art. 36) will be the double of the angle which their direction makes with the paincipal direction. And in the same supposition, *of length. “the unit of the principal or prime direction. *“De Morgan’s Double Algebra, page 116. ae OO sot the product of OA.OB and of a line OD which has the same direction as OA and OB, will be found also in the same plane, and will make an angle three times greater than the angle which the direction OA and OB make with the principal direction. The generality of this fact is evident. 39. It follows from the preceding case that if the diréction of these lines OA and OB is perpendicular to the principal directian, their product will fall on the negative side of principal direction. 40. The product of a line by fear will be called the square of this line; the pro- duct of the square of a line by the line itself, its cube. It follows from Art. 38 that the square, the cube and the other powers of a line OA will be found in the same plane, as this line OA and the principal direction OX; and that the angles which the square, the cube and the- other powers of this line make with the principal direction will be respectively twice, thrice and so forth greater than the angle which this line makes with the same direction. Al. To indicate the different powers of a line we shall employ the same mode used in Numerical Algebra. The square of OA=(OA)?*, the cube of O0A=(OA)§, or the square of «==a2, the cube of «==? and so on. If ¢ is a unit perpendicular to the principal direction, and if ¢ is the unit in this di- rection (Art. 39) 62 — 4, 103 —— — 88 64 9 ao ee = We now know that the angle which «°, for example, makes with the principal di- rection is five times greater than the angle that « makes with the same direction. And in general, n being a positive number, whether integral or fractional, «" will designate a line which is in the plane iz and makes with the principal direction an angle equal to n times the angle between « and ¢. We may consider also the negative powers of lines. n being a positive number, #” is defined to be a line which, being in the plane ¢#, makes with the principal direction a negative angle n times as great as the positive angle between « and 7. Therefore the symbol «7? will denote a line that is in the plane 7%, on the opposite side to «, and making an angle equal, in magnitude, to the angle which is between « and 7. Obviously wat or ahaa? j (Art. 36). It is easy to see that ® being the angle between 7 and ®, we must have Qqr 56) (6) =2 where 4 is a unit line. Also (29 =i. { For, as the line (¢)" approaches nearer and nearer to the principal direction, the number n becomes greater and greater and at the limit, where n= Or will coincide with 2, Ala. Let « denote a line in space. am Dh mee We can write ary ett CLIN (& )o en (ND) aN oe, For, if we designate the unit of « by the symbol U«, we may write =«.Ue; therefore a2 = (x)? (Us)?, and (a2) = («)? (Ue)?. But U« being a unit, (U«)® will also be a unit, and the absolute value of it is 41; therefore (U«)*=1, hence (2?) = («)?. To illustrate this, (a + 6)% — (a-+f)2 and N (« + 8)? = N2 (a +8), 42, It results also from Art. 40 that if the units « and are in the same plane with the principal direction, and that the angle which « makes with this direction is represented by ®, the angle which § makes with the same direction by ¢, and lastly i or * represents the ratio between the number of the degrees of these angles 9 and 9, we shall have Q y) g— >? or Se or if 9= one degree I Ba? and «==6r, Here ? represents the ratio of the angle ¢ to an angle of 1 degree which « is supposed to make with the principal direction. A3. If the lines 8, 3 and a unit « are in the same plane as the principal direction, and if the angles which these 8, ® and « make with the principal direction are respect- ively 9, 9 and one degree, and the length of ® is b; that of ®, -d; we shall have thus (Art. 42) ey B.3 on 3 eA lc 44, It is scarcely necessary to say that when 6*—=—di, this % could represent each one of the linear units which are perpendicular to the principal direction. a 65 = A5. The unit of the principal direction being é, if we employ the symbol V—i, or j, to represent the unit of a direction OY which is perpendicular to the principal di- rection OX, and if we adopt the symbol |, or k, to represent the unit of the direc- tion OZ which is perpendicular to the plane XO Y and is the positive axis of the rota- tion from OX to OY (Arts. 24). We can write (Arts. 13, 14) — pe=avityVit+el, or pavityjt+zk. We must bear in mind that according to our definition of Multiplication — VS xt. KY Sp 7 ATI); (v>)2=—4, (L2=— 5 (Art. 40); or WS 1; ej =a ts and j2=— i, k*=— i. A6. If X=0 then p=yV—i+z_1 or p=yj+zk will represent a line perpendi- cular to the principal direction, and if z=0, then p=wi+y V—i or p=awit+yy will represent a line situated in the plane XOY. AY, If the line OA is in the principal direction (Art. 24) the plane XOA will be- come indeterminate, But in taking it in no matter what position, it will be seen that OD, will always coincide with OD. Therefore in this caae OA .OB=OB.OA=OD. It follows therefore that if the absolute length of OB=b, the unit of OB=f, and the principal unit==7 OB) b==4 x Dhaai te ie eet eae And ne bf represents ® added b times to itself; and bi xf or # xX bz, represents # multiplied by bi, or be multiplied by 8. 48. If OA and OB are both in the principal direction, their product will also be found in the principal direction. Therefore X¢t ands X*xX% 47 ele; or 4° = 1 and 23 ==7 etc. It is obvious that, ¢ being the unit of the principal direction, if 7, 7 and k& are units along the axes OX, OY and OZ, respectively, (Art. 45), we must have as defining equations, jij, kj —jk=—7, ea kde hy tee jk=—kjy=1, 2 = k*=— 7, i ES om These results can be arranged in a tabular form, or multiplication table, as follows ce 4 7 k AY. Hence we see that the units and the lines found on the principal direction have the same properties as the ordinary units and numbers represented by lines. Besides, in the results which we find whether by addition, subtraction or multiplication no abso- lute or abstract term enters. Therefore if in our calculations we replace the principal unit z by 1 no complication or mistake can arise, but on the contrary a great simplicity in the calculation will result from it. Thus, we can write (Art. 45) p=otyV-1+2 1; in this case (V-7*=—1, (1)?=—-I1, and Vaae bee eh Von The symbol Y= represents then the unit of a line which is perpendicular to the principal direction as defined in Art. 45; therefore it has no relation with, and must not be taken for Y-4 of Numerical Algebra which is the symbol of an imaginary quan- tity (see Art. 67, Ex. 1). 00. To show the manner of linear multipltcation of two lines, we have supposed them coinilial; nevertheless to apply our difinition of multiplication to any two lines whatever, it is merely required to add to this definition the following. To multiply a line QR by a line MN, it is necessary to trace from the point M a line equal to QR (Att. 3, 11) or from the point Q aline equal to MN. The definition will be applied to the line MN and to the equal of QR which passes through M or tothe line QR with the equal of MN which passes through Q. We can readily convince ourselves that the line which represents the product of the line MN by the equal of QR which passes through M is equal (Art. 3,11) to the line which represents the product of the equal of MN passing through Q by OR. This line will be found equal to the line which will represent the product of these lines MN and QR transported to any common origin O, without changing their respective directions. It is clearly understood that in all of these three multiplications the principal directions must be the same. Multiplication of Polynominals. 51. It now remains to be shown that, « and # being transported to the same origin — 2 — | (Art. 50) if in «.8 we put e=y+4, and f=A-+y», that is to say, if the lines « and 6 are each the sum of two other lines, we shall have as in numerical algebra (y+?) . Ate) —y. Atay ease, 52, Let us commence by proving that, OA.OM+0A.ON=O0A .(OM+0ON)=OA. OB, OB being the sum of OM and ON. We readily perceive, that after having mulliplied OM, ON and OB by the abstract number of the length of OA, the extremities of these products and the origin O will form a parallelogram, as well as the extremities of OM, ON, OB and the origin O. The feet of the perpendiculars, left fall according to the definition (Art. 24) on the plane which passes through the principal direction and OA, also form with the said origin, a paral- lelogram. Thus we are able to show that the extremities of the productions OA. OB, OA.OM, OA.ON, and the origin O must form a parallelogram whose diagonal and two sides contiguous to the point O, are respectively OA .OB, OA.OM and OA.ON. Therefore OA.OB=OA.OM+0A.0N or OA.(OM+ON)=OA.OM+0A.0N., 538, Let us show now that OR.OB+0S.O0B=(OR+0S8).OB=OA. OB, OA being the sum of OR and OS. Let us suppose for the moment that OR and OS are found in the plane XOA. In this case also we shall easily see that the extremities of the products OS. OB, OA.OB, OR.OB and the origin O form a parallelogram, the diagonal of which starting from O is OA.OB and the two sides starting from the said point O are OR. OB, and OS.OB. Therefore OA.OB=OR.OB+0OS. OB, or (OR+0OS).OB=OR.OB+OS. OB. 4. It results from the preceding two propositions that if OR, OS and the princi- pal direction are found in the same plane, we shall have (OR+0S) (OM+ON) —(OR+0S) OM+(OR+0S8) ON —OR.OM+0S.0M+0OR.ON+0S.ON. 55. Let OA and OB be two lines whatever in the space (Fig., Art. 24) and let us suppose OA=—axvityB, OB=a2,1+y,8,, ES, nace 8, 8, being two unit-lines perpendicular to the principal direction, along which the unit-line is represented by ¢ (Art. 28); « and y, the projections of OA on the prin- cipal direction and on that of 6; w and y,, the projections of OB on the principal di- rection and on that of f,. It is evident that ® is in the plane XOA; 6, in the plane X OB; and the angle which is between f, ~, measures the dihedral angle which these two planes form. Therefore according to the last proposition we shall haye OA xX OB=(at+ y 8) (w, i+ y,8,) en, +yu,Prt+aoy, 1B +y y, BB, =xeri+yoi+aoy,B +yy,b6, (Arts. 47, 48). We shall have also, by multiplying OA by OB (Art. 26), OBxOA=(x,i+y,8) (vit+y 8) =a vity cbB+x,ybB+y,y8,8 (Arts. 47, 48), consequently OAXOB—OBxOA=y/y, (88, —6, 8, or OAXOB=OBXOA+yy, (88,—8 8). Let ® be the angle which is between # and —,; ®, the angle BOX; 9, the angle AOX. Therefore y=QA.sin 9, y,=OB sin 9, and : 68 —f B—=2 sin 9.7(Art. 35), consequently OAXOB=OBXO0A+20A,0B.sin®. sing. sind. 4, We have already seen that (Arts. 28, 29) OAXOB.sing.sin®.sind,¢=FD,, or by taking GF=FD, (Fig., Art. 24), GD,=20A.O0B.sin. ®sin.¢.sin®.7, Therefore OAXOB=OBXOA+GD,. 56. This last relation shows that to have the product OB x OA (Art. 26), we could act exactly as if we would want to find OAXxOB (Art. 24), with the only change that instead of turning D’D, from left to right through the angle AOX, it would be neces- sary to turn it from right to left again through the angle AO X. Thus we should have OG=OBXOA. BT. Now let O0A=OS+OR, without OR and OS, being in the same. eo with OX, We have already seen that (Art. 52) vue. \ eR oe OBxOA=OBx(OR+0S) =0B.0R+0OB.OS. Let us indicate by » the angle that OA makes with its projection in the plane X OB; by » and vy, the angles that OR and OS, make respectively with their projections in the same plane. Therefore (Art. 55) OBxXOA=OAXOB—20A.OB.sinw, sin®. 2, OBxOR=ORXOB—2O0R.OB.siny». sin®, 7, OBxOS=OSxOB—20S.0OB.sinv, sin®. 4; and by substitution we shall have OAXOB or (OR+0S8S)xXOB —OH OB OS. 0B: because the perpendiculars running down from A, R, 8, to the plane X OB are respectively OA.sine, OR. sing, OS. sin; and by a well-known theorem OAsino=OR,.sinvu+O8. sinv or OA. OBsin®, sinw.% =OR.OB.sin®.sinz.1+OS.OB.sin®. sinv.z, 58. Let in general OA=OR+08S, OB=O0OM+ON., Therefore OAX(OM+0ON)=OAXOM+OAXON (Art. 52); OAxOM+OAXON =(0R+0S)OM+(OR+0OS)ON —OR.OM+0S.0M+OR.ON+08.ON (Art. 57). Therefore, whatever may be the positions of the planes ROS, MON, relating to the principal direction, we shall have as in numerical algebra, (OR+0S) (OM+ON) =ORXOM+0SxXOM+ORXON+OSXON, expression which indicates the manner to multiply a binomial by a binomial. It is easy to apply this rule to the multiplication of any polynomial by any polynomial, an HY om Example — Let a=aitbj+ck and B=a,it+b,j+c,k (Art 45). We have af—=[aa,—bb,—cc,+bc,—b,clit+[ab,+a,b]j) +[ac,+a,c] k, and Ba—[aa,—bb,—cc,+b,c— be, |t+[ab,+a,b]j+[ac,+a,c] k. Then | «8—Ba—2[bc,—),cl, and if c=0 and c,=—0, that is if the lines « and $ and the principal direction are in the same plane, or parallel to one plane, we will have, as stated in Art. 36, eh=—6 a, 58a. We have seen that the Commutative Law does not in general hold true in the multiplication of lines. Let us find out whether the Associative Law holds true; that is, if «, B and y are three lines in space, does («. 8). y=. (8. )? Let a—at+bj+ck, B—=a,i+b,j+c¢,k, Yagt+b,j+Cek. It is plainly seen that the equality (». 8). ya. (8. %) requires the conditions b,b,¢—b b,c, +5466, =) 6, Co— DD, 6g +5 gC, , b Dy Cg— 56, Cat 0,6 Cg Dd g6 — DC Cat Dg Cy. For example, as the values Deel tierra hy, Cae Cash 0g satisfy the above conditions, the lines a=ai+tAj+2k, b=a,i+j+3k, Y=a,i+j—2k, will give ton tet (ci. Pp atone, (Sony) =[aa,dg—10a,]i+[aa, tad, +4a,09]j +[20,4g+ 3a d,—2aa, ]h tis OS acs 586. By putting the above conditions in these forms, Cy (C. bg —b . Cg) — bg (0. cy — 9c) +B (b,6,.— bgc,) =0, b, (bCg—D gc) — Cg (6 0,— b,c) +6 (5, 6g— bgc,) =0; it is seen at once that if OR: =—, —=-, and -~=—; 2 jo ata geste b that is, if the lines «, 8 and y and the principal direction are in the same plane, or are parallel to one plane, the above conditions will be satisfied. Then Argand’s Algebra, or De Morgan’s Double Algebra (Ars. 36), is in all cases associative. Reverse of Multiplication. 59. We have seen (Fig., Art. 24) that OD, is the product of OB by O'A. Now let us suppose that the lines OD, and OA being given, we want to find aline which, mul- tiplied by OA will produce OD. This is the Reverse of Multiplication. The figure of Art. 24 shows how this inverse operation should be performed to find the required fac- tor, which in this figure is OB. In some particular cases, for example, in the case where the given lines OD,, OA upc the principal direction OX are in the same plane, and the angle XOA is equal to =, we shall find by this inverse operation certain different lines, each one of which if multiplied by the given faclor will produce the given product; and but one of these factors will be in the plane which passes through the principal di- direction and the given multiplier, the others which are innumerable are in another plane which passes through the factor which is in the plane of the principal direction and of the given multiplier, and also through the biseetor of the angle which is- between OA and the positive axis of rotation from OX to OA. Consequently, if we have «.p—=«a.B for example, « being perpendicular to the principal direction, it would not in general be correct to conclude that =8. 60. When « is perpendicular to the principal direction, we shall be able to have such a relation as «. $2.8, without {= 8 being necessarily true. Therefore from the equality apa, Bf, we shall have a,.B—a«,f—=0 or «(8—f)=0 (Art. 52); therefore if B—B =, a, $0; see Art, 30. this will take place when f—8, is parallel to the bisector of the angle jok or coincides with it (Art. 35), 60a. Wecan also find sucha relation as« .f—«,. 8, without «—=«, For, if «=e-+y, j being perpendicular to the principal direction (See Art. 45), we can write: a!) «bo (u rj) B=a,. B-+7.)8 (Art. 52); and if $ is the bisector, or parallel to the bisector, of the angle 7 o k, then jo (Art do) and % ° p=—«¢, y B, « and «, being unequal to each other. 606. Now it is obvious that, when the multiplier is 7 and the multiplicand is the bisector of the angle 7 o & or parallel to it, the result of the reverse of multiplication will be ambiguous in both cases, where the multiplier being given the multiplicand is required; and vice versa, where the multiplicand is given and the multiplier required. 60c. When one of the factors is perpendicular to the principal direction, the pro- duct will always be in the plane which passes through this factor and the principal direction. For, if erit+tyj+zk, then . Waa jr—(s—y)i+aj, and oj = —(Y + 3)t+ 2). Thus, such a supposition as Deo crOlne |< h will be absurd and impossible. : 60¢c. If to designate the reverse operation of multiplication, we employ the symbols a je and 2 and write by definition, B |x ees and 8X i Ip ae t then where the equality «.—y is true and possible, we may transform it to this form It is obvious that we cannot write in ‘general In order that the symbol is or Fi can signify a line, it is necessary that there should exist a line such that in the first case $ multiplied into it will produce «, and. in the second case, that the line multiplied into 8 will produce « — For example, e being a line out of the plane of 7 and 7, the symbol 7 cannot de- signate any line. For, if —=eityj +2 k. >. PO.) eae then jxBaj—yirai or p==(s—y)t+a) which is impossible as pe is not in the plane of ¢ and 7. Conjugates. 61. When any two points A and B are on a line perpendicular to the principal direction, and are on different sides of this direction, and are equidistant from it, we shall term A a Conjugate of B and B aconjugate of A. We shall also term any line MN the conjugate of the lme PQ and vice versa whenever the point M is the conjugate ot the point P and the point N of the point Q. In this case the lines MN and PQ will be of equal length. We shall designate the conjugate of a line MN, for example, by (MN) and the con- jugates of a, 8, 5 etc. by @’, #, & etc. 62. Conformably to the definition of the addition of two lises (Art. 15) we shall have a+ x= 2a. Ccos¢. t, dee e being the angle which is between any line whatever « and the principal direction. 63. The product of a line and its conjugate will be on the principal direction. For, if from the origin O which is on the principal direction, we draw two lines respectively equal to the given line and its conjugate, these two lines will be found in the same plane as the principal direction, and if one of them make with this direction, an angle 9, the other will make with the same direction an angle 2*—¢; consequently this product, (Art. 37) will be along the principal direction. The length of this product will be equal to the square of the length of the line given. 64. Thus in designating this line and its conjugate respectively by « and @’, and its length by «, we shall have «.«’ or @ , «=a%q, If we replace ¢ by 4 (Art. 49), m 0” or ono = 4%, Example — Let aatt+bj+ck then | aa bj! +k and by combining % o' = 24 7-4 527 7'+ 2k kh’ + ab (17+) 7) +ac(th'+k7r)+bc(jkh'+ky’). As 4v=)j'=kk=7, and obviously ‘=1, j=—), and k=—k, ~S} therefore p+jis—tj+jiz—ij+ij—O (Art. 47); tkh+kim—tk+khiz—ik+ik—0; jk+kjya—jk—kjme—jpk+jk=—0 (Art. 34); a% a! or %%4—=(a2-+ b?-+ e*) 4, It follows that Cerca hae oe 65. oY Le, me 2__ R2__2)2 Lay or by squaring a at i Tee Dane 4v . 4v ag and as @ i 9 OOS Osteo eee een A2—B2% —(:2)2 | Be (Ai eae Ung Sea i290 ieee Pi eOe 2. Br (2\2__9 Pe C2 a (A B C2) 2B ee eae ag SN hence BG? — | (b) Then of the equalities (a) and (b) we have (A2 —B? —C2)2—2B2.C2__ Qn2— Ae ES IIT a and consequently r2(4 B2, C2 —(A2 —B2 —C2)2]=A2. B?. C2; ney A? B2, C2 KB? . 0?—(A2?—B?—09)2° Uh eh ae Besides 4B*,C%—(A2—B? —C2)?=(2BC+A2—B?2 — C2) (ABC—A?2 +B? +C?2) = [A? — (B*? —2BC+C?)] [(B2+2BC+C2)—A?| = [A? — (B—C)*] [(B+0)2—A2] =(A+B—C) (A—B+C) (B+C+A) (B+C—A). Now if we make A+B+C=2S, we shall have are A? .B2.C2 Mow GS(S-- Ay(S— Bb) (S—-C)y: nie ALB.C 3: "=FV S(S—A) (S—B) 8— 0) } ©) Ex. 7. In any quadrilateral prism, the sum of the squares of the edges exceeds the sum of the squares of the diagonals by eight times the square of the straight line which joins the points of intersection of the two pairs of diagonals. Let OA=«, OB=8, OC=y, OD=8; the sum of the squares of the lengths of the edges (Art. 64) = 2 [a o'+ BB+ (y—a) (Y—«') + (y —8) ('— 8) + 288), 2 [Raat 2A 4 277238 —(ay't ya!) —(18'+6y)]. The sum of the squares of the lengths of the diagonals (Art. 64), = (7-42) (142) +01) (1) +@-+a—8) Cre —F) + G-+h—a) +f —2) = 2 [a a’ + BB+ y y'+- 258'— (a B+ Ba’) J, Also s0G=$C+y)= the distance from O to the point of bisection of CD, and therefore to the point of in- tersection of OG and CD; and the distance from O to the point of bisection of AF, like- wise to that of BE, and therefore to the intersection of AF and BE —_— ny 4: _ +-%==} (S++), hence the straight line which joins the first point of intersection with the second =$(*+8—y); eight times the square of the length of this line (Art. 64) =2 (x +B—y) (2'+8'—1’/) = 2 [a a'+ BB+ yy'+ (x B+ Ba’) — (a y+ ya’) —(By'+78)], which, added to the sum of the squares of the lengths of the diagonals makes up the sum of the squares of the lengths of the edges. F Ex. 8. To find the cosine of an angle of a spherical triangle in terms of the cosines and sines of the sides. Let « and 8 in the formula af'+Bo'=2 (ee, +yy,+2,)% (Art. 66) be two unit-lines. A In the figure let O A be denoted by «; OB by 8. Let OA be in the plane X OY, then as z=0 af'+Ba=2lea,+yy,)t. In this supposition we have w==cosb, ~==cosa, y=sinb, and «f'+fa=2icos c. Let BD be perpendicular to the plane XOA and DE to the line OX; then ED=y,, EB=sina, and ED=EB cos ©, then y=sin a cos C. Therefore 2% cos c—=2(cos a cos b+sin a sin b cos C)i or cos c==cosa cos b+sin asin b cos C. 68. Let « be a line in the plane 7 VY —, and its projections on the axes 7 and VY =i, x and y. We have %—=eityy or «evityV—i (Art. 45). If ® designates the angle which is between « and 7, obviously w==«cos§ and y= sin 0, Therefore a—=4 (i cos 9+V —j sin 9), onl ¢=> 1, a= cos 9+-Y —{ sin 9, If the angle which is between « and the principal direction is —® then %—=2(¢ cos 9— Y —i sin 9). 68a. Let « be a unit-line making an angle of one degree with the principal direction OX, and situated , 10 the plane « Y=. We can write Via! (Art. Al) fre = (Vo) a ‘ter But cr): e being a line in the plane « Y—i and making the angle 4 with ¢. Therefore 26 26 pp (V—1)™ or pe (V=-4)”. Let p=OA, then e==OD+DA. If OD=« and DA=y are rectangular co-ordinates of the point A which is in the plane t V—=i, p=wi+yV—i (Art. 45). If ® denotes the angle XOA, e==p (cos 9+ VY —i sin 9) or pe (cos0+Y-1 sin %) (Art. 46). 26 From this and pe Waste, we get 26 i cos 0-++ Wisin 9=(Y—=)* 26 or cos 0+ W214 sin 9= a4". Ex. To prove De Moivre’s formula. Let a—=%cos9+Y—isin® (Art. 68), « being a unit-line. Obviously at = (2 cos 9+ VY —i sin 6)". We know that the angle which the line «* makes with the principal direction is n times the angle which the line « makes with the principal direction (Art. 41). Therefore a2? —=7 cosnd+YVY —i sin n 4 Pee Hh. pee Hence (¢ cos 0+ V — sin §)"=7 cos n 6+ VY — sin n 4, or (cos 9+ Y 4 sin 9)"=cosn 4+ Y-4 sin n ® (Art. 46) which is De Moivre’s formula. In this formula the number n may be integral or fractional, positive or negative, (Art. 41). 26 It is obvious that the symbol (Y—i)* multiplied into a unit-line which is in the / plane 1V—i, will give to it a positive rotation through the angle 9. ( 68. There is another important expression to designate a unit-line ep when it is in _ the plane 7V—i. If in the expression e==cos 9+ VY —4 sin 9 we put ; i 92 04 06 CON Aid ai’ 1 a 1) ee en rd 9 98 95 97 aD 125 12a Dl ee eS ee ee we get Dy Ved C808 Vood 04 pI Tha nando aA RD al or Sy OVE A a OV 24) IOV = 1) See Vie lt Gat inl ae ee 1 oe The second side of this equality is just what we would get by substituting 9Y-1 for « in the expression Now if this series is taken for the definition of e*, the same definition applied to the case where « is replaced by 9Y=4, will give as the result, p. Thus it is seen that we can designate the unit-line ep by the symbol - Views that is | \ ih Vero DA ace 5 eee a _ and —- ° i) 7} cos 6+ Y-4 sin § =e uf ay or ae ov x ico 0+ =isind =e V if a ws RE no CHAPTER Ill. SPECIAL PERPENDICULAR. 69. Let us suppose that « represents a line OA; , a line OB, and that OA=a, OB=b. A perpendicular erected on the plane which passes through the line « and 8, and having a length equal to a.b.sin AOB performs a very important part in the calculations of Linear Algebra; we shall call it the Special Perpendicular of these two lines « and 8. It is scarcely necessary to say that this number abd sin AOB is equal to the area of the parallelogram which has OA and OB as two adjacent sides. 70. The Special Perpendicular of « and $ will be considered as having such a direc- tion that in placing ourselves at the opening of the angle* which is between OA and OB, in such a manner that OB or ® may be at our right and OA or « at our left, and on regarding the point O, we shall see agreeably to our ordinary conceptions that this perpendicular is raised on the plane AOB; if we suppose ourselves placed in the opening of the angle AOB in such a manner that « will be at our right and 6 at our left the Special Perpendicular which we should have conceived as rising on the plane BOA, will be that of § and «. 71. We will represent the Special Perpendicular of «, 8 by Il or JHE and that of B, « by Il or II é it ig evident that - f be IT =—II , w IL +IT =o. DOs (REAL Be It follows from the very definition of the Special Perpendicular that the absolute number or length of Il is ab sin 9%, ® being the angle which is between «, 8 transfer- a, B red to the same origin. Tt is evident that a, being an abstract number, the Special Per- pendicular of a« and f, or II will be=a I ; the Special Perpendicular of —« and is—Il that of «, —® is ‘ate II that eee —, is I] tl G = and. we all a, Pratract number, the Special Decale ofa« and §, or that of «and «8 will be=o; or II =o, I] =0. De, & w, 0% 72. If « and are given, we shall be able to determine Il , and consequently | | . a, B B, a *This angle is supposed here for simplicity to be loss than two right angles. When it is greater than two right angles the Special Perpendicular of OA and OB will be directed downward on the plane AOB. In other words if the angle between OA and OB is less than two right angles, ll ary be along the posi- a) tive axis of the right-handed rotation from 6 to «; and if that angle is greater than two right angles, then II will be along the negative axis of the right-handed rotation from 6 to «. a, pag eee Let us first find the absolute length of I . which is obviously equal to that of II : Oy »% If we represent by ® the angle which is between « and 8 transferred to a common ori- gin, we shall have af'+Ba'==2abcos?.¢ (Art. 65) where «=a, 8=b. From this (a B+ B a’)* = 4q2h2 (1— sin* 6) .¢ (Art. 48), or 4a%b% sin? 6. ¢—Aaa’, BR —(aB'+Ba’)2, (a) As « and 8 are given lines we may write them in these forms— am=oe,t+y+s,k (Art. 14) B= @ot+Yol + Soh. These give a+ B a’ 2 (©, Got Y Yo F120) 1%, oa —= (G2 +y,%7 + 2,7) 4, BP (ae +ye+22)t. Now, substituting in (a), we have a*b* sin? 9 = (2, Yo— Y 430)? + (Uy Sg— 2 %e)* + (YLo— 41 Yo)? which gives, according to the very definition, the square of the length of Ul or of Ul ; a, B B, 73. Now, let us try to find an analytical expression for Il , « and 8 being given. x ? Let LL 2,04 yf Pek (By - a, 8 As the direction of Ul is perpendicular to the lines « and $, we must have (Art. 65) a, 8 cal Wh al pt Sent %,B «a,f eI] alll ica ? on replacing in these two equations «, «’, 8, f, ll , and Il by their equivalents, we a, 8 a, B } shall have these two numerical equations : f 2 ©1%3 + YiY3 +2,23—0, VoVs + YoY + 2923 —D; and (b) gives x +yi+e2—=N CI ‘ lhe: b? gin? 6 a,B «8 =(21Y SSP + (012 sere rp (Y1%o a Ye): > (2) From equations (1), we find — 71420 — 91% 9 a ay Y34%e —LYo __ © 1 Fg — 2X0 ms) Mi Pome 235 Y1%g—%1Yo 3 and, putting these in (2), we get By = (y 1%. — 2, Yo). Therefore equation (b) becomes II po eave —Y 13g) 0+ (©, Sq — %_31) J+ (Y1%e —21Yo) A]. a, Thus equation (b) gives two values for Ul . The reason of it is this: Equations (c) a, B contain only the conditions of perpendicularity of ll to « and ®, which condition will a, 8 be satisfied also by Il . Now, it is necessary to know which of these two values be- B,% longs to Il and which one belongs to II : a, B B, It is obvious that the direction of Mili and that of Il do not depend on 7, 7, and &k. a, B, % Then, let us suppose that 7 be along 8; 7 in the plane of « and 8, and, therefore, k a- long tH . By this assumption we have a, 8 FU eg —— Une ge Ue B= wat, and OI HU Ys); therefore the value of Il will be one of the expressions a, f Yiok, —yYy,%k, and for the sake of consistency with the definition of Il (Art. 70) the first expression a, 8 will be considered the value of Il and the second expression that of II . THUS; eho a, B B, general, LL =(e,y.—y 150) i+ (0,32 — F440) J + Yq — 0, Yo) by a, 8 and ut = (Y 124 — 31Ya) t+ (21% —%, Fg) J + (OY eo —Y1%q) k, B,« 7, j, and k being any three rectangular unit-lines. Remark. If ® be greater than two right angles, y, in the expression y, #7, & will be negative; then, obviously, the direction of ll will be opposite to the direction that Ul a a, 6 would have if 6 were less than two right angles. TA. In a pyramid CAOB let OA=e, OB=f and OC=y. We know that, if the hight CD of the pyramid COAB is H, three times the volume of this pyramid will re =H x the area of the triangle AOB; and if the angle AOB=4, the | area of the triangle AOB= j«8.sin 9%. Therefore if V represents the | B volume of this pyramid, we have el / 6V—«f sin? x H; AG and if the angle DOC= 9, then H=y sin ¢; consequently 6V—«8ysin 9. sing. But according to Art. 65 x Il aE Il 1 =? NT ) cos (;—9).2 (Art. 4), a, B a, B a, f and as N (I[ ize sin® (Art. 69), a, 8 then 8) ela y= 28ysin®.sin¢. 2, a, B a, B ny 8 $e aki Shea a, B a, B In the same manner we can also find that ell ts Ul Bi 2a Y,% Y,% EL Le eaoyee BY B,Y ren 75. From this comes the following important theorem. When three lines represen- ted respectively by «, 6, and y are not in the same plane, we have AD +I yoo W +H r=oI +I « ty Ys % Y From the inspection of this formula we shall see by what law «, $ and y there change their places on quitting the bottom of a to range themselves by its side, or vice versit. It is very necessary to remark that in this formula we assume that in placing ourselves between « and # leaving # to our right and « to our left, and in regarding \ the point O, we shall see y above the plane passing through « and $, Also if we place ourselves between y, «, or between 8, y, leaving « or y at our right and y or $ to our left and in regarding the point O, we shall see 8 above the plane which passes through y, «; and « above the plane which passes through 8, y. 76. If «, 8, and y are in the same plane, p=0; therefore 2%8ysin9, sin ¢—=0, and Y II +11 y=8 II SLL Bia at +I] % = a, B a, 8 1, % 1, % 8, Y By Y Conversely, if y IT ze alll y=0, none of the lines «, 6, y being themselves 0, we a, a, B must have either 8==o or 90; hence in either case the three lines are coplanar. V7. Since ll is perpendicular to the plane AOB (Fig. of Art. 74) and ll ig per- a, 8 By pendicular to the plane BOC, ll and II are both perpendicular to OB the line along eae a, B B,Y ; which is 8; OB is perpendicular to the plane which passes through Ul and ll , and therefore is in the direction of a, B By =mB. II, I ; hence, m being a number, II, II a,B By a8 By If OA=e«, OB=f, OD=5, and OE==«; and if the planes AOB, DOE intersect in OP; it follows as seen above, that, Il and ll being both perpendicular to OP, a, B 6, € II, I is along OP and is therefore—=n. OP, 7 being a number. a@,B d,¢ 78. Formule — We haye already seen (Art. 65) that af + Ba'—=a' B+ P'a==2a8 cos. ¢—(a* + b%* —d?*) 4, (1) » I ——|] CArtist). (2) \ a, B B, % Let the angle which is between the lines « and § transferred to the same origin, be 6, Evidently we have adh Prammer : Bb? cos? 0 + a9 Be sin? 6; but a2 B2 cos? 6= 4 (a f'+Ba')* (Arts. 4, 65), “iti «2 62sin20—=N (II ) (arts. 4, 69); as a, hence a? B24 (a f'+Pa')? + +N (I }. (3) a, Let ay+s, P=p+o. From these we can deduce the following numerical equation afte yltey tie tee tre'toysieted; ——— (H) for we have aB—=(y+8) @to)myp top tyo' too, Ba'—= (e+e) (+8) ey +pi toy +od; therefore by adding these two equations one to the other, we shall have, (2B + Ba!) = (yp +ey) + (yo toy) + bp +pe) + (oF +8e), This theorem still remains good when « and $ compose more than two lines. We know that when «=8 er ie a, B au, % and («f+ Ba’) or aa’ +a a = 2a? 4. Let OH t+Y,J+2,k, BPegit+yYoji +2ehk. We shall have af + Bo’ —a'B+B' 02 (UG +YiYo t+2130)t (Art. 66) (5) or & B’ —Ba' == 2 (2,Yo —Y 120) 41 +2 (Y lo —2, Yo) J +2 (2,%o —%,2_) k. (6) We have also seen (Art. 73) that U Peau —Y 142) t+ (%, Sq — 2, %e)J + (Y Le —2, Yq) k. (7) The frequency of the application of this last formula or theorem in what follows, allows me to recommend to the reader to notice particularly, and to keep constantly in mind the law according to which the terms of this formula are formed from the terms of « and 8. 2a ae If (Art. 14) =H IY IHS K, B= Heit Yol +Zek, Yet t+ Ygj+3sh, and SH yt + Ya) +3, 4K; by addition we shall have &-+B—= (w+ We) it(yy~HYo) J+ (2, + 20) hk, YS (wet 0) t+ (Ygt Ya) J+ (Zg+ 24) A; and by formula (7) ee teresa ere) at 24) ]$ + [(@,+ &q) (23+ 24) — (41+ Ze) (a+ 4) ]J + (Yi. Ye) G+) —(@,+%o) (Yat ys) |; after having performed the multiplications and arranged the terms, we shall have I Seay aa ws Ut hag 91 05) ) (0g g) (x +8), (y+8) . + (ZY 3— Yas) t+ (%oF3— Fo%5)J + (YoUg— Vays) st (2, Yg— Y134) + (© 154 — 2184) J + (Y%y— 2, Y,) & + (Zo Y4—YoZ4) t+ (Gq F4—F 9% 4) J + (Yo%s— LeG4) f, —Il +I eA +I. (3) (7 +8), (y +3) a, a, or again by formula (7) It is well to observe that the formation of the terms of this formula follows the same law as the terms of a polynomial multiplication. If «, 8, and y are three lines (not coplanar) we shall have (Art. 75) vl +H v=ell +i val ell (9) a, 8B a, B T, % BLY By and by formula (2) AM th oe —tH +Il ; (10) B, B, % If GUI Y IAS K, Bagi t+ Yoj +2qk; we have I TAC ETE a (@y2o— 24g) J + (YrTo— 2, Yo) hs a, : and again if {Ut + Ygj +35h, Uy, Ruy *Rs, 4 Ry aL ec we shall have by formula (3) Hl p +H Y= 2 [es (21Yo— Y1 34) + Ys (U1 2—— 21 %q) + 35 (y ,Vo— % Yo) | 4 a, a, 41 Y1 % ZZ, Yi 41 2 ie, ig ea | be 2 eg ee eee A! (11) 4, Ys se Tz; Ys 83 We have seen that the volume of the pyramid OABC (when OA=«, OB=6 OC=~y) is one twelfth of the above. If Ge t+y J+5,k, Bait yeitszgk, ymustt+ys7+23h3 in replacing ~ by U in formula (7), we shall have B,y TT fea 2005) ys Yate —eayall Phy + [ay (Yots—®eY 3) — 21 (40Y¥s—Yora)l J + [y1(%0Ys— Yo2s) — 4 (%_33— Fos) Jk = (1 Ls+ YY gt 21 55)Vqt— (©, Lo+Y Yo 3159) lg? + (2123+ Y Ya 24%5) Yo) — (Vet Yr Yor 2120) Yad + (Ug + YY gt 3423) Zokh— (w, Lat Y1Yot 2120) 23% = (G1 Vet YY gt 2433) (Got + Yoj +2eh) wes (0, %o+ ¥ Yor 2120) (v0 +59 +235), then by formula (5 : 1en by formula (5) Il se ewe eae eee a (12) a, [I | B, Hence — 1 (a B! co 8 a’) 1 ae (a ee 1 a’) 8, (13) II, « 8, ¥ | Whether we employ formula (12), or whether we make use of the method which gave it to us, we can have ie =f (ay! +70) 8— 4 (By +78) =. (14) II, ¥ a, 8 Be TE —-UL =s6v+16)2-(e+pe)1, a, a II, s By a, 8 oe TL —se@v+re)e= TL —sep reer. (15) a, 11 Il, y B, Y a, B We have by formula (12) TL =slev +10) p—4 (06 +82')y, «, II By TL =s(@e +a) 1—4 Gy +18), 8, II Y, % Il —=alyP +By)o—$ (yo +ay')8B, Y, a, B therefore by adding II 519) «|| afi (16) “, I] B, II ¥; B, Y Dick a, B By formula (42) Il = (ay +ya’') B—2 (a8 +Ba’) 7; a, II By by putting here ll instead of « we shall have a, {I =3(II veri ees (II e+ell das YI, Ul a, B a, B a, B a, 8 a, BB, y but the lines ll and ~ being perpendicular the one to the other, ll B+ 6 Il Ai) a, B a, B a, 8 (Art. 65), consequently =} GIT +E ema Gl +I] «)p=a (ell all a’) 8 (17) B, Y By Y 1, % 1, % If in formula (12) we had replaced « by ll we should have found ao TL =:00 v++ID )e—3C0 eo +e Ds. (18) fi, 0 a, 6 %, 6 a, 6 a, 9 a, 8 By y By this last formula —(]] y+31] )e—31 I tae lar je By Boy BY By By %, 6 but by formula (2) II +I] ea): I, I I, 0 By a, 6 a,5 By then é : é (II v+r1] )e—(I +61 )r+(I v+31] «—(I] a, 6 a, 6 a, 6 a, p B BY Boy reall Jao; “o BY Peg fg but by formule (9) and (10) i vil yell oll rao If fall », Ar Aig Seyi aq alii “IL. B,a Ba a, B «LT +II my] +I] e3 Fey Bory ¢ %,B «4,8 consequently CI +I] Pea Ul aslEh )a+s IT call ie 7 emo: b] B, Oar if % aig a, B a, GIT Sie pean +I sad -T>);s4I0 4 a, a, Boy ni a, Y a, a, %, or GIT +I Mab IT +H Ja +1. ll s)erGIL +H», (19) a, a, ’ BLY Y,% a, e,} If «, 8, and y are coplanar GT] + s)a+GIT Alt e+GT +H rao. BY BY X, We can ey the last formula in this form SS + a + (3 tee eo yB+ \6 a a), “Na, IL)! OME +ULs)-+6IL + ede+ GIL +My This Piva! expresses a line in terms of three other lines. Let C= 2 eee ed k, B= Get + YoJ +2ghk, YH Ut + Yu] +25h, Sat +Yy,i+2,k; by formula (11) IT + = 2 [@q (ZeY eu — YoFe) + Ya (WeFZ3—Zq%y) +24 (Yo%s —Xe¥5) | *, “ig »v sil +I] S—= 2 [4 (2341 —Ya21) + Ya (G53, —%5%,) +2, (y ge, me PURY) LE 1, % Y,% SIT ll : [4 (21Yo— Yi Fe) + Yq (@1 Sq —34%q) +34 (Y ye — 2 Ye) |t. a, a, Now by multiplying the first of these three equalities by «—=a,i+y,j+2,h; the second by Pa t+yej+%_k; the third by y=a,t+y,j/+2,k; and finally in adding the products we shall have GIT +I] vex TT +I] oe reace oi vl Ys Biy ORs %,% , 0 a, f ny es which is the second member of formula (19) and = 24,0, (FeYg— YoF3) T+ 2Y 1 Y 4 (WoFg— Zq¥q) J +22134 (YoUg— Vays) ie +2 os (ZyYi— Yg23) PHY aYa (UgF1— 2901) J +2242, (Y3%,— V5Y,) ke +2504 (2, Yo—Y 120) T+ 2 YgYq (U1 Fg— 21 Vq)J +2252, (YLo— 21 Yo) ib = 2 (weet Yr Yat 2124) | (ZoY3— Yo%3)t + (GeFg— Ze%a)) + (YoVg— LoYg) hl — 20,04 (VeFZ3— %o%3 )J —2U1%q (Yo%g—VoYg) K—2y 4 (Z0Yg— Yor)? — 2 Y Ya (Yolg— Vo 5) K—22, 34 (ZeYg—Yo%s)4— 22124 (XaFZy— Zo%y J +2 (to Gat YoYat 203) | (Z3Y1— Y 921) t+ (W354 — 2381) 7 + (Yg%,—%sY;) J] — 2 Ho%q (%3%4— 23%) J —2Gq%, (Y3¥,— L341) K—2 YoY a )33Y1— 321) 2 — 2 YoY (Yg%s1— B31) K— 2 5934 (Z3Y1— Y 921) 4 — 2 29% 4 (%gF4— 23%) ) +2 (504+ Y5Y4+%52a) [ (21Yo—Y1F0) 1+ (© F—— 3, %9) J + (Y1Vo— H Yo) hi] — 20504 (Uy Fq— 31% 9) J —2H5%q (YyLe— 2, Yo) K—W@ygYq (71Yo—Y 132) 2 —2y sYa (Yr%o— TY a) K— 2A gq (21Yo— Y1F2)t— 225%, (@,F9— 340) J = 2 (Gg Y Yat F124) [ (FeY3— Yes) t+ (WoFZ3— FU 5) J + (Yo%3—LoYs) hl +2 (LoS, YoY gt 0%,) [ (23Y1—Y.321) t+ (% 534 — 2 5%1) J + (Yg%1— %3Y1) fe] +2 (@ 504+ Ya Yat 2524) | (Z1Ya— Y 120) b+ (Up Sg— 3, He) J +(Y Lo— 1 Yo) fi] Berea ye ea coe LE ciety) LL ba Y,% a, 8 or GIT +I {8 (a8'+5e) I] +(69+3¢) 1] G3 +37) 1 : (20) a, B a, B By Y 1, % a, B By formula (7) Il ve (3, Yo— Y 120) OH (%1Sg— 2, %q)J+(Y 1 %o— Yoo) &, &, I a (23Y4—Y 334) U+ (@554— 2584) J + (Yg%g— 25Y4) he; Lie consequently by formula (5) act Beney,.5 46 =2(2,Ye—Y 120) (23Ya—Y 324) 1-42(% 12g —2 1 %q)(U324—F 5 ee Ya) (Y svg — Yn) 1; by performing the multiplications, and adding to the second side these three zeros: ©, %_ 3 V4 Hy Go Ge Gq, Ys Yo Ys Ya— Yr Ya Va Yas 21 20 25 Z4— 2) Zq 3 %4 WE Shall readily find | If + II IL 2 (UG 3+ Y Yat 2133) (So Vet Yo Yat Ze 2) t ap y,8 y,o 4,8 2 (gH YY YgH 514) (Votgt YoYgt ZZ) ¢, lS el or again by formula (5) TOL See ey eS 7,5 4,8 = 4 (87 +88) (2 y+ ya) —§ (ae +2a/) By +78), (21) 79, Examples. Ex. 1, The squares of the sides of any quadrilateral exceed the squares of the dia- gonals by four times the squares of the line which joins the middle points of the diagonals. Let P and Q be the middle points of AC and BD; and let AB=z<, AC=8 aud AD=y; then BD==y—z«. Therefore ae Fast eae PQA Oa nh Ro 5) 2? or 2PQ=c 7) Hence the numerical equation that we can deduce is A(P.Q)%*=a2-+ 724+ B24 (ay + ya!) — («BR +Ba')—(yB +8’) formula (3); but pha Komi ht Sea) hE aff telat +9" — (BC) 78 +B y' = y+? +82 — (CD); therefore 4, (PQ)? = (AB)? + (BC)? + (DC)? + (AD)? —[(A.C)? + (BD)? ]. Ex. 2. Let C= o,I+Y, J +2,k, Bot + Yoj +2ek. We know that atest yp ts, Beaty? +2; And af + Bal’ =2 (eGo +Y1Ys +2129), Nd i ca (Art. 73). a Therefore, let us replace «?, 8, «$'-+8, and Nal ? in formula (3) by their a- boye equivalents, we shall at once have (P+ ye? +2") (we tye +24’) (0, Wet ¥ Yo 2120) 2 (Yi Fa S140): 4 ae (3, %g—- © 129)? + (7, Yo— Y1%q)*. ey oo Ex. 3. Let again @—=0,t+Yy,j+2,k, Bat + Yo] +2ehk, x and oaetm ULL a, o,=—m, ULI ; m, and mq being any two numbers. Then o =m, Ull +EttYy tok; a, B a, These give oP =MyP role ryyp+sy, OF = MP Hae HY + 2e5 and W1Oq +40, =—2 (mM, Meg—XVo— Y1Ya— 2 170)t. Also Il —m,Va2 +Ye +35 II — meV «2 +y 2 +2? Il + ; 1 %o ull, v8 ua, ull a, 8 a, 8 a, 8 and Il ll =mPZ(ef+ys+ze)it+m? (aery2+27)t M1 O10 no Veron Verve If It OE | +I. UlULMte PUSCite Us, ole voll, vp a, a ,8 a, B a, B a, B a, 8 2 But by formula (21) [Art. 78] II I] +I] We ull, v8 ve,vull Ue UlL | oll, uP a, B a, B a,B «,8 =) {oeU IL +UIL ve] (UI v-+0- UI a, 8 a, 8 a. a8 “gi JUTE UT sul ull | [UBU'a+UaU'S]; eB +Ba! OS *The unit of a line is denoted by prefixing U to the symbol which represents the line. Thus ull is m, read unit of Il ; Ua unit of «. a) 8 Le BG os which may be easily obtained without the help of formula (24). Therefore N’ (IL) crite (et+Yatso) + me (a; +Y i +21) + MMe qrree) +I] aut M1, qd, = mé (we -+y2+22) + m§ (wt+yf+z7) + 2m, Me (By VoAY 1 Yo+2 120) + (Y12o— 71 Yo) 2+ (3, Le—B 1 Zo) + (Lp Yo—Y 4 %e)?; (Art. 73) = (9%, +, Le)? (MoY HM, Yo)*+ (Moz,+M, 20)" + (YS —3 1 Yo)? + (5, Va—¥ 4 Fa) *+ (U1 Yo—Y 1 Lo). Now adding to the right hand side the expression 2 (Mo% +, Le) (YySo— 71 Yo) H2(MeY HM, Yo) (51 %g—L 1 Fe) +2 (M43, + 120) (4 Y2—Y 124) which is vanishing, we get N I] = (19%, HIM LoA+Y 1 Fo—F Yo)? + (May +MY gtF1V_e— 420)? ® + (192, Zo+%Yo—Y1%3)*. Therefore substituting in the formula of of =} (ordg-tood.)2 +N UL) , (Art. 78) Mo we have (mi +a?+y i +27) (me+a+y3+z%) = (1M g—H 1 Be—Y 1Yg—F 13g) ?+ (Mg VAM LoAY 1g —F Yo)? + (Mey 1 +N YotF1Vo— 120)" Z + (Mo3,+M,Zo+X Yo—Yi%e)?, a formula of numerical algebra due to Euler. (Tait’s Quaternions, Art. 103). More generally put Be feomt (1 U if +Et+Yy,7+2,k, a, Mg —=— Mz, U II Maree Uitte a; x,=o,+2,Ull .: @® and to — ny Ll — 59 — Then m= Ni tmiteit+yi+si , Ro Ng tmetaeet+Yyotse s and Rete Tet, ——— 2 (2, Ne+M,Ma— %Vo—Y 1 Yo— 212) N’ (1) =n5 (meta tyg+2g)+ng (m§+a2+y%+22) Ti sTo — 211 N%q (M,Mg— 4, %e— Yi Yo— 2120) + (9M ye Y Fo — 2, Yo)? + (Mey, +M Yat, Co—%1 20)? + (193, Z +2, Yo— Yi1%e)?3 = (Ny Me— NqMy)? + (Ny LeANg®y)*+ (Ny YotNeY 1)? +(NyFo+924)? + (19%, +, Ve +Y 1 Fa— 2, Yo)? +(Mgy + Yo +1 Lo— © 3q)? + (1192, +Zg4+%,Yo— Yi1%9)?. Therefore (nfm? +a +y?+22) (nz2-+-m2+02+y2-+22) = (1, Ng+M,Me— %1Lo— Y1Yo— 2120)” H (2 Ma— NgMs)?-+H(Ny Ha Me®1)*-+ (My Yogi)? + (11 F9+921)° H+ (1q®,+M Vo t+Y 1 Fq— 31 Yq)? +(e Y HIM Y gHF 1 Lq— Gy Fy)? +( 193M Fg +H, Yg—Y 1 Ve)*} =(04 Ng +m, Mg— Ly L—— Yr Yo— 2132)” +(14Me— NgM,)P+(Me%, 1M, Lo)2+(M4Y 1 HM Yo)? +(M%92,+M 120)" t (1D Meh +Y 15q— 31 Y 2)? + (My Yoo Y +21 %q— HF)? (My Fq+NgZy +0, Yo— Y1%e)?- Ex. 4. To find the volume of the pyramid of which the vertex is a given point, a and the base the triangle formed by joining three given points on the rectangular co-ordinate amis. Let A, B, and C be the three given points; OA=a, OB=b, OC=c; «, y, and z the three co-ordinates of the given point. Porihens OA = a7, OB=—by7 OC—ck.. and OP=ai+yj+z k. Let PA=«, PB=§, PC=y; V= the volume of the Y pyramid PABC. Hence «=0 A—OP=— (r#—a)i—yj—ak x f=—OB—OP=—a2i—(y—)d)j —zk, Y=O0C—OP=—2i— yj — (z—0) k. By formula (7) II =) 9a] 4 (ea) sa) |i+fey— es) y—B)) a, \ =— bzi—azj—(ab—ay—bda) k. And (Art. 74) vl == Il speed: PANES. a, B a, 8 Nes (—wi—yj—azk+chk) (—bszitazjtabk—ayk—bak) | +(—bsi—azj—abk+ay k+bak)(—ait+tyj+zk—ck) { 42 as after having made the two multiplications which are in the first side, we shall find 6V=—abzs+acy+bce—abe, or Vejabe (244421), eae) mae . If V prove to be negative, it will indicate that the pyramid PABC is below the plane ABC. Ex. 5. To express the relation between the sides of a spherical triangle and the angles opposite to them. Retaining the notations and figure of Ex., Art. 68, we shall have, II sin ¢. %,, Il sin @ . to, a, B ByY wu, is the unit of the special perpendicular of « and $8; and we» is the unit of the special perpendicular of ® and y. It is readily seen that the angle which is found between w, and uw, is the supplement of the angle B of the spherical triangle ABC; and that OB or 8 is perpendicular to the plane which passes through w, and u,. Therefore II —=— sinc. sina. sinB.&. EE dik a8 fy By formula (1) | aut + IT e’==— 2sina.sin¢.#, »¥ an % represents the angle which is between the line OA and the plane COB; therefore a(a I] rea EE «' \=—2sina. sine . 8. B, y B,Y But by formula (17) II esi (al [ mea Gl! «’) B—=— sina. sing . B; II , 1 eS ds | “PB By et Gh aces os sinc.sin.a.sinB=sina. sin¢, or sin 9==sinc. sin B. Similarly sin 9 = sin b. sin C, Therefore sine . sin B= sin 6. sin C, or Sim Usa G ca Sih 3 sin CO. Ex. 6, To find the condition that the perpendiculars from the angles of a tetrahedron on the opposite faces shall intersect one another. Let OA, OB, and OC be the edges of the tetrahedron, and OA=z, OB=, 0C=y. Let the perpendiculars from A and B on the opposite faces, be AN and BM. We know that reel Al and rancor 10) 1,8 oey (Art. 70), m and n being two numbers; if these perpendiculars in- tersect in G, the three points A, B, and G will be in one plane, and the special perpendicular of AN and BM will be perpendicular to the line AB=8—e; and ald! which is the special perpendicular of BM and AN LE a, 7,8 (Art. 71) is found perpendicular to AB. Hence (Art. 65) ll, U , U “,y 7,8 a,y 7,8 but by formula (17) Pee Ses e ll y: Tee Ll a, Y a, B a,y 7,8 therefore Geert ermal CL Ty pede LL a) @— 290, a, Y a, ¥ a, ¥ a, Y or (IL v+sIl ) t@—2)y+1e—2)]=0. a, Y $7 It is evident that, ll 8’ +8 ll cannot be =0. Therefore, my a, Y (B—a«)y+y(b—2#’)=0; By + yi ay +ya’; but by formula (1) By’ +yf'==(0 B°+O0C°—BC*)¢ and my yt (OAs O Cl A C*) 2; a OB'+0C*—BO=0 At 0C—AC?, wes 1, a or OB*+ A C= BOt4 OAS. Consequently the condition that all three perpendiculars shall meet in a point is that the sum of the squares of each pair of opposite edges shall be the same. Ex. 7. Any point Q is joined to the angular points A, B, C, and O ofa tetrahedron, and the joining lines, produced if necessary, meet the opposite faces in a, b, c, and 0; to prove that 00 0:700 — 42.0 0a, cof) a) Ono eel DCW AB=$—«, AC=y—«, OB=f—3, OC=y—%, aQ=ws—az«. Therefore peer till +1] (3 —a«’')=0, (B= 6), (7-2) (Ba 9) 17) (§— ae) Oh a +IT ) 8, B, 3 6,6 +([[ Fadel ul +IT )@—a0) =o, or by formula (8) By 5,7 ,° ; eo (IT +H il Pp eae ett +I daw bill +H J 61 VID abE AT 7,8 «(11 +olT voll +I all all 3 (GIL “I = AN B, Y 150 ‘es We can find in the same way, a ae +eLT el +H e+ TL e+ ID oY ~GID, +I =o; {eH +l ll il vl rl t= — (11 +H, «')=0; sale pale GT + IT ae — GIT cell b)=0; i apIl +l el sal: y+] v+I] 3 = (alk = ‘ge a, B BY BY 1, % By Y BLY ) Now if we write eA Bh +I] t= t oll sildic GY t, IT +H easy, BLY 8, Y 7,8 8 lI qin ! 7,0 hy? , and apply formule (9) and (2) we get agztraytaz— w=—0, —ba—y—bz+bw=0), cuetcy+s—cw—), —a—dy—dz+dw=—)0, which give ake Meakign 7 v=0: a aah vce hi p=, Cc b eer: peat as C En T pee oe and therefore tle | og, CA ge NE or a—i b—1 c—1I Wis eg 0 A Cat pee Oe eit oo SPR ys CM a a ha OSs but by supposition Qa Qb Qe 0 a— QA ? Oa C— QC ? Ca QO ? consequently Fa Faso Ra Qa a1 a= Oy etc. ; therefore Qa Qb Qe ae = =I Qa—Qa™ Qb—QB Qe—Qc™ Qo—Q0 \ Ex. 8. Prove that the arc a 6b of a great circle which bisects two sides of a spherical X triangle ABC, intersects the base AB at the distance of a quadrant from its middle point, ~ a Let O be the centre of the sphere; C—O 0, eee Us y——() A; no=— and cos. ‘Let EE, be the line of intersection of the planes passing through «, 8 and y, 4; f therefore containing, respectively, the arcs ab, AB. a 1,0 ll =4(1] veo] )y—4 (I verl[ )a; Oh el a, P a, 8 a, 8 a8 a%,B y,8 putting a—=a(p+%) and B=y(p+y), War I +31] | oa. el es ¥ Ps Y a%,B y,6 Thus we see that y—¢ is parallel to the intersection EE, Besides, the expression (y iP lalbhsiaes) SA yee ah yeas where y+° is a line passing through the middle point c of the arc AB, is vanishing, since y==%. This shows that the line Oc is perpendicular to OF, that is, the arc cE ig a quadrant. ~ Cor. When a great circle bisects two sides of a spherical triangle and one side of a second spherical triangle having the same base, it will bisect the other side also. Ex. 9. To find the cosine of the angle between the planes passing through «, 8 and y, 5, respectively (Fig., Ex. 8). Let Oc=o, Evidently the angle which is between Il and II is equal to the < ma) a, B ? ¢ angle of which the cosine is required. Therefore, if » designate this angle, we have re. >. N(II) .N (1D) 00s 2. = ill Ah all IT a, B 30 30 at, =} noes (Bot oB') —4 ll (2 ol wa) (Art. 78, Form, 21). But S+y=(yo'+oy') o; Ye (by +y7P')B; 3S pm (x +30) a; ‘Ke (10 +0) ox (rf +BY) b+ (2 +32) a—2e, and 6 (%8' +30) a—p=(yo'+oy)o—y. From these, Bo’ +38’ = (ad +5.a') («8+ 8a’) —(Bo’ +8’), =(y0'-+ 07) (08 +80) — (78 +8y); and (yo toy’) (Boi +oB)=2 (B+ By) + (ao +2a'/) (28 + 8a’) —2 (of +o’); (yo! +o 7’) (x0! +0 a) = (78 +8 7’) («@B +8 a’) +2 (a8 +50') —2 (pa +a’): or pt io ee: MRA Ae He Saal ag b ‘i Ww’ —-+- w 7 (% 8’ + 6 a’) (% B' +B a’) wy’ 1 i Foes p + wf = yo +07 ’ Now, substituting, a, 8 3, my EEE [lett bal) 8— (at 400!) Gr’ +B) (4+ Bo!) + (Bye) But (10 oy) © (YB + By) Bt (28 +32) x—2¢ and (yo + © 7’) = (7B +B y') B+ (a8 +8a/) a — 29, give (ro +07) — (P+ BY)? + (YH +BY) (28+ 8a’) (2h! +82!) — (09 30)? +i, te (RA Ba (rH + By) (a2 +00) (2H 48a!) + (1 +BY PAI (yo toy), Again substituting N (1D) .N (IT) Foose i yee [Ai— (yo! +0 ¥)2]; a, ; nn __ «bh +8a’ = Bi 1, @ eth y where 9 denotes half of the angle AOB. Putting N (LT) for sin? 4 Nil). N(ID) cosy. t= 22*#* N’ at ), ae yo boy’ NUD) NOD WaRe el errremr then tanaOb.cosg¢=tanAOc tan AOc or C8 2 ao ire Cor. This formula shows that if EF be made (Fig., Ex. 8) equal to the arc a b, and the are passing through D— pole of the arc AB— and F, cut AB in G; then the are EG shall be equal to half of AB. For, obviously, the angle fT EG=g and the angle FGE 5: | Ex. 10. As a corollary of the two preceding propositions (Exs. 8, 9) may be given the following geometrical theorem. Let ABC be a spherical triangle and a, b, c the middle points of the sides BC, AC, AB (Fig., Ex. 8). Let the ares PAS’, PB« be drawn from P, the pole of the great circle through a, 0b. Prove that ‘ 2/ PAB=2/PBA=x—(spherical excess of ABC). Comparing the triangles EBe« and E,A®, we see that E«—=E,8, Be =A8. Join C and P, m being the intersection of C P and a b. Since P is the pole of a b, the angles at m are right angles; and, therefore, by the equal triangles Bea, Cma, and Abd, Cmb, we have coin, 10 =) Pend se —— Om) oan, Produce CB and CA to meet in H (on the opposite side of the sphere). H and C are diametrically opposite, and therefore CP, produced, passes through H. Now PA=PB=PH, for they differ from quadrants by the equal arcs Af, Ba, and mC. Hence these ares divide the triangle HAB into three isosceles triangles. Then /PAH=/PHA and “-PBH=/PHB. But 7.PHB+ZP HAZ AT BS ABCA; /BCA=ZPAH+ZPHA. Again LPAB=x1—/CAB—/$Ab=r—LZCAB—ZPAH; /PHAC-EP A B= 5 ZCBA—ZP BH, = 67 = Adding and substituting, 2/PAB=27—LCAB—LCBA—ZBOCA; =x —(spherical excess of ABC). Further, let the circle EDE, pass through the points of intersection of the arcs AB, a b and perpendicularly to the plane of the circle EcK, Let ED be a quadrant, then D will be the pole of AB, ¢ of DE, and DE will be a quadrant. Make EF=ab. Let DF cut EE, in G. We know that EG=Bc=cA,. Join D, A and A, F. Obviously, as D is the pole of EE,, DA is a quadrant; and, since EG=cA, GA=KEc is a quadrant also. Hence A is the pole of DG; then AF is a quadrant and the angle DAF is measured by the arc FD; and as ASF is aright angle, AF is a quadrant and FA®8 a right angle. But, as ZFA$ and ZDAE areright angles, we have AGRA LEAR = 7 PAB: Therefore egal) oe A) A —x— (spherical excess of ABC). Cor. All spherical triangles having the same base AB, and their vertices being in a small circle passing through C and parallel to the circle EabE,, will have equal sphe- rical excess; therefore equal areas. CHAPTER IVY. Equations of a Straight Line. 80. Let 8 be a definite line having the same direction as that of the indefinite line drawn through a given point A; « the line from origin O to the same point A; e that to any point P in the same indefinite line and from the same origive; then AP having the same direction as $ is a multiple of this; let AP= 26, where @ isa variable number; the equation OP=OA+OP gives oa +08 (! as the equation of the indefinite line, of which the line AP forms a part. This equation can be put under the form (GI ==) 8, (0 — «) for, Il — Us If the line passes through the origin, then «0, and its equation will be eRe. OF ll <= lp B, In general the equation @—= Pye + Potgts 20s. = 2Po apg where «,, %), etc. are given lines, and P,, Py, etc. functions of the first order of one indeterminate numerical quantity, represents a line. For, any function of the first order of one indeterminate number may be designated by «m-+n; then let p==(wm,+n,)%,+(@ Met Mq)%q 1... from this p= @ (Ne, Me%at . « » -) (0,4, + Netot .. . -) or pulling {0,4 et, =e wa ANd On eae ee lee, pa + wB. Another form in which the equation of a straight line may be expressed is this: instead of the direction of the line and the position of a point in it being given, let us suppose two points in the line to be given, and Jet OA=s, OB=y; then AB=y—« and AP=a2(y—+2), p==a+a(y—za). (2) We can readily deduce the second equation from the first: we have only to suppose that Bmy—«a, Equation (2) may be written p—a%—a(y—a)=0, or e+ (e—1)e«—ax2y=0 which is a special case of the more general form apt+ba+tch=0 subject to the condition a+b+c=0 (Art. 22), A third form may be exhibited in which the perpendicular on the indefinite line DC from the origin is given. If « is perpendicular to in (1), the equation can be written ap’ oa —Qaa’—2Qa%4, (3) where a =<’ ig a constant (Art. 64). From (3) we have Um pet a e204, where the symbol U« designates the unit of «, and U« the unit of «’ (see Art. 41a). 80a. Let pa +ap' C2, « being a determined line, C a constant number. From it we get 0, UatUa, faci; _— thus we see that the given equation may reprsent aline perpendicular to «, and cutting from it a portion whose length is 5-. ay Fr req 80 b. Obviously any two parallel lines may be represented by ep——a+a8, p,——%, + a, 8; or by Il =U; Ul =: 8, (p —«) 8, (p, — «,) or by pata —=Ot, pa +a, =0,2. 80c. To find the equation of a line passing through the extremity of « and meeting 8 perpendicularly. Let ¢ be parallel to this line. Then the equation of the line is p—a+yd, e being a similar line to that in Art. 80. As 6 is perpendicular to $, we have pa +36" =0, and, «, being any line in the plane of $ and 4, we have sl] | Sf es On (Artce70). 8, a B, a, That is 8 and II are perpendicular to %; therefore these two equations give By a =u II ; ae b 8, e, whence we haye p= a+eu ll ’ 6, II >] %, when « is substituted for yy,. This equation gives (p— a) +B (p’ —2') =0, which can easily be obtained directly. 80d. To find the perpendicular line from the extremity of « upon the line p=y7+ 4,8. Let e be the line connecting the origin with the intersection of the lines P—Y + &, 8, and p= a+e Ul (Art. 80). B, &, ey Then oI] 8, II B, x, will represent the required perpendicular. These two equations give y+to,Bme+e FE 6, 1 B, e or YB +B y+ 2a8 af’ + Ba’; fp co Aer) Bae are 26 f Therefore z || =2,8 — («—y); 8, I “he, __ («8 +B a’) —(yp'+By’) ; ee TTR TTY Ph aR = yy [He—8 +6 vi]! 883 nl SM ATLAR(O, eR OrMres 2): Tears, a B, (a — y) It is obvious that, if «=0, the formula will give the perpendicular from the origin to the given line e=y+~a,8. Let this perpendicular be OD, then —arlt (By +yB)B—BP. x]} —=y—4(UB.y'+y. U'$). UB. | This gives OD=3(7> +87), 6 being the unit-line perpendicular to 8. This result can also be obtained directly. 80e. To find the shortest distance between any two lines. Let the equations of the lines be p= a4 +a8, pa +08. Obviously p—pe, represents a line connecting two points, one of which is in the line e—«+af; the other in the line p.=«,+72,6,. eae x __If p—p, be the required shortest line, it must be perpendicular to both lines, there- fore it is parallel to at , and we have BB, ieee cere h ll be ar) or (@e)rap—of ull : (1) 4 @—a) IT +U @—4=eyIL IT. B, 8, BB, BB, 8,8, This determines y, and the shortest line required is oecy EAN bees pee ee ee el] I ona BB, BB, =3(@—«)U TT +0 .«—4)).00 we B, B, pp Consequently the shortest distance required is the number of afe—a)U TL +ULL @—«)1. B, 8, B, 8, To find the extremities of this shortest distance, we must determine @ and z#,. From (1) we have (7 —a) B+ 8 (o'—¢) + 2a B+ aw, (8 B+ 8B) —=0, (a — a.,) B+ 8, (a'— d.,) 2 uo (8 B’+ B, 8’) at 2 v, 8 B = 0. These will give w and @,. Equation of a plane. 81. Let P be any point in the plane of which the equation is required; OD per- pendicular to the plane; and let CP Lesa aU Cil> 6: _ Then e—a—DP, which is in a direction perpendicular to OD, and therefore a (o'—«a') + (p —a} «’—=0), or apt pa = 2424, As this equation is satisfied by the line drawn from the origin to any point in the plane through D and perpendicular to OD, is the equation of the plane. This equation las the same form as equation (3) of Art. 80. In that equation it was tacitly supposed that e lies in the plane ofthe figure while in this equation the extremi- i 9 fi Pome ty of p is supposed to be in the plane through D and perpendicular to OD. Then the line equation (3) represents is in the plane which the above equation designates. If the plane pass through O, e can have the value zero. a p'-+ p «== () is the equation of a plane passing through the origin and perpendicular to «. Sia. Let Be+pp=Ci, 8 being a definite line and C a constant. (See Art. 80a). t. roalQ From it we have US eee ea Thus we see that, if we are not obliged to suppose ¢ to lie in the plane of figure, the given equation represents a plane perpendicular to 8, and cutting from it a portion whose 26 81b. Let oa’ +ap'== C7 length is represent a plane. Assuming that this plane contains 8 and y, we may put «=n ; B 7 n being a definite number, positive or negative. Thus the equation of the plane which is perpendicular to « and parallel to 8 and y, may be put under the form : ree SHE 21 POT edit C anu it +ul 67 GN > or If C=0, then etl +I] pe By. BLY which is the equation of a plane passing through the origin and parallel to 8 and y. 8lc. It is evident that, « and y being two indefinite quantities, p—azatys represents a plane passing through the origin and parallel to « and 8. From it we have al aalil Or “,B «8 81d, The equatian p—=a+tabt+yy ly os ibaa obviously represents a plane passing through the extremity of « and parallel to ® andy. From it we have ell +H e—«l] +I] a, ¥ ’ ; ’ Boy BY or @—) IT +11 @—«)=0; ’ + (p—2«), 8 (p—«),8 Putting nf = Il =o and aul 4 i « ==C7, we have the same forms as a, 8 8, Peele h Pat in Art, 81, In general the equation eee Sel 6 Gen Seer where «,, %,, elc. are given lines, and P,, P., etc. functions of the first order of two indeterminate numerical quantities, represents a plane. For any function of the first order of two indeterminate numbers may be designated by Pa+Qy+S, P, Q, and S being constant numbers. Let then p= (P,e7+Q,y+S,)e, + (Peet QoytSe) tot. ... = (P14, +Pe%ot... )+Y (Q1% + Qe%gt....) +(S,%,+Se%g+....), Or POU ety ate. oa, 4, + Qo%g+s...— f, and P.¢,+ Paap coy, p—a+ab+yy. as in Art. 81d. 81e. We have seen that pa + ap —=2aa’ may represent a plane perpendicular to « and passing through the extremity of a line drawn from the origin and equal to «. Let y~—<-+6, i where y is a line between the origin and any point in the plane, while © is a line in the plane; we then have 6 ok ee oe a theme PAI for Sa’+ai—=0 Therefore pe +ap—ya'+ay’, oF (e—y) a + (p—y') =0 represents a plane drawn through the extremity of y and perpendicular to «. 10 BE Pe 817. To find the equation of the line of intersection of two planes. Let the equations of the planes be (p— 1) a! +a (p’ —y')=0, 4) (p faba y) dt a (Cee Y,) = 0. Obviously the line of intersection contains all the points whose values of p satisfiy both equations. But we may write, (Art. 78 Form. 20), since «, «,, and Il are not coplanar, Qa, a eel +H =H (free) +L (eres) + 1] ell il 2) f % Oa, se) ihe or, by (1) and Art. 75, 2LI IT =I eters) dL nr ead ell Qa, a, a, a ee he where x, an indeterminate number, is substituted for N (11 alii e’) which may have any value. Cp toh When both planes pass through the origin, we have y=y,—=0, and obtain at once : uw or putting « for ————~, alt Il & me, Xa, we obtain =ell, a, % as the equation of the line of intersection of two planes. 81g. To find the equation of the plane passing through the origin and the line of intersection of two given planes. Let Bo -+p B==0 be the equation of a plane passing through the origin. 8 is to be determined so that the plane will contain the line of intersection of the two given planes. | Let the given planes be Mihm ee and a(p—y/) + (p—v/)%=0; or («p+ p%)—(a7/+ 7,4) =0. Obviously f is perpendicular to the line of intersection of the two planes; therefore we may assume = oe ees Brewery a; for « and «, are perpendicular to the line of intersection which is along I] : ’ ‘ Therefore (ratya)e bp (we'+-y a)—=0 a(ap ea’) + y (ae +p e,)=0. (2) It is evident that if a value ofp satisfies this equation and one of the two given equations, it must also satisfy the other equation; that is, if e be such that wap +a!) +y (a,c +e¢)—=0, and ao + pe ay bya’, also a, 0" + pd, ==, y/+ Y, a Then a (ay +ya)+y (2,7 + 7,4) =0. (3) Therefore eleminaling « and y, we have (ap +e’) (2,4, + 7,4) — (a7 +7) (2,0 +e%)=0 or [(a, 7, + 7,4) %—(ey+ya)a}e +e[(ay/+y7,4) e'— (ay +y2')¢]=0 (4) as the required equation, which shows that (a, hy, %,) @— (a y' + yo’) is a line perpendicular to the plane, and therefore to the line of intersection of the two planes (1). 82. To find the line from a given point perpendicular to a given plane. Let appa —=C 4 be the equation of the plane, 8 the line to the given point from the origin; then let the perpendicular line be w«, and we have p——B+a4; then pe == Pai +aaa, and ao aR + 7% 2%’; ix pa tap aba’ +of' + 2ya2 ¢yo=C, i, which gives Qe%2,i—=C.i— (Pa +28") or Py arate ae Are CE © ce GY Qe U %, TG Or putting Ci=ay'+ya’' (Art. 81 e), wag ee a _% (y’—B) + (xy — 8) & a 2a which is the perpendicular line required. Its length therefore is Attemd Es cet 2d 2 & oo =i 82a. To find the equation of a plane which passes through three given points. Let «, 8, and y be the lines connecting the origin to the three given points. Ife be the line from the origin to any point in the plane, then p—«, «— §, and B—y lie in it, and therefore (Art. 76) e—a IT +I] (e+ «') =0 («—§),(2-—y) © (a8) (By) LIL all Pal: 1+ lll el pat Il +I] a’) = BLY By or is the ates equation. The perpendicular line from the origin to the plane containing the extremities of «, 8, and y, is «If wll 3 BY Pr a ~ a), NTT =I quiet tH Jan % BLY 19% Meat MM +I +1 NUL +I +I Sy ute and the length of it is NGI +H a!) BY By 2N° Lil +1] elf ] ye Eee ners Thus remembering the result of Art. 74, we see that NI ol + ] By y es is twice the area of the base of the pyramid, which has «, 8, and y for edges. This may \ OP=2+AP—=8+2a(y—y}. \ Ley) But « is perpendicular to AP, Ne, therefore P 0 = 5m = 4B +B a’ +o [(a y' + y x") — (xf +8 @’)]; el oe) (ea), ee en hehe EE a), and Cae et aenyesaeceny 6 __ [ley tye’) —2aa']p —[(a 8’ +B a!) —2aa'] y ar (a y+ y a’) — (% B+ Ba’) or [(w y+ ya’) —(«8'+Ba’)], OP=[(a 7 + ya’) —2a0']B— [(xB' +8 a')—Qaa']y, Similarly [(# B+ B a!) — (By +y78)]. OQ—=[(%P’ + Ba’) —268] y—[(B y+ 78) — 286] a; [Py +yB)— (ev +7e)). OR=[By' +7?) —2yy]e—[@r +12’) — 277] 8. Hence, remembering that ««’=6$f'’=yy’', we have [(% v'-+ y «’) —(« P+ Ba’)], OP +[(« B+ B a’) —(By'+78')].0Q + [(P v’+ 78')— («7+ 1%')]. OR); while obviously [(ey' +a!) — (@B +8 a')] + [(% B+ Ba’) — (By +78) + [By +78) —CGy +74) =0. Consequently (Art. 22) P, Q, and R are in the same straight line. The Equation of a Conic Section. 85. We will define a Conic Section as the locus of a point which moves so that ils * distance from a fixed point bears a constant ratio to its distance from a fixed slraight line. (Kelland and Tait, Art. 43), oer: } ee Let F be the given point; DQ the given straight line; FP=ePQ the given relation; F, P, Q. D being all in one ane. | Let DF=«, FP=p, DQ=y 8, 8 being the unit-line along the direction DQ, and QP= ae; then Eger et and p—ve—FD+DQ=——2+y8. (a) We have 0% — 6% &'—=— aa + YyBa’s &—'— GH — Ha Ya B's thence ap pe —Qex%e—=—Qaa’, for, «%'+86'=0 (Art. 65); and Qana—=Qaa’+(ap'+pa’), or fia? (20) [Deal + (xp pe!) ]®. () But from the given relation or, as p24==pp’ and «2% —=aa’, (eee 2 2 ? Pit ae ae Ie S c and wr aa’ Ek e then from (d) aa! = [2aa'+ (ap -+pe’)]?, f or pe. wa’ =e? [aa + 3 (xe tea’) ]?; (1) which is the required equation. | 85a. If we had assumed that the point P is above the direction DF, as in the pre- ceding case, and at the left of the direction DQ; or below the direction DF and at the right of DQ; or also below the direction DF and at the left of DQ: we should have written p+oa——a+tyB; (a,) p—e% a——a—y 8; (dg) e—e «== a—yB, (a3) The case (a) will comprise these three cases if we assume that the values of # and y are posilive or negative according to the position of P referred to the directions DF s : + 7 > 7 ere . AF ioe ee ger ns » ee.) and DQ. Besides, these three cases treated as case (a) has been, will also give the equa- - tions (b) and (1). Thus it is seen that equation (1) will apply to any one of the cases (a), (41), (42), (as). The equation (1) shows that the curve which it represents is symmetrical with refe- rence to the direction FD. * 855. If the angle DFP (Fig. Art. 85) be designated by 9, then 4 (a o’-+oa)—=—«p7i cos 9; (Art. 65), P ae ep. au’ ==" [a a’—% 07 Cos 6]%, . or | pp —e* [x —e cos 6]%%, or * =e [x —e cos 6]?. Therefore } Srey (c) and ee, (c) It is obvious that if p can coincide with the direction FD, produced indefinitely; that is, if the curve represented by (1) pass through a point of this direction, it must be with the supposition that 9=0°, 9=180°, or 6= 360°. ip d= 03,40) ives Slee This supposition requires that the point ofthe direction FD, through which the curve (1) passes, must be at the left of F; then for this point, a Ser gL mar e——heA A being the point. (c,) gives, when 60°, p==——"_2; en eT ae then for such a point, p= ex— Uerm— %xX¥—Usem ga HEA 1—e | A, being the point. But i than 1: if e is less than 1, (c¢) will indicate that there isa point at the right of F through which the curye passes; while when ot, this point being at the right of F, is at an ‘infinite distance from it. — « shows that such a point cannot be at the left of F unless e is greater When e is less than 1, both : % and « are real and the curve (1) will meet, 1+e 1 in two points, the direction FD produced indefinitely. The first point (Fig, Art, 85) is ren é e 1+e- 1—e are also real and the curve (1) will meet the direction FD in two points, both of them e I—e therefore the curve (1) will meet the direction FD only in one point which is (Fig. Art. 85) at the left of F. Y at the left of F and the second at the right. When e is greater than 4, being at the left of F. And, finally, if e=1, the value of « will be infinite and p | eee : 85c. Retaining the notation of Art. 85 we have, ! e—FM+MP. (d) i y Let FM=a« and MP=yf8; M then p=ava+yB, (e) 2 and y being two variable numbers. When the relation existing between « and y is known, the equation (ce) may designate a special curve. The values of # and y will be positive or negative according to the position of the point P with reference to the di- rections DX and DY produced indefinitely. (See Art. 85a). 85d. To find the relation between x and y when the curve represented by the ge- neral equation (e) is a Conic Section. Substitute in (1) p= aa + y B; or ep a ataa’+y2BR’ and ap’ +o0' —=2Qeaa’, Thus we see that paar yf represents a Conic Section when m2 02 +9262 — e2 at (1+ a)?. (f) If we put in (f) FM=e+1=e, andMP=y8=y, (Fig. Art. 85c), that is w= and @,°+- be =e? (4 +-2,)%, or, omitting the accents, w? + y* 0? (x -+a0)?; which is the corresponding equation in terms of the Cartesian rectangular co-ordinates of a Conic Section, the origin being at F. — 87 — The Ellipse. 86. When e is less than 1, the curve which equation (1) represents is called the ellipse. A, We have seen that in this case FA=7— DF and FA, =~—— FD (Art. 853). — +e— — + =1—e— Consequently ee fe. e e meee ; AR= (Tg tis) ED= aa EDs AA, the major axis of the ellipse shall be, as usual, designated by 2a. If C be the centre of the ellipse, that is a C, then FO=CR.=FA—CA,=(7—— jp) FD: aT F 4{— a ee ===(2 60) 86a. Changing the origin to the centre of the curve, let CF, be designated by «,, CP by p, (Fig., Art. 86); we have e? . fem peer and p,==—«,+ , eels or ' — a, and p—=p,+«,; then 4 : — 2 ; CSE GS tere oly ae epee Ueal = [(%,p/-+ ep %,) +20, a], ag peal 1—e*)? ) ee ad, and | pe’. aa’ —=(p,-ba) (p+ dé (12S ee rater ep = e4 -[2,¢,. P, p,+ (x, «,) + (p, a + &%, p,) %, é,. We have by substituting in (1), and remembering that «,¢—=e*%a*7, the equation aS (2, 0,-+e, 4)? —4.a%, p—=—Aat (1—e*) 7; which we may now write, «, reprsenting CF, and pe, CP, («p' + ¢a')?—4.a%p p'=— Aas (1+-¢?) 4, (2) which is the equation of the ellipse in terms of the major axis, the origin being at the centre. If e is perpendicular to the direction of «, then «p'+e«'=0, and p =a2(I—e%); that is the curve meets the perpendicular from the centre to the major axis AA, in two points of which the distances from the centre are aV 1—e?. Let b* =a? (1—e?), 2b being the minor axis of the ellipse. ¥ prety s A Xbe the equation of the ellipse, the centre being at the origin; ® perpendicular to « (Art. 86a). We have pp atac’+y2BR and ap +pa' = 2rac’, which substituted in (2) gives a? (a «')2?— 2 q%e «'— y2q2B B’ —=— a2, where b2*—=a* (1—e?), But as «« ==¢%a% (Art. 86a), then ote? a2 — aes 48 ay: therefore w%e2b%7 + YB B' == 5%, giving the relation between @ and y. If we put o==e2%=CM and y,=ys=MP, ©, or Cras and yes we will have 9 67044 BB v,~ at -- os meas or, putting « for x, and y for y,, oo y? which is the equation of the ellipse in terms of the Cartesian rectangular co-ordinates, the centre being the origin. 86c. The equation of the ellipse (a 9’ +p a’)? —Aa%o p’ = —A at (I—e?)¢ (Art. 86a) may be put under a different form. From this equation we have pe PR a creel Hel dial al ee ae 2a* (I—e*) ‘ __ 20%) P'— h(a p+ pa’) (ap + pa’) Say 2a* (l— e?) ; __07pp +a*pp— § (xp +p a’) a p'— § (ap +p a’) pat an, 2a4 (1—e?) ss Ae I on, Th toed ACT cl A = UO Sora ee ae ea a*p— § (xp +p’) a 2a* (1—e?) e only in those cases where « is coincident with ep or when «p’+p2’=0; the equation of the ellipse reduces to If we now write ®o for where ®p igs a line which coincides with ep. DP o+o. =i, (3) From a simple inspection of the value of ®e we see the following properties of ®¢. E (9 -+s) = 0p + Hc, 1a. (ep) =a. Pp, Il. «,@p+p.c=p,s+9c, >, 86d. Let CM=a2.U« and MP=y. U8, (Fig., Art. 860), then ee. Ua+y. U8; ao’-+ea—a(Ua.e+tae. Ue) 2ee2Ue, Var2Qwe, 4; therefore cpt eed AEB 2a* (1—e*) a%(e.,Uet+y.U$)—sx2aae, — 2a4 (1—e*) { __wa*Ua—wvet*Ua+ya2Us. _ eatti—e*).Uc+ya?.U8. om 2a* (I—e?) ‘ _«.Ue UIE. ene 2 2 ee oa 62): <2 9G = Oa UD ripe Ue aa) Gian DESI where b*=a? (1—e?). Therefore, by assuming *p=za. Ua+ry. UB, a oa | (9) we find : uh 4a F0 eee eye ae Pee pte ey ae vee toa ti) a2 and y being positive or negative according to the position of P with reference to the axes CX and CY. 86e. The equation (g) gives Ua.ote., Vere, and /JUB. p +e. U B= 2y29, peg and yore eee Therefore i i Ee alisrea Rami eo gg re tat ard Wo j Sti) 2 2 and Ueto tp. Ue US.e+e.U8 eT PPS ay ee i Sa pa ry aan (7) If we assume by definition b45— (Pp), { we shall have ne Ute 2 pin Dep. s Us At PoE Be I Di? ae haa.) G3 ae but from (J) ab | Shik ey : Hie We) Bap ripe) ie Ua. 2 Ue Se ear an and UB. Po+0, fg eee ie Ear mk ie aes 2 b* | gi wa __ Uo. pp. Ue UB.’ +e. U'B Ie cae ETE TY. OS rere eee UB. (A) By observing the law by which the relation (h) may be derived from (9), (j) from (7), (k) from (7), and denoting by ®7? a-function such that ©! (®p) shall give e, we can write Le @1o—=a%(Ua.p' +p. U'a)Uae+b? (UB. o' +p. U'B)UB, (2) @%) = 1 (1) = Qat (Ua. p'+e. Ua) Ua+2b4 (UB. p’ +p. U'B)UB. (m) These two equalities are easily verified.(*) We can also prove that TOE he— te) eee * og 4 (")We find OD a and doa. \ ‘ r = ae ON ce: . 867. If, further, we write Ve for Ue.p +e. Ura UB. pe. US 5 da iy Peau hh ae ar UB, (n) we shall have V wip Ww (up) — os Tete Oe ya + Piet salen, But from (n) f Ua’ + Vo ; Ue= ofipteUe oes .U'8 and UB . Vo + Vo 4 oe ; Ua.p+toe.U'« UB.’ +e. USB 2,n— eR eh a a 2 ee ot game) a a eect eed UY mghies } (0) By observing the law by which the relation (n) may be derived from (¢), (0) from (n), and by the definition Y-!(¥p)—=p, we can write b(UB. p+. U'8) Ay eS AU a : UB, (p) : y” Yy : ' f y"! W ' "8 pol ME AE Fy ot a It is evident that the properties of ®p (Art. 86c) are possessed by all these functions. 869. Now oD'p-+Do, p71 gives | eo. (W2)' + (W2p) , p’ = 27, or Pee | Pio) eae (pC orate But since | oe. o+o, pmo, W+Vo.c, this becomes Vo. Wop+Wp., Wp 21, or Yo. wp, or . Naw pas which shows that Ve is a unit line; also that the equation of the ellipse may be ex- pressed in the form of the equation of a circle, the line representing the radius being itself of variable length, deformed by the function Y. (Kelland and Tait, Art. 47). 86h. Let «, and 8 be two lines such that 2.08 +08 .¢ 0, This gives iis #,. (02) B+ 028 .¢=—=0 or «,. WV (VB)+U (YR). 4 =0, or Wf Baer as EW ono: WY Be (), Therefore V2, and Vf, are lines at right angles to one another.(*) (*)We find W aa = and We=—+ SIS) ad 86%. p is a line along the perpendicular to the tangent, that is, ®p is a normal, or parallel to a normal, at a point of the ellipse, p being the line from the centre of the ellipse to this point. Let CP=p and CQ=,p,;_ then r PQ=PC+CQ=CQ—CP=p—p=y, ¢ is there- 1 fore a line along the secant PT, and p,=ep+y. Now P, Pe A Pe, . p= (p+ y) 0 (p+ 7) +4 (p +y) (e+), = (e+ 1) (Op + 'y) + (bp + Hy) (o'+7), (Art. 86c, I), =p. P othe. pte bytdy. ty Oot Op. yy +y. Py+oy.y, But P,P p+ Op, p’=im=p. O' p40. 7, pT yt Oy. pty. Op+ Oo yy, Wea by 7 —0, or (Art. 86c, III) 2(y. Pp + Op .y)+y. Py+oy, 7=0, or 27 (Pp +4 Oy) +2 (be+ 467) y=. Thus we see that, however small the length of y may be, Pe+ePy or P(o+hy) (Art. 86c, I) is always perpendicular to the direction of y, (Art. 65 ). But as the length of y gets smaller and smaller, its direction comes nearer and nearer to that of the tangent, and obviously the limit of ®(p+4y) is ®p, therefore #p is a line along the perpendicular to the taegent at the point p. 86j. To find the equation of the tangent to the ellipse. Let T, be any point in the tangent, and CT,=z, then (Fig., Art. 86%) mem p-+ PT, ee lx), and PT,.®p+p.(PT)’=0, (Art. 862), or (= —p) ®'p +p (x'—¢') =0, = mT, Do+dp. nop, p+ .p, then m,.@D'p+dp, n=7% (7) is the equation of the tangent at the point ¢. The equation of the tangent may be written in the form ep. On+br, o'=4, (Art. 86, Il). (s) on ls Let # and y be the rectangular co-ordinates of T,, a point in the tangent; «, and y, co-ordinates of the point p. We will have tae. Ue+y. UB, Gaia ye UB 2a? Flsees By substituting in equation (r) we get and Do ee eds at BF omy which is the Cartesian interpretation of (r). 86k: To find the equation of the normal to the ellipse. Let = be the line from the centre to a point in the normal which passes through the point e in the ellipse. Obviously the equation of that normal is nm==p-+z.p, (Art. 863), (‘) z being a variable number. The equation (¢) gives (= —p) ®'p— Pp (x’—') = 0, or mz. @p—o, 7 =p, DB—Op, (’, (vw) which is another form of the equation of the normal. Let # and y be the rectangular co-ordinates of a point in the normal; 2, and y, the co-ordinates of the point e. We have me \Ue+ty. UB, ; re oe tre and ? p= o be These give Rey LA 1 ; , p. ®p—oe poet (ss—-2)(U=. UB—UP. U's), and FD pa DB, n= (S54) (Ue aU PUB. U'a). Therefore the equation (w) gives x a ae or : cat yb? age wv y, : which is the equation of the normal in the Cartesian form. 861. To find the locus of the middle points of parallel chords. Let all the chords be parallel to y; = the line which is directed from the centre of the ellipse to the middle point of one of them, which may be represented by 2ay; then — 9h — TN+OY, T—LY. are lines from the centre to two points in the ellipse; (x+ay). O(n +a) +0 (x+ay). (7 +07) =i, (x —ay). ® (x—avy)+0(x—wvy). (7®— ey) =. Multiplying out and observing that (Art. 860. I) O(xtay)—Oortye, oy, we get by subtracting mr, Pytoy .n+y. Ur+O7,.y7=)0, or, (Art. 868, IT), | nm. OP y+oOy,.7'=0; t.e., the locus required is a line perpendicular to ®y and passing throngh the centre. ®y is a line perpendicular to the tangent at the extremity of the diameter, parallel to y. (Art. 867). Therefore the locus of the middle points of parallel chords is the diameter parallel to the tangent at the extremity of the diameter to which the chords are parallel. If «, be the diameter which bisects all chords parallel to the diameter $,, and since a OB +08 .¢ =O, we have (Art. 860. ITI) Ba +e .B 0, In the latter equation 8, is perpendicular to the normal ®«, at the extremity of «,, and is therefore parallel to the tangent at that point: hence the diameter 8, bisects all chords parallel to «,. . Therefore the properties of «, with respect to 8 are convertible with those of 8, with respect to «,, and the diameters which satisfy the equation a DB +B. oO are said to be conjugate to one another, and W«,, VB are lines at right angles to each other. (Art. 86h). 86m. Let «=CP and B=CD be a pair of conjugate semi-diameters of the ellipse DQP. » Let MQ be parallel to CD; CM=a2; MQ=y8, and CQ=p; then ' ewe +y8. Equation (3) (Art. 86c) gives (wa +y8) 0 (wx+yf) +0 (vat+yf). (wa! -+yp') =i, j ‘ \ » 7 GQ, = multiplying out and observing that (Art. 867) aD B+8 ,c’—0 and PORa+ha,i' =), we get eS boca Therefore, if we designate y by sin 9, # will be cos 4; then p—=cos9.a+sin 9,8 (v) will be the equation of the ellipse referred to the conjugate diameters 2« and 28. If CM=za, and MQ=y,, then ‘Substituting these values in #*+y*%==1, we get (2) +() = or, omitting the accents, and putting a, for « and b, for &, it becomes o% a3 sa tha which is the Cartesian form for the equation of the ellipse referred to conjugate diameters. 86n. From a given external point two tangents are drawn to the ellipse; to find the equation of the chord of contact. Let # be the line from. the centre to the given point, and oP ar+On, p= the equation of a tangent; then, since it passes through the given point, we have pPR+ OB, pi. As this equation satisfies both points of contact, and since it is the equation of a straight line (Art. 86a) it must be satisfied by every point in the straight line which passes through. these points of contact. Therefore, denoting by « theline from the centre to any point in the secant, the required equation will be oP B+08 , oti, 860. Assume that, e and pe, being two semi-conjugale diameters in an- ellipse, we-haye (Fig. Art. 86m), TU Gea e,==%cos9+8 sin 4, and . p==a cos +6 sin 0, where « and § are a pair of semi-conjugate diameters. “o@'e’ +p’. o' 0 (Art. 86). gives («P'a+ Da , «')cos 9 cos6+ (BO'R+8 . f') sin 4 sin 0,0, or | cos.9 cos #,+-sin4sin®—=0 (Art. 86), 05 COS (0,— 0) = 0, TT or ey ta Therefore TT . cos 9 = cos (5+° )=—sin9, : 5 wT and sin 9, =sin (5+) =cos9; «cos 9+f8sin® and —« sin 9+8 cos 0’ are semi-conjugate diameters. 86p. If 2« and 28 designate any pair of conjugate diameters, we have by retaining the notation of Art. 86m, p—aa+ yf, and P= — ya wB; as the equations of a pair of semi-conjugate diameters (Art. 860), and —y«+a8 or y%—af will be a line parallel to the tangent through the extremity of ». Hence the equation of the tangent will be naatyB+X (ya—ah), where X is a variable number. But y=o and mS oid as 2g eee Hee ° gee aes ae: 6 (y,U« a8); Therefore the equation of the tangent may be written moa het BK (y, U«—mza, w8), ; sd ity : 7 where X is a variable number and m= (=) =a Making —-=Uz, | Us, and ro ne) omitting the accents, the equation of the ellipse may be written Pi aU wi) eu Be and that of the tangent m=a.Ue+y.U8+X(y.U«—mze. U8), where P Ie oragh Bas see AP RAE and (rare or 2h 272 2 pb ip y? = — a (a,2— «?) 9 (a,2—?). 86q. Example — If tangents be drawn at three points P, Q, and R of an ellipse, intersecting in R,, Q,, and P,, prove that PR, .QP,.RQ=PQ.RP,. QR, it a Te ae | > ee v ee? as If Gy Yi %,, Ys Ves Yq belong respec- & tively to the points P, Q, R; we shall have CR,=#2+y8+X (y«—~w§); Q R P =2,%-+y,8+X,(y,« —2,8); ink i Gta Ne eee and GaN aN ee | DAG Gr teas (pr) Namen ap ae pe mee on Tica Rey ey) Sy but Ge yee la ys (ATi SO m); then X (ye%,—@y,) =X, (vy,—y,), X—=—X,. Again CQ =ae+yB+Y (y«— x8); = ae Yoh —\, (Yae —VaP); and CP tie yb Liye a,b); == Hig% + Yah Z, (Yot — Hoh); giving A Ve Or C) > and Z—=—Z, for P,. Therefore PR, =X (y«—a$), RQ—=X(y,«—~«,8); Q, R=Y (yg% — #98), PQ=Y (y «— a8); QP,—Z(y,«—2,P), PR 2 Zit, —= 058). From these we see at once the proof of the proposition. The Hyperbola. 87. When e is greater than 1 (Art. 850) the curve which equation (1) (Art. 85) re- presents is called the Hyperbola. ; 1 malls Fi In this case we have e cA FD — — ee +1 — and : FA =—;FD (Art. 85d); consequently i Nadi == s—7 FD. If C be the centre, that is —~—=AC, we have e2 FC=CF=FA—CA,=- Deo Sas ene PRE 1s ee e2— | —— 87a. If the centre be the origin and CF=«, we shall find, by the same process given in Art. 86a, (a 6’ +p «')2—Aa% p' Aas (e2—1)2, (1) which has the same form as (2) (Art. 86a), e being greater than 1. If, with the origin at the centre, e can have a position perpendicular to AA,, then we must have «p’+oe«’=0; therefore e%—=— a* (e*?— 1) and p= =hay — (e?—4). As e is greater than 1, aY —(@?—4) is not an absolute number, which ¢ was assumed to be. This indicates that the assumption is incorrect; that is, the curve has no points for which e will be perpendicular to A A,. | 87b. The equation of the ellipse, p.P p+p .p —2, is, of course, applicable to the hyperbola, e being greater than 1; and most of the pro- perties of the hyperbola are the same as the corresponding properties of the ellipse. Let CM—2Ue and MPa=y UB (Fig: SArt.eo7): then p—a2Ua+y UB, ele sy UR where 6*=a?(e?— 1). 87c. The equation of the tangent at the point ¢ is x, Do+p, x =17, or ep. PO n+On, p=? (Art. 861); and ®po is a line along the normal at the point e. 87d. Retaining the notation of Art. 861, we have r—=a2Ua+y UB, 6 Sa ee ae . Sts an > ay UL, YY, ‘ therefore Mie ar I, : Oswa ys and Fa Pa ei b* sa2—a? ; . Hence y= — and y,?=— («#,?—a?), a y, or o Bs b xx—a* gute v, Y —— = i « a [x,2*—a?]3 a #43 oe a? cT— b x rao U et — a ary a ar foe oot then which is the equation of the asymptotes. The asymptotes are the lines towards which the tangents continually approach as the tangential points recede from A and A, to infinity. The plus sign belongs to the asymptote above the axis AA, and the minus sign to that below. The asymptotes are equally inclined to the axis AA,. (Fig. Art. 87¢). Obviously the Cartesian interpretation of the above equation of the asymptotes is b ymat—xe. y a 87e. To find the equation of the hyperbola refer Y eee § aS P, F, ed era, +yP. By substituting in p. ®'e+p. =i (Art. 875), red to its asymptotes. If «, and 8 are unit-lines parallel to the asymptotes C X and CY res- pectively, and CM=a, and MP=y; then the equation of the hyperbola may be written we .have — 100 — or (a Patho. é)+y? (6. OB +08 B/))+2ry (e,. 08 +08 4) 7 But a .Pa+he.¢—=0 and 6.08 +08 .8—=0 (Art. 87c); therefore 2ey («PB +68 .¢)—=7, But op SCR a—a% gives 2a*b* My yen yp as eek a, a’ + a &,) —a? (a, +B, 4.) Also, we have 2070 a % +ad% —=2% cos 02; f af’ +8 a’ —= 2% cos 62; a,8'+8 ¢ —=2cos 20¢—=2(2 cos? 6—1) 4; 6 designating the angles FC Y and FCX. Consequently a ; ree v Pad ead Aye owe 79 ¢ Therefore Nay a4; b* h 3 aoany | or LY — which is the relation between # and y, and the Cartesian equation of the hyperbola re- ferred to its asymptotes. It appears that the rectangle under the co-ordinates w and y is constant. Let the area of this rectangle be represented by C*; then we have i pel Ca Therefore the equation of the hyperbola may be written C2 e— My , AR oC B, or, omitting the accents, C2 « and $ being unit-lines along the asymptotes. If « and f are not both unit-lines, we may write the equation under the simple form e164 £ 3 (2) 87f. Let CP=p, CP,=,, (Fig. Art. 87e), then the line along the secant PP, a B p—patatt tat; =(—-1) (@— 4); Phe \4 X — 101— ie Ft’ p ee i (igor ge): 4 Therefore the line jet is along the secant, and obviously when #, becomes equal to ¢, the direction of the secant coincides with that of the tangent, and therefore jo is a line parallel to the tangent. Hence the equation of the tangent is rtatt palia— +), (1) 87g. Let Q and R be intersections of the tangent through P by the asymptotes CX and CY (Fig. Art. 87¢); CP—=p; MP parallel to CY and NP to CX. By putting e=1 and «=—1 in the equation of the tangent we see that CR=2te== 2 C Me and co=t=20n, Hence, the triangle which any tangent forms with the asymptotes has a constant area and is equal to double the area of the parallelogram formed by the Cartesian co-ordi- nates of the point of tangency referred to the asymptotes. Again PQ=cQ—cP==— (t+) —— (te—4), and PR=OR—CP=2ia— (ta++)=ta—+ =—PQ; that is, the part of the tangent limited by the asymptotes is bisected at the point of contact. 87h. Any diameter CP (Fig. Art. 87c) bisects all chords which are parallel to the tangent at P. Let CP be jane, then the tangent at P is parallel to tat (Art. 87/); therefore be 0 5 420 Pg CIP ye, PQ=X(ta+5)—¥(ta—*). But as P, is a point in the hyperbola, this equation must have the form 6 CP,=te4+—. ee 8 8 Therefore COE rei he oe) ok Loe an), or t,%a—=Xi«—YVia, Betta Br NS 8s and er: +Y ai i= (X— Y) #; ih X+Y t, ii! and xX?— Y2=1; an equation which gives two equal values of Y with opposite signs, for every value of X. Hence all chords are bisected. — 102 — 877i. If in X*=1+ Y? (Art. 87h) we put X=sec 6, Y will be tan, then p——«sec8+B6 tan 6 (1) will be the equation of the hyperbola referred to the axes along CP and CQ; CQ being parallel to the tangent PQ; CP=« and PQ=6. 877. It is obvious that, if we take CQ=—=PQ and CR,=PR, the locus of Q, R, will be a hyperbola of which the equation is ao ae P= t% + t ’ —« and § being along the asymptotes CX, and CY. This hyperbola is represented by the dotted curve in the figure of Art. 87e, and is called the conjugate hyperbola. CP and CQ, are a pair of conjugate semi-diameters. The Parabola. 88. When e=1, that is, when FP=PQ, the curve which equa- tion (1) (Art. 85) represents is called a parabola. We have pp. aa'——[aa'+$ (ap +pa)|?, (1) Hence pe. aa (xa')2+aa, (xp -+oa) +1 (xp +pa')?; and, since ««’—«%; (Art. 64), -__ Ape, aa —Aaa' (ap'-+pa2') —(ap+pa')? | t Tea : Qe’. a'a—(ae’+pa')a'—haaw', a po, a a’ —(ap’-+oa')a—h Pe ee ner PSR E Ruane Tt If now we write , : ope teest op teede, a to which the properties of ®p in Art. 86 evidently apply, the equation becomes , fem a a , ° p[Pe— Fal + [Pe— ale =, (3) . pp’ Q’ 5 | teas i\ eee } or oP p+ pp — ae (ap tee) = 0. (A) 88a. If we make e=1 in the last equation of Art. 85d, we will have y*= 2+ Qan, ibs( 4) 5 — 103 — or, if we put —s+e instead of w, that is, change the origin to the vertex A (Fig. Art. 88), we will have y?=2«a, (2) which is the equation of the parabola in terms of Cartesian rectangular co-ordinates, the origin being at the vertex. 88b. If we put FM=a2.Ue, MP=y.Uf$ and FP=p (Fig. Art. 88), therefore e=a.Uaty. U6 and «p’'+pe’=2wai. (See Art. 85c). Then (2) of Art, 88 gives* p= 545 UB, which shows that ®p is a line perpendicular to the axis A X. This result may be easily deduced from (2) of Art. 88. Obviously we have 2 (ap -tp a’) a a’—2 (ap +p a') a a’ Aad hence ®p is a line perpendicular to «, which is along the axis 2. We also have a, P'o-+ Pp ,¢— —0: o@'a+Pa,o'—=0 (Art. 86c. II). 88 c. des is a line along the perpendicular to the tangent at a point of the pa- rabola, e being the line from the focus F to this point. Let FP =e and) Qe, then ER 2 hee eR ERE Bo prs y is therefore a line along the secant PT, and ‘ox=e+y. Then equation (3) of Art. 88 gives thus we see that however small the length of y may be, a Y a Poot 8 tomee ga or 1 oy ener is always perpendicular to the direction of y. But the limit to which the direction of *da=0 and 66 =5% 6, BT) oe approaches is the direction of the tangent, and, in the same time, the limit of ® (p+ a) sa is bo; therefore i @p ck er is perpendicular to the tangent at the point p, or, in other words, is parallel to the normal line at the point p. If » be parallel to the tangent at the point ep, we must have ‘ a a ‘ 88d. To find the equation of the tangent to the parabola. Let T, be any point of the tangent (Fig., Art. 88c), and FT,=7; then PT =2z—p; / ox % ‘ ‘ si Grr et irre (a Gp pe ; eet a or (0 Pears span Vigws ya)" Daz (%p'-+ pa’) = 4% (1) is the equation of the tangent at the point e. Let # and y be the rectangular co-ordinates of T,, a point in the tangent; #, and y the co-ordinates of the point e. We will have To—fUS+y. UP, P=-2e Uat yaa p, and Demy U8. Therefore equation (1) gives yy, ae (ata) 2 Q) If we put 2, —F and z— for «, and #; that is, change the origin to the vertex A, we will have yy, =(e+a,) (3) If ¢ be a line joining the origin F with any point in the normal-line, the equation of the normal-line will be o==p-+a(%e— 55), & iO, Be is a line in the direction of the normal. (Art. 88c). a ar 88e. The equation (2) of Art. 88 may be written under the form : 1 2 Oe ot eae ea a a We have (Art. 880) 2a ,Pp-—-MP, and FP=p, (Fig. Art. 88); sek MP=p— sry (of tox), =O — But MP=—.—FM, EM aa (tp pe) a, 88/. The subtangent MT, is bisected at the vertex A (Fig., Art. 88c). If we put w« for = in the equation of the tangent (Art. 88d), the resulting equation will be Ul ox a 4 y ‘ / ° wa (d Agree AEST Spee Aer Gl, +pa')=1 or een | ; ; ES er atl: +p«')==—7 (Art, 880). Multiplied into «— 7 ; Pea Ch Srey ACG a Reeinte j Ge 7 y : F or OR yee TAC: +p a’) a, But obviously Ce —— KT. a a AF; 1 ' and aa (#P + o')ao—=FM (Art. 88e). Therefore FT,+AF=—(AF+FM)=—AM, or AF—T,F=—T,A=—AM; DP AM} 889. The distance from the focus to the point of contact and the intersection of the tangent with the axis are equal. We have seen that w« substituted for « in the equation of the tangent gives F I F ‘ : aoa Tact +p a')—4, or ©, wa —=— [aa + 3 (xp' +pa')]; ae > waa’, ou’ lau’ +d (ao'+pa')|%, or, by observing that wee’, eae’ =eama’, aa’, 1 ee lao i hae eon 12 wa, ga —~ [aa + § (xp +p a')|%, But Goa hohe. and [xo’ +2 (a p'+pa')]2—ppo', aa’ (Art. 88); Therefore FT,. (FT) =e¢, which gives FT,=e=FP. 14, y (aD eee) — YD gate Ree aia) eee or —yi=2zua' (Art. 885), or Sane) i aye cuit therefore y Pp — aa’, Dp, Yt — 20.0’; and 2% OE QRZae, from which gant fa therefore a= Wi ey or MG=DF, a constant. r : 7 “4 ke oF oe ye ART : v : Ore rapa — 106 — Whence the tangent bisects ae angle FPQ, and FQ is perrentivalar to. ‘and biscote at N by the tangent. 88h. The subnormal MG is constant (Fig. Art. 88c). We have (Art. 88) —=PM+MG; ——2a0', Po+ze (Art. 88 b), where s«—MG; therefore | . 2 a a 88%, Obviously (Fig., Art. 88¢) | FQ=>FD+DQ—>FD+MP =—a+2a0',@— (Art, 88d) = 2a! (de — 57) —=—PG=GP, {(Art..88h) and FQGP is a rhombus. 88). The perpendicular from the focus to the tangent at aay point, intersects it, where 3 it is intersected by the tangent at the vertex, and AM=4MP (Fig., Art. 880). , We have rt 1FQ=4(FD+DQ); ors hae tp (Art. 882); =FA+AN; 4 A — 107 — Therefore AN=aa’', Oo; =4MP (Art. 886); or AN is parallel to, and equal to one half of MP. 88k. To find the locus of the middle points of parallel chords. Let 8, be parallel to the chords; = the line joining the origin to the middle point of one of the chords; then T+20B =p, and — x—aB =o, The equation (4) of Art. 88 gives (n+2B).0 (n-+08)+0(x+06). (w-+09/) ——5[(n-+08) a! +2 (x +06)] =, and (x— a8). 0 (x +08) +0(x—o@B). (x’— 28) —-5 [[» 28) wa (x —@)] =; therefore by multiplying out and subtracting, we have 1 20? a straight line perpendicular to ©§8,, that is, parallel to the axis (Art. 88). This equation may be written (x. OB +08 , x’) — (7B’+8 «')—=0; I ‘ & a i p [e's — 53] + [oe — 5] 8/0; which shows that the chords are perpendicular to the normal (Art. 88c) at the point where p=, or at the point where the locus of the middle points of the chords meets the curve: in other words, the chords are parallel to the tangent at the extremity of the diameter, which bisects them. 881. To change the origin to a point in the parabola. Let » be the line joining the focus F to the new origin O, and let, Oo We have p==wo+ Pp, Substituting in (4) of Art. 88 and omitting the accents, there results : f y ’ ‘ ¢ fi , p. Pp + hp. (ap tea) +2(p. Potroo, p)=0 (1) which is the required equation. If the origin is at the vertex, then o=a« and p.®%o+%o ,°=0 (Art. 880), and U ; y , / ph p+ Po, p—— (ap +pa)—=0. (2) — 108 — 88m. If the unit-line along any diameter of the parabola be Us, and the unit-line parallel to the tangent at its extremity be U8; we may write the equation p—=a2Ua+y UF, (1) where *=OM and y=MP. Let » be the line joining the focus F and the origin O; — then do—>5 is perpendicular to (Arts. 88c¢, 88k), and Uy p (0 o — 5-5) + (bo — 5) 8 —0, (2) or < ia (B. Pot oe. B)—5 (ah +e) =0, (3) Now to find the relation existing between «# and y, substitute cUe+yU6 for e in equation (1) of Art. 881, giving y? (UB. ® UB+0UB. U's) —*7i=0, or 7 a CUE AE EA ea Y =EN(US. PUR+OUB. US” = 200, (A) y where O=—IN(US. © UREOUP.UE (UE PUSS PURE (5) Therefore 2 p= 2 Ua+y UB. (6) 2a The equation (4) is in the Cartesian form, the parabola being referred to a diameter and a tangent at its vertex (See Art. 88a). When the origin O is at the vertex of the axis, equation (4) will become identical with (2) of Art. 88a. Therefore a must be equal to e=FD; that is, when U® is per- pendicular to the axis containing «, we have oe Us. ® US+0US. UB=— a2 * —_ This may easily be verified by (2) of Art. 88. The equation (6) may be put in a somewhat simpler form. We have qy% p= 52 204, a+ 7b N Te. > — 109 — where § UB==8 and a,U«=z«. \ Let t= 7, then ; z {2 62 e aes ie Now by making p Q we get t2 | where « is a line parallel to the axis and # is along the tangent passing through the origin. In the figure 2 OM=—2 EG ENG Mice 88n. To find the equation of the tangent from (8) of Art. 88m. the secant joining the points where ¢ has the values ¢ and %, is, obviously, represented by the equation t* i e—e— ae eerie rs = (tt) a + 8 Then tat is a line along the secant, and its limit is a line along the tangent: hence any line along the tangent isa multiple of ¢#+8, which is the limit of Hae, Therefore the equation of the tangent may be written 2 nay attb+a(la+6), where x is a variable number. If we put s—=—?t t iS SS 2 which is the intercept of the tangent and the diameter passing through the origin, and is evidently the — abscissa of the point of contact, that is —NM. (See Art. 88/). PPP PBF 566 ¢ ¢ CHAPTER V. THE SPHERE. 89. It is evident that there is nothing in the demonstration of Art. 83 which obliges us to assume that the circle APD (Fig., Art. 83) is necessarily in the plane of the figure Bae oor in any particular plane; from which it follows. that the equations there obtained are — 1410 — also applicable to the sphere of which C is the centre and » the radius. Therefore, in general, the equation pp'—(3p'-+ po) = Cz (a) represents a sphere of which the centre is at the extremity of y and VC +8? will give its radius; that is the above equation will be satisfied by putting for p any one of the lines connecting the origin O to the points on the sphere of which the centre is at C and the radius is equal to VC +82.* But if the extremity of p is necessarily in one plane, then the locus represented by this equation will necessarily be a circle. 90. Further if we assume that the tangent plane to a sphere is perpendicular to the radius at the point of contact, the conclusion in Art. 8306 is applicable also. — The equation of the tangent plane to the sphere is therefore Te +on 2724, the centre being the origin. 91. If the tangent plane pass through a given point B, at which the line from the origin (that is, the centre) is 8, we have Bo’ + pB' = 2727, This is the equation of a plane perpendicular to 6 (Art. 81) and cutting from f a por- tion whose length is It is obvious that this plane passes through all points of contact whose tangent planes contain the given point B, therefore, the equation of the plane which passes through the points of contact is (Art. 83) Bo’ + of’ = Qr%4 the centre being the origin. This plane is called the polar plane tothe point through which the tangent planes pass. 92, Let P—=—TOrey et 2s, and Y=ao,+ bo ,+Co3; ©,, g, and », being three rectangular unit-lines, therefore w, y, and z are rectangular co-ordinates of a point in the surface ofthe sphere and a, b, c that of the centre, hence we have pp — (a? + y2+ 27) 4, ( 60’ == (a?-+ b?-+ 62) 4, p'+Se—=2(aa+ by +cz)t; and ” being the length of the radius of the sphere, C= r— 09= r2— (2+ 62+ c?2), “If C is negative and its absolute value >6* the circle will be imaginary. . — 111 — Therefore by substitution in (a) Art. 89a, we have e+ y%+ e2— 2 (aut by +c2)=1r?— (a?+ 52+ ¢?) or (a—a)?+ (b—y)*+ (¢c—z)*=r?, which is the equation of the sphere in Cartesian rectangular co-ordinates. 93. Examples: — Ex. 1. To find the locus of the feet of perpendiculars let fall from a given poiut on planes passing through the origin. Let 8 be the line joining the origin to the given point and let ap -+pa ==() be one of the planes; then (Art. 82) the line-perpendicular is 4(U«.B'+8.U'«) Ua, and for the locus of its foot, e==P—$(Ucx. +8. U2) Us, This equation involves the indeterminate line U«. To eliminate it we have phi +B’ = 288— 3 (Ue. Ph +8. U'a)® and ep PR —4(Ua .f'+8. U'«)?. Therefore ' ; pe'—4 be +p8)=—0 which may be put under the form or Thus it is evident that the locus is the sphere of which 8 is a diameter. Ex. 2. To find the curve of intersection of two spheres. Let the equations of the spheres be Ppa des) 0,0.) a 1, De, gh As the values of p at the points of intersection of the two spheres will satisfy both of these equations, we can write, by substitution, e ee S) + (5 — 3) = 724 — 724 +-8,8'— 88, Any two of these three equations taken simultaneously will represent the curve of intersection. But the third equation represents a plane; therefore, if the given spheres intersect, the curve of intersection must be a real circle. The plane which the third equation represents is perpendicular to 4—4, the line joining the centres of the two spheres. It is always a real plane whether the spheres intersect or not, and it cuts the line joining the centres of the spheres so that the diffe- rence of the squares of the parts is equal to the difference of the radii of the spheres, — 112 — For, let this plane cut CC, through D. We have, C and C, being the centres of the spheres and O the origin — CD=p—é and DC,=—p+4,, where p==OD; therefore INNA CD2i=pp'+00—(Ce' +02), and DC*i=¢ 9’ +8,8/— (3,9'-+ 9 /); or (CD2—D,2) é==88—8,8/+ p (/—2) + (3—8) 9. But p (5/— 8’) + (6 — 8) o’ = (r2— 77,2) 4 4+8,8/—88; 2h CD¢—DC2=r2—r’, This plane is in fact what is called the Radical Plane of the spheres whose centres are in C, C, and whose radii are 7, ”,. Ex. 3. To find the locus of a point the ratio of whose distances from two given points is constant. Let the given points be O and A, the extremities of the line « Also let P be the required point in any one of its positions, and OP=p, Then, if e be the ratio of the lengths’ of the two lines, we can write pr-&==bie This gives (p >e.0:) == e252 oF (ep —«) (c'— a’) =e pp’ (Art. 64), From this pp — (app a’) + aa’ e7 p's pp’ (1—e?) —(%p' +p%) +ae’=0 ; 4 ; ; ax Oy’ or OP 7st KTP EPR) ce eteg. 5 3 5 e% ; Thus the locus is a sphere whose radius is N Toe? and whose centre is at B, where OB=;—,, B being a definite point in the line OA. Ex. 4. If in any line OP, drawn from the origin O to a given plane, OQ be taken such that OQ. OP is constant, to find the locus of Q. Lebigif vintage ap -+pa'—at be the equation of the plane; OP=p, and OQ=7; therefore Up—=Uz. Then, by the conditions, | =. ep==constant = b? (suppose). From this therefore 113 — Substituting in the equation of the plane, we have 2 aq (+40) =at or b2 mw — —-(a n+ ma) == 0; which shows that the locus is a sphere, the origin being situated on it at the point farthest from the given plane. Ex. 5. To find the locus of points, the sum of the squares of whose distances from a series of given points is a constant quantity. Let the lines from the origin to the given points be «,, #),.....%, then eye pee | (aera) (pa. | oa eee + (p — ap) (0 — on)’; ==pp +a,¢,— (#19 +e¢,)+pp +%9¢,—(%90 ted%e)+,... «3 =npp—[p (og. . on) HHH eet 2. op) Payoh gds o 06. andy} =nep—[p. 2a +Ea.¢'] +5 (x2); or , SS aay __ 44 — = (x a’) Ria eee age 9 eee SS . . . . . =_—% which is the equation of a sphere the line of whose centre is Fx and the square of whose radius is Re a a ee n? pie n Therefore the equation denotes a real surface if nor—nN(xo’)+N(Za, Ze’) is positive, that is if ncet+N (la. ra’) >nNZ(a’) or, n being an integer ge nieesee) >NE(«e’), If the origin had been placed at the mean point of the system of the given points, and the lines «,,% ,.....%n were from this point to the given points, then ©«=0 and o> N2(x0') or e8>> Sal, and when these quantities are equal, the locus becomes the mean point of the system. Ex. 6. ‘To describe a sphere, with its centre in a given line, so ‘as to pass through a given point and touch a given plane. Let w«, where x is an indeterminate number and « a unit-line along the given line, be the line joining the origin to the centre; 7 the radius of the sphere; § the line to the given point; and ye +ey at —114— the equation of the given plane. The line-perpendicular from the point a« on the given plane is (Art. 82) ai—wl(ay +ya')y oy ; ) Hence to determine « we have the equation NOE) Sr SN (oa—f) 7 Oral oe pa en | | ? AI— Alay rye Aat—alay tye) , iy alee TE 2 | eee eR a a which gives” hh aks aii) i In fie Srey wel (ay ya’) 2— hy?) — 2a [(e y+ 7a’) a — 2? (af! +8 2) = 46 22— ad; or more simply, by assuming that the origin is taken so that « and 8 are perpendicular to:each: other, . w2[(a y+ y «)2@—Ay?] —2a (ay +y«')a=4b2y2*— a; therefore there are, in general, two solutions. Ex. 7. To describe a sphere whose centre C is in a given line and which passes through two given points P and Q. Let the line of the centre be w, as in the last example, and let rite lines of the given points be ® and y. Then we have CP=(r—24) and CQ=(@—24), ee eee: 2) (y'— a2!) = (6—aa) ('—a2’) or aw [(ay + ya’) — (2 8'+ Bo) = y— Be which gives only one value for «, therefore there is but one sphere, except in the par- ticular case when we have Therefore (v y=6, and «ay +ye’=af'+ fa’ in which case there is an infinite number. The Cone of Revolution. 94. Let the vertex of the cone be the origin. 7 Suppose «, where «1, to be along its axis, and e the cosine of its semi-vertical angle; then, if e be the line from the origin to any point of the surface of the cone, and Ue the unit of e, a, U'ep+Ue.e¢ ter AT Oe | a. p +o, a —+tQePt, therefore (a p' + pa')*—=Ae®, pp’ is the equation of a cone of revolution whose vertex is the origir O, f a 95. In order to change the origin to the point in the axis whose line from the vertex ig a«, we have to substitute «+e for pe; then the equation becomes (a 0’ +p a’)2— 4 e% op’ +A (~2*— e242) i +4 (w—ave?) (xp’ +90’) —=0. Let the radius of the section of the cone made by ao’ +p a —=() be 0b; this necessitates x | V be+ 0% Deo a +4 a which gives Ht—e7h*-+ 99,3 and U—o er==% chad pe as b2+ 2 2 +-9,2° Substituting, we get A b2 b2 — +2 G (% 0’ +p %')®—4 e250 +4 62h%4 + Let the radius remain constant while « changes. If we put woo, then e=1, the locus becomes a circular cylinder of radius b, whose axis is the line «, and whose e- quation is (ao -+00')2—hoo +4b%=0. 96. We can obtain directly this equation thus. Let ¢ be the line joining the origin in the axis « to ecu of the cylinder, and 0 the angle between « and ep. We have NIL =esne and ek aah %, or ae le b. a, 8 From this : NID =, a, B and by formula (3) of Art. 78, N It = 4 (9 ea)? = b? Oras «2=—=], (%p' +p %')®—Ap op’ +4597=0. The Cone ore e a Circular Section. 97. Suppose the vertex to be the origin, and, let the circular section be the inter- section of the plane | — 116 — or ap’ -+pa' <= 24 with the sphere (passing through the origin) pp = 4 (Be +08), where He is the perpendicular let fall from the origin on the intersecting plane. These ‘equations may be written thus e(a. Up+Up. a) 2 et 4 (8. U'p+Up. 8); therefore, eliminating ¢, we find the following equation which Ue must satisfy, £(B. Up+Ue.B)(e. Up+Uup.@)= 22. Hence e2 (8. U'p+Uep. fi) (a. U'p+Ue, «) 40%; he (Be +8) (ap +pa') Ape, or - Ao p'— (Bp + eB’) (a p’ +e a) 0. Now if wp be written in place of , the equation is not changed; it is, therefore, the equation of the Cone. As « and 8 aresimilarly involved, the mere form of this equation proves the existence of the sub-contrary section discovered by Appollonius. (Tait’s Quaternions, Art. 239.) 98. The above equation may be put under the form “aBe2(UB. U'p+Up. US) (Uc. Up+Up. Ua) Apr, es (UB. U'p+Up. U'B) (Ua. Up+Up. Ua) mar. Assuming that «, and 8, respectively correspond to another section parallel to the first, it is easy to see that therefore the expression N[(UB. We+Ue. Uf) (Ue. U'p+Ue. U’2)] ig a constant number. . 99, For a cone which hag a circular section, or, using another term, for a cyclic cone, we have (Art. 97) Ap p'— (xp +p a’) (B p’ + 9B’) ==0 which gives | 2p p'—ap'(Be'-+ eB’) +2 p’—pa’ (Bp +8!) =0 and 2pp'—fo' (a p’ +p a’) + 2p p'— pf’ (a p’ +p a) —0; adding, [4p—«(Bp’-+ 6B’) —B (a p’-+a")] p’-+ 5 [4 p’— a’ (Be’ +98’) —B’ (a o' + a’)| =0, Let Pp—=Ap—a(Bp'+ef')—f(ap+pe’), ae — M7 — which has obviously. the same properties as ®p in Art. 6c. Therefore the equation of the cyclic cone, that is the cone which has a circular sec- tion, may be written in the form 0. Po+4+p , =O. 100. To find the equation of the tangent plane toa cyclic cone at the extremity of. Put p+Ap for po, then (@ + Ap). P (p+ Ap) +@(p+Ap). (p+ Ae) =O or (0. p+ Pp.c)+ (Ap. Pe+ dp. Ap) +(p,WAp+ Ap. o') + (Ap, PAp+PAp. Ap) =O; a 2(Ap. Pp+ho. Ae) + (Ap. DAp+ A, Ap’) =O, From this Ap. (®p+4 Ap) + (Op +i @Ap), Ao 0, Thus we see that, however small the length of Ap may be, Pp+hPAo or P(p+$Ap) (Art. 86c, I) is always perpendicular to the direction of Ap. But as the length of Ap diminishes, its direction, no matter on what side it 1s, ap- proaches coincidence with the tangent plane. Therefore ®¢ is perpendicular to the tangent plane at the extremity of ep. The equation of this plane is therefore (* being the line of any point in it) ®p . (n—p') +(x —p) . P= 0 or, by the equation of the cone, mr. Po+p, x =(), 101. The equation of the cone of normals to the tangent planes of a given cone p. Pe+ he. =O can be easily deduced from that of the cone itself. For we may write this equation in the form (P1 De), Po+ hp, (bth) =0 ? where ®!@p=p, Then, if we put ®p=s, a line of the new cone, we get for the new Cone O15 a-+ec, (ots) == () b the origin being at the vertex. RPascal’s Theorem, 102. Let O be the origin; OA, OB, OC, OD, OF five given lines on the surface of a cone, and terminated in a plane section of the cone ABCDEF, not passing through O; OX any line lying on the same surface. Let OA=«, OB=$, OC=y, OD=%, ims, OX =, — 118 — If we write el ie a ae Sie ne / Parlay sh eg Tee ies Peal i miu =0 (I) II II II Trl) AY gL ere Leiter et % ; ; B,Y £5? 7,0 Ps” : Bey £50 : 1,8 Pot ; ‘ ’ we haye the equation of a cone whose vertex is 0; ‘for the equation is not altered. by putting wp for ep. Moreover the lines a, B, y, 8, and « are sides of the cone, because if any one of them be put for the equation is satisfied. This is obvious for « and « (Art. 71). The equation, by formula (21) of Art, 78, may be put under the form a —_ vy JT le IL) a1, AT ll /iles f ‘i | qT, as ; a ea 3 os nay, p,m Bit, ae ee (TL a ees aie a I )=0. op TL a Tes ee ead u. ie th Se 15 P,% Ys py Br ep BY ee And, applying again the same formula, we have hts Ca es “) (IT lll. ‘) (ell ol, a) (« eI +IL «) FUP hy: BY (IT +10 2) (+10, 1 +I + el. IL s)=0 OF, SOY SATE), nes GIT +I ‘) GIT + +I (M1 El ) ID + ally a’) % Bia BLY 1 »& 0, ? GIL + Gl “IL GIL. “II, ‘) ( I at ‘=O. a, a, B, oe Thus we see that if we put ~ or § for p both terms vanish an BY If we put y for ep, we have ‘identically GIT +I YOGI +IT, ell ule OGH, All, oe ’ Bea Chee ae 7,8 GIT il OOlr +I DG If elke GI +H a’) =0 (Art. 75), 1,8 A ff Hence Rte (1) is satisfied if coincide with ‘any one of the five coinitial lines «, B, y, 5, and «; that is OA, OB, OC, OD, and OE are lines -on tlie surface of the cone represented by (1). “As is remarked by Hamilton, this is a very simple proof of Pascal’s Theorem — for (1) is the condition that the intersections of the planes of «, B and 6, e; 8, y and ¢, 9; y, 8 and p, «; shall lie in one plane; or making the statememt for any plane section of the cone, that the points of intersection of the three pairs of opposite sides of a hexagon inscribed in a curve, may always lie in one straight line, the curve must be a conic section.” (Tait’s Quaternions, Art. 247.) t 4 — 119 — CHAPTER. VI. SURFACES OF THE SECOND ORDER. 103. We have seen that the equations pp-—(rp' +ey)—C#+=0, (1) 4 erp p'— (a p'-+pa')? =0, (2) hp p'— (ap +pa')2—4 b%=0, (3) Ape'—(xp'+pa') (Bp -+e8)=0, (4) represent, respectively, a -gpherical surface (Art, 89), a cone of revolution (Art. 94), a cylinder (Arts. 95, 96), and a cyclic cone (Art. 97). ' Every one of these equations has this peculiarity that, an expression in the form («o’-+«') being considered as one term, the unit-line of each term is 7, therefore de- noting 7 by 1 (Art. 49), ee” will become ¢? and ap’ +a’, ae’+pe'; thus these equations will have numerical form though they contain directed quantities: Moreover defining ap’ +o of the first order, pp’, («p’+e2')*, and («p’+ 2’) (Be’+ 68) of the second order in e; the above equations will be considered of the second order in e and the TLE they represent are called surfaces of the second order. Before beginning the study of surfaces of the second order in general, we shall con- sider one or two more illustrations. 104. To find the equation of the surface formed by the revolution of a conic section about its axis (Fig. Art. 85). It is obvious that in the demonstration of Art. 85 there is nothing which obliges us to assume that the line DQ and the point P are necessarily in the plane of the figure or in any particular plane; besides ® which was assumed to be the unit-line along DQ disappears in the final equation; therefore it may be supposed to be in any direction in the plane through D and perpendicular to DF. It follows that the equation of the sec- ond order ah | pp . aa-—e% (a a’) 9— @2 (a p' +p a’) aa! — 5 (ap' + pa)*=0 obtained in Art. 85, represents, in general, a surface of the second order which is formed by the revolution of the conic section about DF. But if the extremity of p is necessarily in one plane, then the locus represented by the above equation will be a conic section which is the curve of intersection of the sur- face. with the plane ell +I] p=. a“, a, i 105. In Art. 85 we had assumed that DQ and FP were in one plane, then PQ being parallel to FD, we were able to write p=ew#%. But if P is not in the plane of DF and DQ, this equality cannot exist in general. In this case we may proceed thus: — We can write PQ=ete.sinPDQ | — #20 — and N I] e+e, sin PDQ, (Art. 69). B, (e+) Therefore N Il Silay, B,(p+o) ~~ and o—eNIT B, (e-+«) which gives pp =e? II II ; or, by formula (3) of Art. 78, 2 pe’ ete’ +etaal+o%(ap/-+ pa’) —T (Be + pB)® (2) another equation of the second order. This represents the locus of a point which moves in space so that its distance from a fixed point bears a constant ratio to its distance from a fixed line. It is obvious from its definition that the curve of intersection of this locus with the plane of « and & is a conic section: an ellipse if e>1, and a para- bola af ¢=<32 It is easy to deduce from equation (a) that of Art. 104. For this we must assume that in (a) «, 8, and p are coplanar; and, as in both cases 8 is perpendicular to «, by this assumption we can find («p+ pa')? hater | (Bp +e f)P= Ao? [2 which, substituted in (a), will give the special case of Art. 104. Putting the origin at D (Fig. Art. 85), that is p—« for p, equation (a) will have the simpler form f ‘ ‘ 4 ‘ e2 4 , A Pia tora AN Loa gid ed a mh tak (2) 106, Let us change, in (b), the origin to D, where DD,=4; then pep becomes p+3, and equation (b) takes the form pp + (Sp +5) +55 — (a p' +p a’) — (% 0+ da’) +a a’ —=e%[p p+ (5 p' +98) +38— 4 (Bo + 08) 2— 4 (Be + pf) (83° -+ 58") —4 (82 -+38')2]. (c) Thus we perceive that if g (Bp + p2)— (xp +p a’) eGo +08) — 5 (Be +08) (BO +58) (2) or p[(°— a’) —e* [8— § (BY +38) Bf] ] + [(8—a)—e* |8— 3 (83+58') B{]p'—=0, —() the terms of the first order in e will disappear from (c). In order that (e) be true in general, we must have 5—a—e? |3—2 (60 -+56)8| =0, (f) — 121 — To solve this we have a? (82° +28) — (28 +B") —e2 [(B2'-+28) —(62'-+26')] =0 (B2' +88!) — («B+ 8a’) =0; then, remembering that «8’+f«'=0, B3' +58’ —=0, Hence (f) gives .—— t= eo OU \ " 6 (1—e*)—=« a = ame (See Arts. 86, 87). Thus we see that —if e is not 1— there is a definite point such that if we take it as the origin, the terms of the first order in pe will disappear. Referred to this point as the origin, the equation of the surface becomes ‘ a or 2 a 0 ; : 0% ox ely a, ab OE Yr pam ee en Md ea mm ere ded or | eet , C7 a (i—e*) pp + (eB Bp)*—-— = 0. (9) 4", This equation, if satisfied by +e, will also be satisfied by —p; then it represents a central surface, the centre being at the extremity of 6, which is also the origin. 2™*, This surface is a surface of revolution (of the second order) whose axis is parallel to 8. For, the equation, if it be satisfied by pe, will be satisfied by any p, where e,=e and the contained angle of p and & is equal to that of pe, and 8; that is the curve of in- tersection of this surface with a plane perpendicular to 8 is a circle. 107. For the potnts where the surface represented by (g) in Art. 106 cuts the direc. tion of «, we must have p$’+8e—0, Then . er a’ re — ((—e2)2 or eu e% a — c fo), Sacer! = Z 4{—e* and 1{—e? Qew ™ AA, = 7— == 2a, A and A, being points where the locus cuts the direction «. (See Arts. 86 and 87), For the points where the surface cuts the line parallel to 8 through the centre, we must have pRB +Bp = 28 pr 2oei, Substituting in (9), 16 — 122 — or eo? =(1—e*) a*=5*, (See Arts. 86a, 87 a). 108. If e=1, the case excluded in Art. 106, (a) or (b) of Art. 105 gives ap +pa'—aa'+t (Bp +pB')2, (h) again a surface of the second order whose intersection with a plane perpendicular to f is a straight line. For, let the equation of the plane be Bo'+pe$'=2czt; by substituting, (h) gives ap + pa aa’ + 627 which represents another plane, so that the section is a line perpendicular to the plane of « and 8. Hence there is a case in Art. 105 where the locus will be a surface which may be considered as a cylindrical surface whose base is the parabola in the plane of «. and ® and whose generatrix is perpendicular to this plane. The existence of such a case in Art. 105 is almost obvious. General Equation of the Second Order. 109. The general equation of nwmerical form of the second order in pe must evidently contain a term independent of p; terms of the forms B(«p'+e8'), B, (Be +ef),..... of the first order in p; and terms of the forms App’, A, (~p’+p')%, Ag(Be +eB)*,..... C (« p+ ea) (Rieust-0 8.) Bra ae of the second order in ». A term of the form aul +I] «') being equivalent to aul +I e) is of the form («,p’+p4,) where PB p,B Bro = Bm a — B,« Now the terms B(ap’+pa'), By(Bp'+p8'), .. 2... may be written in one term having the same form as (ype’+¢7') where ¥— Be Bip. eee The terms A (ap +pa,)%, A, (Be + 6')2, a Boe have a form which is merely a particular case of the form C (a e'+ pa’) (Be +f’) which may be written as (x p’-- pa’) (Bp + eB), « being substituted for C«. Thus we may write the general equation of numerical form of the second order in ¢ as follows: — S3 X, (e+ pa’) (Bp +68) +2Ape +(e +ey)— Ct. (a) The numerical factor 2 is introduced for convenience. By changing the origin to D where OD=8, pe becomes p+, and equation (a) takes the form — 123 — S, [a (o' +8) + (0-+8) a'] [f (o’ +3’) + (0+) fT] ‘ + 2A (p +8) (e+ 8’) + [y (p' +8) + (e+ 8) y= Cz or =. [ap +p a’) + (28 +8a/)] [(B e+ 8!) + (88 +38) + QAP +ZATT+ARA (pF +30) + (ye +ey) +i +3 7)= C7; from which we get = [(a p+ pa’) (Be + pf’) + (28 +8a’) (Be + pf) + (a p' +p a’) (8 3'+ 38") + (0 8' +5 a’) (8 3'+68')] \ H2A PP H+ QASH + IA (p8' +30") + (yp toy) + (yF +8 y') = Cz, or X. (ap +pa') (Bo +o8') +2, (a8 +6’) (8 3' +68’) +2 [[xp +p 2!) (8 +58!) + (28 +32) (Bo +08')] HRA PP HLACT AIA (CT +50) + (ye tex) + (ye +27) = Cr, (0) 110. Now if = [(% 9’ +p a’) (B28 +68') + (a 8’ +a) (Be +8’) H+ QA(CT +30) + (ye +e 7')=0; (c) then, the terms of the first order in p disappear and Z. (ap +pa’) (Be +oB) +2. (a8 +50’) (83 +38) + 2A oe +2A80' + (yi +87) = Ci (d) will be the equation of the surface of the second order referred to its centre. Equation (c) can be put under the form [a o’ (B38 +28) +0 a’ (83438) + (08 +30’) Bo’ + (a8 +82’) of] +2AT0+2A 03 +70 +07'=0 or Slo] a’ (83 +28) +8 (as +30) | +] a (88 +38’) +8 (ad +3a') | o'] +2Q2AT9+2A pt + ye +ey'=0 which gives eX [a (b3'+58') +8 (2d +30')] +5 [a (8 8’ +58') +6 (ad +4.a')| 9" 2 Mae ey OK Sp pO: -—- : e[=,o' (Be +3f) +B (ae +5e)(+2A0+y] + [S]a (BX 426) +8 (22 4+34')[+2A5-+y]/'=0. In order that this equality be true in general we must have S[a (89 +5B') +8 (2d +5e')]+2A5+y7=0 (c) an equation of lines of the first order in, which in general gives a single definite value for 6. [But in the cases where equation (e) represents a line or a plane, any point of fy these will be a centre of the surface]. With this definite value of 6, and putting D i=Ci—(yX +87) —QASI—S, (a7 48a’) (88 +38) equation (d) becomes =. (ap + px) (Be +08) +2A pp =D, (f) 110 lie 0. E. (ap +a’) (Bo -+ eB) + 2A p=, (9) which is not altered by putting ap for e; therefore the surface represented by it is conical. (See Art. 97). Equation (g) may be put in the form X.p a (Bo +oB')+ Apo +S. ae (Pp'+of') +A pe’ =0 and Teh (ap tea')tApe +=. Be (ap +p’) +Ape'=0. Adding | | e[Z{a’ (Be +e B')+B (ap tea) {+A +[S]a(Bp+eB)+B(ap'+e’){+Ap] Yom I And putting De s(x Bp +h) +h (ap +ex')]+Ap, we have 0. Po+p, 7°00. (h) The function @ hag the same properties of the function ® in Art. 86c. A function which has the property that Art. 86c, I., exhibits is called a Self-Conjugate function. From above function ® we have o.Pp+Po.c' == 3 [(ao'+aa') (Bo + ef) + (oh +Bo') (ap +ea’)] +A (se +c) and 0, Dota, Some > |) a’ -+-% 9") (Bo +-oB’) + (p B’ +B 0’) (ac’-+oa')|-+A(cp’+ 0), ¢,Po+0p,c—=p , Po+c ,(, as in Art. 86c, II. This property is of great importance. eS 112. If D in (f) does not vanish the surface designated by (f) is an ellipsoid or hy- perboloid. By dividing by D, and thus altering only the length of the constants, we see that the equation of central surfaces of the second order, referred to the centre, is (excluding cones) =. (49+ po’) (Bo +p P') +29 pp = 7; which can be put under the form p» Pp+Pp. =, (*) where Pp [a (Be +p 8) +B (ap +e a] +90; ®o being a self-conjugate function. 113. To find the equation of the tangent plane. Let a secant plane pass through the point whose line from the centre is p; and, let ep, be the line from the centre to any point of section. Put p’—=p-+y, where y is a line along the secant plane; then — 125 — P, Bet Pe, . p= (p+ y) (p+ y) + P(E +8) . (e+) which gives (Art. 867) y. @ (o+ dy) +0 (0+ dy). 70. Thus however small the length of y may be ®(pe+ 4y) is always perpendicular to the direction of y. But as y gradually diminishes in length when the secant plane approaches the tangent plane, its direction, no matter on what side it may be, steadily nears coin- cidence with the tangent plane. Therefore ®p, being the limit of (p+ 4y), is perpendi- cular to the tangent plane at the extremity of p. The equation of this plane is therefore (x being the line of any point in it) Pp. (n—p') +(7—p). =O or, by the equation of the surface, m=, Pp+Pp , n=, 114. If OY is perpendicular from the centre O (origin) on the tangent plane, then, since ®p is a line perpendicular to that plane, we can write OY=a.p and putting OY forex; oPp.Pp+aoPp, Pps or 2405. Pp—7 ane (2p) which gives 1 OY=N(a® =F 6, 115. If tangent planes all. pass through a fixed point, the curve of contact is a plane curve. Let T be the fixed point; « the line joining the origin toT; e the line from the origin to a point of contact. Then (Art. 113) a, Dp+Pp . a= or 0, Pata. ome which is the equation in ¢ of a plane perpendicular to ®«. Now ®« is the normal line of the point where OT cuts the central surface; ihaverane the curve of contact lies in a plane parallel to the tangent plane at the extremity of the diameter drawn to the given point. The plane of contact is called the polar plane to the point. 416. To find the enveloping cone whose vertex is T. Let « be the line from the origin to T, = tothe point of contact of one of the tangent planes. Then «—- will be along the tangent line passing through T and the extremity of x. Let pe be a line from the origin to any point of the tangent line. We have pata (r—a«), — 126 — which gives 1 | 7 —p—— a-+-a@. 1 Substituting this in a,On+On,ae—=71, (Art. 113) and x, On+bn, ni; (Art. 112) we have (7, Po+dp, hoe ~Pe+da,a/)+(a, Pa’ +da , a’) 7 w 8si=— and f 1 Ul ’ ‘ tT ’ U] ° sage al es (LG. Wat oe -P)+(p. Patda. pli, Therefore, eliminating 2, (2. Pata, a’) (p. Pp+p . 6’) —(p. Mp+p, p')—(a, Pa +a, a’) = (p. Pata, p')*?—2(p, Data . 0’) or, adding 7 to both sides, [x Matha, a'—il le. Pp+he. g—i]—[p. Pata, p—iPP—=0, (2) which is the equation of the cone required. If the origin be transferred to the vertex T, that is if p+ be put for pe, the result will be [~. Pata , a/—2i] (0. De4-Pp, o')—(p. Pata. ’)*=0, If we put wA for «, A being a definite line parallel to «, and afterwards assume that w is infinitely great, we get (A. PA+A . A’)[p. O'pt+op. o]—(o. A+ . (2 =0 the equation of the cylinder formed by tangent lines parallel to A. 117. To find the locus of the middle points of a system of parallel chords. Let them be parallel to «, « being the line joining the origin to the middle point of one of them; then =+a« and s—g« are points in the surface. From the first, (s+a«) . O (n+p a)+ (x+ma) , (n’+a'x) 2, (Art. 112) or (x, On+Or , n')+2 (x, Da+Oe , nm’)? (a Peta , a’) =o) From the second, (x Orton, x)—2Qa (x ,Pe+0a, m) +a? (a .Potbdea, arog; therefore, subtracting, nm, Detde , n'—0, (a) This equation shows that the locus of the extremity of z, the middle point of a chord parallel to «, isa plane through the centre perpendicular to ®«; that is, a plane parallel to the tangent plane at the extremity A* of the diameter which is drawn parallel to «, “We assume here that a line drawn from the centre and parallel to « shall meet the surface in a point A. But if the surface be an hyperboloid the point A can not be always real. s AS — 127 — If we call this the plane BOC, B and C being any point belonging to the curve of its intersection* with the surface; and if CB=8, OC=y, we shall have ’ 8. Datda, p=), and, therefore, a, B48 , o’ =), or, as « satisfies the equation x, 2 8+08 , x’ —0 * which is the equation of the plane which bisects all chords parallel to OB. Let AOC be this plane, and, since OC or y is a line in it, Yi? BOB: y= 0, he B, Py+oy , i =0, But, y being in the plane BOC, we have y. Pata, y'=0 or a, Py+Py, a0, Therefore by equation (a) «, 8 both satisfy the equation of the plane x. Py+oy7. 70, which is the plane bisecting all chords parallel to y; that plane is therefore the plane AOB: We are thus presented with three lines OA, OB, and OC such that all chords parallel to any one of them are bisected by the diametral plane which passes through the other two. These lines are termed conjugale semi-dia- meters, and the corresponding diametral planes, the conjugate diametral planes. (Kel- land and Tait, Art. 61.) It is evident that the number of conjugate diameters is unlimited. 118, «, 8, and y being a system of conjugate semi-diameters, we have a, DP+OB . o' = 8, Datda , =O; B. Py+Oy. Py. OP+68 , =; a, Dy4+Oy a’ —y. Pa+Oa.y 0, They show that y is perpendicular to both ®« and 8, and is therefore a line per- pendicular to their plane; hence —el] (a) \ *This curve of intersection will be always real for ellipsoids, 488 — e being a numerical coefficient. In the same way, since ®y is perpendicular to both « and 8, we have ey=yll. (0) a, B thus we see that y, 8, and « are, respectively, parallel to Peal b) Da, OB a, Py OB, by iHE HL ost BSB. F987 eB y y=yo |] ; a, f and Py, ®8, da to Putting (b) in this form and, substituting in (a), we have yor] | or ’ a, B Pa, PB upon which Hamilton founded his solution of linear equations. (Kelland and Tait, Art. 61.) 119. We have seen that any line whatever may be expressed in terms of any three non-coplanar lines. Let then «, 8, and y be any three non-coplanar lines, and e=eUae+yU$+2U y designate a central surface of the second order referred to «, 8, and y. Substituting in (¢) of Art. 112, | (wUaty US+2 U «) (wx U«+y U 8+2U y) +0(¢Ua+ty U8+2U x). @Uat+y U b+2 Uy) =; or 22(Ua. ®Uc+OeUa, Ua)+y2(UB. & Ub+OU 8. U'$)+22(U y. & Uy+-eUy. Uy) +oy (US. Uc+oU«.Us)+a2(Ue. 8 Uy+oUy. Ue)+y2(U8.& Uy+eU x. Up) 7. Now let «, 8, and y be a system of conjugate semi-diameters, then Us. UctoeUa«. UB), Uc. ®Uy+"Uy.U«=0, USB.®%Uce+oUy.U' 80; therefore o2(Ua.P' UetoU«e.U'a)+y2(U8. & UP+O UB. U'B)+22(Uy. ® Uy+euy. Uy) =7; the equation of central surfaces of the second order (cones excluded) referred to its conjugate diameters with Cartesian Co-ordinates. Let «, 8, and y be real, and ¢==a,,-B=6,, and y==c;. x — 129 — We have Uc. ®Uc+OUe, Ve=—, ete. t a," Therefore the ordinary equation of the ellipsoid referred to conjugate diameters. 120. Principal axes of the ellipsoid.— Let us conceive a point on the ellipsoid so that its distance from the centre be a maximum or a minimum; that is, the tangent plane to the ellipsoid at that point be perpendicular to the line joining the centre and the point. Let this line be ,, then, Art. 113, Po, = 91%: which, by (¢) Art. 112, gives td ——- v 11 Ra. - Let a plane passing through the centre be parallel to the tangent plane at the extremi- ty of »,, that is perpendicular to ,, it will bisect all chords parallel to , (Art. 117). Therefore the curve of intersection of this plane and the ellipsoid will be such that all tangent planes to the ellipsoid at the points on this curve of intersection will be necessarily parallel to ,; therefore, on this curve of intersection must be at least a point whose distance from the centre is a maximum or minimum. Let ©, be the line joining the centre to this point. Obviously , and , are perpendicular to one another, and (Art. 113) 10) PO p=GJo%e? which gives BER sass a SiS oe Das A plane passing through the centre and being parallel to the tangent plane at the ex- tremity of »,, will bisect all chords parallel to o,, and pass through »,. The curve of intersection of this plane and the ellipsoid will be, necessarily, so that all tangent planes to the ellipsoid at the points on this curve will be parallel to o,. Therefore the tangent plane to the ellipsoid at the common point of the two curves of intersection will be parallel to wm, ando,, that isto the plane of», and ,. Let , be the line joining the centre to the common point of the curves. Obviously », will be along the section of the two planes passing through the centre and perpendicular to ©, and ,. Now it is evident that the plane of ©, and , will bisect all chords parallel to ©,; and , is perpendicular to the plane of , and ., therefore to the tangent plane parallel to that of , and ,; consequently Ow, = 93% which gives W_@ Seale ae oes > 29s Thus we see the existence of a system of conjugate diameters which are at right angles to one another. These diameters are called the principal axes of the ellipsoid. 17 130 — 421. To show that there is only one system of principal axes. Let ©,—4, heh, and ©3—=C, and e—«rUo,+yUo,+2U, DW, YWs. ZH, or = + HH + p a b Ae x z then, Pp Kho, pr Pug t= 005; therefore (Art. 120), oP Fis YIaMe F930 5 | a b % Now if for any point on the ellipsoid which is not in the curves of intersection of the ellipsoid and the planes of ©, and 9, , and ,, ©, and 3, we have De=Xo, we must have also TI YGJaa , FJgg_ XG, XY AZo; a b G a b C or z idan Oy “a Y eos De Fa — 3 =0, which requires (Art. 20) 9, 7X0, g-—X=—0, 9,—X—0 or J1—IJo— Js» and consequently (Art. 120), @'O = O50, O20. or No, =—Nog=N ®,, which will be true only if the supposed ellipsoid be a sphere. Then, in general, there is only one system of principal axes in the ellipsoid. 122. Let ,, ,, and », be three unit-lines along the principal axes of an ellipsoid; and let P=LO FY Vets, we, y, and z being Cartesian co-ordinates referred to the principal axes. We have Pp=—27Po0,+yPo0,+-2Pw,; but P = G11, POZH=Jo%, POZ—IJ 3% then, substituting, P P= HOH Y Jog HF 3%5- Therefore equation (7) in Art. 112 gives 20°99 .o,o,+ 2 Y 79g % 9% gt 2579 5%%%g=1. But (Art. 120) a therefore, the ordinary equation of the ellipsoid referred to the principal axes. 423. Let PLO +¥Y¥ +753, \ 1, g, and ©, being unit-lines along the principal axes of the ellipsoid. This will give O90 +p, 2a1, Substituting We have Po == HG, HY Jot SGgg (Art. 122). But, remembering that ©,, ©,, ©; are unit-lines at right angles to one another, y 1 | Jia At? Coot pe Pus 5 a! v y . oe Pia as £1) pe he Wapeai 8 (2) or observing that OrP +90, 2H, Ogi’ pHs 2yt, Wp + pH, 221; 0,0 pw ,0 +OW 020 FOO be cae An ft = 9+ a Fs. (3) Let us write We for OP +0, Wg0 +P My Wap +P, ed SS ea SN. iy) oe yy (4) la ; 26 . Dele e we shall have as in Art. 86f, 2p W (Vp) = 20p. (5) We can write o,. Pp+Op.0 ay .Po+Pp . a, Og. Dp+Po. ws b% — b dp 1 dh 4 Se? 9, “ae Abe h o2 ye +Pe Wop +E, 20 PM, eee Bat A WO, ie 3 haan b4 alien chara cf Os ; (6) — 132 — nie y % A or ON igs eter pa Coes coe ") We have, from (3), { : , __ Op +pe O,.Pp+Pp,0, = “Za2 i; etc., etc. Substituting in (1) p= a*(o,. Ppt Pp. &,)0,+5%(og. Pp +h, De) gt C%(, bp +p, a5) 5. Therefore / D-lp = G2 (4 0' +961) © + D8 (woe’ +p bg) Og+ 67 (M50 +p Hs) O5, (3) ri because ©! produces p. Also from (4) o,.Ve+WVe. é,—-1 ain : etc., etc Substituting in (1). a t , b ’ 7 C “| 7 Wp 5 (49 P41) OFS (Mee +P Hq) Oot 5 (Mgr +P Hs) 5. (9) 2 2 2 This will give 5 a t 7 ? } b / , , C v ¢ r poh I Wp = 5 (0. Ve +P. 54) O45 (Og. Vet Vp. bg) Og 5 (0g. Ve tHe. 4s) 05. (10) It is evident that the properties of ® in Art. 86c, apply to all these functions. 124, If ®p is, in general, a self-conjugate function, that is if o,Pp+p,c=p, Oso+0c,0, where « and ep may be any lines; 2p, 30, . . . . are self-conjugate also. For, putting ®o for ep, we have 6 (©2p)'+ 2p 6 = 00, Do+hc, Oo, = 0% , p' +9 (08s); which proves the proposition for ®%¢ The generality of this is evident. 125, We have seen that P85 == 2a Arto) Therefore, by (¢) Art. 112, we can write p. We +p, = 2% or, Y being a self-conjugate function, Q Yo. Wet Wp , Wp 21, Ye. p=, or NWp=1 which, if we put s= Wp, becomes Ait the equation of a sphere. eens v7 Hence the ellipsoid can be changed into the sphere by the deformation of each line, the operator being the function VY. And, as Y1Wp—=p, it is obvious that, reciprocally, the sphere will be changed into the ellipsoid by the operator ¥Y-. By writting Pe 5 PV, we have (Art. 118) ore USB) EAR eas OS VER Wat Wa eR etc., etc., so that Va, "8, Wy, the lines of the unit-sphere which correspond to conjugate semi- diameters of the ellipsoid, form a rectangular system. 126. We have from (3) Art. 123 » (Ore He 45,)* | (oe tee)? | (Mgr! +e 45)? p Po. 6 et tt. eo. Pp+Pp.p ha? + 7b? + het Let «, 8, and y be any rectangular unit-lines; we have, by the above formula, (o,¢' +a0d,)? (9% +269)? (w,2’ +%0,)? Uthat cba we wh cae G6C7,. CLC. &, DetPda.o — But putting P=7e-y P+ z,y which gives eS A pone ead ta dd oh ene 5} a ++ 5) B+ 5) Y +, (ee +2e)* (Be ek)? (re +er}?- or ig ee i + rn + j : Thus we have 5 lone ba) (ob BG)® (Opt ty)? 4 4, A etc., etc. : Therefore ; (x, Pa+tda,a)+(8, DB+O8, fo) +(y. Py + hy, y) Ee Again if «, 8, and y be any rectangular unit-lines, oa] +I] ba, Gee Yi > Baan 0,0 + aw ' ‘ = [Sar at oeeoe Lal eat +P, ha? eh — We Fad, ; +]] rn ree 8 - Js o,p'+ 8 Dy OY HY, + e e e e es ® es [ ha 1 [al ha? 1 | We +a, Wo% + ab, Ose +40, hg tae A ba Nc Q! of , 0 y H =B 0 — 5 | OP FRO, ob +BoQ of8' +80, (Art. 78, formula 11); tn ier Nh b® / hc OP HYO, Sov +Y%, OF +1% ha® if 4 b? : hc? Oe +A0,, Wg% +%0, : 3% +20, = 32 a bter Barre, 8 +Bo,, 58 +B, ; we YO, » Oat PYSse MOET eE Suk = 14 vA , =Gag7p7 l(a +4%0,)O,F eee, AT] |(o, B+ B4,)o,+ es oes As foyy+144) oy+ Bae: FY «| [1 . x |(oa’+ad,)d4+....f]; {(o,P'+Bo,)o,+... A, f(oyy+y4,) 0+ fe I | ’ ' F ~ Wh a2? Belt a He 24 = —s.5(L] +IT »); 2h2;,2 8a%btc By By Qa']; ? — + , arbre with the sign + if the rectangular unit-linégy «, 8, and y are so situated that 7 is the axis of positive rotation from 8 to «; or with the sign — if y is the axis of positive rotation from « to 8. For in the first assumption «LT +|] eee 5 By ByY and in the second raul +f] a = — 25, By ByY 127. Remarks:— Is. It is evident that, g being a numerical quantity, if we write (0+ 9)e for ®e+gp, considering ®+g as a functional symbol, ®+4 will have the same properties as ® (Art. 86 c). We may write (P+ 9)?p for (P+) [(® +g) ]. Also (+g) for (® +9) [(®+9) (+9) p]. 2’, The symbol © may be treated as a coefficient; that is, if ®>—=y, we may write p= oly. — 135 — Thus (3) Art. 123 gives et | ©, ee 9 ogp as wl). 1a f 1c To verify this put w o +90 Wop + po ,0 +O vere) er aoe Wee ee a for p in (8). We can write also p=o*y, poy, for d%p—=y, P3p=y, respectively. 3°, We assume that p, (+ 9), and (6+ 4)%. are not generally coplanar. But if these lines be generally coplanar, so are (®+ 9)p, (©+¢)%p, and (®©+ )%»; since, putting ¢ for (®+ 9), they may be written s, (®+ 9)s, and (®+ )%s; then (®+9)%. will be in the same plane. If ina particular case we should have, for some definite line p, (®+9)e=hoe where / is a numerical quantity, we shall obviously have (® + g)*=h%o and (® + g)§p = ho; then in this particular case also p, (P+ 9), (®+ g)%e, and (®+ )% will be coplanar. Therefore (®+g9)%: can be expressed in terms of p, (P+ )p, and (+ )%; and we can write (D+ g)®p=Ap+B(P+g)e+C (P+ 9) %. («) Tt will be seen that in this equality A, B, and C are quantities independent of the line ge. Then if any three lines «, 8, and y be substituted for e, they will in general enable us to assign the values of the three coefficients. The equation may be written —A p——B(®-+-g)o 4 C (P-+ g) 9p — (9 -t g) 8p. (D) Substituting (®©+g9)-'s for ep, we have —A(®+ g)"p =Bep +0 (®+ g)p—(® +g) *e; (c) thus the unknown inverse function (®+g)"!9 is expressed in terms of direct operations. Let (P+ 9) e=1. We can write (P+ 9)*p= (+9) 1, yt ates (P+ 9)%e = (2 + g)?x. Substituting in (0d), =Apg== Br+C(O+9)y— (2+ 9)*¥. But by (c) By+C (+9) y—(®@+9)*y=—A (®+ 9)", therefore (D+ 9)" =e, that is the symbol (®+ 9) may also be treated as a numerical coefficient. A™, If ®o is self-conjugate obviously ®p+gpe or (®+ )¢ is self-conjugate, and we can see that (&+¢)%, (®+ 4), etc., are self-conjugate, therefore (®+ q)-'p is self-con- jugate also. — 136 — 428. Examples: — Ex. 1. To find the point on an ellipsoid, the tangent plane at which cuts off equal portions from the axes. Let ep be the line from the centre to the point of contact; #, y, and z the co-ordinates of it referred to the axes; P the length of the portion cut off; ©,, g, and », unit-lines along the axes. Obviously the extremities of Po,, Pos, and Po, are on the tangent plane; therefore putting Po, for z in the equation of the tangent plane, we have P(o,. Pp+%p.4,)=7% or substituting : , CGO, Y|Y M+TOg | for p, eP(o,. PotOo. d,)+yP (oy. Pogt Pog. d,)+2P(o,, Mos+Po,.5,)=4, bard muh fa 1 Atal | ; au Por Faia Bag paler OMe T ees y therefore coe ia th oa, A Similarly y Pos 3 1, 4 aP.—g=; 2 2 pe ’ em 4 or Pisaygtw tenes? a a Derek aki b? mg Dua V a+bi+e. Ex. 2. To find the perpendicular fiom the centre of the ellipsoid on a tangent plane. We have 1 | bs (OY)%=O0Y {OX} =F Domne (Art. 114), | x Therefore At —A > , 2 OY)? ria =" 44+, (art. 123, 2): Ex. 3. To find the locus of the points of contact of tangent planes which make a : given angle with the axis of z. wh —A37 — _ We know that ®p being perpendicular to the tangent plane at the point pe, the con- tained angle of », and ®p will be the complement of the angle which the tangent plane makes with the axis ,. Therefore, by supposition, Oar U'Pp+UPp . Oe == sie) is constant, from this ®,.Pp+Po,o,—=P. Np or (o,. Pp+p, 6.) =P? Do, Wo, the equation of a cone whose axis is that of z. The intersection of this surface with the ellipsoid is the locus required. To find the Cartesian equation of the above cone put ny y Zz DEEL Ea arraee for @o (Art. 123, 2). Thus c at " bt * c# which gives - oe Yy 2 eS pe NY ii: 62. P4 The section of this cone by a plane parallel to that of ©, and , will be (n a*)* om (qo?) 2 - where 4—P? Pisa cap st 7) This section is an ellipse whose semi-axes are na*® and nb?. Ex. 4. To find the locus of a point when the perpendicular from the centre on its polar plane is of constant length. Let = be the line connecting the origin to the point, then the equation of the polar plane is ep. OP n+On, p=? (Art. 115), and the length of the perpendicular on it is (Art. 114) 4 ; 2NOn- then Nz is constant. Hence the required locus is Nec 2 or 12 nr, Pn+Oor, ere which gives t(b2r)' + O29 x , n= 5 a, — 138 — a concentric ellipsoid, with its axes in the same direction as those of the first. By Art. 123 its Cartesian equation is oo ye 2% Ex. 5. A sphere, passing through the centre of an ellipsoid, is cut by a series of spheres whose centres are on the ellipsoid and which pass through the centre thereof; to find the envelope of the planes of intersection. Let (op —«) ('—2«') =a’ (Arts. 83, 89) ‘ ‘ U f or pe'— (xp' +e a’) =0 é by the first sphere. One of the others is pp—(np' +p x)=), where a, On+On,n'=7% (Art. 112). The plane of intersection is (x —a)e +p (n—a’)=0, (1) For another point on the ellipsoid the plane of intersection will be (= — a) p' +e (7/—a)=0. (2) Then obviously these two planes pass through a line situated on the plane (=— t)e +e (7,—7)=0 (3) which is perpendicular to the direction of the chord ~—-7; then the intersection of the two planes (1) and (2) is perpendicular to z—7. Therefore as the direction of the chord x,— approaches the tangent plane at the extremity of =, the plane (3) approaches per- pendicularity to the tangent plane; that is as =, comes nearer and nearer to = the in- tersection of the planes (1) and (2) approaches more and more perpendicularity to the tangent plane at the extremity of =. And as ®x is also perpendicular to this tangent plane (Art. 113), ¢ being a line along the intersection, we must have ®r—=20 or nao Plo (Art, 127, 2); « being a numerical quantity. Hence wade . [® (a b-p)]'+ [0 (wb p)] . w (1p) =A ™ or v2 [po . (P-lp)'+ Blo , o'] = 3, and (a ®-"p —a) p+ p (w@ bp —a)'—=0 or ale. (P%p)'+ bp . Jap’ + pa’, Now, eliminating a, ep. (Pp) blo , p’ = (ap -+pa')2, a cone of the second order. J — 139 — - Ex. 6. From a point in the outer of two concentric ellipsoids a tangent cone is drawn to the inner; to find the envelope of the plane of contact. If nr, Onr+On, rt be the outer, and | p. Wp+ Vp-p 4 be the inner ellipsoid, ® and ¥ being any two self-conjugate functions of the first order in e, the plane of contact is nm. Wot Vo! x9, For another point on the outer ellipsoid the plane of contact is Mota toe oly t. —— 7, The line of intersection of these two planes is situated in the plane (x— 7). Wp+Wo. (x—x)'=0 which is perpendicular to the chord s—7: therefore as in the preceding example, Pr—ZYo, or TG Pp. This gives a[P1Wp . Wp+Wp, (o1Wp)']=7, and w2(@-1 Wp. We+Wo, (1 ¥ p)'] = 7, Therefore, eliminating a, PIV. Wo+ Wp, (PTW o)'=7, or AAR NER LA: AE where ¥®!Wo or W[b-!(¥)] is a function of the first order in p, and it is involved in this equation in such a manner that its non-conjugate part is necessarily absent. There- fore, by designating by we the self-conjugate part of ¥@!¥p, the above equation may be written ep. up +ap.p'=t, which shows that the required envelope is another concentric ellipsoid. Remark:— It will be seen that, ®p being any function of lines, in which the variable e is of the first degree, if the function xp is the self-conjugate part of ®p, we can write op—ep+II ; P,& and therefore 0, Pp+Po, =o [we + 11 j+{ee+ Ll le fade Pee =p.Up +ap.p'. Ex. 7. To find the locus of intersection of tangent planes at the extremities of con- jugate diameters. If «, 8, and y be the lines to the surface along the semi-diameters, the tangent planes are x. WB4+08. 7, i me Oy+oy rhs A nemecy | or, (Art. 123, 5], ; Eee 2e) Wee Re 24, T Ae 2p) O28 21, me yey iit with the conditions that Ye, 8 and Vy form a rectangular system of unit-lines (Art, 125). Hence, by the formula (19) Art. 78, (W a IL + Il Ya) Wx 2 yee py —(vx|[ A eee tal Hl +... )up+(erll ee fe dc WBOWVy Vy, va Ya, vB or, since Va, YB, and Vy form a rectangular system of unit-lines, 20 5 (Un Vo ee For (Ur W B+ Bos. JERE (Ea ey aoe Ae y [x (2a) +... J Vat fa (UG) +... |e b+ ie (ey +... ey, hive VYraVae+W¥64+Vy, Therefore Wa Mr aVe, Pot VG, w+ Vy, vy, that is Wn. Wrst or Nasa Vas: This may also evidently be written Ua Wren Wa 2.3; or mi (Em) ah Ware ee. 2, Therefore x, On+Orn, 731 which shows that the locus is similar and similarly situated to the given ellipsoid but larger in the ratio, ¥3:1. | Ex. 8. To find the locus of the intersection of three spheres whose diameters are semi-conjugate diameters of an ellipsoid. “ If « be one of the semi-conjugate diameters, a, Pata , aay or putting 4% for a (Art. 123, 5), | > a (W22) 40% ie oY And the corresponding sphere is | pp— 4 (ap’+ pa’) 0, tae ee A ei or, as a eed ~yfa, (WWAy! + (WHAp), =O —4 (Wp. Wat Ve, (¥-p)J=0, Also —$ [to B+ ws, (Go) |=), Get eee ee Ee. (Up) | 0, 8 and y being the other two of the semi-conjugate diameters. Hence by formula (19) ATLeTO; wall Pram) Se v6, Vy — vw, |] be einen ll +o. a) ¥8+ ("PLL cr kOe LN ae Wy, Wa oe MG or 20-16 (To Wat... Wat (Ww b+...) B+ (yt... Wy Wip==(Wa+¥ 6+ Vy) «pp therefore eee Cet 00-06; Ne Oe or Ne pe Veo 02 This, evidently, may also be written Wolo (Wo) Wola gy, (Wt) 2 O4e or, finally, | p .{P-2p)-4- W-%9 , o == Ot, “This is Fresnel’s Surface of IE in the Undulatory Theory.” Tait’s Quaternions, Art. 263. Ex. 9. To demonstrate that if any rectangular system of three lines be drawn from a point of an ellipsoid, the plane containing their other extremities passes through a fixed point. To find the locus of the latter point as the former varies. Let = be the line of the point of the ellipsoid; «, 8, and y a rectangular system of three unit-lines drawn from this point. We have xc, On +Oon, n= (a) and (x+aa), O (x+aea)+0(n+a0). (7 +00’) = 7; ‘ Qa, On+On, a’ therefore Ct—-— 2 se, : a,PatPa.a Hence: the line of the other extremity of the chord along « is a Or ED ro! i) 2 a, Datda, ao” and the lines of thé extremities of the chords along # and y will similarly be determined. Let ; — 442— Bd’ + xf’ uted A rrer een Sin B.D'B+08.8 : yOn+ Ory’ and fo eee {og ey oy The equation of the plane passing through the extremities of «,, §,, and y, will be (Art. 81 c) ep—Ac +B +Cy, with the condition A+B+C=1. Now putting 2 @p ji ‘ ‘ : Qo’ @ - ‘ A ee PATER ee ere reece aaa m m mmoES mim—[@.Patha a+b, PR+OB.B)+(y. Py + oy. 7), and, therefore, to satisfy the condition A+B+C=1, m we get a(a. Pron, e)+8 (8, d'x+O7.8')+y (y. ®'n-+0 7. ¥’) m pt+2 But, «, $, and y being rectangular unit-lines, we can write a(x. Or+On,a')+6(8B On+on, i)+y(y. Pr+Or. y)=—2or; AOn therefore em=on+ which proves the first part of the proposition. This may be written Imep=Or+imr=(S+4m)z, 7™—=}M(P+ 4m)" which, substituted for = in (a), gives EO+t) [te xe) q+ ett Org) dlls a) d= Te (8+ ) | (0+) | A E (e+ tr) | (e+ wy p’ (ee) or which may be written (Art. 127, 3°.) whel(on tye By] [looky ee Bye] ela the equation of a concentric ellipsoid, — 143 — Ex. 10. Given two surfaces of the second order, which intersect each other, there exists in general a set of three non-coplanar lines, whose directions are those of con- jugate diameters in every one of the surfaces of the second order passing through the intersection of the two surfaces given. Let the given surfaces be eo. Po+Pp, =i and (op — a). ¥ (op —a) +0 (0 —a) . (p’— a’) = er, One of the surfaces of the second order passing through their intersection is f.(p. Pp + Pe. o'}) —(p—a). 8 (p — a) —¥ (p—a@) . (('—2') = fi— er where f and e are numerical quantities. This may be written e. (fPp—Wp)’+ (fPp—Wo) . p' +2(p. Vata, ¢’) = (f—e)ita, Watta.a’; or p. (fP—P)p+(fP—V)e.p+2(p Vara, o') =(f—e)ita., Vata, a’, Let «,, 8, and y, be along a set of semi-conjugate diameters. Then (f#—)«, is per- pendicular to 8, and y, (Arts. 117, 118); (f®—)8,, to «, and y,; (f®—)y to «, and 8. Assuming that another surface of the second order e(fP—¥)'p+ (fP—Y¥)p. +2 (pWa+ Va) =(f—e)ti+a, Pao+tWa, a’, which is passing through the intersection of the two given surfaces, has a set of semi- conjugate diameters in the same direction; (f/@—‘Y)«, will be perpendicular to 8, and 1,3; (F®—Y)8, to «, and y,; and (f®—Y)y, to «, and 8. That is (f@—)«, will be pa- rallel to (fF @—W)«a,; (f®—¥)B, to (fO—Y¥)8; and (f@—¥)y, to (f;2—¥) y,. Therefore we must have ==) (fo—¥)«,, (f,o—¥) %, or ¢—fll =o, @ eh Oe ae Il iy Pa,, Va, and in the same way aos (e = () OB, WB, oy, Vy, Hence the set of conjugate diameters which are the same in all are parallel to the roots of Hb ae (a) Po, Fe as we might have seen without analysis. Let VYo=s, then p= W-lo: NE eeslll alll Pp, Vp OWlo EW-le -) O¥-lo,6 Wetec | substituting in (a), or ? ' OW-le—= Ge =) (b) where g is a numerical quantity. Thus we see that the solution of (a) may be obtained by that of (2). : To find the locus of the centres. We have fle. Spt he. p)—(p. Pot Vo. pyr 2p. Var Wa , o') =(f—e)ita, Da+ha a, Putting p+°é for e, Jb BY f(p. Pe+ Op. pale. +03. e')+ fe, H3+08, 3) —(p.\W pte. pf) — 2 {pi Ws wh oie) (0. oS 2) +2(0. Pata, o)+2(6.Wao+Wa, 3’) =(f—ejitaPa+Da, a’, Assuming that the extremity of ¢ (OD=8) is the centre of the surface the above e- quation represents, we must have Qf(p. PEAS, p')—2(o. WS5+ WS. o')4+2(o, Wa+Wa.o') =O or e [fs —WS+Wa]'+[fos—WS+Va], 0, which to hold true in general we must have fes—W¥5+Va=0, or (fo —W)8+V a= 0; 3 the equation of the locus of the centres. Ex. 141. To find the generating lines of a central surface of the second order. Let the equation of the surface be p. Pe + Pp. pi; then, if « be the line from the origin to any point on the surface, and = a line parallel to a generating line, we must have (=A oT for all values of the numerical coefficient wx. — 145 — Hence (®+on). DO (xton) +O(x+on) , (x’ +a") 7 or ot(n. Orton. n)+Qa(«, Pxr+Oorn, «')=—O which must give a,Pnr+Oon, a0, z,Pnr+Oon,7—)0, for @ is assumed to be an indeterminate numerical quantity. The first of the these two equations represents a plane through the origin parallel to the tangent plane at the extremity of «, the second is the equation of the asymptotic cone. The generating lines are therefore parallel to the intersections of these two surfaces, as is well known. From these equations we have c=|] a, 7 where y is a number to be determined. Now, $ and y being any two lines not coplanar with «, we can write nino ey JOE ea Gat | ha! ylt.x+n. Y—=y. 11 +II Y a, 7 a, 7 or y(n. 06 +08, 2) =a] +IT x, Ba = B, Uy (attty ody nln b +i] bite it 9, % Ti which may be put, respectively, under the forms pci lla +yerrll ea? and Teli shame Oo Abas From these equations we have 4 (ab BIT) po LL) a, B 15% St +ylI Syl lH r) OP, Dy Tie fy 8,11 Thae i a, f 1,% a, y,% c=y2]] + hy [eer ore)p—@up+opayy—s(el] +I «)., oS, dy 7; 7,8 19 — 146 — or gr=ytl | yl ata di +II au’) a; OB, Dy Pa, I] 7,8 7,8 B, y rOa+da7r’=0, where «, §, and y being given lines anil +I Mx and AL +1] a are determined numerical quantities. 08, Oy 1 | OR, Ox bs Baettactete Let col] +][ weemell + », (a) OB,Py DR wy B Wiay Psat belonging to the two generating lines. Equation (a) gives ao’ | | EDA a’ =m (« Il +I] a’) Dey Deby BLY By y or « {oJ —mlI t+{el] ld fica) dB, dy Byy OB Ox BLY which being true for every value of 8 and y, must give ol] en Ll ; OB, DY BY from this we haye Il —me|] a OB, Py B,Y and by substitution gr=y2mo| | +yll +4eL] +I] a") &, BLY ba, IT] B, Y BY BLY This, according to the sign of y, gives one or the other generating line, which is real if m is negative, or imaginary if m is positive. Since «, 6, and y are assumed not to be coplanar, Il may be any line whatever, “= B,Y provided it is not perpendicular to «, and we may write for it 6. Substituting the value of y before found, we have — 147 — =e TT +4 (x0 +00’) a; = 4 (2 [Doo] + [007g] a) x — 3 (2D a+ ae!) om \/ = 4 (910, Da+be, [@-16]')«— 3 (a Da + ba a’) bol Oe 416 2m @ x, 0 at == || +\/ — (Art. 78, form. 12), 2m Go, I] ' 2,9 a, p-1 6 or Jee ’ , } , —| = 40 0[(Pa , (-16)'+ 010, Pala—i(a® a+ Do «') d-1 6] = hl ‘ 2m Pa, 9 =! fa (Da , (P16) + o-16, @' a) Da — f(a Pa +6 a’) 6] = vi Da, — [ (a 0 +60’) da — 4 ( P' a + Po «’) O]= al : 2m ® «, 0 =) =|] oe aealil «{] Pa, 0 Pa, 4 Putting -=Il_ Pa, 4 we have —| sro] | + Im” a, Yr with the condition t+. Da+tbda,. 70, for II .V2+Da, lf aoa 4 Pa, 4 Pa, 0 ———r>p phi ggeeee——— CHAPTER VI. ON SOME ADDITIONAL APPLICATIONS. 129. Let usedesionate by f, (i), fa:(t) fa (fiena. + diverse functions of one indeter- minate number, and by f, (t, vu), fo (,u),...... diverse functions of two indeterminate Ber | joe numbers. Let «,, %s, %3...... be given lines; following the nature of the functions Pee PGi or AC OP PRRs Glens « the equation ea. f, (t)+%fe(+...... may represent a right line or a curve, and the equation em a,f, (t+u)t+% fo (ttu)+..... a plane or a surface. The first equation is often written under one of these forms: p= Zaft) and p=%(%); ( the second equation, px tef(tt+u) and p=(t+4u). 130. If P is a point on the curve that p= a f(t) p represents; (0 Pe, es 64 9, deiea a eee and similarly, if Q is any other point on the curve OQ=p.=4,f, (+A +e,f, (-+At)+...... i where A¢ is any number whatever. The line PQ is therefore Po —P, Ap, =a, [fy (+A) —f, Ol +%e [fe (t+ AD —fy (H]+..-.-. pies Ali+s4)—fAl) fo (t+ 4i))~fel) At ; At 1 AtEm qa fo ee ; dja, Gh), tht) and pot at af owe pred Ora We can represent this result by de__, af{t) |, dp__d®(i) ta apd ay, Chard eamemede Let OP be P, —(%), and OP Pg = P(t+Ad), . Let us suppose that é represents time, and At is the interval of time that a point moving along the curve takes to come from P to Q: pg—p, or ®(¢+At)—t, represents the line PQ. It is evident that a or ve can represent the average velocity of a point which passes from P to Q on the line PQ; in the interval of At; it is also evident that when Q approaches nearer and nearer to P, that is to say, when At becomes smaller, the average gens A aie velocity represented by > approaches also more and more to the actual velocity at the ee A ome ‘ s . * A . ti) point P of the point which describes the curve; therefore a or a He. this point when it is at P; that is to say, if the point which describes the curve arrived at the point P ceases to be accelerated in departing from the point P, it will continue is the velocity ot : aah to move on the tangent at the point P with the velocity an; this expression represents not only absolute value of velocity but its direction as well. 131. Example. Let us suppose that a point moves from O to X on the line OX, conformably to this equation «= f(t), % being the space passed over, and i the time; let us also suppose that another point moves along the curve OBC in such a manner that the line which joins this point to the point which describes the line OX, shall have constantly the 0 B, x same direction as OY; we will represent the length of this line by the function F (a). To find the velocity of the point which, comformably to the conditions above laid down describes the curve OBC. The equation of this curve may be written thus, p—xa+F (x) 8, « being the unit of the direction OX; 8 that of the direction OY. Therefore dp dw dF(«u) dz c= oho ee This equation shows that the course travelled over on the line OX being OB,, if we describe the line BM equal to the velocity at the point B,, and the line BN which has the direction OY and the length one) =, B=BM+BN=BL. we shall have the velocity at the point To find the absolute value of this velocity cf we have only to multiply it by e, the product will be the square of its absolute valuexz (Art. 64). Suppose that OY being perpendicular to OX we have F—V;5 Cos9 . t, J 2 —*, tal 9 ———-———— 9)°, and F (x) =a . tan oF cos?0 Then the equation of the curve will be e—=a71+ (x tan 9— ba te ba a?) 4 2 v6 cos® 6 ; OX being the principal direction. — 150 ~ These equations give, ae au.. 9% — <= — f — —-~___ dt ait Kien V_ cos* 5) and ae COS 9. Consequently, aaa cos 9. ¢-+(v, sin9—g. t)j is the velocity at the end of the time é. To have the absolute value of this velocity, let us multiply it by ot cos #9. 4—(v, sin §—g t)y, we shall find de de 2 de ; ie ita Gi) is = [v2 cos? 6+ (v, sin 9 —g t)?|z =[vi—2v, sin. gt+g%t*| 1. It is unnecessary to observe that an example of this nature is given for illustration merely. Q 132. Let us suppose that a curve is represented by p= Sa fii), and that s being the length of this curve measured from some fixed point is t= (s); then p——Zafo(s)) or p= =a F (s} or p==®(s). The last equation gives Ap—=pa—py = (s+4s)—$(s) = PQ, OP being p,; OQ, pg. Thus dp __(s-+4s)— (3) As As i Aep__. F(s+As)—F(s) or Ree? Ms 3 de_-. dF(s)__d ®(s) sta ie ee Fas must be a linear unit in the direction of the tangent at the extremity P of p,, for evi- Ap dently a q A At the proximate point, denoted by s+As this linear unit tangent becomes d(e+Ae)_. dF (s+As)__d&(s+As) Cee OR yee age cae But ®(s+As) is by Taylor’s theorem equal to hence d(p+Ap) ds) d2%®(s) CAE ATI ee ge ge ed ds ds £. ds* ha ds? 1,2 d ® (s) ds ds a a <> Now if we designate the conjugate of tion by its conjugate, we shall have d{p+Ap) d(p'+Ap’) ds ae a Os __a4®(s) d's) d*(s) d ®' (s) — ds Pr ds? As, ds +, ® ® a®(s) d* & (s) a? & (s) a? &' (s) ds ‘Tg Nee ads eft ds% aoc : +. S&H ee AO OD Witenes Ot chy WE RCE Col eee ee eae But we know that ad (p+ Ap) d(e+Ae)_. ds : Gia ae and oi Rae Seo (Art. 64); hence aes d? & (s) Gd smamnd s* 1 eee 30s d? ®(s} d®' (s) dst * > ds by dividing this by As and afterwards by making 4s=0 we shall find d? ® (s) d#'(s)__ d®(s) d? ®'(s) __ cs? ees Ce i oa 2@ ; : : ; 3 : : Hence ss which is a line in the osculating ‘plane ofthe curve, is also perpendicular s ad®(s) de to the tangent dee det d ® (s) 133. If A® is the angle between the successive tangents Ts and d®(s) d* ®(s) ds to Rhee ° AS-i woe ®(s) . ® A§ and if d ao is represented by PP,, ss by QQ,, and lastly if QP,=PP,, a thy = we shall have P.Q,=QQ—QP, __d 0 (s) d ® (s) (As)? Sante se apse Gstaerlace Besides QQ, =QP,=—1; the angle Q,QP, being extremely small we can suppose P,Q,=A9%. Hence —Ap—Nf ee d® @(s) (As)? BP =AI=N[— A Agia Pee es sista be ] Ad. d%&(s) d3@(s) As or aN ds? Ss ds3 ime ope cereals i Therefore Ao eas) Asgard Suaamus® If C is the centre of curvature at P, PQ or As being infinitly small, the triangles Q,QP,, QCP at the limit become similar, and a7/0(s) a Hence N as? —R poe 1 d s* d* ®(s) . : so that the number of 7 38 the reciprocal of the radius of absolute curvature at the point P, to which point s is corresponding. : 2 ; ; ’ 134. We have seen that the line i il is perpendicular to the tangent, and that it d2 ri) s) M s% is in the osculating plane, thus — must be on the same line as R. Then d a d? ® (s) R=R.U 732 Besides d? ® (s) yu?) ne ee Te jot Mt Ds hot 1 Ce ds* . f — 153 — Consequently d? & (s) ds* see d? ® (s) * ds? Thus, if OP=(s) is the line from the origin O to any point P of the curve, and if C d? (s) is the centre of curvature at P, we have PC=R, andOC=OP+PC=®(s)+ a Dy ase is the equation of the locus of the centre of the curvature. 135. Hence also lI is perpendicular to the osculating plane; and the d®(s) d* &(s) Cae ds? unit of this line may be represented by d®(s) ,.d* ®(s) ds ’ ds® then or NI is the tortuosity of the given curve, or the rate of rotation of its osculating plane per unit of length. (Tait. Art. 283). 136. Ex. 1. Let 12 e—=2l+8-,, where ¢ is an indeterminate number, and «, 8 given lines. The curve which this equa- tion represents is evidently a parabola. See Art. 88m. d dt Here Fee tt) as a7 pee a? ¢ adty* and Teh) +8 (-) ; whence, if we assume «8'+6«'=0, dt? ; ee : .) (a a +6 B' t?) 4, , “y= v . : (as) Faxeere 20 — 154 — a Cee, stee and aie +8Bt?) *. ; ‘ TDS BBE ate BB t FO Se ae ae me as* (wo’+BB't2)? ds (a a’ + 6B’ 1)? ; d* o BB t a “e ds? (@+8). Coa ep ey? + aw See JOR aan, BRe 8, a%—e, orn 7 (ee eee eet a i ,d2 9 b% a4 +02 64 72 62 ‘ and N dst (a? 6i2)@ («2 oar (2)3 } 2 62 1 2 od 52 42 hence Roe ee = oe Therefore, for the line to the centre of curvature we have (Art. 97), ae ta Wen oe Eisen 1 OG be te aa, Ba 9 (ot? , xa! PP es Bluey which is the linear equation of the evolute. 137. Ex. 2. To find the curve whose curvature and tortuosity are both constant. Tait. Art. 284 Son aes) 2 We have curvature = N= ry : = ING ees and tortuosity = N +TI == 6}, d ® (s) U d? ® (s) ais ds* : N&f=e ma N&T 9 = tN Gig? ae ds a ve* PP Geb ds’ ds* Let « be a unit line perpendicular to the osculating plane, that is =I] dp ate as.) aise . Hence I] — cbt. dip seaA% ds’ ds* This gives a _ pax de dtp mL =e ap =ou Uae a ai as ds* A Ss 2 a — 155 — Integrating we get l If ty pce, de G 2 6 as ds’ ds* where 8 is a constant line. From this dp dp’ 2 — 4 — Cue oo. i+ Bp +c,(—*p'+8— | : Ut | eke and VIA A Tame a yet then 6%#—c%@+e2, And also I Sill 1 =i aul =II vale ce io aes de Vials me ely vil ia aol lils 6) lis’ds ie ass de d*o ds’ ds? but I] Ea SM aS as" Ghee) 4 is II ; a ds’ : : I or by integrating —— I pies @ e, 2 : ea d where » is a constant line. Eliminating oe Soe iil sales ds? (—[I+ ©), 8 II, 8 o,B ps8 e,8 But I = e%sin%,e, from this lI — 6% sin 6. &, IT, 8 p,8 where 9 is the angle between pe, @ transferred to the same origin; and € is a unit line perpendicular to 8 and being on the plane passing through e, %. Evidently cos 9 1 = — — —— Up. sin 9 sin 9 dp de : We have SELLY: Ae, lin) SL ds ny Ae — 156 — or by integrating ef’ + Be’ ——2¢,si+2ai=—2ph coss.z, where 2a is a constant number; hence eT a ee cos ey: A, re 2 aie l = 56% (— iar ) Ub— 262 U¢ 99 Pp fice Sle ee 2,2 7 J =— 6,56 +ah— 6%; ; dl? cee —f——esp+as—s%—[] 5 Pidas o, B d* 4 or ae ae ee Ot ; ds* <, 6 6 The complete integral of this equation is 3 | p= § cos 8 soy sin. 2s —z5(csp—ab +1] ) (a) ae (0) b $ and 4 being any two constant lines. From this we have Bo’ + pf’ (62 +28) cos. Bs+ (ba +8) sin. Bs —2e,sit+2at or (62' +28") cos. @s+(8 a +n8')sin. 6s=0, which requires that By+eR=0, Ba+nh=0; also we have d . 2B ~f—= #6 sin. Bs+n8 cos. bs— a, ds = = 6 de’ c. 8 and r designate, respectively --(a+«) or —a—« and —(b+6) or —b—8, By formula (2) of Art. 140, et a eb (ee Pec) be Dap—FI | /, a, B ——[Aab+BG@6'+60)+Cba+Dap+Ell iF a, f =— (qa); also (—g)ar=— (qa), and (—q)a(—r=qar. Thus we see that the rule of signs applies to the complex multiplication as well as to the multiplication of lines or numbers. 444. Let q=a+« and r=b+8. We have gar=Aab+Bf'+62)+Cba+Dap+Ell : a, If «=0, then anr=Aab+Dag=a(Ab+D8). And if 8=0 qab=Aah+Cba=b(Aat+Cz), — 160 ~- Alsoysit &— ONG =—=0); anb=Aab. We see that if A—=C—=D=1, we shall have as in ordinary multiplication, anr=ar, Gnb= bq, and aa0 =a by a, b being any two numbers, q and 7» any two complex quantities, 145. Putting a=0 and b=0 in (2) Art, 140, we get anB—Boh +6e) seal (a) ee (Art. 70) which shows what is the result of the complex multiplication i a into 8. We see at once that we cannot write a rvB=—Bae unless E=0. (See Art. 441) Then, with the assumption E=0, GAP Pine —— Be Eee): by which a great many problems may be solved. 146. If « be perpendicular to 8, (a) of Art. 145 will give For B@ 6 +6«)—=0, (Art. 65) The geometrical interpretation of this is as follows: « and $ being any two lines in space perpendicular to each other, if « operates on # and is considered only as an operator, it will bring 8 in the direction of the line EK II and stretch it till it becomes equal to E lI 0,8 a, B This action of bringing 8 in the direction of EI| may be considered as giving to a, 8 8 a positive rotation, if E is negative, or negative rotation, if E is positive, around the axis « through the angle 5 in the plane perpendicular to «. (See Art. 70). 147, Putting I for Boor ll for « in (a) of Art. 145, we get r¥ B, Y a, %, kon tf “* — 161 — and Il resol A : BY Tie se B, Y For AML +I] coe IT evel = (Art. 65). a Y 4,7 B, Y 148. We have as a special case of Art. anh or directly from (a) Art, 145, aa(—f)=—(@n8), (—2)n8=— (a8), and (—«) a(—8) =a nf. 149. We have from (a) Art. 145, . GAG 2B AP; (« +8) A(% +8) = 2B (+8) (e+ 6) = 2B («+8)8; (2 +8) a(@—8) = Bl@+4)—6)+@—He'+e+ELL (« +8), (« —8) =2B (22 82) 22] And in general (+P) a(r+8) = Ble +0 +5) ++9@ +5 +E] (+), (y+) Bey +7rye)+Be8+60e)4+BG6y +76)+BG5+8 6) +al] +e] +EIT «ell am, B,Y a, 8 —aayteanitbayt had. 150. Distributiveness. — Let para, —b-+8, == O-4- vs sd+6, designate any four complex quantities, a, b, c, and d being any numerical quantities; «, B, y, and 3 any lines, We can write [(a+a) + (b+8)]a[(e+y)+(d+9)], =[(a +b) + (x +B)]a[(c+d) + (y+ 9)], =A (a+b) (c+d)+Bl@+$)('+2)+ (y+) @ +69) +C (c+d) («+8)+D (a+b) (y+8)+E II , (Art. 140, formula 2); (EFA (+) — 162 — —KacteBlay +re)+Cee Day HE LL a, +Aad+Bae (o3'4+50)4+Cde+Das+ElL a, 6 +Abct+BGry'+16)+Cc8+Dby+E] | BY + Abd+BG8+56)4+Cd8+Db8+E] T° B, 6 = (a+a)a(e-+7)+(a+2)a(d+3) + (b+ $)A(e+y)+(b+8)a(d+94). Therefore (pt q)a(r+s)=partpastgqart+gqas. Thus we see that the complex multiplication is distributive without depending on the values of A,B, C, D, and E. 151. Let p designate a complex quantity which operating on « will produce 8, that is, Bp ne, Put pc -Fy ethen B—=pae=(c+y) ae, =cane+tyaz, (Art. 150), =Dcea+tyaz, (Art. 144); =DeatBey+10)+ELI Y,% Therefore RE Dice +EI] (Art. 139), and Bay +yo)—0. [Hence « is perpendicular to y and 2 is also perpendicular to y.] Therefore we have af +B oe = 2Dca?, y : j ODT) at (a 6 +8 «’) or po ae aD . 6 being the contained angle of « and 8, Also It =I —rI] a, B «,(De«+E ) «, II Y,% YX % & vet — 163 — —=Ee2y—1E (ay +ye’)e (Art. 78, formula 12) ' Therefore 1 1 Ipe seem eel x, \ Ri 3 a. \ Or p= [qos sino ULL 1 2 aw | p= eres ULL denotes the unit along the line ll a, B a, B Hence we can write 6 seme = [qcosd-# sin. Ull na > =f cose sino. ULL jae fart. 142). D E a, B & In fact / 4 : ; por = [soos + sesino. UL ]ae Ele NIULED gil yo “D . 6 : === ¢o9 0: Sie [- js (Ariss 4g4, 142; 147); oA oa Ul = = cos 0, %+—~ [42 B—4 (a 8'+Ba’)a], (Art. 78, formula 12) . Thus we perceive that Ye { | 3 { | : pe (A 6 Bo’ — poset gesin. ULI or spep elteoteaall : : ss ) operating on « moves it until its Schon eatin na that of ® without altering its length; that is pte ine, ULL ) woe 8, us h: mele ’ 0 being the contained angle of « and 8. — 164 — 152. It is necessary to observe that in aes) etn 0.U Jae D E a, B the action of altering the direction of « may be considered as giving to « a rotation through the angle 4 in the plane perpendicular to I] . The result of this rotation is a 8 evidently independent of the definite values of D and E; and, it is a positive rotation around the axis — II (see Arts. 24, 70). We must bear in mind that if C C, is perpen- a, B dicular to the plane passing through OA, OB; and OA=«, OB=8, In the first figure OC will be along ren , OC, along +I ; and in the a, B a p b second figure OC will be along ee || a ’ and OC, along ==y fi . Then the axis a of positive rotation from « toward # is AL | in the first figure as well as in a, f the second. 153. Let cos 9-++« sin 0 be a complex quantity and « any line; ® being any positive angle and « a unit-line. We have - (cos 9+ sin 6) ne = cos 9Aa+sin 4 (enc), —Decosé. a +sin 0[Blew+ae)+E]] 1, E,% —Dcos.a4+Blue’+eo)gin 6+sin 6, EI| - &, a Which is a complex quantity. If « be perpendicular to «, (cos9+sin 4, ‘)ae—=Doosd.e+Esine. 1] ; £,% which is a line in the plane passing through « and lI , and therefore perpendicular E,% to «. To determine the position of Dcos#.«+Esin?0, Ul we must have the definite £,% values of D and E. + oo — 165 — Let OX and OZ be perpendicular to each other; « 7 along OX and « along OZ, and OY along iT : ee If d and —e be, respectively, the values of D and E, d and e being positive numbers; and if OA=dcos®.«—=«dcos®. Ue, OB=—esin 0 Al —=—esin?#, UII : £, % =, 0 OD will represent the expression Ayah eer ae All gE, % or dacos?. Ue—exsine. ULI : Ey & We see that by making D positive and E negative, « becomes the axis of the positive rotation from « to the line yee ecasenee dl al : 7% Moreover if we make d=e=1 we shall have (cos 9+ sin 9) nx«=cos 0. a+sind. 1] ae =«cos?,Uas+esin®, Ull : a, € then putting QDA=«cos9 and OB=z gin 8, OD=«cos§.U«+e¢sin0. Ul ‘ = (cos 9+ sin 6) na, es and (O D)* = 22 cos? 0 +4? sin? 6a? oD=« Therefore the assumption d=e=1, that is D=1 and E=—1, will give to the complex multiplication this property: that whenever cos§+<«sin® operates on a line perpendicular to the unit-line «, it turns it, without changing the length, through the angle 9 in the positive direction of rotation around the axis « and in the plane perpen- dicular to «. It may also be seen that, with the conditions D=1 and E=—1, cos(— 64) +¢ sin (— 9) or cos 9—« sin § operating on a line perpendicular to the unit-line «, turns it, without — 166 — changing the length, through the angle ® in the negative direction of rotation around the axis « and in the plane perpendicular to «. 154. Let c+y be acomplex quantity, c being any numerical quantity and y any line in’ space. — ia) | . ] To determine two lines « and 8 such-that we can write GAB CY, suecaae We have (Art. 145) ang=Bei'+se+E]] . . a, 6 Therefore the equality anb—=e+y { must give (Art. 139) Bes + pa) c and ELI — 5 a, or, designating by ® the contained angle of « and §, © 2Be6 cos §=c ; . ; (a) and ; + Hep ain oye ne.be Ane = tan ps which gives the angle 9. 7 As to the lines « and 8, obviously they are in the plane perpendicular to y, their contained angle being ® while their lengths are variable and depend upon each other; and (a) gives WS a 155. Associativeness.— We have seen that (Art. 141) the commutative law does not in general hold in complex multiplication. Let us inquire whether the associative law holds. That is, if p, q, 7 be three complex quantities, have we i Pats pal(qnr=(pag)ar. Let peate,. g=b+8, yoect+y. » By Art. 140 we have 2 LA fale paqmAab+B@h+6«)+0be+Das+E LL a, B and (pagvara[Aab+Ba'+62)+0be+Dap+ELL In(e+y),. \ ¢ = z a, ; , =A[Aah+B@h' +62] ¢ . 1 i sae fé if — 167 — ( h +BN[(Cbe+Das+EL] }i+r(coe+Der+ElT \ | , +Cc[Cbe+Das+E ll | a, B +D[Aab+B@b'+62)] 7 +E . [(Cbao+Da8+ET ],v ‘. a, B \ =Atabc+tABG@h+8ee +BCb@y +12) +BDaGy+76)+BEN(LL verti ) a, 8 a8 Motciee Chere Oke LL a, B +ADaby+BDG@6'+62)y +cEoll +DEall +L E% (ey +r eB —2R2 Gy +y6)a, a, Y BY Again by Art. 140. garaAbc+BGr'+16)+Cc8+Dby+EL] . Ber and Pale arleaia ve\a(Aber BOy+76)+Cc8+Dby+E LI) t | By =Atabc+ABGy +76 a +BCc#+$2)+BDb@y+y29+BEN(@1] +I] 2) st Plea Pak +ACbc«a+BCGy +yr6)« +CDach+D%aby+DEal] : PY ieee! +CEcdt peli + 2hE2 Gy ty o)B—LE% (26+ 6e)y, a, a, ‘Now we see that the equality c (pagar=palqar) requires (Art. 139) the two conditions Ac@@+hoy+Cb@r+y2)+Dalby' +78) =AaBy+yb)+Cco@s+60+Db ay +72), (1) and ; C2bcatADaby+BDG@l+6x)7+CEb1] —1E2 Gy +78) ps a, — 168 — —ACbcat+BCGe'+u6)a+D2aby+DEbI] —4H2@6'+6e7. (2) *, % To satisfy the first condition we must have (A—C)c@f+6e)+(D—A)a Gy +y7$)+(C—D)b@y+r“)=—0. But e(uf' +62), aly +yv6) and b(y+y#) may have any values whatever, then this equation cannot be true unless A—C=0, D—A=0 and C—D=0 or (Nims Be Az De -and Ci that is AG 1), only with this condition the equality (1), in general, will exist. And with this condition the equality (2) becomes A2bca+A2aby+AB@b'+6e)y+AEbOL | —2 Be Gy +16). «,Y —A® boat ABGy+yS)a+A2aby+ARbL| —pE2@s+8ey, : af | A or AB@B +B a) y—LE* Gy +yh)e=ABGCY +yh)a—LE%@6 +62) y, or (AB+4E?*) @6' +60) y= (AB+4E*) Gy +y6)«. Since «, B and y are any three lines, this equality will in general exist, only with the condition 2 2 AB+--=0 or pent 2 2 which gives K2% A=— pp: Thus we see that by making =e eel’ A= Os Deana oR we have pagqar=(pagqgar=pa(qar). Also with the same conditions the associative principle holds for any number of com- plex factors. For, obviously, pagaras=[(paq)ar|as =[pa(qar)]as =pal(qar)as] =palqa(ras)] = (paq)a(ras) Plt CLC. — 169 — F, j since we may consider yn as a single complex quantity and the above proof applies { directly. 156. Making in the above article a=0, b=0, and c=0, we get arnkhay=(«a8)ay=aa(bay) which is true with the same conditions. And also anBayani=(«ak)a (ra) =ete., ete. 157. Obviously, with the conditions \ A= C=) and a. the general formula (2) (Art. 140) gars Aab+ Bap +se+OberDas+E] (1) %, where g==a-+«, r==b+8, will have the form qara— 55 ab+ Bas + B a’) — Fa be —aerell (2) and exp nieseee) cul © (See Art. 145) (3) %, 158. If we designate a+« by q, we willdesignate a—« by kq, a being any number and « any line. Let g=—a-+u, r—=b+8, kq=a—z, kr=b—8., By the formula (2) of Art. 157, we have id Ce NE te al sae+E ll , 2B 2B aay Re g aie; oar aes which is a numerical quantity; and Z eee wy 08 — 2B 62 also a numerical quantity. oe Therefore K* kK? (qakq)a(rakr) =[(— x0" —2Bz22]a al 55? 2 BE? | E* pee 2B 28 E? | ES — ap lap? >, From =A+G& and r=b+8, : 22 2 a? + 2B a4] [55 b? +2B 84] (Art. 144) a7b® + H%b 2a? + H%q262 + A B%26], — 170 — we have EK? We : qar=—pp 00+ Bb tha) — [By bat Saas ell 4 and by definition of kq, i? ; , (gar) =— So ab+Bll'+$0)+[F batons ell 3. Therefore Gene han eae ES ab+B@f'+6«)|? 2B 2B Sgr aay we Ss ak, Rese 2BN [aq ba tie ae—El] ] But by Art. a 2 Ne [ag ab x5 e—Ell =a! te greece I )+ pa ab 6629, and by the formula (3) of Art. 78, zen LT il D5 Bab +80) +6 0)? =_2* B22 62) Therefore (qar)ak(qar= 55 es ae a? b2+- H? 69 o24- K* a? 62 +4 B% «2 62], Thus we see that the conditions E2 A=D=—C¥w and A=—x give also (qakq)a(rakr=(qar)ak(qar). 159. We have seen that qakq=kqng=— 55 02—2Be", (Art. 158) where | q—at+e, kqg=a—a, If ine, th 5 oe I= Jpae bee Bethy (Art. 151) and k= ana eftee—sall Therefore bee y ; ; 4 HOODY “eat * 7 haga ee ae 2 Bia wll Putting E=—1, B=--3, and D=1 we have 2 inkg= a vo —171— If gen, | then g=B@6'+54)+EL] , @, kq=Bes+6u)—E]] . Therefore of qnkq=—s Be ee reo) —2pEeN* [I . or putting again E—=—1, B=—}, qakg=2* B?, 160. Before proceeding further in the calculus of complex quantities we shall give a résumé of the foregoing articles on the complex multiplication. 1. We have seen (Art. 150) that in the complex multiplication of g—=a-+« into ry—=b+8, defined by the formula (a+«)an(b+$s)=Aab+Btu e'+60')+Cba+Das+ELL (Art, 140) a, B the distributive law holds independently of the values of the numerical co-efficients A, B, C, D, and E. 2. We have (Art. 142) mqaanr=mn(qar) as in ordinary multiplication. Also the rule of signs holds (Art. 143) independently of the values of A, B, C, D, and E. 3. To be able to write as in ordinary multiplication, anr=ar, qab=b9q, anb=ab, anb=—ab, aab—ba, ank—Baa, we must assume the equality A=CS DSA 4(Art. 144) 4. To retain the commutative law of ordinary multiplication we must make (Art. 144) GD anda b= The condition E=0 will deprive qar of its last term EL . a, B 5. We have seen (Art. 153) that if E has a negative value and D a positive value, ey) the result of the operation of cos§+«gsin® on «, which is perpendicular to the unit-line e, may be considered a positive rotation of « in the plane perpendicular to «, « being the positive axis of the rotation. And, if E=—1 and D=1 this rotation will be through the angle ® and will not alter the length of «. This property is very important in some geometrical interpretations. 6. To retain the associative law of ordinary multiplication the required conditions are A=C=D and A=— Art. 154) ai 2B" 7. Thus it igs seen that to retain the commutative law as well as the associative law 2 we must first assume E=0; 2', a= will give, if B be a definite number, A=0; 3’, A=C=D will give C=D=0. Therefore in this case the general formula will be reduced to qar=—B@h+6e) which obviously does not depend at all on the number parts of g and +. If in this we make a=0, D=0; arnB—BC@P’ +6 «')—=— 248 B cos 0, which is the 2B multiple of the product «, 6, and the cosine of the contained angle of a and 8. We will also have in this case ang—0, Baa=0, anb=0, Crt 2 Bee, aabay—)d, and taj—0; ¢ and jy being mutually perpendicular lines. This forms a system* of calculus which, if applied properly, may be used in the solu- tion of a great many problems. 8. Finally, by making K2 9B? A, B, C, D, and E being definite numbers, we retain the distributive and associative principles and lose the commutative principle. In this case the general formula is A=C=D and A= — 2 2 72 (+e) ab+8)=—seab+Beb+6e)—E po Papel , a which will give anb=Bab+buy+EL] , a, p *Commutative System. —173 — ‘ ~ Ge E2 E2 K2 ) an(b+8) =— 55 4b— a eh=— 54 (b +8), \ E2 (a+a)nb= op O(a +4), E2 anp= opt? 2 a b=— 5p ba, \ _# anb=—p,ab, AA%—=—2Bae, 2g ¢ (a-+4) a(a+a)=—Z, t+ 2Bat— a, hh? - A (a+2) a (a— 2) =—5e a —2B«9; and inj=El] =-Eh, i,j FAb=EL —— Ei, j,k kai=Ell =— Ky, k,@ tajak=liaj)ank=—Ekak =—E (kak) t, J, and k being a system of three mutually perpendicular unit-lines drawn from one point and situated so that a po- sitive rotation (see Art. 152) through a right angle about 7 as an axis brings 7 to coincide with k; that is ae fy lI =1, and II Ani: kyg a j This forms a new, complete, and general system of calculus. eee St eee — 174 — Quaternions. 161. By making A>C=D=1, E=—1, and therefore B=—4%, we do not break the law of consistency at all, while much simplification and great advantages in geo- metrical interpretations (Art. 153) are gained in this new system of calculus. Therefore in the rest of this investigation we will make A=C=D=1, B=—4, and E=—1. Then the general formula of Art. 140 will be reduced to the form (a+e)n(b+i)—ab—4 Oe +80)+he+08—[1 ) a, 8 which gives — 1. Pes ee sell a, B 22, anf=baa=a8. a. anb=ab. Nh, a A% = — 22, 5. «nlf =—II a, f a, II a, B 6. bf pate a) > [For aL | +I] dy 1} a, iim a, 8 a, B a8 fk anh+Bawm—lC«i'+6e’), 8, Py vat yen AE a, 8 And if « be perpendicular to §, 9, yee ai El e a, B For («' + 8a’) =0. We have also iaj=k, Ink, KAt=—), tajak=(inj)ank=kak=—1, [hey ia Fs ti} or tajakz=ia(jak)=iniz=—I|; t, j, and k being the same unit-lines as in 8 of Art. 160. 162. This particular case also forms a complete system of calculus called Quaternions by Sir William Rowan Hamilton who discovered it. 163. The following quotation (Notes by Prof. A. S. Hardy to Argand’s book) from a letter of Hamilton on the discovery of the symbols 4, 7, and k, assumed to have the above properties, may be read with interest: —175 — “Oct. 15, 58: P. S.— To morrow will be the fifteenth birthday of the Quaternions. They started into life, or light, full grown, on the 16™ of Oct., 1843, as I was walking with Lady Hamilton to Dublin, and came up to Brougham Bridge, which my boys have since called Quaternion Bridge. That is to say, I then and there felt the galvanic circuit of thought to close; and the sparks which fell from it were the fundamental equations between 1, J, k; exactly swch as I have used them ever since. I pulled out, on the spot, a pocket-book, which still exists, and made an entry, on which, at the very moment, I felt that it might be worth my while to expend the labor of at least ten (or it might be fifteen) years to come. But then, it is fair to say that this was because I felt a problem to have been at that moment solved —an intellectual want relieved — which had haunted me for at least fifteen years before. Less than an hour elapsed before I had asked and obtained leave of the Council of the Royal Irish Academy, of which society I was at that time president, to read, at the next general meeting, a paper on Quaternions, which I accordingly did on Nov. 13, 1843.” 164. Before proceeding further we will give, for the sake of illustration a few simple examples. Ex. 1. Let paatysb+sy, «, 8, and y being three unit-lines perpendicular to each other. To find the lenght of p. By Linear Algebra. We have op = (watyP+sy) (we +yi+zy), p* =? + 2 + 2%, By Commutative System (’*). We have pnp (catyB+sy)a(vatyh+zy) =a anaty*bab+sz2 yay +Qry%t*aB+azanytyshay. But Sap 2 Dee OA 2 Bo 2 Ree 6p 2b 6228; Tate b iby) andea nl) aay 00k ay); therefore 2B6%*=2Bae2+2By2+2B2%, or fee Ua By Quaternions. From p—watyb+zy we get enp(matyiP+sy)a(aatyi+zy) *Arts 141 and 145. — 176 — = a2anatyBab+styay +ay (anbB+haa)+az(anytyane) +y2(Baytyas) But BN PS Gn ee eh ee ee Yay=— Vs; and, anb+Pac—O0; anytya«cO0; Baytyabk—0. Therefore e% =a + y* + 3%. Ex, 2. On the sides AB and AC of a triangle are constructed any two parallelograms ABDE and ACFG: the sides DE and FG are produced to meet in H. Prove that the ° sum of the areas of the parallelograms ABDE and ACFG is equal to the area of the parallelogram whose adjacent sides are respectively equal and parallel to BC and AH. Let AB=«, AC=§, AH=e; then BO=b—a; , hH=ttee, and AG==e+y8; aw and y indicating } two numbers. Let the parallelograms AEDB and IN ha ACFG be respectively designated by A and B. D By Linear Algebra. The formula of Art. 660 gives @oo-o ee eae ewe ba —t At= 5) [(e+- a %) a’ —«a (e'-+ @ @’)| = 5) (ea'—ae'), and Bix Vp +yt)—le+ ys) = ee therefore (A-+B)i VS few ae pe eB] whence the proposition. If one of the parallelograms, as AF,, be interior, it is easy to see that the area of the parallelogram on AH, and BC is the area of AD minus the area of AF,; and, the sum of the areas of AD and AF, is equal to the area of the parallelogram on AH, and AB+AC. The required solution may be exhibited also in the following form. That is es til —[{ =—I] , (arts. 69, 70, 81, and 78 form. 8) ee a,c and be - B(e+y8) Be j f —11T— Therefore X a+Besll.+T =H... Bye a, € (@—a),« which is-the proposition. Note. The solution by the Commutative system or by Quater nions, is the same as the above. Ex. 3. In any quadrilateral, the product of the two diagonals and the cosine of their contained angle is equal to the sum or difference of the two corresponding wane for the pairs of opposite sides. Let OABC be the given quadilateral ONY AB, and BC «, 8, and y, jesvecuren Then OC=a+ 8-+y, OB=-«+8, VAC=6 + By Linear Algebra. We have (Art. 65) | (+8) (8+ 7’) + (B+ 7) (2+ 8) = (@ B+ Ba’) + 2B Bi + (a y+ ya’) + (B y+ 78) | —= (a B+ BB+ 7B) + (3 a'+ BB+ By’) + (a y+ ya) = (%+ B+ 7) B+ 8 (a+ B+ 7’) + (27+ 7), which gives thé proposition. . By Commutative System. We can write . ! (+8) A(B+y)anBt+Baktranyt fay =aenb+hab+ynbtany (Art. 145) = («+B +y)Ab+any. From this we have (Art. 145) ) (2 +8) (By) + (B+ y) (2+) = (2 + 8+ y) BB (et B+ y') + (a y+ yo); whence the proposition. i By Quaternions. rf From (a +B)A(B+Yy)San8+8AB+any+ Bay aes UN [(a-+8)(8-+7) + (+7) (e'+8)] +11 ie i B), (8.+ ’) =<. ores] weer sete TD Meoeerga lit. Pa Therefore (Art. 139) (x +8) (8+ 7’) + (B-+ 7) (2+ 8) = («B48 x’) + 2B B+ (x y+ yx) +8 ee ie — 178 — = (« B+ BB+ 8) + Ba'+ BB +B y') + (a y+ 7a’) = («+B +y) 848 (e+ B+ 7) + (27+ 72’), the genes Uy Ex. 4. Let in a tetrahedron an edge be equal to the edge opposite to it. Prove that the lines which join the points of bisection of the other opposite edges are at right angles to each other. Let OABC be the given tetrahedron, the length of the edge AB being equal to the letgth of the--edge OC; and OA=¢,: OB=6, and OC=y: ‘therefore A B— Poa, AC=1—2, ann BC=y—8. . The assumed equality AB=OC wives y= 8—« or y?=($—4)2, By Linear Algebra. From the assumed equality y*=(%—2)? we have 17 =(6— 2) P—2@) or Qyy— Qa a'+ 2a B'+ WBa’—2868'—0, Adding to it By'+ yB—(By'+- 78) +2 y+ ye'— («y+ ye) —0 will give (y+ B—«#) yor (y+ B — 8) ee oe © 7 (y+ B— a’) + (7+ B'— a’) — 8 (y+ B’— a’) = 0; (YB — a) (y'+ e'— 8) + (y +a —§) (7+ B'— «') = 0, which gives the proposition. For the line joining the bisection points of the edges OA and BC is + (8-0) += 4 6+1—2); and the line joining the bisection point of OB to that of AC is =4(«+y—§). By Commutative System. From the equality y*=(%—+)* we have TA1T= (P—4#) a (Be) =Bab—anB—Baatana =BaAb—2Zanhtane, or YAY ane +2Qanh—PAS—D, Therefore adding Bay—yab+ya%z—any=0 (Art. 145) will give - (y+B—a)ny+(y+8—2) Ax — (y +8 —2) n8=0 or (y-+@—«)a(y+«—8)=—0, The proposition follows from this, be ep — 179 — By Quaternions. The equality y?—=(&—*)* gives Yay = (P—4) a (b—a) or TAY 4anetBae+rangB—Basb=. Adding to both sides Bary—yra8—4any+rn.e, Bivesee P=“) ay + (y—-2 +f) a2 — (y—4 +8) AB Bay—yaba—eanytins or (y+ B—a) A(y+%—$)=Bay—yah—aartyas, from which we have AN [fy-+6—») (y'+e'—8) + (r+a—9) (+ 6—«)] +11 | (y+8—«),(y +2 — 8) =167'+70+1[ —r1y¢+6n—Il[ —16y+re)— |] +ieerep+Ll ? Pat sae Cs 1, % Therefore (Art. 1389) N[(y+ 8 —e) (+ @—8) + (y+ —8) (+ B'—@/)] =0, which is the proposition. Cor. The above proposition may show that if in the given tetrahedron each edge is equal to the edge opposite to it, the lines which join the points of bisection of opposite edges are at right angles to each other. It is also easy to prove that in the above mentioned case the lines which join the points of bisection of opposite edges are at right angles to those edges, N. B. The solutions by Quaternions may be shortened considerably by using Hamil- ton’s notations of abridgment which will be given hereafter. Some analogous simplification may be easily introduced in the Linear Algebra also. 165. Let p and q denote, respectively, the complex quantities a+«, b+68, where a and b are any two numerical quantities; « and ® any two lines. We have seen (Art. 161) that pag will be, in general, also a complex quantity of which the number part is ab—4(@6'+6«') and the line part be+a8—J]] . a, B Hamilton calls Quaternion any complex quantity; scalar the number part and vector the line part of it. Aud, he writes q=Sq+Vq, where S is read scalar of, V vector of, q being any complex quantity or quaternion. Using the same notations we may write paq=S(paq)+V(paq), where, if p=a+« and qa bs, S(paq) =ab—1@b' +62), ’ Vipag=be+as—I] : — 180 — Also x AB=S(2n8)+ V («A8), Sy gS aed Yeates sd 3 . and V (anf) =—IT area 2 6 sin 0, ( where 9 is the contained angle of « and : and <« a unit-line along {I : =, 166. Hamilton denotes the multiplication of quaternions as in the ordinary multipli- cation and he writes | p.q or pq, S.pq, V-pq, «8, #%, SaB, and Vas I instead of / peg Sipod, Veg: tat, @ oa ola t) aaa a In the following. pages we will continue to use the sign «a as before. TOC aL las 100 be carefully noticed that the definitions of the symbols S and a red the equalities S(Vq)=0, V(Sq)=0; and S(S¢)=S49,, Viva For the number part of Sq cannot be other than Sq itself and the line part of Vq must similarly be Vq itself. we Therefore S[S@ns]=S (ns), S[S(anBayJ=Slenboy), V (x 8), y V (a ABAY). VIV (eas) = | | VV (eabar)]= 168. From- | anp—pas'+bay—]] a, B and | Bax=—hoou+ae—]] ’ scl B, o we have 1 S (ung) =8 Baa), V(zn8)=—V (Saez) or V(anf)+V (bax)=0. ww ar8+-Bra=2S$ (a8), %AB—BRAa=2V (zn8), & co ee Je (2 +8) A(2+6)manatanb+hnat+ Bas -m — 42 +258 (x \8) — 6, b. (2—8) n(x —f) = — 42928 (« nB) —62, | — 181 — [S (#48) + V(en8)] [8S Bae) +V bas)] [S (#8) + V (2 n8)] o[S (2 nf) — V (2 98)] [S | HT (« v8)]? —V(anh)av (x \8) S (x nf)]? + [Vann], which is the formula (3) [Art. 78]. When « and 8 are at right angles to each other: 8. S (« v8) =0. \ _ And «, § being any lines, a any number: 9. Sa —¢, 410. Sa Oe tae Vr ar, 42 S(a+e)—=a 13 V(a+e)=e 4A. San il\= | | a, B M5. S[aaV(ans)|= 16, call Sie lB mare jog eal } a, B a, B a, B a, B 17. . anV («ab)=Vl«eanV («8)]. 18. slf = 19. vil al 169. Obviously the symbolic expressions — $(@6'+6) and —|J of Linear Algebra a, f designate, respectively, the same things as the expressions S (« 78) and V (zaB), or SaB and Va, as wrilten by Hamilton. Therefore we can write: >, ae ViVentaveny=—ll _. Ti Ut a 8 5? BLY «, IT BAY De vill ayn=—II —|][ . a, any y, II — 182 — 4. snl I ye NT UL ie Uk a’. 7 BY By or sl Seat LL +f] eb B, 7 B,Y py 170. Obviously we write S (pq) =Sp8q; V (peg) — ¥ == Ver; p and q being any two quaternions or complex quantities. We can also write V(page=V (par) =V[(pag#lp.o7)] =V[pa(g=r)], and S(pag=S (pan=Slpaqg=(par)] =S8 [pa (qr). For example: V (@ v8) +V (any)=V[«anl(B+y)], V (x ABAY) +V («A8) = Vl[aan(bay+ )]; S(«ak) +S (eazy) =Slzean(b+y)], S (7 AB AY) +538 (2ns3=S [zn (BAY +8)]. (71. If gq and » be two quaternions, that is two complex quantities, we may write dS aa PSD IeE NV a Therefore FATSHSGSST SOE Voto. Voy Gia —=S8q.Sr4+8q.Vr4+8r.Vqg+S(VqaVr)+V(VqaVr), and PAGS TSG FO. Va aS Oey Ty =Sr.8q4+S8r.Vq4t+8q.Vr+S(VraVg)+V(VraVq) . Hence S(gar=Sq.Sr+SVgave), | V(qar)=Sr.Vq4t+Sq.Vr+V(VqaVvn); and S(raqg)=Sr.Sq+S(Vravg), V(raqgv=Sr.Vq4+Sq.Vr+V(Vravq). By these we get S (q an r) =e (r nan 9q)3 V(qar)+V (raq)=2(S¢q ~Vr-esr. Vq). 172, We can write tell «IL «) (arts. 163, 169). But “ll ell vail wil for +I a A 78, Form. 9), B, a,b «, Therefore Ji anes 4 eae Ae This shows that a change of order amongst three lines produces no change in the number part of their complex product, provided the cyclical order remains unchanged. The same formula of Art. 78 shows that S(aanbay)=—S («anyn$)=—S bascay), that isa cyclical change of order amongst three lines changes the sign of the number part of their complex product. 173. We have V(znbary=V(ya8as). For V(aanhary=Vl[enl—E © metal TE ese ) (Art. 170) BLY E foyer i+ Lh) (Art 169) “, II Boy —=— 1G +yH)athlay+ye)B—LEb+he)y =— pb Hbay + ger tye)p— py +16 a cote A Mina 2 i II (Art. 78, Form. 12) ee Be =—426'+50y—vyall ) cart. 169) B, @ =Viya}—1@¢+62)—]] B, =Vlirabae)] =V(rabaz), — 184 — 174. We can write V («n8Bay) + V(yaen$)e— eb +8e)y = 27S (a8). For | Vientan= Vil ,erse)— pL Jay] : a, 8 =—1 6+ 62')y—V a spt and Viynens)=V (yal §@b +62) 2-H J Ls chs home e; V (Anbar tV (tnans)=— Ge + bay —=27S(«n8) (Art. 165). 175. The formula (12) of Art. 78 may be exhibited under the form V[zav Gar]=r1S (108) —BS (any); -and the formula (16), Vea (Bay) + VB AV (i ne)] + V[rAV (@n8)]=0 or V [av (BAY) +BAV (74%) HYAV (« A8)] 0, (Art. 167). 176. We have V(enbary)=2%S bar)—BS (zany) +78 (xn$). For VEnbary=V[zalj—s@6r +76) eet ) = joysrnes I a, TI B,Y —=—4by'+y Sat sy tre)B— £6 thay = 28 (3A7)—BS(zar) +78 (28). (77. The formule (17), (18), (19), (20), and (21) of Art. 78 may be exhibited, respec tively, under the following forms (see Arts. 169, 172): 4. VV @AB)aAV Gary)J]=—BS («abay) —=— BS (yneaf) | ee x VV (@n8) AV (y.08)]=2S (an 8Ba1)—1S@ahad) : sr seen ase aa) 3, 5S (anBary)=eS Bara’) +8S (taxa) +718 (anbad). A. 8S («abar)=V(eank) . SAd+V (BAY). S(zan8)+V (yane) .S (BA). \ ( Ashe S[V (2B) AV (7¥A3)]=S (eas). SBAY)—S (ear). SBA). 178. We can write Yor S(«abaynd)/=S («af) .S(yn8)—S (zn1) .S(Ba2) +8 («n’) . Say). S(2nbarat)=S[|S (zabar+VEnbar) {ae | =S[]V (#ABA8)|A8], (Art. 167) =S[]«S Bay) —8S (zany) +7S («a8){ a4], (Art. 176), =S(#n?).SGay)—S (baa). Seay) +S8(142). S@ a8). 479. We can also write For S(anbarat)=SCnanso). S(enBarnt)=S[|S(zabay)+V abaya SP. S(anbavj]+S[V(enbarat =S[F.S(en8ay)]+SBAV nba], (Art. 168), S[A|S(anBar)+V @abay){], (Art. 170), SEA(eafay)], (Art. 165), S(3AaABay), (Art. 156). ———_ > P Pt 34.4. ¢ ¢ —__—_- lated “ Ass eee 2 sage ene PP?