Irving Strin^han 
 
 *th. 
 
-4^ .C 
 
 f 
 
NOTES 
 
 ON 
 
 ELEMENTS 
 
 OF 
 
 (ANALYTICAL) 
 
 SOLID GEOMETRY 
 
 BY 
 
 CHAS. S. VENABLE, LL.D., 
 
 PROFESSOR OF MATHEMATICS, UNIVE-RSITY OF VIRGINIA. 
 
 NEW YORK: 
 UNIVERSITY PUBLISHING COMPANY. 
 
 1891. 
 
Math. 0ot. 
 
 IN preparing these Notes I have used the treatises of Gregory, 
 Hymers, Salmon, Frost and Wostenholme, Bourdon, Sonnet et 
 Frontera, Joachimsthal-Hesse, and Fort und Schlomilch. 
 
 C. S. V. 
 
 V 
 
 COPYRIGHT, BY 
 
 UNIVERSITY PUBLISHING COMPANY 
 1879. 
 
NOTES ON SOLID GEOMETRY. 
 
 CHAPTER I. 
 
 1. WE have seen how the position of a point in a plane with ref 
 erence to a given origin O is determined by means of its distances 
 from two axes (Xr, Oy meeting in O. In space, as there are three 
 dimensions, we must add a third axis Oz. So that each pair of axes 
 determines a plane, CXr and Oy determining the plane xOy ; O.v 
 and O2 the plane xOz ; Oy and Oz the plane yOz. And the posi 
 tion of the point P with reference to the origin O is determined by 
 its distances PM, PN, PR from the zOy, zOx, xOy respectively, these 
 distances being measured on lines parallel to the axes CXv, Oy and 
 Oz respectively. This system of coordinates in space is called The 
 System of Triplanar Coordinates, and the transition to it from the 
 System of Rectilinear Plane Coordinates is very easy. We can best 
 conceive of these three coordinates of P by conceiving O as the 
 corner of a parallelopipedon of which OA, OB, OC are the edges, 
 and the point P is the opposite corner, so that OP is one diagonal of 
 the parallelopipedon. 
 
 2. If PM = OA = a, PN = OB = b, PR = OC = c, the equations 
 of the point P are x = a, y = b, z = c, and the point given by these 
 equations may be found by the following construction : Measure on 
 OX the distance OA = a, and through A draw the plane PNAR 
 parallel to the plane yOz. Measure on Oy the distance OB = 3, 
 and draw the plane PMBR parallel to xOz, and finally lay off OC 
 c and dnuv the plane PMCN parallel to xOy. The intersection 
 of these three planes is the point P required. (Fig. i.)* 
 
 3. The three axes Or, Oy, Oz are called the axes of x } y, and z 
 respectively ; the three planes xQy, xOz, and yOz are called the 
 
 *. 
 
 * For Figures see Plates I. and II. at end of book. 
 
 814023 
 
4 NOTES ON SOLID GEOMETRY. 
 
 planes xy, xz and yz respectively. The point whose equations are 
 x a, x = b, x = c is called the point (a, b, c). 
 
 4. The coordinate planes produced indefinitely form eight solid 
 angles about the point O. As in plane coordinates the axes Ox 
 and O> .cUvicieUtte plane considered into four compartments, so in 
 space coordinates the planes xy. xz and yz divide the space con- 
 slcjttfed ia.;o .eight Compartments four above the plane xy, viz. : 
 Q-xyz, Q-xy z, Q-x y z, Q-x yz ; and four below it, viz. : Q-xyz , 
 Q-xy z, O-x f y f z , Q-x yz . By an easy extension of the rule of 
 signs laid down in Plane Coordinate Geometry, we regard all x s 
 on the right of the plane yz as + and on the left of yz as ; ally s 
 in front of the plane xz as -f- and those behind it as ; all z s above 
 the plane xy as + and those below it as . We can then write the 
 points whose distances from the coordinate planes are a, b and c in 
 the eight different angles thus : 
 
 In the first Octant, Q-xyz P l is (a, b, c) 
 
 In the second Octant, P 2 is (a, b, c) 
 
 In the third Octant, P 3 is ( a, b, c] 
 
 In the fourth Octant, P 4 is ( a, b, c) 
 
 In the fifth Octant, P 5 is (a, b, c) 
 
 In the sixth Octant, P 6 is (a, b, c] 
 
 In the seventh Octant, I\ is ( a, b, c) 
 
 In the eighth Octant, P 8 is ( a, b, c). 
 
 The signs thus tell us in which compartment the point falls, 
 and the lengths of a, b and c give us its position in these compart 
 ments. 
 
 1. Construct the points I, 2, 3 ; o, i, 2 ; 0,0, i ; 4, o, 3. 
 
 2. Construct the points i, 3, 4 ; 2, 3, o ; 3, o, i ; 2, o, o. 
 
 5. The points M, N and R are called the projections of P on the 
 three coordinate planes, and when the axes are rectangular they are 
 its orthogonal projections. We will treat mainly of orthogonal pro 
 jections. For shortness sake when we speak simply of projections, 
 we are to be understood to mean orthogonal projections, unless we 
 state the contrary. 
 
 We will give now some other properties of orthogonal projections 
 which will be of use to us. 
 
NOTES ON SOLID GEOMETRY. 5 
 
 6 DEFINITIONS. 
 
 The projection of a line on a plane is the line containing the 
 projections of its points on the plane. 
 
 When one line or several lines connected together enclose a plane 
 area, the area enclosed by the projection of the lines is called the 
 projection of the first area. 
 
 The idea of projection may be in the case of the straight line thus 
 extended: if from the extremities of any limited straight line we draw 
 perpendiculars to a second line, the portion of the latter intercepted 
 between the feet of the perpendiculars is called the projection of the 
 limited line on the second line. 
 
 From this we see that OA, OB and OC (coordinates rectangular) 
 are the projections of OP on the three axes, or the rectangular coordi 
 nates of a point are the projections of its distance from the origin on the 
 coordinate axes. 
 
 7. FUNDAMENTAL THEOREMS. 
 
 I. The length of the projection of a finite right line on any plane is 
 equal to the line multiplied by the cosine of the angle which it makes with 
 tht, plane. 
 
 Let PQ be the given finite straight line, xOy the plane of pro 
 jection ; draw PM, QN perpendicular to it ; then MN is the projec 
 tion of PQ on the plane. Now the angle made by PQ with the plane 
 is the angle made by PQ with MN. Through Q draw QR parallel 
 to MN meeting PM in R, then QR = MN, and the angle PQR 
 = the angle made by PQ with MN. Now MN = QR = PQ cos 
 PQR. (Fig. 2.) 
 
 II. The projection on any plane of any bounded plane area is equal to 
 thut area multiplied by the cosine of the angle between the planes. 
 
 i. We shall begin with a triangle of which one side BC is parallel 
 to the plane of projection. The area of ABC = - BC x AD, and the 
 
 area of the projection A B C = - B C x A D. But B C = BC and 
 
 A D = AD ccs ADM. Moreover ADM = the angle between the 
 planes. Hence A B C= ABC x cos angle between the planes. (Fig. 3.) 
 
 2. Next take a triangle ABC of which no one of the sides is pa 
 rallel to the plane of projection. (Fig. 4.) 
 i* 
 
6 NOTES ON SOLID GEOMETRY. 
 
 Through the corner C of the triangle draw CD parallel to the 
 plane of projection meeting AB in D. Now if we call 6 the angle 
 between the planes, then from i A C D = ACD cos 6 and B C D 
 = BCD cos 0. . . A B D -B C D =(ABD - BCD) cos or A B C 
 = ABC cos d. 
 
 3. Since every polygon may be divided up into a number of 
 triangles of each of which the proposition is true it is true also of 
 the polygon, i. e., of the sum of the triangles. 
 
 Also by the theory of limits, curvilinear areas being the limits of 
 polygonal areas, the proposition is also true of these. 
 
 8. The projection of a finite right line upon another right line is 
 equal to the first line multiplied by the cosine of the angle between the 
 lines. 
 
 Let PQ be the given line and MN its projection on the line CXv, by 
 means of the perpendiculars PM and QN. Through Q draw QR 
 parallel to MN and equal to it. Then PQR is the angle made by 
 PQ with Ox, and MN = QR = PQ cos PQR. (Fig. 5.) 
 
 9. If there be three points P, P , P" joined by the right lines PP , 
 PP" and P P", the projections of PP" on any line will be equal to 
 the sum of the projections of PP and P P" on that line. Let D, D , 
 D" be the projections of the points P, P , P" on the line AB. 
 Then D will either lie between D and D" or D" between D and 
 D . In the one case DD" = DD + D D" and in the other DD"= 
 DD - D"D = in both cases the algebraic sum of DD and D D". 
 The projection is -f or according as the cosine of the angle above 
 is -f or . 
 
 In general if there be any number of points P, P , P", etc., the pro 
 jection of PP" on any line is equal to the sum of the projections of 
 PP , P P", etc., or, the projection of any one side of a closed po 
 lygonal line on a straight line is equal to the sum of the projections 
 of the other sides on that line. 
 
 10. USEFUL PARTICULAR CASE. 
 
 The projection of th? radius vector OP of a point P on any line is 
 equal the sum of the projections on that line of tlie coordinates OM, MN, 
 NP of the point P. For OMNP is a closed broken line, and 
 the projection of the side OP on a straight line must be equal to 
 the sum of the projections of the sides OM, MN, and NP on that 
 line. 
 
AZOTES OA> SOLID GEOMETRY. 7 
 
 11. DISTANCE BETWEEN Two POINTS. 
 
 Let P and Q, whose rectangular coordinates are (x,y, z] and (x , 
 y, z ), be the two points. (Fig. 6.) 
 
 We have from the right parallelopipedon PMNRQ of which PQ 
 is the diagonal, PQ 2 = PM 2 + MN 2 + QN 2 . But PM = x - x, MN 
 
 yy y NQ= z z. 
 
 Hence PQ 2 = (x - xj+ (y -y Y + (* - z )\ 
 If one of the points P be at the origin then x o, y = o, 2=0, 
 and PQ 2 = * 2 + / 2 +z" 2 . 
 
 12. TO FIND THE RELATIONS BETWEEN THE COSINES OF THE ANGLES 
 
 WHICH A STRAIGHT LINE MAKES WITH THREE RECTANGULAR 
 AXES. 
 
 Take the line OP through the origin. Let OP r, the angle 
 POjt: = of, POy = /3, POz y, and x { , y , z the coordinates of P. 
 
 Then by Art. 1 1, r* = ,v 2 +y 2 + z -. 
 
 But, Art. 8, x = r cos a ; y = r cos ft ; z = r cos y. 
 
 Hence r 1 = r 2 (cos 2 a -f cos 2 ft -\- cos 2 ^) or 
 
 cos 2 <? + cos 2 ft + cos 2 y = i. (i) A very im 
 portant relation. 
 
 Cos #, cos /?, cos y determine the direction of the line in rectan 
 gular coordinates, and are hence called the direction cosines of the line. 
 We usually call these cosines /, m and respectively. So the equa 
 tion (i) is usually written P + m^ + n? = i, (i), and when we wish to 
 speak of a line with reference to its direction, we may call it the line 
 (/, m, ii). Only two of the angles a, ft, y can be assumed at pleas 
 ure, for the third, y, will be given by the equation 
 
 cos y = A/ 1 cos 2 a cos 2 ft. 
 
 13. We can use these direction cosines also for determining the 
 position of any plane area with reference to three rectangular coordi 
 nate planes. For since any two planes make with each other the 
 same angle which is made by two lines perpendicular to them respec 
 tively, the angles made by a plane with the rectangular coordinate 
 planes are the angles made by a perpendicular to the plane with the 
 coordinate axes respectively. Thus if OP be the perpendicular to a 
 plane, the angle made by the plane with the plane xy is the angle y ; 
 with xz is the angle ft ; and with yz is the angle a. So cos a, cos 
 ft, cos y, are called also the direction cosines of a plane. That is, the 
 
g iVOTES ON SOLID GEOMETRY. 
 
 direction cosines of a plane with reference to rectangular coordinates are 
 the direction cosines of a line perpendicular to this plane. 
 
 14. The relation cos 2 a + cos 2 ft+ cos 2 y = i enables us to prove 
 an important property of the orthogonal projections of plane areas. 
 For let A be any plane area, and A x , A y> A z its projections on the 
 coordinate planes yz, zx and ;ry respectively. Then Art. 7, II., A z 
 = A cos a ; A y = A cos ft ; A a = A cos y. 
 
 Squaring and adding we have 
 
 A Z 2 + A,, 2 + A. 9 = A 2 (cos 2 a + cos 2 ft -f cos 2 y) 
 
 or A/+ A/ + A 3 9 = A 2 . 
 
 That is, the square of any plane area is equal to the sum of the squares 
 of its projections on three planes at right angles to each other. 
 
 15. To FIND THE COSINE OF THE ANGLES BETWEEN Two LlNES IN 
 
 TERMS OF THEIR DIRECTION COSINES (cos <*, cos ft, cos y) 
 AND (cos a , cos ft , cos y ). 
 
 Draw OP, OQ through the origin parallel respectively to the given 
 lines. They will have the same direction cosines as the given lines, 
 and the angle POQ will be the angle between the given lines. (Fig. 7.) 
 Let POQ = 6, OP = r, OQ r , coordinates of P(x,y, z\ coor 
 dinates of Q (xy z }. 
 Now by Art. n, 
 
 PQ* = ( x -x>y + (y -yj + (Z - Zj = X* + f + Z* + X* + / 
 
 2}>/ + 2ZZ }. 
 
 And from triangle POQ, 
 
 PQ 2 = r 2 -f r * 2rr cos 0, 
 
 hence r 9 + r 2 - 2rr cos B = x* +/+ z* + x* +/ 2 + z* 
 
 -f 2yy + 2zz ). 
 
 But r*= .r 2 +j 2 + ^ 2 and r 2 = ,r 2 + y a + ^ 2 . 
 
 Therefore rr ; cos 6 = xx + yy + zz , 
 
 a x x y y z z 
 
 or cos B = .+.--+-.. 
 
 r r r r r r 
 
 Hence cos 6 = cos cos <* + cos ^cos ft + cos y cos y (i) 
 which we write cos B // + ww + # . (2) 
 
NOTES ON SOLID GEOMETRY. g 
 
 Cor. i. If the lines are perpendicular to each other cos 6 = o or 
 // + mm + nri = o (3). (3) is called the condition of perpendicu 
 larity of the two lines (/, m, ), (/ , m , ). 
 
 Cor. 2. From expression for cos 6 we find a convenient one 
 for sin 2 6. 
 
 Thus sin 2 0=1- (// + mm + nn Y = (/ 2 + m*+ n*} (l z +m* 
 
 + n 2 ) (// + mm + nn Y 
 whence sin 2 6 = (Irti I m}*-}- (ln r / ) 2 + (mn m n)\ (4) 
 
 1 6. To express the distance between two points in terms of their 
 oblique coordinates. 
 
 Let P (xyz) and Q (x y z ) be the two points. (Fig. 8.) 
 The parallelopipedon MPQN is oblique. Let the angle xOy 
 = A, xOz = //, yOz = v, and the angles made by PQ with the 
 axes respectively a, ft and y. Project the broken line PMNQ on 
 PQ. This projection is equal to TQ itself. Hence we will have 
 
 PQ = PM cos a + MN cos ft + NQ cos y. (a] 
 
 Now project the broken line PMNQ on the axes xyz respectively. 
 We obtain thus the three equations 
 
 PQ cos a = PM + MN cos A. + NQ cos /* j 
 PQ cos ft = PM cos A + MN + NQ cos v V (b) 
 PQ cos y PM cos /* + MN cos v + NQ ) 
 
 Now multiply the first of equations (3) by PM, the second by MN 
 and the third by NQ and add them taking (a) into account and we 
 have 
 
 PQ 2 = PM 2 + MN 2 + NQ 2 + 2PM . MN cos A + 2 PM . NQ cos 
 
 /* + 2MN,NQcos v (c) 
 
 or PQ 2 = ( X - X J + (y -/) + (z - z Y+ 2(x - x ) (y -y ) 
 cos /I + 2(x x )(z z ) cosfii + 2(y -y )(z z ) cos v. (5) 
 
 Cor. If one of the points as Q be at the origin then 
 
 PO 2 x* + y* + z* -f 2xy cos A+ 2x2 cos jn + 2zy cos v. (6) 
 
 17. Direction Ratios. In oblique coordinates the position of a line 
 
 PM MN NQ 
 PQ is determined by the ratios -j-y ; -^ - ; ~p^-, and these we 
 
 call direction ratios. We may name these /, m, n respectively, 
 
IO NOTES ON SOLID GEOMETRY. 
 
 taking care to note that \ve are using oblique coordinates and call 
 the line PQ, the line (/, ;//, n}. To find a relation among these 
 direction ratios, we divide equation (c) Art. 16, by PQ 2 . We thus 
 have 
 
 i = I 2 + m* + ;/ 2 + 2//T2 cos A + 2ln cos /* + 2mn cos v, (7) the 
 desired relation. 
 
 1 8. The coordinates of the point (xyz) dividing in the ration : n 
 the distances between the two points (x y z 1 ) (x"y"z") are 
 
 mx" + nx ! my" + ny mz" + nz 
 
 x =. , y = . z= - . (8) 
 
 m + n J 7?i + n m + n 
 
 The proof of this is precisely the same as that for the correspond 
 ing theorem in Plane Coordinate Geometry. 
 
 19. POLAR COORDINATES. 
 
 The position of a point in space is also sometimes expressed by 
 the following polar coordinates : 
 
 The radius vector OP = r, the angle PO0 = 6 which the radius 
 vector makes with a fixed axis Oz, and the angle CO^tr cp which the 
 projection OC of the radius vector on a p ane yQx perpendicular to 
 O0 makes with the fixed line Ox in that plane. (Fig. 9.) 
 
 We have OC = r sin 6. Hence the formulae for transforming from 
 rectangular to these polar coordinates are 
 
 x r sin 8 cos <p } 
 
 Y r sin 8 sin <p\ (9) 
 
 z = r cos 8 j 
 
 and these give r 1 x^ +.V 2 + 
 
 tan cp = 
 
 a * z 
 
 cos 8 = = 
 
 (10). 
 
 Conceive a sphere described from the centre O, with a radius = a 
 and let this represent the earth. Then, if the plane zOx be the 
 plane of the first meridian and the axis of z the axis of the earth, 
 
 Q latitude, cp = longitude of a point on the earth s sur 
 face. 
 
NOTES ON SOLID GEOMETRY. Ir 
 
 20. Distance between two points in space in polar coordinates. 
 
 Let P be (r r , 6 , q> } and Q (r, 6, <p). Project PQ on the plane xy, 
 MN is this projection, draw OM and ON the projections of OP and 
 OQ respectively on that plane. Through P draw PR parallel to MN, 
 then PR = MN. (Fig. 10.) 
 
 And we have 
 
 PQ 2 = PR 2 + RQ 2 = MN 2 + (QN - RN) 2 . 
 
 But in triangle MON 
 
 MN 2 == OM 8 + ON 2 - 2OM . ON cos MON, 
 or MN 2 = r 2 sin 2 & + r 8 sin 2 6 2rr sin sin d cos (tp <p ). 
 
 Moreover QN r cos and RN = PM = r cos 6 . 
 
 Hence PQ 2 = r 2 sin 2 + r* sin 2 - 2rr sin sin cos (q> <p ) 
 
 + (r cos - r cos )* 
 or 
 
 PQ 2 =r 2 +r 2 - 2 rr (cos 6>cos^ + sin sin 6 cos (<p - q> )). (n) 
 
CHAPTER II. 
 
 INTERPRTTATION OF EQUATIONS. 
 
 TRIPLANAR COORDINATES. 
 
 21. LET us take F (x y y, z] = o, that is any single equation con 
 taining three variables x, y and z. This may be considered as a 
 relation which enables us to determine any one of the variables when 
 the other two are given. Let these be x and j . So the equation 
 may be written 
 
 *=/(*, jOi 
 
 in which we may attribute arbitrary and independent values to x and 
 y. And to every pair of such values there is a determinate point in the 
 plane xy ; and if through each of these points we draw a line parallel 
 to the axis of z, and take on it lengths equal to the values of z given 
 by the equation, it is clear that in this way we will get a series of 
 points the locus of which is a surface, .and not a solid since we take 
 determinate lengths on each of the lines drawn parallel to z. Hence 
 F (x, y, z) = o represents a surface in triplanar coordinates. 
 
 22. If the equation contains only two variables as F (x,y) = o then 
 it represents a cylindrical surface. 
 
 For F (x, y] = o is satisfied by certain values of x and y inde 
 pendently of 0, and x and y are no longer arbitrary but one is given 
 in terms of the other ; to each pair of values corresponds a point in 
 the plane xy, and the locus of these points is a curve in that plane. 
 If through each point in this curve we draw a coordinate parallel to 
 2, every point in that coordinate has the same coordinates x andj/ as 
 the point in which it meets the plane xy. Hence F (x,y) = o repre 
 sents a surface which is the locus of straight lines drawn through 
 points of the curve F(.r, v) = o in the plane xy and parallel to the 
 
 12 
 
NOTES ON SOLID GEOMETRY. 13 
 
 axis of z. This locus is either what is called a cylindrical surface 
 with axis parallel to z or a plane parallel to the axis of z according 
 as the equation F (x,y) o in the plane xy represents a curve or a 
 straight line. 
 
 For example, x 1 4- y* r 2 = o in rectangular coordinates is a 
 right cylinder wiih circular base in plane xy (since jv a +y = r* is a 
 circle in plane xy) and its axis coincident with the axis of z. 
 
 And ax + by c = o is a plane parallel to the axis of g, intersect 
 ing the plane xy in the line ax + by = c. 
 
 Similarly F (x, z) = o represents either a cylindrical surface with 
 axis parallel tojy or a plane parallel to_> . 
 
 F (y, z) = o represents either a cylindrical surface with axis parallel 
 to the axis of x or a plane parallel to this axis. 
 
 23. An equation containing a single variable represents a plane or 
 planes parallel to one of the coordinate planes. 
 
 Thus x = a represents a plane parallel to the planejyz. 
 
 And as _/"(.#) = o when solved will give a determinate number of 
 values of x, as x = a, x I, x = c, etc., so it represents several 
 planes parallel to the coordinate planers. 
 
 Thus also F(jy) = o represents a number of planes parallel to the 
 plane xz. 
 
 And F (z) = o, a number of planes parallel to xy. 
 
 24. Thus we see that in all cases when a single equation is inter 
 preted it represents a surface of some kind or other. 
 
 The apparent exceptions to this are those single equations which 
 from their nature can only be satisfied when several equations which 
 must exist simultaneously are satisfied. As for example 
 
 (x of + (y b)* + (z c) z = o. This equation can only be 
 satisfied when (x a)* = o, (y b}* = o, (z c} z = o, or x = a, 
 y b, z = c. 
 
 Now these represent three planes, but being simultaneous they 
 represent the point a, b, c. 
 
 So also (x a) z -f ( y b}* = o is only satisfied by x = a, y = b, 
 and hence though x = a is a plane, and y = b is a plane, the two 
 together must represent a line common to both of these planes, that 
 is their line of interseciion, which must be parallel to z. 
 
 25. In general two simultaneous equations as 
 
 f(*,y, *) = F (x,y, z) =o 
 
14 NOTES ON SOLID GEOMETRY. 
 
 represent a curve or curves, the intersections of the two surfaces 
 represented by the two equations. 
 
 Thus _ 7 r taken simultaneously we have seen represent a straight 
 
 line parallel to the axis of z, the intersection of these two planes. 
 
 F (x) = o ) 
 
 p ) \ _ f represent a number of straight lines parallel to the 
 
 axis of z, the intersections of the several planes parallel respectively to 
 the planes yz and xz. 
 
 F (x) = o ) 
 
 p/j) of re P resent a numbe r of straight lines parallel to the axis 
 
 of y, e;c. 
 
 f. represent the curves of intersection of the two cylin- 
 P (.v, z) = o j * 
 
 ders F (x, y) = o and F (x, z) = o, e c., etc. 
 26. Three simultaneous equations 
 
 F (x, y, z}=o\ F (x, j>) = o\ 
 
 as f(x, y, z) = o I or F (A-, z) = o I etc ., 
 
 represent points in space or the intersections of the lines of intersec 
 tion of the surfaces. 
 The simplest case is, 
 
 y ^ > representing the point (a, b, c], 
 
 z = c] 
 So also 
 
 2z \ 
 
 x + y = 2z v. represent points which can be found by 
 
 solving the three equations which themselves represent different sur 
 faces. 
 
 Interpretation of Polar Equations. 
 
 27. i. r = a represents a sphere having the pole for its centre. 
 Hence the equation (r) = o which gives values for r as r a, 
 r = b, r c, etc., represents a series of concentric spheres about the 
 pole as centre. 
 
NOTES ON SOLID GEOMETRY. jcj 
 
 2. 6 = a represents a cone of revolution about the axis of z with 
 its vertex at the origin of which the vertical angle is equal to 2a. 
 Hence the equation F (6) = o giving values 6 = a, 6 = ft, etc., 
 represents a series of cones about the axis of z having the origin for 
 a common vertex. 
 
 3. cp = (3 represents a plane containing the axis of z whose line 
 of intersection with the plane xy makes an angle /3 with the axis of 
 x. Hence the equation F (cp) o which gives values q) = (3, cp 
 = ft , etc., represents several planes containing the axis of z inclined 
 to the plane zOx at angles ft, ft , etc. 
 
 4. If the equation involve only rand 6 as F (r, 6) = o, since 
 F (r, 6) = o gives the same relation between r and 6 for any value 
 of cp, ii gives the same curve in any one of the planes determined by 
 assigning values to (p. Hence it represents a surface of revolution 
 traced by this curve revolving about the axis of z. 
 
 Example, r = a cos 6 is the equation of a circle in the plane 
 xz, or in any plane containing the axis of z. Hence r a cos 6 
 represen s a sphere described by revolving this circle about the axis 
 of*. 
 
 5. If the equation be F(<p, 0) o for every value of cp there 
 are one or more values of 6 corresponding to which lines through 
 the po e may be drawn, and as cp changes or the plane fixed by it 
 containing Oz revolves, these lines take new positions in each 
 new position of the plane, and thus generate conical surfaces 
 (a conical surface being any surface generated by a straight line 
 moving in any manner about a fixed straight line which it inter 
 sects. ) 
 
 6. If the equation be F(r, (p) = o, for every value of (p there are 
 one or more values of r, thus giving several concentric circles about 
 the pole in the plane determined by the assigned value of cp. As (p 
 changes, or the plane through Oz revolves these values of r change, 
 and the concentric circles vary in magnitude. The equation thus 
 represents a surface generated by circles having their centres at the 
 pole, which vary in magnitude as their planes revolve about the axis 
 of z which they all contain. 
 
 7. If the equation be F (r, 6, cp) = o, it represents a surface in 
 general. For if we assign a value to <p as cp = ft, then F (r, 8, ft) 
 = o will represent a curve in the plane (p = ft. And as cp changes 
 or the plane revolves about Oz this curve changes, and the equation 
 will represent the surface containing all these curves. 
 
!6 NOTES ON SOLID GEOMETRY. 
 
 28. Two simultaneous equations in polar coordinates represent a 
 line, or lines the intersections of two surfaces. And three simulta 
 neous equations represent a point or points the intersections of three 
 surfaces. 
 
 Thus 
 
 r = a j 
 
 6 a v taken simultaneously represent points determined 
 
 9 = ft) 
 
 by the intersection of a sphere, cone and plane. 
 
CHAPTER III. 
 
 EQUATION OF A PLANE. 
 
 COORDINATES OBLIQUE OR RECTANGULAR. 
 
 29. To find equation of a plane in terms of the perpendicular from the 
 origin and its direction cosines. 
 
 Let OD p be the perpendicular from the origin on the plane, 
 and let it make with the axes O,r, Oy and Qz the angles a, ft and y 
 respectively. Let OP be the radius vector of any point P of the 
 plane ; OM, MN and NP the coordinates of P. (Fig. n.) 
 
 The projection of OM + MN + NP on OD is equal to the pro 
 jection of OP on OD. 
 
 The projection of OP on OD is OD itself, and the projection of 
 OM + MN + NP on OD is x cos a +y cos ft + z cos y. 
 
 Hence we have x cos a +y cos ft + z cos y p. (12) 
 
 30. To find the equation of a plane in terms of its intercepts on the 
 coordinate axes (coordinates oblique or rectangular). 
 
 Let the intercepts be OA a, OB = b, OC = c. The equation 
 (12) may be written 
 
 x y z 
 
 p sec a p sec ft p sec y 
 
 But since ODA, ODB and ODC are right-angled triangles, we have 
 p sec a = OA a, p sec ft = OB = b, p sec y = OC = c. 
 Therefore the equation becomes 
 
 X V Z 
 
 the equation of the plane in terms of its intercepts. 
 
1 8 NOTES ON SOLID GEOMETRY. 
 
 31. Any equation Ax + By + Cz = D (14) of the first degree in x, 
 y and z is the equation of a plane* 
 
 For we may write (14) 
 
 x j^ _z_ 
 
 Jl J>/ _D/ 
 
 X "B" ~C~ 
 
 D D D 
 
 And putting -^= a, -g- = ^ --= A W e have the form (13). 
 
 Hence (14) is the equation of a plane in oblique or rectangular 
 coordinates. 
 
 Hence to find the intercepts of a plane given by its equation on 
 the coordinate axes, we either put it in the form (13) or simply 
 raakej> = o and z = o to find intercept on x ; z = o and x = o to 
 find intercept onj/ ; x= o andjy o to find intercept on z. 
 
 "Example. Find the intercepts of the plane 2x + $y 52 = 60. 
 
 32. It is useful often to reduce the equation AJC + By + Cz = D 
 to the form x cos a + y cos ft + z cos y =/ in rectangular coordi 
 nates. We derive a rule for this. 
 
 Since both of these equations are to represent the same plane, we 
 have 
 
 cos a _ cos (3 _ cos y _ p _ Vcos 2 a. + cos 2 (3 + cos 2 y 
 A B C D yT7 A 2 + B 2 + C 2 "" 
 
 Hence cos <* = 
 
 + B 2 + C 2 -v/A 2 + B 2 + C 
 
 + B + C 2 A/A 2 + B 2 +C ! 
 
 = -vWWc* (I5) 
 
 it is in the perpendicular form (12). 
 
NOTES ON SOLID GEOMETRY. ig 
 
 Hence the Rule: If we divide each term of the equation Ax + By 
 = D, by the square root of the sum of the squares of the coefficients 
 ofx, y and z, the new coefficients will be the direction cosines of the per 
 pendicular to the plane from the origin, and the absolute term will be the 
 length of this perpendicular. Give the radical the sign of D. 
 
 Example. Find the direction cosines of the plane 2x + $y 40 
 = 6 and the length of the perpendicular from the origin. 
 
 Result. 
 
 2 2 3 4 
 
 cos a = === 7=, cos ft = T=, cos y =, 
 
 A/4 + 9 + 16 v 29 V 2 9 V 29 
 
 P = 
 
 V 29 
 
 33. To find the angle between two planes (coordinates rectangu 
 lar). 
 
 If the planes are in the form 
 
 x cos a + y cos ft + z cos y = p 
 x cos a + y cos ft + z cos y p , 
 
 then since this angle is equal to the angle of two perpendiculars from 
 origin on the planes the cosine will be (Art. 15) cos V= cos a cos a 
 + cos /3 cos /3 -f cos y cos y 1 . 
 If they are in the form 
 
 A.v + By + Cs = D 
 A .v + B> + C z = D . 
 
 Then cos a = ==. r, cos /? = 
 
 B 2 4- C 2 A/A 2 + B 2 + C 2 
 
 C 
 
 cos = 
 
 A al B 
 
 COS flf = ~, COS P = 
 
 A/A 2 + B /f + 
 
 And cos V 
 
 cos y = = 
 
 AA + BB + CC 
 
 AA 2 + B 2 + C VA /f + B 2 + C 2 
 
2Q NOTES ON SOLID GEOMETRY. 
 
 From this 
 . Q _ (A 2 -f B 2 + C 2 ) (A 2 + B 2 + C 2 ) -(AA + BB + CC ) 2 
 
 (A 2 + B* + C 2 )(A 2 +B 2 + C" 2 ) 
 
 d . v - (AB -A B) 2 +(AC -A C) 2 +(BC -B C) 2 
 (A 2 + B + C ) (A 8 + B a + C 2 ) 
 
 Cor. i. If the planes are perpendicular to each other, then cos V=o. 
 . . AA + BB + CC = o (18) is the condition of perpendicularity of the 
 planes. 
 
 Cor. 2. If the planes are parallel sin V = o. Hence 
 
 (AB - A B) 2 + (AC - A C) 2 + (BC - B C) 2 = o 
 or AB - A B = o AC - A C = o BC - B C = o 
 
 ABC 
 A 7== B 7= C 
 
 or the condition that the two planes shall be parallel, is that the coefficients 
 ofx,y and z in the two equations shall be proportional. 
 Ex. i. Find the angle between the planes 
 
 x + 2y + 32 = 5 and $x 4y 4- z = 10. 
 
 2. Show that the planes 
 
 x + 3y 5 Z 20 anc * 2X + V + z I0 are perpen 
 dicular to each other. 
 
 3. Write the equation representing planes parallel to the plane 3* 
 + 2y 6z = ii. 
 
 34. To find the expression for the distance from a point P (x y z ) to a 
 plane (coordinates rectangular). 
 
 i. Let the equation of the plane be of the form 
 
 x cos a + y cos ft + z cos y = / when p OD. 
 
 Pass a plane through P parallel to the given plane, and produce 
 OD to meet it in D . 
 
 The equation of this plane will be 
 
 x cos a +y cos ft + z cos y = p when OD =/ . 
 
 Now let PM be the perpendicular from P on the given plane. 
 Then PM = OD - OD = p 1 - p. 
 
NOTES ON SOLID GEOMETRY. 21 
 
 Hence PM = x cos a 4- y cos /3 + z cos y p. 
 
 And (x 1 cos a + y cos /? + cos y p) (20) is the expression 
 for the perpendicular from the point x y z on the plane x cos or + _y 
 cos ft + cos y =p, the sign being + or according as P is or is 
 not on the side of the plane remote from the origin. 
 
 2. Let the plane be in the form Kx + By + Cz = D. 
 
 A 
 
 Then cos a = - etc., etc. (15) Art. 32. 
 
 V A 2 + B s + C* 
 
 Hence the expression 
 
 (y cos -FJF cos /3 + z cos y p] becomes 
 -D 
 
 Ex. Find the length of perpendicular from the point (3, 2, i) on 
 
 the plane 
 
 3Jt- + 4_y 6z = 24. 
 
 9 4- 8 - 6 24 13 
 Result. p 
 
 - - 
 
 A/9+ 16+ 36 
 
 35. The equation of the plane in the form x cos a + y cos (3 4- 
 cos y = p may be used to demonstrate the following theorem in 
 projec ions. 
 
 7% volume of the tetrahedron which has the origin for its vertex and 
 the triangle ABCy2?r its base is equal to the three pyramids which have any 
 point (x, y, z) in the plane ABC for their common vertex and for bases 
 the projections of the area ABC on the three rectangular coordinate planes 
 respectively. 
 
 For let A be the area of the triangle ABC and 
 
 x cos a +y cos /3 + z cos y = p 
 the equation of its plane. 
 
 Multiply this equation by A. 
 
 Then A cos a . x -f A cos /3 . y + A cos y . z = Ap 
 or -JA cos a . x + ^A cos ft .y + -JA cos y. s 
 
 But A cos a, A cos /3, A cos y, are the projections of A on the 
 planes yz, xz, and xy respectively, and x, y and z are the altitudes of 
 the tetrahedrons which have these projections as bases and the point 
 (x, y, z) as common vertex, and -JA/> is the volume V of the pyramid 
 
22 NOTES ON SOLID GEOMETRY. 
 
 which has the origin for vertex and A for base. Hence the theorem 
 is true. 
 
 Calling these projections A x , A y , and A x , we may write the equa 
 tion of the plane A^v + A y y + A^ = 3V. (22) 
 
 36. To find the polar equation of a plane . 
 
 Let OP = r, POS = 6, P OM = cp be the polar coordinates of a 
 point P of the plane. (Fig. 12.) 
 
 Let OD = a = perpendicular on plane ; angle DOS a, D OM 
 = jff, and POD = GO. 
 
 Then yy p - = cos POD = cos GO, or = cos GO. Now in order to 
 
 express GJ in polar coordinates conceive a sphere about O as centre 
 with OP = r as radius. Prolong OD to D" on the sphere. Draw 
 the arcs of great circles SPP , SD"D , MP D and D"P. 
 
 The triangle SD"P has for its sides SD" = a, SP = 6, D"P = GJ 
 and angle D"SP = D OP = /3 <p. But 
 
 cos D"P = cos SD" cos SP + sin SD" sin SP cos D"SP. 
 
 Or 
 
 cos GO = cos a cos 6 + sin a sin 6 cos (fi cp). 
 
 Therefore = cos a cos 6 + sin a sin 6 cos (f3 cp) (23) is the 
 polar equation of the plane. 
 
 37. The general equation of the plane Ax + By 4- Cz = D may 
 be reduced to the form 
 
 A x 4 B f y + C 2 = i (24) by dividing by the absolute term D. 
 
 And also to the form 
 
 z mx + ny+ c (25) by dividing by C transposing and putting 
 
 = m, = n and -^ = c. These two forms are very useful in 
 
 V^/ V^ v_/ 
 
 the solution of problems and in finding the equations of the plane 
 under given conditions. 
 
 Plane tinder Given Conditions. 
 
 38. i. The equation of a plane through the origin will be of the 
 general form Ax + By + Cz = o, for ihe equation must be satisfied 
 by ^ = 0,^ = and 2 = 0. 
 
 2. The equation of a plane which contains the axis of z is of the 
 
NOTES 
 
 SOLID GEOMETRY, 
 
 form Ax + By = o ; a plane containing the axis of y is Ax + Gar 
 o ; one containing the axis of x is By + Cz = o. 
 
 3. The equation of a plane parallel to the axis of z is Ax + By 
 = D; of one parallel to the axis ofjy is AJI* + Cz = D ; one parallel 
 to the axis of x is By + Cz = D. 
 
 4. The equation of a plane parallel to the plane j/s is A^t: = D ; 
 parallel to ^0 is By = D ; parallel to xy is Cs = D. 
 
 These equations we have had already in the forms x= a, y 3, 
 
 39. To find the equation of a plane containing a given point (a, b, c) 
 and parallel to a given plane Ax -f By + Cz = D. ( i ) 
 
 First, since the required plane is to be parallel to (i ) it may be writ 
 ten A* + Bj/ + G&=D (2) when D is undetermined. Secondly, the 
 coordinates (a, b, c) must satisfy (2). Therefore Aa + ~B& + Cc =D . 
 Hence by subtraction we eliminate D and obtain 
 
 Ax + By + C0 = Aa + B3 + O (26) 
 the required equation. 
 
 Example. Find the equation of the plane passing through the 
 point (i, 2, 4) parallel to the plane 2x + 4y 32 = 6. 
 
 40. To find the equation of a plane passing through three given points 
 (x , y , z ), (x", y", z") and (x ", y ", z "). 
 
 Let the equation of the plane be of the form Ax + By + Cz = i, 
 A, B and C to be determined by the given conditions. 
 
 Since the plane is to contain each of the points, we must have 
 
 Ax + B 
 
 Hence 
 
 I,/", 
 
 B = 
 
 X, I, ^ 
 
 JP", i, z 
 x ", i, 2 
 
 C = 
 
 Substituting these values in the equation Ax + By + Cz = i 
 
NOTES ON SOLID GEOMETRY. 
 
 we have 
 
 I,/ , 
 
 #, I, * 
 
 *, I, * 
 
 x + x" r , i, 
 
 (27) 
 
 But from plane coordinate geometry the coefficients of x, y and z 
 in these equations are the double areas of triangles in the planes jy0, 
 xz and xy respectively. Moreover these triangles are the projections 
 of the triangle of the three given points, on these planes. Hence 
 comparing this equation wi h the equation (22) 
 
 we see that 
 
 x ,y ,ss 
 
 x">y", 2 
 
 ,z = 3V 
 
 = 6V. That is = 6 times the volume of the 
 
 pyramid which has the origin for vertex and the triangle of the three 
 given points for base. This equation fully written out is 
 
 ^(yv--yv )+^ (yv-y3 )+y (jV -yv)-6v, ( 2 8) 
 
 41. To find the equation of the planes which contain the line of intersec 
 tion of the two planes Ax + By + Cz = D and Ax 4- By + Cz = D . 
 
 ThisequationisA# + By + Gs D + K(A jc + B> + C 0-D / )=o(29) 
 
 when K is arbitrary. For this represents a plane when K takes a 
 particular value and it is satisfied when A^ + By + C-s: D = o and 
 K x 4- B^ + C z D = o are satisfied simultaneously. Hence it 
 is a plane containing their line of intersection. Hence as K is arbi 
 trary it (24) represents the planes containing the line of intersection 
 of the two given planes. 
 
 42. When the identity KU + K^Uj + KaUs = o (30) exists between 
 the equations U=o, U 2 o, U 3 = o of three planes, then these planes 
 intersect each other in one and the same straight line. This is an 
 easy corollary of Article 41. Also when the equation of the first 
 degree in x, y and z contains a single arbitrary constant all the 
 planes which it expresses by assigning particular values to this con 
 stant intersect each other in one and the same straight line. This 
 line of intersection may be at infinity and then the planes are all 
 parallel. 
 
 Example I. The planes represented by the equation 6x+T&y + 2z 
 =. 3 (M arbitrary) all contain the line of intersection of the two 
 planes 6x+ 22 3 = and y o. 
 
NOTES ON SOLID GEOMETRY. 25 
 
 Example 2. The planes represented by 2x + ^y 42 = n (n 
 arbitrary) are parallel. 
 
 Example. The planes $x + \v + 60 = 2 } 
 
 X + 2V + 3 I ? 
 4.v + 4T + &z = 2 J 
 
 intersect in one and the same straight line because 
 
 2 = o 
 
 is an identity. 
 
 43. Wfo between the equations of four planes in any form U o, U, 
 = o, U 2 o, U 3 o the identity 
 
 KU + K 1 U 1 + K 2 U a -hK 3 U 3 = o (31) ^/j/j, then these four planes 
 intersect each other in one and the sam* point. For then any coor 
 dinates which satisfy the first three U = o, U, = o and U 2 = o will 
 satisfy the fourth U 3 = o. 
 
 44. Example i. Find the equation of the plane passing through 
 the origin and containing the line of intersection of the two planes 
 Ax + By + C0 = i and A x + B> + C z = i. 
 
 First we have Ax + By + Cz i + K (A .r + B> + C z 1)= o 
 for all the planes containing the line of intersection of the two given 
 planes. But as the required plane must contain the origin, the 
 equation must be satisfied by (o, o, o). Hence we have i K=o. 
 .-. K = - i. 
 
 The required equation is therefore 
 
 Ax + By -f Cz i (A x + B> + C z i) = o 
 or (A - A ) x 4- (B - B )^ + (C - C ) z = o. 
 
 Ex. 2. On the three axes of x, y and z take OA = a, OB b, 
 OC = c and construct on these a parallelopipedon having MP as the 
 edge opposite parallel to OC, and AR in the plane xz the edge 
 opposite and parallel to BN. 
 
 Find the equation to the plane containing the three points M, N 
 and R. 
 
 \? V 
 
 Now NR is the line of intersection of the two planes +-7- = i and 
 
 a b 
 
 -= i. Hence the plane containing this line must be of the form 
 
26 A OTES ON SOLID GEOMETRY. 
 
 OC V 
 
 "*~ ^ I -- l ) - To determine K \ve impose the 
 condition that this plane shall pass through the po nt M (a, b, o). 
 
 Hence we have + -r i + K ( i ) = o. . . K = i 
 at \c J 
 
 Therefore the required plane is 
 
 x v z x v z 
 
 --t-V H ---- i o or -- (-^-_j-__ 2. 
 a b c a b c 
 
 Ex. 3. Find in like manner the equation of the plane containing 
 the points P, B, and C, in the same figure. 
 
 Result, 
 
 45. If two given planes be in the normal form as 
 x cos a +> cos ft + z cos y=p and.v cos a +y cos /3 + zcos y = p . 
 The plane containing their line of intersection is 
 
 x cos a+y cos /3 + z cos y p-\-K (x cos a +y cos ft + z cos y 
 
 -/)=o 
 
 And if K = i the equation becomes 
 
 x cos a -\-y cos /3 + z cos y p (x cos a + y cos ft + z cos ;/ 
 
 -/> )= o 
 
 which represents ihe two plane bisectors of the supplementary angles 
 made by the given planes. 
 
 That is to find the equations to the plane bisectors of the supplementary 
 angles made by two given planes, put their equations in the normal form 
 and then add and subtract them. 
 
 Example. Find the two planes which bisect the supplementary 
 angles made by the planes 2.v + 3.y Vz = 5 and 3^ + 4^23 =-- 4- 
 
 Result, . .. = . 
 
 A/14 V29 
 
 Remark. If we place A = x cos (Y + y crs ft + z cos y p and 
 A = x cos a + v cos /? + cos y /> . 
 
 Then A A = o is the plane bisector of one of the angles be 
 tween the planes A and A and A -f A = o is that of the supple 
 mentary angle. 
 
 46. The three planes which bisect the diedral angles of a triedral have. 
 a common line of intersection. Let A = o, A = o and A" = o be three 
 
NOTES ON SOLID GEOMETRY. 
 
 planes in the normal form, and let the origin be within the triedral angle 
 formed by the three of which P their point of intersection is the vertex. 
 
 Then the plane bisectors of the angles made by these planes is 
 A A o, A" A o, A A" = o. And as these when added 
 together vanish simultaneously, it follows that these three planes 
 have a common line of intersection. 
 
 We can give this theorem another form by conceiving a sphere to be 
 described about the vertex of the triangular pyramid as a centre. The 
 three planes A= o, A = o, A"=o cut the surface of the sphere in 
 arcs of great circles which form a spherical triangle and the three 
 planes A A =o, A" A = o and A A" = o cut the sphere in 
 three arcs of great circles which bisect the angles of this spherical 
 triangle and their common line of intersection pierces the sphere in 
 the common intersection of these arcs. Hence the above demon 
 strates the following theorem, namely, The arcs of great circles which 
 bisect the angles of a spherical triangle cut each other in the same point 
 (the pole of the inscribed circle of the triangle). 
 
 47. To find the point of intersection of the planes AJC + By + Cs 
 = D, A x + B> + C z = D , A",r + B > + C"z = D". 
 
 We have by elimination 
 
 D, B, C 
 D , B , C 
 D". B", C" 
 
 A, D, C 
 
 A , D , C 
 A", D", C" 
 
 A, B, D 
 A , B , D 
 
 A", B", D" 
 
 (32) 
 
 A, B, C [A, B, C A, B, C 
 
 A , B , C A , B , C A , B , C 
 
 A", B", C" A", B", C" A", B", C" 
 
 Hence the condition that one of these shall be parallel to the line 
 of intersection of the other two, or that the planes shall not meet in 
 a point, is 
 
 A, B, C 
 
 A , B , C 
 
 A", B", C" = o, that is 
 
 A(B C"-B"C ) + A (B"C - BC") + A"(BC - B C) = o. 
 47. The condition that four planes 
 
 Ax -fBy +Cz +D = o 1 
 
 A x + B> + Cz + D =o 
 
 A"x +B > +C"z +D" =o 
 
 A "x+B "y + C "z + ~D " = o shall meet in a point is 
 
28 
 
 NOTES ON SOLID GEOMETRY. 
 
 A, B, C, D 
 
 A , B , C , D 
 
 A", B", C", D" 
 
 | A ", B ", C", D " 
 
 = o. (33) 
 
 49. We have seen that the equations of two planes Ajr 
 D = o and A. x+ Bjy + Cz D = o added together one or both 
 of them multiplied by any number give the equation of a plane 
 which contains the line of intersection of the two given planes. If 
 we combine these two equations so as to eliminate x we shall obtain 
 a plane parallel to the axis of A\ containing this line of intersection. 
 If we eliminate y we obtain a plane parallel to the axis of y contain 
 ing the same line ; and finally if we eliminate z we obtain a plane 
 parallel to the axis of z containing the same line. 
 
CHAPTER IV. 
 
 THE STRAIGHT LINE. 
 
 50. The equations of any two planes taken simultaneously represent 
 their line of intersection. 
 
 re P resent a strai S ht lme the c - 
 
 ordinates of every point of which will satisfy the two equations. 
 
 If we eliminate alternately x and y between these equations we 
 obtain equations of the form 
 
 x m p\ , x ^ planes parallel respectively to the 
 
 y ~ nz + q \ 
 
 axes Oy and Ox which represent the same straight line as equations 
 (35). These non-symmetrical forms (35) are very useful. The 
 planes x =. mz + a, y = nz + b are called the projecting planes of the 
 line on the planes of xz and jar, and these equations are also the 
 equations of the projections of the line on those planes respectively. 
 
 If we eliminate z we get - ~= -- or y = x + g -- p the equa- 
 n m J m m 
 
 tion of the projection of the line on the coordinate plane xy. 
 
 The equations (35) of the straight line contain four arbitrary con 
 stants, m, n, p, q, to which we can give proper significance by com 
 paring these equations with the equation y mx + b in plane coor 
 dinate geometry. 
 
 The equations (35) may be thrown in the form 
 
 ^_^ =z _^ = , 
 
 m n i 
 
 which gives us an easy choice of fixing the line by the equations 
 of any two of its projecting planes. 
 
 51. To find the equations of a straight line in terms of its direction cosines 
 and the coordinates a, b, c of a point on the line : (axis rectangular.} 
 
 29 
 
30 A T OTES ON SOLID GEOMETRY. 
 
 Let a, ft, y be the angles made by the line with the coordinate axes 
 respectively. Let / be the portion of the line between any point 
 (x, y, z) on the line and the point (a, b, c). Then / cos a = a- a ; 
 / cos ft yb ; / cos y zc and eliminating / we have 
 
 (37) 
 
 cos ex cos ft cosy 
 
 This form (37) of the equation of a straight line is symmetrical 
 and is therefore very useful. It contains six constants but in reality 
 only four independent constants, since the relation cos 2 o + cos 2 ft 
 4- cos 2 y = i holds, and of the three a, b, c one of them may be 
 assumed at will, leaving only two independent. 
 
 We have seen that the equation (35) may be thrown into the form 
 (37). So also (37) may be thrown into the form (35) by finding 
 from them expressions forjy and x in terms of z. 
 
 52. To find the direction cosines of any straight line given by its equations. 
 
 If the equations be in the form 
 
 x a vbzc T HT i XT i 
 
 = = r^ . L, M and N are proportional to the 
 
 direction cosines of the line. 
 So that we have 
 
 cos a _ cos ft _ cos y __ -\Xcos- a + cos 2 ft + cos 2 y 
 L ~M~~ N ~ 
 
 Hence 
 
 N 
 
 Hence to find the direction cosines of any straight line 
 
 A.r + By + Cz = D 
 MX 4- B> + C z = D 
 
 we throw the equations into the form 
 
 .va _ yb _ z c 
 L "IT" ~N~ 
 
 by eliminating y and x, and. then write out the direction cosines as above 
 equal to each denominator divided by the square root of the sum of the 
 squares of all three. 
 
NOTES ON SOLID GEOMETRY. 
 
 Thus to find the direction cosines of the line 1 
 
 y = nz + q \ 
 
 X p V Q Z 
 
 " 
 
 we write it 
 
 m n 
 
 Hence 
 
 m 
 
 cos =^+*+r. cos *> 
 
 (39) 
 
 Ex. i. Find the direction cosines of the lines 
 x 5 y ~ 2 z + 
 
 ,, . _ = 24 . v 
 
 ., T ^ 3^-1 j \ > > 3x 4 y+2z - 10 U; 
 
 53. To find the cosine of the angle between two lines given by the equa 
 tions 
 
 xa yb z c , x a vb zc 
 and ri = : 
 
 L M N L M N 
 
 We have shown (Art. 15) 
 
 cos V = cos a cos a -f cos /3 cos /? + cos ^ cos y. 
 
 LL + MM + NN 
 
 *+ N s VL *+ M" 2 + N ~ 2 
 
 Hence cos V = 
 
 TT.U i- u i r .\ mz-\-p] x = mz-\-p | 
 
 If the lines be m the form . , \ 
 
 y nz + q } y = n z + q \ 
 
 wm+nn +i 
 
 Then cos V = ^= =.. (41) 
 
 2/ 2 , 
 
 Ex. i. Find the cosine of the angle between the lines 
 
 X** 2Z + 6 
 
 Ex. 2. Find the cosine of the angle between the lines 
 xy z =4) x+y + z =2p 2 
 
32 NOTES ON SOLID GEOMETRY. 
 
 These equations may be put in the forms 
 
 = _ = _ (I ) and ___ = 
 
 4-3 -5 
 
 , T 5 + 12 
 . . cos V = - 
 
 V26X38 
 
 54. The condition of perpendicularity of two lines given by the 
 equations in last article is LL + MM + NN ^ o. (42) 
 The condition that they shall be parallel (see Art. 15) 
 
 is (LM / -L M) 2 + (LN / -L N) 2 +(MN / -M N) 2 i=o 
 
 or -=^=-(43). These two conditions when the lines are in the 
 
 1 A IV J. IN 
 
 x = mz+p ) x = m z+p 
 forms 
 
 y nz + q \ y 
 
 m z+p \ 
 n z + q f 
 
 become mm + nn +i =. o, (44) and m == m , n = ri (45) respec 
 tively. 
 
 55. To find the condition of the intersection of two lines 
 
 x = m,+p } 
 
 V nz + q ) 
 
 This is derived by eliminating x, y and z from the four equations. 
 Subtracting the third from the first we have o (m m }z +pf> . 
 
 /0 z _P~P ^ Similarly from the second and fourth z ~~ n r^_ n , 
 and since the lines intersect these two values of z are equal. There 
 
 fore we have -^^= ^ . (46) 
 m m n n 
 
 Ex. Find / so that the lines i ^ ~ 
 
 tersect. 
 
 If the two lines are in the form 
 
 xa yb zc , x a __ y b _ z c 
 
 ~TT TT~ "N~ (I) ~T7" = ~W~ N r 
 
NOTES ON SOLID GEOMETRY. 33 
 
 the elimination of x, y and z can be effected more readily by writ 
 ing (i)= K and (2) = K . 
 
 .-. x-a ^LK ) , 
 
 r Vif . <I =I*K LK. 
 
 x a = L K j 
 
 Similarly b- b M K MK 
 f-^ r ^ N K -NK. 
 
 Therefore eliminating K and K we have 
 
 L, -L , a-a 
 M, -M , - 
 
 ( N, -N , r-/ =ro 
 
 or 
 
 (47) 
 
 The Straight Line under Given Conditions. 
 
 56. The equations of a straight line parallel to one of the coordi 
 nate planes as xy are z = c, v = mx -\-p. 
 
 The equations of a straight line parallel to one of the coordinate 
 
 x = 
 
 axes as z. are 
 
 y 
 
 5 7. 7 o find the equations of a straight line, passing through a given point. 
 If (x, y, z ) is the point 
 
 xx vy z z , ft . 
 we have seen the equation is == =- = -^ (4) 
 
 or if the equations are in 
 
 , r x = mz+p) xx m(zz\} , , TT .,. . 
 
 the form * \ then } ,/ \ (49). Hence if the 
 
 y = nz + q j ^r-y = n(zz ) j v v 
 
 equations of a straight line contain only two arbitrary constants, all 
 the lines obtained by assigning values to these arbitraries pass through 
 a single point. 
 
 58. To find the equations of a straight line passing through two given 
 points (x , y , z ) (x", y", z") using (48) we have 
 
 ~ " = , or dividing (48) by this to eliminate 
 .L IVl JN 
 
 3* 
 
34 NOTES ON SOLID GEOMETRY. 
 
 L, M, and N we have 
 
 xx yy zz . 
 
 :?r=P=-77=7w- <5> 
 
 If one of the points as (x",y, z") be at the origin then the equa 
 tions become 
 
 59. To find the equation of a straight line passing through a given 
 point (x ,y, z } and parallel to a given straight line 
 
 xa yb z c 
 
 L M 
 
 From the first condition we must have -W~ ~ / = -, 
 
 -L M N 
 
 , , L M N 
 
 and Irom the second condition -=r T =-r T -.=-:. Tr . 
 
 L M N 
 
 Hence the required equation is 
 
 x x __ yy _ zz 1 . 
 
 L M ~~N w 2 ) 
 
 If the equation passing through the point x ,y , z be of the form 
 
 A--JV = m (z-z ) \ x == mz+p 
 
 i // /! an d the given line be 
 
 vy ~ n (z z ) ) y nz + q. 
 
 Then = n and / = /, and the line will be 
 
 ^2$$Y 
 
 60. To find the equations of a straight line passing through a given 
 point x , j r , z and perpendicular to and intersecting a given right line 
 x a _ y b _z c 
 
 I m n 
 
 The required line by the first condition will be of the form 
 
 where L, M, and N are to be determined by the conditions 
 
 = o (Art. 54) 
 
NOTES ON SOLID GEOMETRY. 
 and 
 
 61. Ex. i. Find the equation of the line joining the points (b, c, a) 
 and (a, c, It) and show that it is perpendicular to the line joining the 
 origin and the point midway between these two points ; and that it 
 
 is also perpendicular to the lines x =y = z and =-= -. 
 
 a b c 
 
 Ex. 2. The straight lines which join the middle points of the 
 opposite sides of a tetrahedron all pass through one point. 
 
 Take O one of the vertices as origin and OA, OB, OC as the 
 axes of x, j , s. 
 
 Let M, M , M" be the middle points of BC, AC and OC respec 
 tively, N, N , N" the middle points of the edges OA, OB and AB 
 opposite to those respectively. Then to find the equations of the 
 lines MN, M N , M"N". 
 
 We apply the equation ; - = , ,, , -77- to the points 
 
 x x y y z z 
 
 (M, N) (M , N ) (M", N") respectively. 
 Let OA = 20, ; OB = 2b ; OC = 2c. 
 Then M is (o, b, c) and N is (a, o, o). 
 Hence the equation of MN is 
 
 J ^~^ (l _.^_ ^ / T \ 
 
 ~ a~~ b ~ c ^ 
 
 Similarly the equation of M N is 
 
 x __yb _z . 
 a b c 
 
 And the equation of M"N" is 
 
 X V Z C 
 
 (i) and (2) give x , y =, z = and these values satisfy (3). 
 Consequently these lines pass through the point (-, , - -). 
 
36 A 07^S OA SOLID GEOMETRY. 
 
 Straight Line, and Plane. 
 
 62. To find the conditions that a line shall be perpendicular to a plane 
 given by its equation. 
 
 If the plane be of the form .v cos a+y cos fi + z cos y =/ (i) 
 we know that cos <v, cos /?, cos y are the direction cosines of the 
 perpendicular from the origin on the plane. 
 
 And the equation of this perpendicular will be 
 
 cos a cos // cos y 
 
 If any plane A^ + Br + Cs= D be parallel to the plane (r) we must 
 have 
 
 A B C 
 
 cos a - cos// ~~ cos y 
 
 and if the line -~= C - be parallel to the line 
 
 x v z 
 
 -, we must have 
 
 cos at cos ft cos y 
 
 L M N 
 
 cos a cos/3 cosy 
 Hence the conditions that the line = M = -^ shall be 
 
 perpendicular to the plane Ax + Ey + Cz = D will be 
 ABC 
 
 == (54) 
 
 ) . xp \~q z 
 
 If the line be in the form v we write it- =" - = - 
 
 ) m p i 
 
 A B C A = ;;/C) 
 
 And the conditions are - = - = or _, , \ (55) 
 
 m n i B = C ) 
 
 The equation of a line passing through the point .r , r , z 1 and 
 perpendicular to the plane Ax + fy + Cz = D will then be 
 
 xx yv zz 
 
 A B 
 
NOTES ON SOLID GEOMETRY. 37 
 
 or in the unsymmetrical form 
 
 *-* = ~ (z-z) 
 
 yy = ~(z-z). 
 
 Ex. Find the equation of a line passing through the point (i, 2, 
 3) and perpendicular to the plane $x+2j \z = 5. 
 
 63. To find the condition that a straight line shall be parallel to a 
 given plane. Let the plane be Ax + l$y + Cz = D and the line of 
 
 x a y b z c 
 the form ^- ~ . 
 
 Now if this line is parallel to the plane it will be perpendicular to 
 the normal to the plane. Hence the required condition will be 
 
 AL + BM+CN = . (56) 
 
 64. To find the conditions that a straight line shall coincide with a given 
 plane Ax + By + Cz = D. 
 
 x a _ y b _z c 
 L, TT~ : ~N~ 
 
 The line must fulfil the condition (56) of parallelism above, 
 AL -f BM -f CN = o. And also any point on the line as (a, b, c) 
 must satisfy the equation of the plane. Hence we must have the 
 additional condition Aa + Bl> + Cc D = o. (57) 
 
 2. Let the equations of the line be of the form x=mz +p ) 
 
 y=nz + q } . Sub 
 stituting these values of ,r and y in the equation of the plane, we 
 have 
 
 A(ms +/>) + K(nz -f q) + Cz = D, 
 
 whence z = 52 JT t And for coincidence this value of 
 
 Am + B/* + C 
 
 z must be indeterminate, and therefore A/ + B^ D = o ) (58) 
 
 Am + B + C = o f are the 
 conditions of coincidence. 
 
 NOTE. This last method is a general one of determining the con 
 ditions of coincidence of a straight line and any surface given by its 
 equation. That is substitute x and^ of the line in the equation of 
 the surface and since the z in the resulting equation mu^t be inde- 
 4 
 
38 NOTES ON SOLID GEOMETRY, 
 
 terminate if there be coincidence we treat this equation as an iden 
 tity and make the coefficients of the different powers of z separately 
 equal to zero. , 
 
 65. To find the expression for the length of the perpendicular PD from 
 any point P(x , y ,z ) on a straight lin.> AB given by its equation. 
 
 i. Let line be - -= : - where a, b, c are the coordi- 
 cos a cos p cos y 
 
 nates of any point A on the line. Now PD 2 = PA 2 AD 2 . (Fig. 13.) 
 
 But PA 2 = (x - a y + (v -by + (z -cY and AD being the projec 
 tion of PA on AB, we have 
 
 AD (x a) cos a+(yb) cos /S+(z 6) cos y. 
 Hence 
 
 + (z -b)cosy)\ (59) 
 2. If the given line be of the form 
 
 x a __ y b _zc 
 
 A B : C 
 
 Then 
 
 A 
 
 cos a = ____ etc., etc. 
 
 And therefore PD 2 
 
 3. If the given line be x mz + p ) 
 y = nz + q ) 
 Then PD 2 
 
 66. To find the expressionfor the shortest distance between two straight 
 lines given by their equations, 
 
 This shortest distance is a straight line AB perpendicular to both 
 the given lines PB and SR. (Fig. 14.) 
 
 Let the given equations 
 
 x a vb z c . xa yb z c 
 
 = = --- and - 7 -z -,= -- 7 and 6 = the 
 co 5 a cos// cos;/ cos a cos ft cos y 
 
 angle between the lines. 
 
 And L, M, N the direction cosines of the perpendicular AB. 
 
NOTES ON SOLID GEOMETRY. 
 
 Then we must have 
 
 L cos a + M cos ft -f N cos y o } 
 L cos a + M cos ft + N cos ;/ = o. j 
 
 Whence 
 
 L M 
 
 39 
 
 cos /? cos T -cos / cos y cos <* CO3 y cos a cos 
 
 N 
 
 cos <* cos ft cos M cos ft 
 
 M 2 -|- 
 
 [(COS /S COSy - COS /3 Cos y)*-t ^COS a COS > - CO* a COS >) 2 -j- (COS a COS ^ -COBa COS 0) 2 1 
 
 I 
 
 sn 
 (Art. 15). 
 
 Now let P be the point (a, b, c) on the line PB and Q be the point 
 (a , b\ c) on the line SR, Then as the projection of PQ on AB is 
 AB itself, we have 
 
 AB = (a-a )L + (6-6 )M + (c-c )N = 
 
 (62) 
 (a- a / )(cos^cosv _co!ycos/3 )4-(6 6 ) (cosa cosy - cosacosy )+(c -c ) (cosacos/3 -cosa cos^S) 
 
 If the given lines are expressed in other forms we can find cos a, 
 cos ft, etc, from the given equations and substitute them in (62). 
 
CHAPTER V. 
 
 TRANSFORMATION OF COORDINATES. 
 
 67. To transform to parallel axes through a new origin the coordi 
 nates of which referred to the old axes are a, b, c. 
 
 Let OA = .r, AN = r, NP = z be the coordinates of P referred to 
 the origin O and the axes O,r, Qv and Oz. Also let O be the new 
 origin, and OA = a, A N = b, N O = c be its coordinates and let 
 O H = x , HK =_/ and KP = z be the coordinates of P referred to 
 O as origin and axes parallel to the original axes. (Fig. 15.) 
 
 Then .r == OA = OA + A A 
 
 or x = a + x ) 
 
 Similarly y = b -f r ! (63) 
 
 and z c + z \ 
 
 Substituting these values in the equation of a surface we obtain the 
 equation referred to the new origin and axes. 
 
 68. To pass from a rectangular system to another system tlie origin 
 remaining tlie same. 
 
 Let Ox, Or, Oz be the old axes at right angles to each other ; O_v , 
 O/, Oz the new axes inclined to each other at any angle. (Fig. 16.) 
 
 OM = x, MN =.y, NP = z 
 OM =.< M N =.r , NT = * . 
 
 Now the projection of the broken line OM -f MN + N P on the 
 axis Ox is equal to the projection OM of the radius vector OP on 
 Ox. Let cos a, cos # , cos a" be the cosines of the angles which the 
 new axes make with the axis O.* ; then 
 
 x = x cos a + y cos a + z 1 cos a 1 
 
 40 
 
NOTES ON SOLID GEOMETRY. 41 
 
 If cos ft, cos ft , cos ft" be the cosines of angles which the new 
 axes make with the axis Qy, and cos y, cos /, cos y", the cosines 
 of the angles which they make with Oz, we shall have similar values 
 and z. Hence the three equations of transformation are 
 
 x = x cos a + y cos a + z cos a" j 
 j/ = ^ cos ft + y cos /? + cos /S" > (64) 
 z = x cos y + y cos / + z cos 7". ) 
 
 We have of course 
 
 cos 2 a + cos 2 ft +cos 2 y i } 
 cos 2 a + cos 2 ft + cos* y = i > 
 
 (B) 
 
 cos 
 
 For the angles A, /*, v between the new axes of y and z , of 3 and 
 .r , of ..v and y respectively we have 
 
 cos A = cos a cos a" + cos /J 1 cos ft" + cos y cos ;/ J 
 cos fA = cos r"cos or + cos // cos /? +cos y"cos y I (C) 
 cos r = cos cos a + cos y5 cos ft +cos ^ cos y . } 
 
 69. To pass from one system of rectangular coordinates to another also 
 rectangular. 
 
 The formulae in this case are the same as those in the last with the 
 exception that since the new axes are also rectangular cos A = o, cos // 
 = o, cos v =o and formulae (C) give 
 
 cos a cos a" + cos ft cos ft" + cos y cos y"= o ) 
 cos a" cos a + cos ft"cos ft +cos ^"cos y = o V (D) 
 cos <* cos + cos ft cos /T +cos y cos y o.) 
 
 Since between the nine quantities there are six equations of con 
 ditions, (B) and (D) there are only three of the quantities, cos a, 
 cos ft, etc., independent. 
 
 70. In changing from rectangular axes to rectangular, there is 
 another set of equations of condition among the quantities, cos a, 
 cos ft, etc., equivalent to the preceding which result from the fact 
 that the new axes are rectangular. For a, a , a" being the angles 
 made by the old axis of .v with the new rectangular axes, etc., we 
 must have 
 
 cos 2 tf + cos 2 <* + cos 2 a" = i } 
 cos 2 /? + cos 2 yS + cos 2 ft" = i ( (E) 
 cos 2 ;/ + cos 2 7/ + COS 2 y" = i ) 
 
42 A 07v<:s av SOLID GEOMETRY. 
 
 cos a cos /3 + cos rr cos ft + cos <*" cos y#" = o i 
 
 cos a cos 7 + cos <? cos ;/ + cos " cos 7" = o >- (F) 
 
 cos /? cos 7 + cos ft cos 7 + cos ft" cos 7" o ; 
 
 and the new coordinates expressed in terms of the old are 
 
 .v = .v cos a +jcos ft +scos y \ 
 y ^cos a +ycosft +s cos y (F) 
 z x cos " +y cos/?" +s cos 7" ) 
 
 71. In the study of surfaces by sections made by planes it is often 
 necessary to transform the coordinates in space to coordinates in 
 the cutting plane. To do this we must fix the plane with reference 
 to the old coordinate planes. Let the equation of the plane be given 
 as z= A^ + By. Then the angle 6 which this makes with the plane 
 
 xy is determined bv the equation cos 6 = and the 
 
 angle cp which it traces on that plane makes with the axis of x 
 
 ^ 
 by the equation tan cp = , the trace being Ajv + Bj> = o. 
 
 Let x Oy be the given plane, cutting the plane xy in the line 
 Ox which take for the axis of x and let CV a line perpendicular 
 to it in the given plane be the axis of v and OR = .r , RM =y 
 the coordinates of any point M in the plane referred to the axes Ox\ 
 (V ; also let OQ = .v, QP = r, P.AI =a z be the coordinates of M 
 referred to the old axes O v, Or, Os. Then the angle MRP = 9 
 and xOx = (p. (Fig. 17.) 
 
 Then PR =y cos ft PM =j/ sin 8. 
 
 OQ = OR cos (p + RPsin <p y QP = OR sin <p-RP co^ (p. 
 
 .-. z-y sin 8 \ 
 
 x = x cos q>+y cos 8 sin gj V (65) 
 y = .r sin <py cos 8 cos cp ) 
 
 And if these values be substituted in the equation of any surface 
 F(.v, r, z) o the result will be a relation between x andy, coor 
 dinates of the curve cut from the surface by the plane. 
 
 72. If the cutting plane contain one of the coordinate axes, the 
 formulae are simplified and in many cases sufficiently general. 
 
 Let x Oy (Fig. 18) be the cutting plane containing the axis of y ; 
 Ox its trace in the plane ; zx the axis of x ; PM = .v , OM = _r , 
 
.VOTES ON SOLID GEOMETRY. 43 
 
 the coordinates of any point P in the section ; ON =. x, NQ =.r, 
 QP = 2, the coordinates of P referred to the old axes. Then angle 
 PMQ = 0, and MQ = x cos PQ = x sin 0. . , The formulae 
 of transformation are 
 
 x = x cos 6* ) 
 ^=/ (66) 
 
 z x sin } 
 
 That is, we have only to make x = x cos 0, z x sin 0, y -=.y in 
 the equation of any surface, in order to find the equation of the sec 
 tion of this surface by a plane containing the axis of j 1 and making an 
 angle with the plane xy. 
 
CHAPTER VL 
 
 THE SPHERE. 
 
 73. To find the equation of the sphere. 
 
 i. In rectangular coordinates. 
 
 Let , <, c be coordinates of the Centre, and Radius = R. 
 
 The equation is then (Art. 1 1 )(x aY + (y-l>Y + (z c) z R z (67) 
 or if the origin be at the centre 
 
 ^+/ + 2 2 = R 2 . (68) 
 
 2. Tn oblique coordinates. 
 
 Let A, /<, v be the angles of the axes then the equation is (Art. 16) 
 
 x-a)(y-b) cos A-f 
 2(xa)(zc) cos )A+2(yb)(zc) cos r = R 2 (69) 
 or if the origin be at the centre 
 
 x* + .) 2 + -s 2 + 2xy cos A -f 2AT2 cos yu + 2yz cos r = R 2 . (70) 
 
 3. In polar coordinates 
 
 Let r , a, ft be the polar coordinates of the centre then the equa 
 tion is 
 
 2 2rr (cos ^ cos rt + sin 8 sin ^ cos (q> >5)) = R tf . (71) 
 
 If the pole be at the origin and the centre on the axis of 0, the 
 equation is 
 
 r=2Rcos#. (72) 
 
 Since that is the equation of the generating circle in any one of its 
 positions. 
 
 44 
 
NOTES ON SOLID GEOMETRY. 
 
 74. The Sphere under conditions (coordinates rectangular). 
 The equation (67) may be written 
 
 45 
 
 JV a +>* + * 2CLX 2by 21-2 + 0* + &* + * R 2 O 
 
 or .T 2 -f / + ,s 2 + DA- + Ey + FZ + G = o. (73) 
 
 And since this equation contains four arbitrary constants, the 
 sphere may be made to fulfil four conditions (which are compati 
 ble) and no more. Four given conditions give four equations for 
 determining the constants D, E, F, G, and with these determined 
 we know the radius and centre of the sphere, for we have only by 
 completing the squares to throw the equation (73) into the form 
 
 / DV / ^\ 2 / F\ 2 D 2 E 2 F- 
 \x+-\ + ( y + - ) +(* + -) = + + -- C* 
 \ 2/ \ 27 V a/ 444 
 
 / D E F\ 
 to see that the centre is ( -- , -- , -- ) and the radius is 
 
 V 2 2 2j 
 
 ^ F~~ 
 
 7" 
 
 1. The equation of a sphere passing through a givtn point d, e, f, is 
 v 2 4-y + 2 + D(^-^) + E(.y-^) + F(0-/)-^-^-/ 2 = o. (74) 
 If the given point be the origin the equation is 
 
 = o. (75) 
 
 2. The equation of a sphere cutting the axis of z at distances c and c 
 from the origin is 
 
 x*+^+(z-c)(z-c ) + I)x + Ey= o (76) for" 1 ^ 
 
 must give two values for z, c and c , and this equation fulfils that 
 condition. 
 
 3. The equation of a sphere touching the axis of z at a distance cfrom 
 the origin is 
 
 y = o (77) for this gives two coinci 
 
 dent values of z = c when " > . 
 J =o f 
 
46 NOTES ON SOLID GEOMETRY. 
 
 4. 7 he equation of a sphere touching all three axes at distance a, from 
 origin. 
 
 To meet these conditions the equation must be of such a form as 
 
 je=o ) 
 to give equal roots for z when \ the same equal roots for y 
 
 when * > and the same equal roots for x when - I . Let 
 
 z=o } z=o j 
 
 the distance of points of contact from origin be a, then the equation 
 will be 
 
 = o (78) 
 
 as this fulfils the above conditions. 
 
 5. The equation of a sphere passing through the origin and having its 
 centre on the axis of x is 
 
 * s +y + **= 2R.V. (79) 
 
 6. The equation of a sphere tangent to the plane xy at the point (a, b) 
 is 
 
 (;t--rt) 2 -f -) 2 + * 2 + F*=o (80) 
 
 for then z=o gives x=a, and y=b t a point (a, b) in the plane xy. 
 75. Interpretation of the expression 
 
 (x-a? + (.y-b)* + (z-c)*-W. (i) 
 
 i. Let (x, y, z) be the coordinates of a point P without the 
 sphere whose centre O is (a, b, c] and radius = R and let PM be 
 tangent to this sphere at the point M. Then PM 2 = OP 2 OM 2 . 
 
 Now 
 and hence PM 2 = (x-a)*+ (jy-) 2 + (z-<) 2 -R 2 . 
 
 Therefore the expression (i) is the square of the tangent from the 
 point P to the sphere. 
 
 2. Let P (x, y } z) be a point within the sphere. Join OP and 
 erect a perpendicular PM to OP meeting the sphere in M, and join 
 OM. 
 
 Then PM 8 =OM*-OP = R* -((x-ay + (y-by + (z-c)*} 
 
NOTES ON SOLID GEOMETRY. ^ 
 
 That is the expression (i) becomes negative and represents the 
 square of the half chord through P perpendicular to the radius 
 through P. 
 
 76. .Radical plane of two spheres. 
 
 Def. The radical plane of two spheres is the plane the tangents 
 drawn from any point of which to the two spheres are equal. 
 If the equations of the two spheres are 
 
 the equation of their radical plane is 
 
 o 
 
 For this expresses (Art. 75) that the squares of the tangents from 
 point (_r, y, z) to the two spheres are equal, and moreover it is an 
 equation of the first degree in x, y and z and therefore the equation 
 of a plane. If the spheres intersect their radical plane is their 
 plane of intersection. It may be easily proved that the radical 
 plane of two spheres is perpendicular to the line joining their cen 
 tres. 
 
 77. ,The six radical planes of four spheres intersect in a common 
 point. 
 
 Let S = o, S = o ; S" = o ; S " o be the equations of the four 
 spheres. Then the equations of their radical planes are 
 
 S-S =o S -S" =o 
 
 S-S" = S - S " = o 
 
 S-S " = o S"-S"f:==o 
 
 These may be arranged in groups of four equations, which added 
 vanish simultaneously and therefore the planes intersect in a common 
 point. This point of intersection of the six radical planes is called 
 the radical centre of the four spheres. 
 
 78. Examples : 
 
 i. Find the centres and radii respectively of the spheres 
 
48 NOTES ON SOLID GEOMETRY. 
 
 40 = o. 
 
 \x + 5 jy = o. 
 
 2. F ind the equation of a sphere passing through the origin and 
 the points i, 2, 3, i, 4, 5, 3, > i- 
 
 CHAPTER VII. 
 
 CYLINDERS, CONES, AND SURFACES OF REVOLUTION. 
 
 79. CYLINDERS. Def. A cylinder is a surface generated by the motion 
 of a straight line which always intersects a given plane curve, and is 
 always parallel to a fixed straight line. The moving straight line is 
 called \\\Qgeneratrix; the plane curve which it always intersects is 
 called the directrix or guiding curve. 
 
 80. To find the general equation of a cylinder (coordinates rectangular}. 
 Let m, n, i be the direction cosines of the axis. 
 
 And let > (i) be the equations of the generatrix in 
 
 y nz + q \ 
 
 which m and n are constant since the generatrix remains parallel to 
 the axis. For convenience take the guiding curve in the plane xy, 
 
 its equations will then be ^ ~ \ . (2) Now making z = o in 
 
 z o 5 
 
 the equations (i) we obtain x=p y q for the point in which the 
 generator pierces the guiding curve (x, y) in the plane xy. 
 
 Hence we have F(/>, q) = o, (3) and eliminating the arbitrages/ 
 and q between (i) and (3) we obtain 
 
 ?(x-mz,y-nz) = o (82) 
 
 the general equation of cylinders. 
 
 If the cylinder be a right cylinder with its guiding curve in the 
 plane xy and the axis of z for its axis, then in equation (82) m = o, 
 and = o, and the required equation of the cylinder is 
 
 F(*, y) = o. (83) 
 
NOTES ON SOLID GEOMETRY. 
 
 49 
 
 8 1. Cylinders of second order. We shall confine ourselves to cylin 
 ders whose equations are of the second degree. 
 
 i. To find the equation of the oblique cylinder with circular base. 
 
 Here F(x,y) = x 2 + / R 2 = o. Hence (xmz, ynz) = o 
 gives (x mzY + (y nzf R 2 = o (84) the required equation. 
 
 2. To find the equation of the right cylinder with circular base. If 
 the axis be the axis of z, the equation is J?(xy) = o that is 
 
 3. To find the oblique cylinder with elliptical base. Let the guiding 
 
 curve in plane xy be + ^ = i, z = o. 
 a 2 b- 
 
 x* v* 
 Then (Jf t jf) = ^ + J an ^ the equation is 
 
 
 4. The equation of the right cylinder with elliptical base whose 
 
 x* v* 
 axis is the axis of z is F(x, y) = o, that is, s + ^ = i. 
 
 5. The equation of the right parabolic cylinder whose guiding 
 curve is f = ^dx\ z o, is f $dx. 
 
 82. CONES. 
 
 Def. A cone is a surface generated by a straight line which passes 
 through a fixed point and always intersects a given plane curve. 
 The fixed point is called the vertex, the moving line the generator, 
 and the given plane curve the directrix or guiding curve. 
 
 83. To find the general equation of a cone. 
 
 Let the coordinates of the vertex be (a, b, c) the equation of the 
 
 x a y b z c 
 generator ^ ~~n~ ~T~ ^ and take the dircctrix in the 
 
 plane (xy) its equation being then ^ 1 - r ) ~ L (2). Now if we 
 
 z o ) 
 
 eliminate w and by means of the definition of cone and the equa 
 tions (i) and (2), the resulting equation will be the equation to the 
 cone, the locus of the right line (i). 
 
 Making z = o in (i) the values of* and y, namely, x ~ a ~ mc 1 
 
 y = b nc ) 
 
 which result will be the coordinates of the point in which the gene- 
 5 
 
50 NOTES ON SOLID GEOMETRY. 
 
 rator meets the plane xy and these will consequently satisfy (x,j ) 
 = o the equation of the directrix. We have therefore 
 
 F(a me, bnc) = o (3). But from (i) m =^~ , n = - , 
 
 2 C Z C 
 
 and therefore (3) becomes 
 
 zc z c 
 
 az ex bz cy 
 
 or 
 
 zc 
 
 the general equation of cones. If vertex be on axis of z, then a o 
 
 and b = o and equation (85) becomes F( , ) = o. (86) 
 
 \z c z c ) 
 
 84. Cone with vertex at origin. 
 
 If the vertex of the cone is at the origin and the directrix in a 
 plane parallel to the plane xy, and at a distance c from it then the 
 
 equation of the generatrix will be = = , (i) the vertex (o, o, o) 
 
 and the directrix will be v*^J ~ I . (2) 
 
 z = c ) 
 
 To find the point in which the generator meets the directrix we 
 i). We thus get 
 
 Hence we have (mc, nc) = o, but m = , and n - from (i). 
 Therefore 
 
 <?,7 r )= (B7) 
 
 is the equation required. 
 
 The equation (87) is a homogeneous equation in x,y and z. 
 
 85. Cones of second degree. 1. The equation of an oblique cone with 
 circular base. 
 
 The equation of the directrix is ~F(x, y) = x 2 +y R 2 = o. 
 Hence 
 
 zc zc zc 
 
NOTES ON SOLID GEOMETRY. 5! 
 
 or (az -ex)* + (bzcy}^ K\z-c}\ (88) 
 
 2. To find the equation of a right cone with circular base, the axis of 
 z being the axis of the cone and vertex being (o, o, c). The equation 
 
 . F(*,.> )=-* +y-R s =o 
 of the directrix is v 
 
 z c. 
 
 Hence 
 
 ^ )=o is -^ f 7^-R = 
 
 W s <7 f 2 zc 
 
 R 2 
 or jt- 2 +y= (2 r) 2 (89). This is a cone of revolution about the 
 
 axis of z. 
 
 3. The equation of a right cone with vertex at the origin and circular, 
 elliptical, or hyperbolic bases. 
 
 The equations of the circular base (directrix) are 
 
 Hence 
 
 / ex cy\ . fix 1 c^y 1 2 R 2 
 
 \T" ~z)~ *~z* s 2 "" ""?" ^ 9 
 
 The equations of the elliptical and hyperbolic directrices are 
 
 ^ +- _ i = o ) and | 2 --| - 1 = o \ respective]y . 
 z=c] z= c) 
 
 Hence the cones are 
 
 x 9 f z 2 . . 
 ~ i =oor--+--=-5 (91) 
 
 c*x* W 
 
 86. SURFACES OF REVOLUTION. 
 
 To find the general equation of a surface generated by the revolution of 
 a plane curve generator about the axis of z. 
 
 Let SPi=r be an ordinate of the point P to the axis of z of the 
 
52 NOTES ON SOLID GEOMETRY. 
 
 plane curve and OM = x, MN =y, NP = z the coordinates of P. 
 Then SP 2 = ON 2 OM 2 + MN 2 , or r 2 .r 2 +y 
 
 That is, the distance from any point of revolving curve (gen 
 erator) from the axis of z is r==vC?4j? (i). But r being an ordi- 
 nate of the generating curve to the axis of z we must have by the 
 equation of the curve in any position r = (z) (2). Therefore 
 eliminating the arbitrary r between (i) and (2) we have 
 
 V^+y = F(s) (93) 
 
 the required equation of surfaces of revolution about axis of z. 
 If the curve revolved about the axis of .r the equation is 
 
 Vy +s 2 F(JC). (94) 
 
 87. Surfaces of revolution of second order. 
 
 i. Equation of Cylinder of revolution about the axis of z. The 
 equation of the revolving line is r = a. 
 
 V. -**+ = gives x-+f = a\ 
 
 2. Equation of a Cone of revolution about the axis of z, vertex at 
 (o, o, c). The equation of the generating line is r = m(zc}. 
 
 Hence .# 2 + y 2 = m*(zc)* (95) the required equation where m is 
 the tangent of the angle made by side of cone with axis of 2. 
 
 3. Equation of the Sphere. The equation of the generating curve 
 a* or r =\/a* z 2 . 
 
 Hence 
 
 4. Equation of the Surface generated by the revolution of an ellipse 
 about its conjugate axis. 
 
 r* Z 9 2 /70 OX 
 
 The generator is - + -= i or r~ = ~(b z 2 ). 
 Hence the equation of the surface is 
 
 +=- (96) 
 This is one of the ellipsoids of revolution called the oblate spheroid. 
 
NOTES ON SOLID GEOMETRY, 
 
 53 
 
 5 . Equation of the Ellipsoid generated by the revolution of an ellipse 
 about its transverse axis the {Prolate spheroid]. 
 
 Take the axis of x as the axis of revolution. Then the equation 
 
 jv r 
 of the generator is + 75 = 
 
 _ 
 
 or r 2 = >(# 2 x*)> Hence vj * + s* F(,r) gives 
 a 
 
 v _ x 
 
 i r**; (97) 
 
 the required equation. 
 
 88. Hyperboloids of revolution. Definitions. When the Hyperbola 
 revolves about its conjugate axis it generates the Hyperboloid of revo 
 lution of one sheet. When it revolves about the transverse axis it gen 
 erates the hyperboloid of revolution of two sheets. 
 
 i. Equation of the Hyperboloid of one sheet. Let the axis of z be 
 
 r 2 z 2 a 2 
 
 the conjugate axis then -273 = i or r 2 = j3(s 8 + 2 ). Hence 
 
 
 2. 7%f equation of the Hyperboloid of revolution of two sheets. Take 
 the axis of x as the axis of revolution. Then the equation of the 
 
 a: 2 r 2 b* 
 
 generator is ^ = i or r 2 = (j\r 2 ^ 2 ). 
 
 Hence for the equation of the surface we have 
 
 A- ,- 3 
 
 ?--?-= (99) 
 
 89. Equation of the Paraboloid of revolution about the axis of\. 
 
 The equation of the generator is r 2 \dx. 
 
 Hence the equation of the Surface isjy 2 + 2 2 \dx. (100) 
 
 5* 
 
CHAPTER VIII. 
 ELLIPSOIDS, HYPERBOLOIDS, AND PARABOLOIDS. 
 
 89. To find the equation to the surface of an Ellipsoid. 
 
 Def. This surface is generated by a variable ellipse which always 
 moves parallel to a fixed plane and changes so that its vertices lie on 
 two fixed ellipses whose planes are perpendicular to each other and to 
 the plane of the moving ellipse, and which have one axis in common. 
 
 Let BC, CA (Fig. 19) be quadrants of the given fixed ellipses traced 
 in the planes > ; 0, zx ; OC c their common semi-axis along the axis 
 of z, OA = a (on the axis of x), and OB = b (on the axis of y) the 
 other semi-axes ; QPR a quadrant of the variable generating ellipse 
 in any position, having i s centre in OC and two of its vertices in the 
 ellipses AC, BC, so that the ordinates QN, RN are its semi-axes ; 
 also let ON = z, NM = x, MP =y be the coordinates of any point 
 P in it : 
 
 3? y* 
 Then -^ + .- = i. And since Q is on the ellipse AC we 
 
 have . = i -. Similarly -- = i . 
 
 a c u c 
 
 Hence eliminating RN 2 and QN 2 we have 
 
 * ^L 
 
 - r +- 7 ,-+--r== , (loi) 
 the equation to the surface. 
 
 90. To determine the form of the ellipsoid from its equation. Since in 
 
 .v 2 y 2 z* 
 the equation y -f- y- -f - = i , x can only receive values between a 
 
 54 
 
NOTES ON SOLID GEOMETRY. 
 
 55 
 
 and a, y between b and b, and z between c and c, the surface 
 is limited in all directions. 
 
 A 2 V 2 
 
 If we put z = o we obtain -f r= i, for the equation to the 
 
 trace on xy, which is therefore the ellipse AB. 
 
 x* z 1 
 If we put y o we have ~--\ ^- =i, the ellipse AC. 
 
 If we put x = o we have 1 r -f = i, or the ellipse BC. 
 
 These three sections by the coordinate planes are called the princi- 
 pal sections, and their semi-axes a, 6, c, are the semi-axes of the ellip 
 soid ; and their vertices the vertices of the ellipsoid, of which it has six. 
 
 If we make z=h we have 
 
 the equations of any section parallel to xv, which is an ellipse similar 
 to AB, since its axes are in the ratio of a to b, whatever be the value 
 of h, and which becomes imaginary when h > c. In the same man 
 ner all sections parallel to xz 9 and yz are ellipses respectively similar 
 to AC and BC. The whole surface consists of eight portions pre 
 
 cisely similar and equal to that represented in the figure. 
 
 yA ^_ j,2 z z 
 Cor. If =0 the ellipsoid becomes ^ t-- = i the ellip 
 
 soid of revolution about the axis of z, Art. (87), all the sections of 
 which by planes parallel toyz, are circles. Hence the spheroids may 
 be generated by a variable circle moving as the variable ellipse, in 
 Def. Art. (89). 
 
 91. To find the equation to the hyperboloid of one sheet. 
 
 Definition. This surface is generated by a variable ellipse, which 
 moves parallel to a fixed plane, and changes so that its vertices rest 
 on two fixed hyperbolas, whose planes are perpendicular to each 
 other, and to the plane of the moving ellipse, the two hyperbolas 
 having a common conjugate axis coincident with the intersection of 
 their planes. (Fig. 20.) 
 
 Let AQ and BR be the given hyperbolas traced in the planes zx,yz ; 
 OC = c their common semi-conjugate axis coinciding with the axis 
 of z ; OA = a, OB = b the semi-transverse axes ; QPR the generating 
 ellipse in any position having its plane parallel to xy, its centre in 
 
56 NOTES ON 1 SOLID GEOMETRY. 
 
 OC, and its vertices in the hyperbolas AQ, BR, so that the ordinates 
 NQ, NR, are its semi-axes. Also, let MN = x, MP = y, ON 2, 
 be the coordinates of any point P in the generating ellipse ; then the 
 ellipse PQR gives 
 
 "NQ 2 + NlT 2 = I 
 
 NQ 2 z* 
 Also from hyperbola AQ 2 "~=i. 
 
 NR 2 z" 
 And from hyperbola BR -- ^ i. 
 
 Hence, 
 
 f 
 
 A 8 V 2 Z 2 
 
 or - + ^- - i (102) the equation to the surface. 
 
 92. To determine the form of the hyperboloid of one sheet from its 
 equation. 
 
 Since the equation (102) admits values of x, y and z positive and 
 negative however large, the surface is extended indefinitely on all 
 sides of the origin. If we put z = o we obtain 
 
 ~z- + ~i = i for the trace on xy which is the ellipse AB. Similarly 
 
 r 2 s 2 
 the sections by the planes xz and j 2 are respectively ^ - i the 
 
 v 2 z 
 hyperbola AQ, and 1 -^ ^ = i the hyperbola BR. The ellipse AB 
 
 and the hyperbolas AQ and BR are the principal sections. The sections 
 parallel to xy are all ellipses similar to and greater than AB. The 
 sections parallel to xz and yz are hyperbolas similar to the principal 
 sections. 
 
 The semi-axes a and b are called the real semi-axes of the surface 
 and c the imaginary semi-axis, since x = o and y = o give z = 
 c\/ i. The extremities of the real axes are called the 
 vertices of the surface. The surface is continuous and hence is 
 called the hyperboloid of one sheet. The hollow space in the inte 
 rior of the volume of this hyperboloid of which the ellipse AB is the 
 smallest section has the shape of an elliptical dice-box. 
 
NOTES ON SOLID GEOMETRY. 
 
 57 
 
 Cor. If b = a the equation becomes = _ = i that of the 
 
 fl 2 c j 
 
 hyperboloid of revolution of one sheet. Its sections parallel to xy 
 are all circles. 
 
 93. To find the equation to the hyperboloid of two sheets. 
 
 Definition. This surface is generated by a variable ellipse which 
 moves parallel to itself, with its axes on two fixed planes at right an 
 gles to each other and to the plane of the generating ellipse and ver 
 tices in two hyperbolas in those planes having a common transverse 
 axis. 
 
 Let AQ and AR be the given hyperbolas traced in the planes zx, 
 xy, OA = a their common semi-transverse axis along the axis of x, 
 OB = b OC = c the semi-conjugate axes along the axes of y and z ; 
 QPR the generating ellipse in any position having its plane parallel 
 loyz t its centre in Ox, and its vertices AQ, AR so that the ordinates 
 QN, RN are its semi-axes. Let ON = x, MN =j>, MP = z be the 
 coordinates of any point P in the ellipse. (Fig. 21.) 
 
 
 also from hyperbola AQ " i 
 
 c a 
 
 and from hyperbola AR "- = i. 
 
 l> a 1 
 
 Hence 
 
 
 - >-? - 
 
 the equation to the surface. 
 
 94. To determine the form of the hyperboloid of two sheets from its equa 
 tion. 
 
 The equation shows that all values of x between -f# and a give 
 imaginary results, therefore no part of the surface can be situated be 
 tween two planes parallel toyz through A and A the vertices of the 
 common transverse axis ; but the equation can be satisfied by values 
 
58 NOTES ON SOLID GEOMETRY. 
 
 of x, > , z, indefinitely great, therefore there is no limit to the distance 
 to which the surface may extend on both sides of the centre. 
 
 V* Z* 
 
 If we make x-= o we have ~- -\ = 
 
 o c 
 
 for the principal section by the plane yz. For x = h and h > a 
 
 v* z* // 2 
 we have - -\ ^- - 2 i which represents similar ellipses. The 
 
 principal sections by the planes xy and zx are AR and AQ, respec 
 tively. For the sections parallel to xy and putting z = I we 
 
 have ^ = = i + a hyperbola similar to AR with its vertices 
 
 in AQ and the opposite branch of that hyperbola and conjugate axis 
 parallel to Oy. In the same way the sections parallel to zx are hy 
 perbolas similar to AQ with vertices in AR and its opposite branch 
 and conjugate axes parallel to Qz, 2a is the real axis of the surface 
 and its vertices the vertices of the surface. The axes 2b and 2c 
 are the imaginary axes of the surface as it cuts neither y nor z. The 
 whole surface consists of two indefinitely extended sheets perfectly 
 similar and equal, separated by an interval. Hence its name. 
 
 Cor. \ib-c the equation becomes ~ ^ = i the equa 
 tion to the hyperboloid of revolution about its transverse axis. 
 
 95. Asymptotic cones to the two hyperboloids. 
 
 i. The hyperboloid of one sheet has an interior asymptotic cone. 
 
 x* r 2 z* 
 Putting its equation T + ^ -j- = i (i) in the form 
 
 
 4. -=s /I ^ -- - . (2) Now when z is very great 
 
 ,- is very small, and hence the limiting form of (2) for z increased 
 
 without limit is 
 v 
 
 r + ~- = r (3) the equation of an elliptical cone having 
 
 its vertex at the origin and its elliptical section parallel to xy. 
 
 Moreover, this elliptical section is always within the corresponding 
 
NOTES ON SOLID GEOMETRY. 
 
 59 
 
 section of the surface by the same plane. For putting z h in 
 (i) and (2) respectively we have 
 
 ~-?- + ~rr = l + r f r tne section of the surface 
 a 1 // c* 
 
 x * / /r 
 
 ^- + 4^- = 7- for the section of the cone. 
 
 a 2 b 1 c 2 
 
 This cone is asymptotic to the hyperbola. 
 
 2. The hyperboloid of two sheets has an exterior asymptotic cone, 
 
 x~ V* z~ 
 
 Putting the equation , ^-r r = i (i) under the form 
 
 a 2 (j r 
 
 we have as a limiting form of 
 
 this equation when jy and z increase without limit, 
 
 jv 2 j 2 z~ 
 
 = ----- -| --- (2) an elliptical cone with vertex at the origin 
 
 and with an elliptical section parallel to the plane \z. Moreover, 
 this elliptical section is greater than the corresponding section of 
 the surface by the same plane. For putting x .h in (i) and (2) 
 
 i 2 2 h* 
 
 respectively we have -,- H --- -- = 5 i 
 * 2 
 
 * 
 
 ^ "T" 55 ?" 
 
 This cone is asymptotic to both branches of the hyperboloid. 
 
 96. To find the equation to the elliptic paraboloid. 
 
 Definition. This surface is generated by the motion of a parabola 
 whose vertex lies on a fixed parabola, the planes of the two parabolas 
 being perpendicular to each other, their axes parallel and their con 
 cavities turned in the same direction. 
 
 Let OR be a parabola in the plane xy, its vertex at the origin, its 
 axis along the axis of .r, and / its latus rectum ; RP the generating 
 parabola in any position with its plane parallel to zx, vertex in OR, 
 and axis parallel to O.v, and let / denote its latus rectum. Also let 
 ON x, NM = r, MP z be the coordinates of any point P in it : 
 also draw RM parallel to Oy. (Fig. 22.) 
 
60 NOTES ON SOLID GEOMETRY. 
 
 Then z* = I . RM = / .M N and / = I .OM ; 
 
 . . j- + -j- = x ( IO 3) the equation to the surface. 
 
 97. To determine the form of the elliptic paraboloid from its equation. 
 Since only positive values of x are admissible, no part of the sur 
 
 face is situated to the left of the planers. But the surface extends 
 indefinitely in the positive direction of x. If we makej = o, 2 2 = I x 
 is the equation to the principal section OQ, and all sections parallel 
 to zx are parabolas equal to OQ, with vertices in OR ; similarly, all 
 sections parallel to xy are parabolas equal to the other principal sec 
 tion OR, with vertices in OQ. If we make x h we have 
 
 21 + Jl-, 
 Ih + I h ~ 
 
 Therefore the sections parallel to zyare similar ellipses, and hence its 
 name. 
 
 Cor. If /= / the equation becomes j 2 + 2 = /.v, the paraboloid of 
 revolution. 
 
 98. To find /he equation to the hyperbolic paraboloid. 
 
 Definition. This surface is generated by the motion of a parabola 
 whose vertex lies on a fixed parabola, the planes of the two parabolas 
 being perpendicular to each other, their axes parallel, and their con 
 cavities turned in opposite directions. (Fig. 23.) 
 
 Let OR be a parabola in the plane of xy, vertex at the origin, and 
 axis along with the axis of x, and / its latus rectum, RP the generat 
 ing parabola in any position, vertex in OR, axis parallel to O.v, 
 and let t denote its latus rectum, and ON = x, NM =y, MP z, 
 the coordinates of any point P in it ; draw RM parallel to Oy. 
 Then 
 
 z-=! . MR and/=/.OM ; 
 
 but OM - MR = ON = AT. 
 
 V 2 2 2 
 
 Hence : --- - = x (104), the equation of the surface. 
 
 99. To determine the form of the hyperbolic paraboloid from its equation. 
 The surface cuts the coordinate axes only at the origin, and since 
 
 the equation admits positive and negative values of.r, v, s, as great 
 
NOTES ON SOLID GEOMETRY. 6 f 
 
 as we please, the surface extends indefinitely both ways from the 
 origin. 
 
 If we makejy = o we have z~ = I x the principal section, the para 
 bola OQ, with its concavity turned towards the left ofyz, and all sec 
 tions parallel to zx are parabolas equal to OQ with their vertices in 
 OR. Making z = o we have y 2 = Ix the parabola OR, and sections 
 parallel to xy are parabolas equal to OR with vertices in OQ. 
 
 If we make x o we have the principal section in yz, 
 
 z^ I = y +Jl or two straight lines through the origin; and for sec 
 tions parallel toyz making x = h we have 
 
 - -- f = i a hyperbola with its vertices in OR, and con 
 jugate axis parallel to Oz. For h negative the section becomes 
 
 2 y 2 
 
 7- - i a hyperbola with its vertices in OQ, and conjugate 
 In Ik 
 
 axis parallel to O.v. 
 
 The surface has but one vertex, and consists of one sheet and one 
 infinite axis. 
 
 100. Asymptotic planes to the hyperbolic paraboloid. 
 
 v* z* 
 The equation --- -jr = x may be written 
 
 j 2 z* / l x\ 
 z -j~ == ~jr ( T + ~7 ) which has for its limiting form 
 
 / / \ z J 
 
 y l Z <1 
 
 when y and z become infinitely great with regard to x, - = 7- , 
 
 V z v z 
 
 or - : = = -. This represents two planes =. = H -- and 
 
 V ^ 
 
 - = = -- - . through the origin and asymptotic to the surface. 
 
 V? 
 
 These planes contain the asymptotes to all the hyperbolic sections of 
 the surface parallel toj 2. 
 
 101. The elliptic and hyperbolic paraboloids are particular cases of the 
 ellipsoid and hyperboloid of one sheet respectively when the centres of these 
 surfaces are removed to infinite distance. 
 
 Take the equation + ^- H =i, and transfer the origin to 
 
62 NOTES ON SOLID GEOMETKY. 
 
 the left vertex of the axis 20, (a, o, o). (New coordinates being 
 parallel to the primitive.) 
 
 -,. or 
 
 - 1 . v J 1 o 
 
 a 2 
 
 r 2 y* 
 
 or multiplying through by a - -- t- -y^- =2jr (i), 
 
 in which , and are the semi-latera recta of the principal sec- 
 a a 
 
 V 1 c* 
 
 tions in xy and zx. Now make a =. oo and put - and , 
 
 a a 
 
 which remain finite, equal to / and / respectively. 
 .*. (i) becomes 
 
 =y- JT 2x, the equations to the paraboloids. 
 
 1 02. The equations of the surfaces of the second order which we have 
 been studying are of the two forms 
 
 =D (i) 
 
 =kA* (2) 
 
 and we will show hereafter that all the surfaces of the second degree 
 may by transformation of coordinates be included in these two forms. 
 
 The first form (i) includes the sphere, ellipsoid, hyperboloids, 
 cones of second order, elliptical and hyperbolic cylinders which 
 have centres. For if .r, -y, z be written for (x, y, z) in (i) 
 the equation is not altered, therefore for every point P (x, y, z) on 
 the surface there is a point P ( x, y, z) and PP passes through 
 the origin O and is bisected in O. 
 
 Moreover, the coordinate planes bisect all the chords parallel to 
 the axes perpendicular to these planes respectively and are principal 
 planes of the surface. 
 
 The second form (2) includes the elliptic and hyperbolic parabo 
 loids and the parabolic cylinder which have a centre at an infinite 
 distance. 
 
 The planes ^.s: and zx are principal planes of the two paraboloids, 
 the other principal plane being at an infinite distance. 
 
 Also both families may be represented by the equation 
 
NOTES ON SOLID GEOME IRY. 63 
 
 the origin being at the vertex and A = o when the surfaces have 
 no centre. 
 
 EXAMPLES. 
 
 1. Construct the sphere whose polar equation is 
 
 r = a sin 6 cos cp. 
 
 2. Find the locus of the point the sum of the squares of the dis 
 tances of which from n fixed points is constant. 
 
 3. Find the locus of the point the ratio of the distances of which 
 from two fixed points is constant. 
 
 4. Find the equation of the surface generated by the motion of a 
 variable circle whose diameter is one of a system of parallel chords 
 of a given circle to which the plane of the variable circle is perpen 
 dicular. 
 
 5. The sphere can be represented by the simultaneous equations 
 
 x = a cos cp cos 6 } 
 y = a cos cp sin 6 > 
 z = a sin cp 
 
 6. The ellipsoid may be represented by the equations 
 
 x = a cos <p cos 6 } 
 y b cos cp sin 6 Y 
 z = c sin cp 
 
 7. The hyperboloid of one sheet may be represented by the equa 
 tions 
 
 x a sec cp cos 6 \ 
 y b sec cp sin 6 \ - 
 z = c tan gj 
 
 8. The hyperboloid of two sheets may be represented by the equa 
 tions 
 
 x a sec cp \ 
 
 y = b sin 6 tan cp > 
 z c cos 6 tan cp ) 
 
 9. A line moves so that three fixed points on it move on three 
 fixed planes mutually, at right angles. Find the locus of any other 
 point P on its line. 
 
64 NOTES ON SOLID GEOMETRY. 
 
 Solution : 
 
 Let the three fixed planes be the coordinate planes (x,y,z) the coordinates 
 of P. A, B, C the points in which the line meets the coordinate planes of yz, 
 xz, xy, respectively. Take PA=<z, PB=/>, PC=c, ON=.r, NQ=,y, QP = z, 
 <ACA =:<p, <CB .r=:0 (CA being the projection of CA on the plane xy and B 
 the projection of B on the axis of x). 
 
 Then x=a cos cp cos Q,y=b cos q> sin 6, z= c sin q>, and therefore the sur 
 face is an ellipsoid. 
 
 10. Find the locus of a point distance of which from the plane xy 
 is equal to its distance from the axis of z (coordinates rectangular). 
 
 11. Find the locus of the centres of plane sections of a sphere 
 which all pass through a point on the surface. 
 
 12. Find the equation of the elliptical paraboloid as a surface 
 generated by the motion of a variable ellipse the extremities of whose 
 axes lie on two parabolas having a common vertex and common 
 axis and whose planes are at right angles to each other. 
 
 13. Find the equation of the hyperbolic paraboloid as generated 
 in a similar manner by the motion of a variable hyperbola. 
 
 14. Construct the surface r sin 6 = a. 
 
 15. Find the equation to the surface B = JTT in rectangular coor 
 dinates. 
 
CHAPTER IX. 
 RIGHT LINE GENERATORS AND CIRCULAR SECTIONS. 
 
 103. SURFACES of the second degree admit of another division, viz. 
 into those which can be generated by the motion of a straight line 
 and into those which cannot. This property which we have seen to 
 belong to the cylinder and cone we shall now show to belong also to 
 the hyperboloid of one sheet and the hyperbolic paraboloid. The 
 ellipsoid being a closed finite surface does not possess this property ; 
 nor the hyperboloid of two sheets, since that consists of two surfaces 
 separated by an interval ; nor the elliptical paraboloid, since that is 
 limited in one direction. 
 
 104. Straight line generators of the hyperboloid of one sheet. 
 The equation of the hyperboloid of one sheet 
 
 a 2 y 2 z* .r 2 z* r 
 
 v- + ^ 5- i mav be written - -r- = i ~-~ 
 
 a z o* c a 2 c b 1 
 
 fx z\ fx z\ _ f y\( y\ - ( \\ 
 
 \a c ) \a c ) \ b ) \ b) 
 
 Now (A) is satisfied by the pair of equations 
 
 (B) 
 
 Z f V\ 
 
 + 7 =K I + i)J 
 
 and also by the pair 
 
66 A OTES ON SOLID GEOMETRY. 
 
 And m being arbitrary equations (B) represent a system of straight 
 lines, and all of these lie on the hyperboloid as the two equations 
 together satisfy the equation to the hyperboloid. 
 
 Similarly equations (C) represent another and distinct system of 
 straight lines which also lie on the hyperboloid which is the locus of 
 both systems, and we shall see the lines of either system may be used 
 as generators of the surface. 
 
 105. No tivo generators of the same system intersect one another, 
 For example take two of the system (B), 
 
 / /-v A v 1 
 
 m i ) = i - -. 
 
 \a c J b 
 
 (0 
 x 
 
 a 
 
 (2) 
 j 
 
 a c \ j 
 
 Combining the first equation of (i) with the first of (2) we obtain 
 
 (in m") ( I J = 
 
 Combining the second equation of (i) with the second of (2) we 
 have 
 
 or v ~ . 
 
 These values for y being incompatible the lines do not intersect. 
 
 1 06. Any generator of the system (B) will intersect any generator of 
 the system (C). 
 
 Take 
 
 , (*--*-}= t-Z 
 c / b 
 
 of system (B) 
 
 (x z \ , ( y\ 
 
 t + r) = w ( + i) J 
 
NOTES ON SOLID GEOMETRY. 67 
 
 U (4) of system (C). 
 
 Eliminating x, y, and a; we obtain the identity m m" m m", 
 therefore the lines intersect. 
 
 Hence, through any point of an hyperboloid of one sheet two 
 straight lines can be drawn lying wholly on the surface. 
 
 107. No straight line lies on an hyperboloid which does not belong to 
 one of the systems of generating lines (B) or (C). 
 
 For, if possible, suppose a straight line H to lie entirely on the 
 hyperboloid, it must meet an infinite number of generating lines of 
 both systems (B) and (C). Let two of these (one of B and one of 
 C) intersect H in two different points, \ve could then have a plane in 
 tersecting the surface in three straight lines, which is impossible since 
 the equation is of the second degree. Hence no such line as H can 
 lie on the surface. 
 
 1 08. The hyperboloid of one sheet may be generated by the motion of a 
 straight line resting on three fixed straight lines which do not intersect, and 
 which are not parallel to the same plane. 
 
 In the first place it is necessary that the motion of a right line 
 which is to generate a surface should be regulated by three condi 
 tions. For, since its equations contain four constants, four condi 
 tions would fix its position absolutely ; with one condition less the 
 position of the line is so far limited that it will always be on a certain 
 locus whose equation can be found. 
 
 Take then three fixed generating lines of the system (B), these do 
 not intersect, nor are they parallel to the same plane. Now, if a 
 straight line move in such a manner as always to intersect these three 
 straight lines, it will trace out the hyperboloid of which they are the 
 generating lines. 
 
 For the moving line meets the hyperboloid in three points (one 
 on each of the fixed straight lines), and hence must necessarily lie 
 wholly upon the surface. For the equation of intersection of a line 
 and this surface being a quadratic equation, if satisfied by more than 
 two roots, it is satisfied by an infinite number. The moving straight 
 line, therefore, in its different positions, will generate the hyper 
 boloid. 
 
68 NOTES ON SOLID GEOMETRY. 
 
 109. Lines through the origin parallel respectively to generators of the 
 systems (B) and (C) lie on the cone 
 
 x* y z- 
 
 r + -73- = = asymptotic to the hyperboloid. 
 For this equation of the cone may be put in the form 
 
 A A A + A = _,vz, 
 
 \ a c J \a c t b b 
 
 which gives two systems of lines through the origin lying on the 
 cone, one system evidently parallel to the lines (B) and the other to 
 the lines (C). 
 
 no. The projection of a generating line of either system upon the 
 principal planes., is tangent to the traces of the surface on those planes. 
 
 The equation of the trace of the surface on the plane zx is 
 
 The projection of the line of system (B) on xz 
 
 f_ i\ + * + =2m . or ^}. * + 1=*. i = I (I) . 
 
 \ a c J a c 2m a 2m c 
 
 "V P 
 
 Now. the condition that a line in the form ~| = i shall be 
 
 p q 
 
 .r 2 z 1 a* c* 
 tangent to the hyperbola ~ F =I IS ~^~ T T - 
 
 This -condition is fulfilled by the projection (i), for 
 
 ( +!) (i-m Y 
 
 Hence this projection is tangent to the hyperbola. 
 
 III. The straight line generators of the hyperbolic paraboloid. 
 The equation of the hyperbolic paraboloid 
 
NOTES ON SOLID GEOMETRY. 69 
 
 =y --- = x may be written 
 
 And hence it is satisfied by the pair of equations 
 
 y 
 V7 
 
 
 
 n + ^) =i _ 
 
 or by the pair 
 
 y z } 
 
 HE). 
 
 Hence the surface has two systems of straight line generators (D) 
 and (E). 
 
 The lines of both systems are parallel to the asymptotic planes 
 of the surface respectively. The equations of these planes being 
 
 z 
 = = o and 
 
 112. We can show in the same manner as in the Articles (34) 
 and (35) that no two lines of the same system intersect ; and that a 
 line of either system intersects all the lines of the other system, and 
 that no other line than the lines of these two systems can lie on the 
 hyperbolic paraboloid. And hence that through every point of the 
 surface two lines may be drawn which lie wholly on the surface. 
 And as in (108) that this paraboloid may be generated by the motion 
 of a straight line which rests on two fixed straight lines and is con 
 stantly parallel to a fixed plane ; also by a straight line which rests 
 on three fixed straight lines which are all parallel to the same plane. 
 
 113. The projections of the generating lines on the principal planes are 
 tangent to the principal sections of the paraboloid. 
 
 The principal section in xy is y* = Ix (i). 
 
NOJ^ES ON SOLID GEOMETRY 
 The projection of any line of the system (D) on xy is 
 
 i 
 
 m 2 2m 
 
 j= mx ^ or y = - jp-fJ . ( 2 ) 
 
 /t / / -WJ O ^-rw * / 
 
 Now the tangent line to the parabola y>= Ix is of the form 
 
 y tx + : and if / = then = -5 -. 
 4^ 2 4/ 2m 
 
 Hence the projection (2) is tangent to the section y = lx t 
 
 114. Distinctions of surfaces of second order generated by straight 
 lines. 
 
 All the generators of the cone intersect in one point. All 
 the generators of the cylinder are parallel. Hence cones and 
 cylinders are called developable ruled surfaces. In the case of the 
 hyperboloid of one sheet and the hyperbolic paraboloid, the gen 
 erators of neither system intersect or are parallel. These are 
 styled skew ruled surfaces. The distinction between these last two 
 surfaces is that the generators in the paraboloid are parallel to a fixed 
 plane. 
 
 115. Plane sections of surf aces of the second order. 
 
 If we intersect the surfaces represented by the general equation 
 
 A* 2 + By 2 + Os 2 + 2h. xz + 2B> + 2Cxy + 2& x+2 B"y+2C"z - D 
 by the plane z = o we will obtain 
 
 A^ 2 + B| 2 + 2C A7 |-2A"jt: + 2B y = D (i) a conic section. 
 
 If we intersect it by a plane z a we have for the curve of inter 
 section 
 
 Aa 2 -h B^ 2 + 2C xy+ 2G x + 2H>=D , 
 
 a conic similar to the conic (i). 
 
 Therefore sections of surfaces of the second order by parallel 
 planes are similar curves, and hence, in determining the form of these 
 sections we may confine ourselves to the discussion of sections through 
 the origin. 
 
NOTES ON SOLID GEOMETRY. ^ 
 
 1 1 6. To determine the nature of the curve formed by the intersection of 
 a surface of the second order by any plane. 
 
 Take the equation 
 
 = 2A r ,r. And in order to get the equation of the 
 curve of intersection in its own plane 
 
 Make 
 
 x x cos q> -f y cos 6 sin cp 
 
 y zz: x sin cp y cos 6 cos cp 
 z =y f sin 6. See Art. (71). 
 
 Arranging the result we have 
 
 x *(K cos 2 r/> + B sin 2 cp) + 2x y (K B) cos 6 sin cp cos cp 
 
 +/ 2 ((A sin 2 v + B cos 2 93) cos 2 +C sin 2 0) = 2 AV cos <p 
 
 -f 2Aj/ cos 6 sin (>, 
 
 the equation to a conic section which will be an ellipse, parabola or 
 hyperbola, (including particular cases of these curves,) according 
 as the quantity 
 
 (A B) 2 cos 2 6 cos 2 cp sin 2 cp (A cos 2 <p + B sin 2 <p)(A cos 2 6 sin 2 cp 
 
 + B cos 2 #cos 2 <p + Csin 2 0) 
 
 or -AB cos 2 cp-AC cos 2 ^ sin 2 BC sin 2 <p sin 2 8, (i) 
 is negative, zero or positive. 
 
 Hence every section of an ellipsoid is an ellipse because A, B and 
 C are all positive. 
 
 The sections of the hyperboloids may be ellipses, parabolas or 
 hyperbolas since one or two of the quantities A, B and C will then 
 be negative. 
 
 For paraboloids A = o. Hence for the elliptic paraboloid in 
 which B and C have the same signs the section is an ellipse ; except 
 when B = o or cp = o in which cases it is a parabola. 
 
 For the hyperbolic paraboloid since B and C are of contrary signs 
 the section is a hyperbola except when 6= o or cp=o when it is a 
 parabola. 
 
 1 1 7. Circular sections. Since the section is referred to rectangular 
 axes it cannot be a circle unless the coefficient of xy vanishes 
 
72 NOTES ON SOLID GEOMETRY. 
 
 or (A B) cos 6 sin cp cos cp o 
 
 7T 7T 
 
 or # or &= - , or cp o 
 
 2 2 
 
 which shows thaty^r # circular section the cutting plane must be perpen 
 dicular to one of the principal planes of the surface, 
 
 1 1 8. Let us now examine the surfaces of the second order for cir 
 cular sections. 
 
 Take first the surfaces having a centre and therefore represented 
 by the equation 
 
 A* 2 + B/ + Cz 2 = i. (i) 
 
 Since every circular section must be perpendicular to a princi 
 pal plane, let the cutting plane contain the axis of y, and make 
 the angle 6 with the plane xy 
 
 To transform (i) to this plane make 
 x = x cos 6 
 
 y=y 
 
 z x sin 6. Art. (72). 
 
 Hence we have 
 
 .v"(A cos 2 + C sin 2 0) + B/* = i (2) 
 which represents a circle if 
 
 Acos 2 #-fC sin 2 B 
 B-A 
 
 or 
 
 tan 2 9 = . (3) 
 
 We must now examine for each of the surfaces which axis it is 
 that coincides with the axis ofj>. 
 
 i. For the ellipsoid A = , 3=-^-, C = 
 
 Hence for a real B b must lie (in value) between a and c or the 
 axis of the surface to which the cutting plane of circular sections is 
 parallel is its mean axis. 
 
 2. For the hyperboloid of one sheet since we cannot have B ne- 
 
NOTES ON SOLID GEOMETRY. 
 
 gative we must put A = ^ B = C - 
 
 . . b > a or the cutting plane is parallel to the greater of the real 
 axes. 
 
 3. For the hyperboloid of two sheets since we cannot have A and 
 C negative, we must put 
 
 .*. b > c or the cutting plane is parallel to the greater of the im 
 aginary axes. 
 
 Since tan 6 has two equal values the cutting plane may be inclined 
 at an angle 6 or 180 6 to the plane of xy. Hence there are two 
 sets of parallel circular sections of the surfaces having a. centre. If 
 the surface becomes one of revolution we have tan 6 = oo or o, and 
 the two positions of the circular sections coincide with each other, 
 and are parallel to the two equal axes. 
 
 119. Secondly. For the surfaces not having a centre, we take 
 equation B_/ + Cs 2 = 2k x (i). 
 
 i. For the elliptic paraboloid, B and C have the same sign. 
 Transforming (i) we have B/ 2 + Or 2 sin 2 # 2K x cos 6\ and hence 
 for circular sections we must have the condition C sin 2 = B, or 
 
 = \ - 
 
 sin 6 = \ - Therefore the cutting plane is perpendicular to the 
 ^ 
 
 principal section whose latus rectum is least. 
 
 2. For the hyperbolic paraboloid, since B and C have different 
 signs, sin 6 is imaginary, and no plane can be drawn which shall in 
 tersect it in a circle. This was evident, too, from the fact (Art. 
 116) that the hyperbolic paraboloid can have no elliptic sec 
 tions. 
 
 i 20. Then, to sum up, all the surfaces discussed with the excep 
 tion of the hyperbolic paraboloid admit of two sets of planes of cir- 
 
 7 
 
74 
 
 NOTES ON SOLID GEOMETRY. 
 
 cular sections. Therefore they can be generated by the motion of a 
 variable circle whose centre is on a diameter of the surface. 
 
 121. The planes of circular section may be found directly from 
 the equations of the surfaces, as follows: 
 
 The equations of the central surfaces 
 
 may be written 
 or 
 
 which shows that either of the planes 
 
 <\/A BJI +V^-^ = o (i) <v/A B.r VB C.s = o (2) 
 cuts the surface in the same line in which it cuts the sphere 
 
 Hence the planes (i) and (2) and all planes parallel to them cut 
 the surface in circles. 
 
 The equation to the elliptic paraboloid may be treated in a similar 
 manner, thus showing its planes of circular section. 
 
 122. Sections of Cones and Cylinders. 
 
 i. The sections of the cones may be inferred from Art. 95. For 
 elliptic cones, sections of the hyperboloids by any plane are always 
 similar to the sections of the asymptotic cone to the surface made by 
 the same plane, as is evident from the equations respectively. Hence 
 the section of a cone of revolution by a plane will give an ellipse, 
 parabola, or hyperbola. But we will examine this case more par 
 ticularly. 
 
 In the equation of the cone of revolution 
 
 A . + /= L 2 ( 2 _, ), or .v>+y = (*-<) 
 
 (when - - J put x=x cos 0} 
 
 tan v J 
 
NOTES ON SOLID GEOMETRY, 75 
 
 And we have for the curve of intersection by the plane containing 
 the axis ofjy 
 
 jr 2 (cos 2 #tan 2 v sin 2 6) + _/* tan 2 v+2cx sin 6c 1 = o (i). 
 
 This equation (i) represents an ellipse, parabola, or hyperbola, 
 according as cos 2 6 tan 2 v sin 2 6 is > = < o, that is according as 
 tan #<=:> tan v. 
 
 2. For the cylinder of revolution about the axis of z, we make 
 x=x cos 0,y y in its equation x*+y* = r*; /, the curve of in 
 tersection is a- 2 cos 2 04-/ 2 = r* an ellipse. 
 
 EXAMPLES (Coordinates Rectangular). 
 
 1. Find the right line generators of the hyperboloid 
 
 *+>_:=, 
 
 9 4 i 
 
 for the point (2, 3 ?) on the surface. 
 
 2. Find the right line generators of the paraboloid 4jy 2 2$z 2 = icxxr 
 for the point (? 2. i) on the surface. 
 
 3. Find the planes of circular sections of the following surfaces : 
 
 36 (0 
 =i44 (2) 
 i2 (3) 
 (4) 
 
 ^ -f. j; 2 g 2 
 
 4. In the hyperboloid of revolution of one sheet f --- r = i 
 
 find the equations of the generating line whose projection on the 
 plane xz is tangent to hyperbolic section in that plane at its vertex. 
 
 5. Find the sections of the cone x*+y*=(z 2) 2 by planes con 
 taining the axis of j/, at angles to the plane xy of 30, 45, and 60 
 respectively. 
 
 6. Find the curve of intersection of the surface 
 
 by a plane inclined at an angle of 30 to the plane xy, and whose 
 trace on that plane makes an angle of 45 with the axis O.v. 
 
CHAPTER X. 
 
 TANGENT PLANES, DIAMETRAL PLANES, AND 
 CONJUGATE DIAMETERS. 
 
 123. Straight line meeting surfaces of second order. 
 We can transform the general equation 
 
 to polar coordinates by writing x = /r, y mr t z nr, (when /, m, n 
 are in rectangular coordinates, direction cosines, and in oblique co 
 ordinates, direction ratios). The equation becomes 
 
 r*( A/ 2 + Bfl* 2 + Cri 2 + 2K ?nn + 2 Win + 2C7m) 
 
 +C" + F = o. 
 
 Hence a straight line meets the surface in two points, and if these 
 two points be coincident the line is tangent to the surface. 
 
 124. Tangent Plane to surfaces of second order. 
 
 Let the origin be on the surface (and therefore F= o) then one 
 of the values of r in (2) is r = o. Now, in order that the radius 
 vector shall touch the surface at the origin, the second root must be 
 o, and the condition for this is A"/+B"w + C"n o. Multiplying 
 this by r and replacing Ir, mr, nr by x, j>, z, this becomes 
 
 = o. (3) 
 
 Hence the radius vector touching the surface at the origin lies in the 
 fixed plane (3); and as /, ;;/, n are arbitrary, A",r + B j + C"2 = c is 
 the locus of all the radii vectores which touch the surface at the 
 origin, and is therefore the tangent plane at the origin. 
 
 Hence, if the equation of the surface can be written in the form 
 + != o (where u. 2 represents terms of second degree and u { terms 
 
 76 
 
NOTES ON SOLID GEOMETRY. ^ 
 
 of first degree in #, y, and z), then u l = o is the equation of the tan 
 gent plane at the origin. 
 
 Therefore, to find the equation to the tangent plane to the surface at the 
 point x y z , transfer the origin to this point. r l he equation may then be 
 writ/en u 2 -f Uj = o, and Uj = o is the tangent plane referred to the 
 point of contact as origin ; then in Uj= o retransfer the origin to the 
 primitive one. 
 
 125. For the central surfaces (origin at centre) take the equation 
 
 and let (x ,y , z ) be the point of contact. Transferring the origin to 
 the point (x , y r , z ) by the formulae 
 
 X = X + X \ 
 
 y y + y V we have 
 
 2= Z + Z 1 } 
 
 Ax* + B/ + Os 8 + 2 Axx + 2 fyy + 2Czz =o. 
 Hence the tangent plane at the new origin is 
 Axx -f By/ + Czz = o. ( i ) 
 
 Now retransfer the origin for equation (i) to the centre by the 
 formulae 
 
 x = x x } 
 
 y = y y \ and we obtain 
 
 *=*-* ) 
 
 + Czz Ax * Bi/ 2 - CV a = o, 
 
 or AAVT + BJ/ + Czz =i (2) the required equation of the tangent 
 plane, at the point x y z referred to centre. 
 
 i. For the sphere A =B = C = -^ . 
 
 a 
 Hence (2) gives xx +yy + zz = a\ (3) 
 
 2. For the ellipsoid A = -i -, B = -1 C= i 
 xx yy zz _ 
 
 3. For the hyperboloid of two sheets A = ~ E~ ^ C = \ 
 
 xx yy zz 
 
 7* 
 
78 .VOTES ON SOLID GEOMETRY. 
 
 4. For the hyperboloid of one sheet, A= , B=: , C= ^. 
 
 xx yv 
 
 126. For the surfaces which have no centre (origin at vertex) by 
 treating the equation By 2 + Cz* = 2A. x in a similar manner we 
 obtain 
 
 Bj/y + Oss = K (x + x ) (7) for the equation to the tangent plane 
 to the elliptical paraboloid and 
 
 Bij/ Czz = A. (x + x ) (8) for the tangent plane to the hyper 
 bolic paraboloid. 
 
 Remark. The same method may be applied to cones and cy 
 linders. 
 
 127. Polar planes to surfaces of second order. The equations (3) 
 (4) (5) (6) (7) (8) are the equations to the polar planes to the sur 
 faces respectively with respect to the point (.v f , y , z ) and these polar 
 planes possess properties analogous to the polar lines to the conic 
 sections. 
 
 128. The length of the perpendicular from the centre on the tangent 
 plane to the ellipsoid is p = \Az 8 cos 2 a + &* cos 2 ft + c* cos 2 y , when 
 cos a, cos (3, cos y are its direction cosines. 
 
 The equation to the tangent plane is "-^ + "-jrjH g- = i. It may 
 
 also be written x cos a +y cos ft + z cos y p. Hence we must 
 have 
 
 / cos a cos /? cos y a cos a _ b cos ft _ c cos y _ 
 
 i ~ x y z x y _s/ 
 
 ~J ^~ ~^~ ~a~ ~T ~7 
 
 Hence calling the direction cosines /, z, w, the equation of the 
 tangent plane may be written 
 
 Ix 4- my -f .s = V^ 2 + ^ 2 ^ 2 + ^ 9 2 ( 9 ) 
 
NOTES ON SOLID GEOMETRY. 
 
 79 
 
 129. To find the condition thai the plane 
 
 3C \f 2 
 
 f- -~ H =i (i) shall be tangent to 
 
 a ft y 
 
 x* / z* 
 
 the ellipsoid 5- + ~-r -\ r 
 
 a u c 
 
 xx yy 22 
 Comparing (i) with ^ + -^ -\ - 
 
 we must have 
 
 I _ X I _ y I 
 
 " o~ j 7T~ /o i 
 
 a 2 ft 6* y 
 
 a x b y c z 
 
 or - , r = -^-, - = ; /. squaring and 
 
 ^f a ft by c 
 
 adding r + ^ H r i is the required condi- 
 
 <^ p y 
 
 tion. 
 
 130. The sum of the squares of the perpendiculars \>, p , \>" , from the 
 centre of the ellipsoid on three tangent planes mutually at right angles is 
 constant. 
 
 Let cos a cos ft cos y\ cos <* cos ft cos j/, etc., be the direction 
 cosines. 
 
 Then p 2 = 2 cos 2 -f <$ 2 cos 2 /5 +<; 2 cos 2 y 
 
 p * = a" cos 2 a +b~ cos 2 /? +t f cos 2 y 
 p 2 = <z 2 cos 2 a" + ^ 2 cos 2 ft" }- c 1 cos 2 T/ , 
 
 and adding we have 
 
 131. Cor. Hence the locus of the point of intersection of three tangent 
 planes to the ellipsoid which intersect at right angles is a concentric sphere 
 of the radius V V + b* + c 1 . 
 
 For 2 the square of its distance from the centre is equal to 
 p*+p *+p "\ and therefore to a* + tf + c\ 
 
 Remark. In the case of hyperboloids one at least of the quantities 
 a 2 . b~, c 2 is negative, and hence their sum may be negative or nothing ; 
 in the former case there is no point in space through which three 
 rectangular planes touching the hyperboloid can be drawn, and in 
 the latter case the centre is the only point which has that property. 
 
8o NOTES OA SOLID GEOMETRY. 
 
 132. Diametral Planes. Definition. A diametral surface is the 
 locus of the middle points of a series of parallel chords of a given 
 surface. Diametral lines or diameters are the intersections of the 
 diametral surfaces. 
 
 133. To find the diametral surface corresponding to a given series of 
 parallel chords in a surface of the second order which has a centre. 
 
 Let the equation of the surface be 
 
 /, m, n the direction cosines of each the parallel chords, and # , /, z 
 the coordinates of its middle point. 
 The equation of the chord will be 
 
 x oc y v z z 
 
 m =: zn i*. 
 
 I m n 
 
 Then for the points in which it meets the surface (i) we shall 
 have 
 
 or 
 
 Imposing on this the condition of equal roots for r, we have 
 Atx + Bniy + Cnz = o (2) the equation of the diametral surface, a 
 plane passing through the centre. 
 
 
 134. The diameter = = is one of the series of parallel 
 / m n 
 
 chords bisected by the plane (2), and is called the diameter conju 
 gate to the plane, and conversely the plane lx + my + nz o is con- 
 
 A.v Bv Cz 
 
 luxate to the diameter - = = -. 
 I m n 
 
 If a diametral plane be chosen as a new plane of xy and its con 
 jugate diameter be taken as the new axis of z, the centre O being still 
 the origin; then, since every chord parallel to Oz is bisected by 
 the plane xy, the equation of surface will contain only the second 
 power of z. Hence, if there be three planes through the centre the 
 intersection of any two of which is conjugate to the third, the equa 
 tion of the surface referred to these planes will be of the form 
 
 AV + By + CV=i, (3) 
 that is of the same form as the equation referred to rectangular axes. 
 
NOTES ON SOLID GEOMETRY. 8 1 
 
 !35- To find the conditions that of three planes through the centre of a 
 surface of the second order each may be diametral to the intersection of the 
 other two. 
 
 Let the planes be 
 
 / v _|_ mv + nz = 0, I x + my + n z = o, l"x 4- m ! y + n"z = o. 
 The equations of the diameters conjugate to the first plane are 
 
 and if this be parallel to the other two planes, we shall have 
 / . m , n . ... I ,, m ,, n 
 
 / A + M B + "c= oandr A + " B+" c= 0; 
 
 these with the third equation I + m + n o, found in 
 
 like manner, are the required conditions. 
 
 These three planes are called conjugate planes, and their intersec 
 tions conjugate diameters. 
 
 Since we have only three relations between the six quantities there 
 will be an infinite number of systems of conjugate planes in each 
 surface. 
 
 136. Equations referred to conjugate diameters. If in (3) Art. 134 we 
 make 
 
 A >_ B _JL c-- 
 _ a , v i - b , v ^ - -& 
 
 Then for the ellipsoid 
 
 X~ I 2 2 2 
 
 + -f = i will be the equation referred to conjugate di 
 
 ameters, and a, b , c will be the semi-conjugate diameters. 
 For the hyperboloids we shall have 
 
 a- 2 y z* , x 1 y z 2 
 
 ^-^- 7 T= i and jr^-^bc r. 
 
 Remark. The tangent planes at the extremities (.* , j/, z) of any 
 diameter to a central surface are parallel to the diametral plane 
 conjugate to the diameter so that the conjugate plane of the diameter 
 through the point (x 1 , y , z } on the ellipsoid is 
 
 xx vy zz 
 
82 NOTES ON SOLID GEOMETRY. 
 
 137. The sum of the squares of three conjugate semi-diameters of the 
 ellipsoid is constant. 
 
 In the first place, any point on the ellipsoid may be represented by 
 the equations x =. a cos A, y = b cos //, z = c cos Y, when cos A, 
 cos ju, cos v are the direction cosines of some line, for the condition 
 cos 2 A + cos 2 //-fcos 2 v i cause these three equations to satisfy the 
 equation of the ellipsoid. 
 
 Therefore if cos A, cos fa cos r, cos A , cos //, cos v are the direc 
 tion cosines of two lines answering to the extremities of two conju 
 gate diameters, these will be at right angles to each other. 
 
 T, xx yy zz 
 
 ror the equation +~j^- + g-= o will give 
 
 cos A cos A + cos yu cos //4-cos v cos v = o. 
 
 Now the square of the length of any semi-diameter ~v 2 +y 2 +V 2 
 expressed in terms of A, /i, v, is 
 
 rt 2 = a* cos 2 A + < 2 cos 2 JJL + C* cos 2 r, 
 and of the conjugates in terms of A , //, v\ A", // , y" 
 2 =rt 2 cos 2 A + 2 cos 2 yu + r cos 2 r 
 <:"*= a* cos 2 A"+3 2 cos 2 // + r> cos 2 v". 
 Adding we have 
 
 a" 2 + 3 /2 + c 2 = a 2 + tf + <r 2 , since the lines A, ju, r, A , // , r x , and 
 A", yu", y" are mutually at right angles. 
 
 138. To find the locus of the intersection of three tangent planes at the 
 extremities of three conjugate diameters. 
 
 The equations of the three tangent planes are 
 
 cos A + v cos u, -f - cos v i 
 a o c 
 
 OC V Z 
 
 cos A + cos v -\- - cos v = i 
 
 (2 O C 
 
 x y ,, z 
 
 - COS A -f COS V -\ COS V I . 
 
 a b c 
 
 Squaring and adding, we get for the equation of the locus 
 > H 
 
 tV~T- 
 
 ~-f-- r = 3 an ellipsoid with the semi-axes a\/~$~, b^/ 3 , 
 
NOTES ON SOLID GEOMETRY. 83 
 
 139. The parallelepiped whose edges are three conjugate semi-diameters 
 of an ellipsoid lias a constant volume. 
 
 Let Qx, Oy, O.2 be the semi-axes of the surface a, b, c\ Oo; , Oy, 
 Oz r any system of semi-conjugate diameters a , b , c\ let the plane 
 of x y intersect that of xy in the semi-diameter OjCi A, and let 
 Oy 2 = B be the semi-diameter of the curve ^ # which is conjugate to 
 CXrj. Hence parallelogram a b = parallelogram AB. 
 
 .-. Vol (a, b\ f ) = Vol (A, B,-O 
 
 for these figures have the same altitudes and equal bases. 
 
 Let the plane z Oj 2 intersect xy in the semi-diameter Oi^ C, 
 then this plane must contain Oz ; for, being conjugate to OA*, in a 
 principal plane it must be perpendicular to that plane ; hence CXv,, 
 Or,. Oz form a system of semi-conjugate diameters, and any two of 
 them are semi-conjugate diameters of the plane section in which they 
 are situated. 
 
 .-. Vol (A, B, c ) =Vol (A, C, c) 
 
 Vol (A, C, c ) =Vol (a, b,-c) 
 . . Vol (a 1 , b , c ) =Vol (a, 6, c). 
 
 140. To find the diametral plane bisecting a given system of parallel 
 chords in the case of the surfaces which have not a centre. 
 Taking the equation of the surface 
 
 r , , , xx vy z z 
 and one or the chords - = - =. ~ - = r 
 I m n 
 
 the equation of the diametral plane will be 
 
 ;;/ BJ/ + W Cz= A 7. 
 
 Hence the diametral planes are parallel to the common axis of the 
 principal parabolic sections. 
 
 We cannot, therefore, in these surfaces have a system of three con 
 jugate planes at a finite distance, but we can find an infinite number 
 such that for two of them each bisects the chords parallel to the other 
 and to a third plane, by proceeding as in Art. (135). 
 
 By taking the origin where the intersection of these two meets the 
 paraboloid, and referring to these three planes, the equation of the 
 surface will be of the form 
 
84 NOTES OiV SOLID GEOMETRY. 
 
 And the third plane is evidently the tangent plane to the surface at 
 the new origin. 
 
 141. The tangent planes to the hyperboloid of one sheet and /he hyperbolic 
 paraboloid at a point x y z intersect the surfaces each in two right line 
 generators through the point of contact. 
 
 The equation of the hyperboloid of one sheet referred to any con 
 jugate diameters is 
 
 and the equation of the section made by any plane y = ft parallel 
 to the conjugate plane of xz, is 
 
 and it is evident that the value /3b gives us the section of the 
 tangent plane at the extremity (.v , y , z ) of the diameter b ; or 
 
 x 3 z"* 
 
 ~~^ 7? > two right line generators. 
 
 Ct c 
 
 For the hyperbolic paraboloid 
 
 B> 8 -CV=2E".,v (i) 
 
 the tangent plane through the origin is x = o, and its intersection 
 with (i) is 
 
 B^ 2 CV= o, two right line generators. 
 
CHAPTER XL 
 
 GENERAL EQUATION OF THE SECOND DEGREE IN 
 
 x,y, AND z. 
 
 142. In order to discover all the surfaces represented by the gen 
 eral numerical equation 
 
 Bj/ 2 
 
 Gz 2 +2C ^ + 2C"a = D (E). 
 
 we will first transform the coordinates to a new origin by means of 
 the formulae 
 
 x =. a + x* \ 
 
 z 
 
 and endeavor to determine the coordinates (a, /?, y) of the new 
 origin in such manner as to cause the terms of the first degree to dis 
 appear. If this can be effected the equation will be reduced to the 
 form 
 
 Ax 2 4- Br + Cs 2 + 2A. zy + 2% zx + 2C xy = F (F) 
 
 in which there is no change when x, y, z are substituted for 
 + x, +y, +z, and which therefore represents a surface having a 
 centre, and the new origin of coordinates is at this centre. 
 
 Now, several different cases may arise according to the numerical 
 relations among the coefficients A, B, C, A x , B , C , A", B", C". 
 
 i. a, ft, y the coordinates of the centre may each have a finite 
 value found from the three equations determining the conditions of 
 the transformation. 
 
 2. a, /3, y may have infinite values. 
 
 3. a, /?, y may be indeterminate. 
 
 8 85 
 
86 A T 07^ES ON SOLID GEOMETRY. 
 
 The surfaces corresponding to these three cases will be 
 
 (A) Surfaces having a centre. 
 
 (B) Surfaces having no centre (centre at an infinite distance). 
 
 (C) Surfaces having an indefinite number of centres. 
 
 143. Making the actual transformation of (E) by the formulae (i 
 we have 
 
 And in order that the terms of the first degree in x, y, z shall dis 
 appear, we must have 
 
 (C) 
 
 which are called the equations of the centre. 
 
 1. If these three equations give finite values for a. /?, y, then 
 the surface represented by the given equation has a centre. 
 
 2. If two of ihese equations are incompatible this shows infinite 
 values for a, /3, y, and the surface has no centre. 
 
 3. If the three equations reduce to two, then the surface has a 
 line of centres. For each one of the equations is the equation of a 
 plane, and two taken simultaneously represent a line, and the surface 
 is an elliptical or hyperbolic cylinder. For, cut the surface by the 
 planes P and Q, P cutting the line of centres (D) and Q containing 
 that line. The section by P is a curve of the second degree having 
 its centre on the line D, and hence an ellipse or hyperbola. The 
 section Q will be two straight lines parallel to the line D, and as Q 
 may revolve about D in all its positions giving two straight line sec 
 tions parallel to D, the surface is a cylinder. 
 
 4. If the three equations reduce to a single one, then the surface 
 has a plane of centres (i. e., the given equation represents coincident 
 or parallel planes). 
 
 Note. The equations of the centre can be found in any given 
 
NOTES ON SOLID GEOMETRY. 87 
 
 equation most readily by finding the derived equations with regard 
 to x, y, and z respectively (i.e., by differentiating with regard to x. y, z 
 respectively), the x, j , and 2 in the resulting equations standing for 
 a, ft, y. 
 
 144. Example i. Determine the class of the surface represented 
 by the equation ,v 2 + ^y 1 + 42 2 + 2yz + \zx + 6.17 2 6x 2$y 322= 26. 
 
 The equations of the centre are 
 
 6> + 22 + 6x 24 = O > . These give y = 2 I 
 82 + 27 + 4^32 = 0) *=3J 
 
 and the surface has a centre. 
 
 Example 2. Determine the class of the surface 
 
 The equations of the centre are 
 
 2JI- + 22+2 V 4 = O ) 
 
 f 
 
 2 y + 22 + 2X 2 = O , x - 
 
 4 Z + 2y + 2X + 2 = O ) 
 
 the first two of which x+y+z = 2, x+y + z = I are incompatible, 
 hence the coordinates of the centre are infinite, and the surface has 
 no centre. 
 
 Example 3. Determine the class of the surface 
 
 2 2 2yz zx-\-^xy+2z = o. 
 The equations of the centre are 
 
 2x z +4y ~ o 
 8j/ 22 + 4^v = o 
 
 2Z 2V X+ 2=OJ 
 
 The first two of these are identical, hence the three equations re 
 duce to two and the surface has a line of centres (/. e. } is a cylinder). 
 Example 4. Determine the class of the surface 
 
 8-v 2 + 1 8v 2 + 22 2 + 1 272 + 8zx + 2^xy 5o.v 757 252 + 75 = o. 
 The equations of the centre are 
 
 1 6x + 82 + 247 50 = o 
 
 = 
 i2r+ 8.v 25 = o 
 
88 
 
 NOTES OX SOLID GEOMETRY. 
 
 which are all three the same, each being 8.r+ i2> 4-4.3 =25. Hence 
 the surface has a plane of centres, and consists of a pair of parallel 
 planes. 
 
 145. Recurring to the general equations of the centre 
 A <*4-C7? + By + A"= o ] 
 
 + ir=o( (c) 
 
 we may find an easy rule for a relation among the coefficients in any 
 given equation by which we can distinguish the central surfaces from 
 those having no centre and those having an infinity of centres. 
 
 The common denominator of the values of a, /3, and y in these 
 equations is the determinant 
 
 A, C , B 
 C , B, A 
 B , A , C 
 
 2 + CC 2 -ABC-2A B C . 
 
 Now, if R be different from zero, the surface has a centre ; but if 
 R = o it may either have no centre or an infinity of centres. 
 
 The value of R may be written out by the following mnemonic 
 A, B, C 
 
 form : 
 
 A B C 
 
 A B C 
 
 the letters to be multiplied by columns for the first 
 three terms, and by rows for the two last. 
 
 146. To find an easy rule for F, the new absolute term in the trans 
 formed equation of the central surfaces when the origin is moved to the 
 centre. 
 
 This complete transformed equation is 
 
 A_v s + By 2 + Or + 2 A zy + zR zx + 2C xy = F when 
 
 F=D- 
 
 Now, multiplying the first of the equations (C) of the centre by 
 a, the second by ft, and the third by y, and adding them 
 
 we have 
 
NOTES ON SOLID GEOMETRY. 
 
 8 9 
 
 Hence F = D (A" + B"/? + C . Therefore the rule for F is 
 substitute for x, y, z in the terms of the first degree one-half the co 
 ordinates of the centre (i. e., ^a, |-/?, ^y respectively and take result 
 from D. 
 
 Example i. Taking the Example i, Art. (144), in which the coordi 
 nates of ihe centre are found to be x =-i,y 2, z = 3, 
 we have F=26 + 26 x J- + 24 x i +32 x f 1 1 1; 
 
 and the transformed equation is 
 
 X* + 3> 2 + 43 2 + 2yz + 4zx -f 6,r> = 1 1 1 . 
 
 Ex. 2. 2^ + 3y* + 4Z z + Syz + 6xz + 4xv 6x SyI4z = 20. 
 Here the coordinates of the centre are x = \>y = 2, s= i. 
 
 . . F==2o + 6 x-J + 8 x i + 14 x 1 = 17; 
 and the transformed equation is 
 
 2 A* 4- 3/ + 4-s 2 + 870 + 6^2- + ^xy 1 7. 
 
 147. Removal of the terms in xy, xz, yz. Reduction of the equation of 
 the second degree to two forms. 
 
 Fora more complete discrimination of the surfaces represented by 
 the general equation, we will now remove the terms in^ry, xz, yz by 
 a transformation of coordinates. So far we have made no supposition 
 as to the direction of the axes. Henceforth, for convenience, we will 
 consider the axes rectangular. 
 
 Taking the equation (E) in rectangular axes we propose now to 
 transform it to a system also rectangular in such manner that the 
 terms in xy, xz, yz shall disappear. The disappearance of these 
 terms can only be effected by taking for coordinate planes either dia 
 metral planes or planes parallel to them. 
 
 We will therefore begin by finding a diametral plane conjugate to 
 a given diameter. 
 
 148. To find a diametral plane conjugate to a given diameter. 
 
 x x yy __ zz 
 
 I m n 
 
 Putting x-= x + lr, y y + mr, z=z -\-nr in the general equation, 
 and arranging with reference to r, we have for the coefficient of the 
 first deree in r 
 
QO NOTES ON SOLID GEOMETRY, 
 
 and this placed equal to zero is the equation of the diametral plane, 
 namely 
 
 ( A/+ B + C m)x + (C7+ Bm + A n}v + (B7+ A m + Cn)z + A 7 
 
 149. 70 determine a diametral plane perpendicular to the chords which 
 it bisects, that is, to find a principal plane. 
 
 In order that the diametral plane shall be perpendicular to the line 
 
 x a v b z c 
 
 m n 
 
 , we must have the conditions fulfilled 
 
 I m 
 
 or putting each of these equal to s. 
 
 At +R n+C m = 
 
 (A) 
 
 and also the condition / 2 + w 2 -f # 2 = i . 
 
 To determine /, w, and n in equations (A) we first find s. Writ 
 ing these equations 
 
 they give the result 
 
 A-J, C , B 
 C , B j, A 
 B , A , C-.r 
 
 = o 
 
 or 
 
 or 
 
 C *-ABC-2A B C f =o (D). 
 
 This cubic has necessarily one real value for s, which substituted 
 in (A) gives one set of real values for /, m, n. Hence there is one 
 principal plane. 
 
 For convenience of discussion let us take this plane perpendicular 
 
NOTES ON SOLID GEOMETRY. gi 
 
 to the axis of z, then / = o, w = o, and n =i. And hence equa 
 tions (A) give B = o, A = o, and the general equation transformed 
 to this principal plane as plane of xy is of the form 
 
 z = D. 
 
 Now we know from the like discussions in conic sections that one 
 transformation is always possible, and but one to a system of rectan 
 gular axes in the plane xy which shall cause the term in xy to dis 
 appear. Hence there are three principal planes, and three sets of 
 values for /, m, n, and the cubic (D) has three real roots. 
 
 The general equation may then be always reduced in rectangular 
 coordinates to the form 
 
 * = D. (E ) 
 which represents then all the surfaces of the second order. 
 
 1 50. The reduction of this equation Lx 2 + My 2 -f Nz 2 + 2L x + 2M y 
 + 2N z = D to two forms. 
 
 i. If L, M, and N are different from o. 
 
 Then we may cause the terms of the first degree to disappear by 
 
 L M N 
 
 transferring the origin to the point x = ~r~>-J = ivf * \f 
 
 1 4 i * L IN 
 
 The surface will then have (# ,.> , z ) for its centre, and the equation 
 will be of the form 
 
 Lr + My + Ns 2 ^ F. (I.) 
 
 2. If one of the three coefficients, L, M, N, for example L = o 
 and L be different from o. 
 
 We cannot then cause the term 2~L!x to disappear, but by trans 
 ferring the origin to the point 
 
 D M N 
 
 x = --., y = z = -- ^p- the equation will take the form 
 2L M N 
 
 M/ + Ns f = 2V x. (II.)* 
 
 The forms I. and II., we have seen, belong to the surfaces of the 
 second order, which we have already discussed. Hence the general 
 equation of the second degree (E) represents these surfaces and no 
 others. 
 
QO NOTES ON SOLID GEOMETRY. 
 
 151. The form I. we have seen represents the ellipsoid, the two 
 hyperboloids and cones of second degree, and includes the elliptic 
 and hyperbolic cylinder, Mv 2 + Ns 2 - F and parallel planes N2 2 =F. 
 
 The form II. represents the elliptic and hyperbolic paraboloids, 
 and the parabolic cylinder. 
 
 152. The complete reduction of the equation of the second degree to the 
 simple forms I. and II. Use of the discriminating cubic (D). 
 
 The resolution of the equations (A) furnishes for each value of s 
 in the cubic (D), one system of values of/, m, n. We have then three 
 systems, /, m, n] / , ??i , n /", m", n , which are the direction cosines 
 of the three rectangular axes {principal axes ] to which the surface 
 must be referred in order to cause the products .%T, xz, yz to dis 
 appear ; the formulae of transformation are then 
 
 y mx + my + m"z 
 z = nx + n y + n"z . 
 
 If we take only the terms in x* in this substitution we find 
 L = A/ 2 + Bw 2 + C;* 2 + 2h. mn + 2BW+ zC lm. 
 
 But if we multiply the equations (A) respectively by /, m, n and add, 
 remembering that / 2 + 7 2 + 7z 2 = i we have 
 
 A/ 2 + B;;; 2 + CV + 2h. mn + 2 BW+ 2C lm s ; 
 
 Hence L is a root of the cubic (D) and M and N are the other two 
 roots. 
 
 For the values of L , M , N we will have 
 
 L = A 7 +B"w +C"n \ 
 M = A 7 +B";;/ r + C V (M). 
 
 The absolute term D does not change in this transformation since 
 the origin is not changed thereby. 
 
 For the surfaces having a single centre after solving the cubic, we 
 have only to calculate F, for which we have given a rule. 
 
 For the surfaces having no centre the coefficient designated by V 
 is equal to L , and is computed by first finding in equations (A) 
 the values of/, m, n, which correspond to s = o. Both in the cases 
 of surfaces having no centre and a line of centres, one root of cubic 
 = o and we have only a quadratic to solve to determine L and M. 
 
NOTES ON SOLID GEOMETRY. 93 
 
 153. For surfaces having a centre, if \ve wish only to discover the 
 particular class of the surface, without making the complete trans 
 formation of the equation to its centre and axis, the sign of the roots 
 of the discriminating cubic will tell us whether the surface is an ellip 
 soid, hyperboloid of one sheet, or hyperboloid of two sheets. These 
 signs we can ascertain from inspection by Descartes s rule * without 
 solving the equation. 
 
 Example. Find the nature of the surface 7# 2 + 6y 2 +5.s 2 4_> $xy 
 = 6. The cubic (D) gives 
 
 s 3 (7 + 6 + 5)s a + (42 + 35 + 30 -4 4)5+28 + 20 210=0; or 
 s 3 i8s* + 99^162=0. 
 
 .-. The row of signs is H 1 , three changes of sign. Hence 
 
 all the roots are + and the surface is an ellipsoid. 
 
 So also for surfaces having a line of centres, the signs of the roots 
 of the quadratic into which the discriminating cubic degenerates, 
 serve to distinguish the elliptic from the hyperbolic cylinder. 
 
 And for surfaces having no centre, the signs of the roots distinguish 
 the elliptic paraboloid from the hyperbolic paraboloid. 
 
 154. Recapitulation of the method of reduction of numerical equations 
 of the second degree and of distinguishing the surfaces represented by them. 
 
 We now propose to give the mode of distinguishing the nature of 
 the surface represented by any given numerical equation of the second 
 degree in x, y, and z, and of finding its principal elements. 
 
 I. Form the equations of the centre, and also the discriminating 
 cubic from the remembered form 
 s *- (A + B + C)s z + ( AB + AC + BC - A 2 - B 2 -C 2 ).r + AA 2 + BB 2 
 
 + CC 2 -ABC-2A B C = o, 
 
 observing that the absolute term is equal to R, the denominator 
 of the values of the coordinates of the centre in the general equation, 
 
 ABC 
 and therefore can be formed by the mnemonic A B C (Art. 145). 
 
 A B C 1 
 
 Then 
 
 155. i. If R be different from o, the surface has a centre. Find 
 
 Note. "All the roots being real the number of positive roots is equal to the number of 
 changes of sign in the row of signs of the terms, and the number of negative roots is equal to the 
 number of continuations of sign." 
 
94 
 
 NOTES ON SOLID GEOMETRY. 
 
 the coordinates of the centre and transform to the centre by the rule 
 in Art. (146). Determine the signs of the roots of the cubic by Des- 
 cartes s rule. Then calling these roots L, M, and N, and calling F 
 the new absolute term on the second side of the equation. 
 Then 
 
 a. If L, M, N all have the same sign as F, the surface is an ellip 
 soid. 
 
 b. If L, M, N all have a different sign from F, the surface is im 
 aginary. 
 
 c. If two only of the roots L, M, N have the same sign as F, the 
 surface is the hyperboloid of one sheet. 
 
 d. If only one of the roots L, M, N has the same sign as F, the 
 surface is the hyperboloid of two sheets. 
 
 e. If F = o and L, M, N all have the same sign, the locus is a 
 point. 
 
 f. If F = o and one of the roots L, M, N has a different sign 
 from the other two, the surface is an elliptic cone (Art. 85). 
 
 156. 2. If R = o the cubic has one of its roots s = o and is 
 degraded to a quadratic, the coefficient of s, namely AB + AC + BC 
 A 2 B 2 C 2 , becomes the absolute term. 
 
 And if the equations of the centre are incompatible the surface has 
 no centre. 
 Then 
 
 a. If the roots M and N of the quadratic (degenerate cubic) have 
 the same sign (i. e.) if AB + AC + BC A 2 B *C 2 >o the surface 
 is the elliptical paraboloid. 
 
 b. If M and N have different signs (/. e.) if AB + AC + BC-A" 
 
 B 2 C *<o the surface is the hyperbolic paraboloid. 
 
 c. If one of the roots M or N be zero (* . e.) if AB + AC + BC-A 2 
 
 B 2 C 2 = o the surface is the parabolic cylinder. 
 
 157. 3. If R = o and the equations of the centre can be reduced 
 to two equations, the surface has a line of centres. The cubic as in 
 (2) has one of its roots S = o and degenerates into the quadratic 
 
 j_(A+B + C)s + AB + AC + BC --A"-B -C" = o. 
 
 Then 
 
 a. If the roots M and N of this quadratic have the same sign (i. e.) 
 
NOTES ON SOLID GEOMETRY. 
 
 95 
 
 if AB + AC + BC A" B 2 C ~> o the surface is an elliptic cy 
 linder. 
 
 b. If the roots M and N have different signs (i. e.) if AB + AC 
 + BC A " 2 B 2 C 2 < o the surface is the hyperbolic cylinder. 
 
 c. If in the reduced equation of the cylinder Mz 2 -f Ny 9 = H, H be 
 equal to o, and M and N both of same sign, the locus is a straight 
 
 z = o } 
 line > . 
 
 J = o \ 
 
 d. If H = o and M and N be of different signs the surface con 
 sists of intersecting planes. 
 
 158. 4. If R = o and the equations of the centre become a 
 single equation, the surface has a plane of centres, and consists of 
 two parallel or coincident planes, which are readily found by solving 
 the equation with reference to any one of the variables. 
 
 159. 5. In the case of surfaces of revolution the cubic has equal 
 roots. To examine the cubic for equal roots in the case of central 
 surfaces of revolution, we simply look for a commoe root between it 
 and its first derived equation (differential). 
 
 1 60. GENERAL REMARK. In any of the above cases we may com 
 plete the reduction by solving the cubic to get the new axes and 
 thus obtain their direction by finding /, m, n from equations (A). 
 And in the case of the surfaces without a centre we may find V, 
 from equations (M). 
 
 161. REMARK I. In the cases of surfaces having a line of centres 
 and of those not having a centre, we can distinguish readily the sur 
 face represented by a given numerical equation through sections by 
 the coordinate planes. 
 
 i. If the equations of the centre show a line of centres, sections 
 by the coordinate planes will tell whether the surface is an elliptic or 
 a hyperbolic cylinder. 
 
 2. When the equations of the centre show no centre, then 
 
 a. If there are ellipses among these sections by the coordinate 
 planes, the surface is an elliptical paraboloid. 
 
 b. If there are hyperbolas among these sections, the surface is a 
 hyperbolic paraboloid. 
 
 c. If all these sections are parabolas, or one of them parallel 
 straight lines, the surface is a paraboMc cylinder. 
 
96 NOTES ON SOLID GEOMETRY. 
 
 162. REMARK II. Again, if the terms of the second degree in the 
 given equation break up into unequal real factors, the surface must 
 be either the hyperbolic paraboloid or hyperbolic cylinder, and these 
 two surfaces are otherwise readily distinguished. We may note also 
 that if the terms of the second degree in the given equation form a 
 perfect square, the surface is either a parabolic cylinder or two 
 parallel planes. 
 
 163. We will now illustrate by a few examples : 
 
 Ex. i. yjt- 2 iy~ + 6z i + 2^xy+ \2yz- 1202: = 84. 
 
 As this is a central surface with the origin at the centre, we only 
 need the discriminating cubic, which is 
 
 j 8 343^4-2058 = o ; or s*os* 343^4-2058 = o. 
 
 The signs 4- -- 1- show one continuation and two changes, and 
 hence the surface is a hyperboloid of one sheet, or two sheets, accord 
 ing to the sign of 84. 
 
 By trial we find that 7 is a root of the cubic, and then by depress 
 ing the equation we find the other two roots are 14 and 21. There 
 fore the equation of the surface referred to its centre and axes is 
 
 7* -f 14^-212*= 84; or.r 2 + 2_y 2 -3S 2 =:i2. 
 
 Ex. 2. 2 5.r 2 + 2 2_/ + i6s 2 4- i6yz4zx 2oxy z6x 40^442 
 
 = -46. 
 
 The equations of the centre are 
 
 250, ioy 22 = 13 
 I o:r 4- 2 2j 4- 82=20 
 - 2X+ 8y+ 162 = 22 ; 
 
 whence we find the coordinates of the centre x \,y =i, * =i. 
 Moreover 
 
 F = 46 4- 26 \ + 40 J + 44 I = 9- 
 The discriminating cubic is 
 
 / 63^+ 1 134^5832 = o. 
 
 Its signs give three changes. Hence all the roots are positive. 
 The surface then is an ellipsoid. 
 
NOTES ON SOLID GEOMETRY. 97 
 
 By trial we find that 9 is one of the roots of the cubic. Hence the 
 other two are 18 and 36. The reduced equation is then 
 
 9jr J -fi8> 2 -f 36s 2 9 ; or A- 8 
 And the principal semi-axes are i - , __ 
 
 V2 2 
 
 Ex. 3. 5 x* + i of + i ys 2 + 2 6yz + 1 8zx + i $xy + 6x + ?> + 1 02 = 64. 
 The equations of the centre are 
 
 5.r+ 7^+ 92 =-3 
 7A:+iqy+ 132 =4 
 
 Multiplying the first of these equations by i and the second 
 by 2, and adding, we obtain the third. Hence the equations are 
 only two independent ones. The surface is therefore a cylinder. In 
 tersecting it by the coordinate plane xy, i. e., making z = o, we obtain 
 
 5^ 2 + 1 4xy + i oy* + 6x + Sy 64, 
 
 which is an ellipse. The surface is therefore an elliptic cylinder. 
 To complete the reduction we transfer the origin to the point 
 
 where the line of centres > pierces the plane 
 
 x, y, that is, to the point z o, y i, x 2, 
 and find F=64 +6 4=:66. 
 
 Also the discriminating cubic is 
 
 .r 3 32,r-f6 > 9 = o, which gives j 8 32^ + 6 = o, 
 
 the roots of which are 16 + 5^ 10 and 16 5\/ 10. 
 And the reduced equation of the cylinder is 
 
 io)jr + (i6 5V 7 io)> 2 =66. 
 Ex. 4 . 5 jp 2 + 5/ J + 8s 2 + 4sv + \zx Sxy + 6x + 6y 30 = o. 
 The equations of the centre are 
 
 ^4-4*= 
 
98 NOTES ON SOLID GEOMETRY. 
 
 Adding the two last of these equations we have 
 $x4y+2z 2j. An equation which is incompatible with the 
 first. Hence the surface has no centre. 
 
 The cubic / iS>r 2 + 8ij= O ; or s*i 8^ + 81 = o, 
 
 which gives two roots equal to 9. The surface is therefore a para 
 boloid of revolution 
 
 2 
 
 To find V, we first determine /, m and n. For these we have the 
 equations 
 
 4/ ^ 272 = o 
 
 o 
 
 U- U 7 2 2 ! 
 
 which Rive / = , m = n -- . 
 3 3 3 
 
 Therefore (Eq. M) L = 3 .-+ 3 .-+-.- = - 
 
 and 2V 2L =9. 
 
 The reduced equation of the surface is therefore 
 
 Ex. 5. 2A70+2B ^v+2C .r^ 4-2A v .v + 2B jv4-2C"2; D. 
 
 The cubic is 
 
 s-(A 2 + B 2 f-C 2 )s-2A B C = o . 
 
 The surface is a hyperboloid if A , B and C are all different from 
 o. If A B C is of the same sign as F in the reduced equation the 
 cubic will have two roots of the same sign as F and the surface will 
 be a hyperboloid of one sheet. In the opposite case it would be a hyper 
 boloid of two sheets. 
 
 If A = o the cubic becomes 
 
 j. 2 _(B 2 4-C *) =o, whose roots are of different signs. Hence 
 the surface 2H zx + 2C xy-\- 2A".r+ 2B V + 2C r z = o is a hyperbolic 
 paraboloid. 
 
 Ex. 6. jc 2 +_y 2 4-9s 2 + 6r0 6xz 2~\y+ 2.v 42 o. 
 
 The equations of the centre are incompatible and the terms of the 
 
O.Y SOLID GEOMETRY. 
 
 99 
 
 second degree form a perfect square, hence the surface is a parabolic 
 cylinder. 
 
 EXAMPLES. 
 
 164. i. Find the nature of the surfaces represented by the follow 
 ing equations. 
 
 ( i ). i i~v 2 -f 5_y 2 4- 2z 2 2o\ z + 4zx 4- 1 6xv 4- 2 2.v + 1 6 v 4- 42 4- 1 1 = o. 
 
 ( 2 ). .X 2 +/ 4- S* 4- 2F + 2X2 4- 2AJ I O,V I OV IOZ-\-2$ = O. 
 
 ( 3 ). 3 jc 2 37 2 1 2J2T +i2zx + Sxv 6x 6>> + 3^ = 0. 
 (4). 4.i- 2 + QV 2 + 97s 2 1 60jt- + 54^ =36. 
 
 (5). 3X*+2f2XZ \- 4VZ 4.V 80 8 = O. 
 
 2. The equation 7^- 2 4-8/ + 4s 2 7^0 nzx jxy = ^ represents 
 a hyperboloid of one sheet. 
 
 3. The equation .r 2 +> ;2 +32: 2 + yz+zx+xy jx 1 4> 252 = 12 d 
 represents an ellipsoid, a point, or an imaginary surface according 
 as d is < = > 67. 
 
 4. The equation _v 2 4-.v 8 4-2 8 -i-yz+zx + xy = a* represents an oblate 
 spheroid. 
 
 5. Find the nature of the surface (y z)*+ (z xY + (x v) 2 = a~. 
 
 6. Find the nature of the surface yz \ zx + xy # 2 . 
 
 7. ^a- 2 + 4v 2 4-9s 2 4- I2yz + 6zx + 4xy+i4x+ 16^+2404-47 = o re 
 presents an elliptic, a parabolic or a hyperbolic cylinder according as 
 a > = < i. 
 
CHAPTER XII. 
 PROBLEMS OF LOCI. 
 
 165. PROB. I. To find the surface of revolution generated by a right 
 line turning around a fixed axis which it does not intersect. 
 
 Let the fixed line be the axis of z and let the shortest distance a 
 from the revolving line to the axis of z lie along the axis of x in the 
 original position of this line so that its equation is x a, y = mz. 
 
 Then the equation of the surface is 
 
 or 
 
 The hyperboloid of revolution of one sheet. 
 
 Prob. 2. To find the locus of a point whose shortest distances from two 
 given non-intersecting, non-parallel straight lines are equal. 
 
 Take the axis of z along the shortest distance between the two 
 lines, the plane xy perpendicular to z at the middle point of this 
 distance 2c, and the axes of x and y bisecting the angles between the 
 projections of the line on their plane. Then the equation of the lines 
 will be 
 
 z = c \ z c ) 
 y mx f y mx } 
 
 ( y mxY xo ( V + mxV 
 
 and we have (z cY + - r L = (* + + ~ ~- 
 
 I + m~ i + /// 8 
 
 or 
 
 cz(i + ?n*) -f mxy = o, a hyperbolic paraboloid since it has no 
 centre and its term of second degree breaks up into two real factors. 
 
 Prob. 3. 7 wo planes mutually perpendicular, contain each a fixed 
 straight line. To find the surface generated by their line of intersection. 
 Take the axes as in Prob. 2. Then the equations of the planes are 
 
 100 
 
NOTES ON SOLID GEOMETRY. IO I 
 
 K(z c) +y mx o ; (i) K (z + t)+y + Mx = G. (2) 
 The condition of perpendicularity of these planes is 
 
 KK +i m* o, and eliminating K + K between this equation 
 and equations (i) and (2) we have 
 
 y _ /;/V + ( i _ m *Y ( i - m*y 
 which represents a hyperboloid of one sheet. 
 
 Prob. 4. To find the surface generated by a right* line which always 
 meets three fixed right lines no two of which are in the^santy fflantj *++ 
 
 For greatest simplicity take the origin at the Centre of & paraljek)-- 
 piped, and let its faces be at the distances a, &, d respectively from- 
 the coordinate planes yz, xz, and xy. Then take three edges of this 
 parallelopipedon as the three fixed lines fulfilling the conditions. 
 
 Assume for the equations of the movable line 
 
 *^. m 2Z* -_=*.> (4) 
 cos a cos ft cos y 
 
 The conditions that the line (4) shall meet the lines (i) (2) and 
 (3) are respectively 
 
 y b _ z +c z c _ x + a x a _ y -f b 
 cos ft cos y cos y ~ cos a cos a ~~~ cos ft 
 
 Eliminate the arbitrages a, ft, y by multiplying the equations to 
 gether, and we have for the surface 
 
 or reducing 
 
 ayz + bzx + cxy + abc o, 
 
 which the discriminating cubic shows to be a hyperboloid of one 
 sheet. The same surface will be generated by a straight line resting 
 
 x = a } y = b } x = a 
 
 on the other three edges > , - , . 
 
 z = C ] z c \ y = o 
 
 Prob. 5. To find the surface generated by a right line which alw ns 
 9* 
 
102 NOTES ON SOLID GEOMETRY. 
 
 meets three fixed right lines, no two of which are in the same plane, but all 
 of which are parallel to the same plane. 
 
 Take one of the fixed lines as the axis of x, and then the other two 
 parallel to the plane of A:; . Then their equations are 
 
 Now, the equations of a moving line meeting lines (i) and (2) are 
 - . ,- (4) (/ and k arbitrary), and the condition that this 
 jine shall also meet (.3) is Ic mk (cb], 
 arid eliminating therV and k by means of equation (4), we have 
 
 cy mx(cb) 
 
 2 z b 
 
 or cyz -\-m(b c)xz ( by o, 
 
 a hyperbolic paraboloid, as its equation shows no centre, and the 
 terms of the second degree break up into two real factors. 
 
 Prob. 6. To find the surface generated by a right line which meets 
 two fixed right lines, and is always parallel to a fixed plane. 
 
 Since the two fixed lines must meet the fixed plane, we can take 
 
 -y " 
 
 Z 
 
 1 1 (i), y [ (2), as in 2, as the fixed lines, and the 
 
 = c \ z = c } 
 
 plane yz as the fixed plane. 
 
 Then the equation of the moving line parallel toyz 
 
 is Z (3), /, A and k arbitrary. 
 
 The conditions that this line shall meet the lines (i) and (2) 
 
 mk = k+/> 
 -mk=-lc+t\ 
 
 or mk = Ic and/> = o ; 
 
 or eliminating /, k, and /, 
 
 y 
 
 mx = c -- ; 
 z 
 
 or mxz = cy, a hyperbolic paraboloid. 
 
NOTES ON SOLID GEOMETRY. Ic> 3 
 
 Prob. 7. Two finite non - intersecting non-faralld right lines are 
 divided each into the same number of equal parts ; to find the surface 
 ivhich is the locus of the lines joining corresponding points of divi 
 sion. 
 
 Let the line which joins two corresponding extremities of the given 
 lines be the axis of z ; let the axes of x andjy be taken parallel to the 
 given lines and the plane of xy be halfway between them. Let the 
 lengths of the given lines be a and b. 
 
 Then the coordinates of two corresponding points are 
 
 z =. c, x ma, y = o ; z = c, x = o, y mb ; 
 and the equations of the lines joining these points are 
 
 + ^-=1 ] 
 
 ma mb 
 
 2X Z 
 --- =1 
 
 ?na c 
 whence eliminating m the equation of the locus is 
 
 a hyperbolic paraboloid. 
 
 Prob. 8. To find the locus of the middle points of chords of a surface 
 of the second order that has a centre, which all pass through a given fixed 
 point. 
 
 Take the given point for the origin and two conjugate diametral 
 planes which pass through it for the planes of zx and xy, and a plane 
 parallel to the third conjugate plane for that of> 2; then the equation 
 to the surface will be of the form 
 
 ax* + by 1 + cz 1 + 2a"x +/ o. 
 
 Let x viz, y = nz be the equations of any chord. Combining 
 these with the equation of the surface, we have 
 
 (am* + bn^ + c )z y + 2ct mz + d o, 
 in which the values of z belong to the extremities of the chord. 
 
104 VOTES ON SOLID GEOMETRY. 
 
 Therefore the z of its middle point is 
 
 a"m 
 
 and the other two coordinates of the middle point are 
 
 x = mz t (2) y = nz . (3) 
 Hence eliminating m and n the required equation of the locus 
 
 a surface of the second order similar to the first, and passing through 
 its centre and through the origin. 
 
CHAPTER XIII. 
 SOME CURVES OF DOUBLE CURVATURE. 
 
 1 66. To find the equations to the equable spherical spiral. 
 Definition. If a meridian of a sphere revolve uniformly about its 
 
 diameter PP while a point M moves uniformly along the meridian 
 from P to P", so as to describe an arc equal to the angle through 
 which the meridian has revolved, the locus of M is the equable 
 spherical spiral. 
 
 Taking PP as the axis of z, PAP the initial position of the plane 
 of the meridian as the plane of xz, the equation of the sphere is 
 
 ac*4y +* = <!*. 
 
 Let MON = 0, AON q), then, by definition 6 = cp, 
 and from polar coordinates 
 
 x = a cos cos cp, r a cos 6 sin cp ; 
 /. x = a cos 2 6, y = a cos sin 0. 
 
 Therefore **+/ = cos 2 (cos 2 + sin* 8) = ax. 
 
 Hence the equations of the spiral are 
 
 .r+y + = 2 (i) #" + v* = ax; (2) 
 
 or the spiral is the curve of intersection of the sphere and a right 
 circular cylinder whose diameter is the radius of the sphere. 
 If we subtract (2) from (i) we obtain 
 
 z~ <? 2 ax (3) a parabolic cylinder. 
 
 And the equations (2) and (3) also represent the curve, which is 
 therefore also the intersection of a right circular and right parabolic 
 cylinder at right angles to each other. 
 
 167. To find the equations to a spherical ellipse. 
 
 Definition. The spherical ellipse is a curve traced on the surface 
 
 105 
 
I0 6 A OTES ON SOLID GEOMETRY. 
 
 of a sphere such that the sum of the distances of any point on it 
 from two fixed points on the sphere is constant. 
 
 Let S, H be the t\vo fixed points on the surface of the sphere 
 whose radius is ;-, C the middle point of the arc of the great circle 
 which joins them. If P be any point of the spherical ellipse, SP and 
 HP arcs of great circles, then 
 
 SP-fHP = 2a = a constant. 
 
 Through P draw PM, an arc of a great circle perpendicular to 
 SH, and let SH = 2;/, CM = y, PM = 0. 
 
 Then, in the right-angled spherical triangle SPM we have 
 
 cos SP = cos (y + q>) cos 0. 
 And in the triangle HPM 
 
 cos PH = cos f cos 6. 
 
 Now, cos SP + cos 
 
 UTD /SP+HP\ /SP-HPX 
 
 HP ? POJ 1 I rns 1 
 
 V 2 / f \ 2 ) 
 
 /SP-HPX 
 = 2 cos a cos 1 ) 
 
 /SP + HPX /SP-HPX 
 
 And cos HP cos 
 
 Therefore, 
 rns f 
 
 X 2 / 
 
 = 2 sin 
 SP-HPX cosy cos cp 
 
 \ 2 J 
 
 /SP-HPX 
 
 * A ( "-2 J 
 
 COS 
 
 2 J cos a 
 
 /SP-HPN sin Y sin q> cos 6 
 
 sin - V ?B A . . 
 
 V 2 / sin r 
 
 Squaring and addini 
 
 cos 2 Y 9 /) s 2 V 2 2/3 
 cos cos c H ; sin 2 <z> cos 1 Cr-xsi; 
 
 cos r.f sin- (y 
 
 or if we transform from polar to rectangular coordinates 
 
 o o 
 
 cos* 
 
 c 
 
 + -r-r^y = ^ (0 
 
 
 
 This equation and the equation of the sphere 
 A- 9 +.!* + ** nr r 2 (2) 
 
 determine the spherical ellipse, as the intersection of a right elliptic 
 cylinder and the sphere. 
 
NOTES ON SOLID GEOMETRY. 
 
 107 
 
 1 68. To find the equations to the helix. 
 
 Definition. Whilst the rectangle ABCM revolves uniformly about 
 its side AB, so that the parallel side CM generates the surface of a 
 right circular cylinder, the point P moves uniformly along CM, and 
 generates a curve called a helix. 
 
 Let AB be the axis of z, and when the rectangle is in the plane 
 xz let P and M both be at D on the axis of x, and let the velocity of 
 P = n times the velocity of M. 
 
 .-. PM = ;/. arc. DM. 
 
 Also let AN = jv, NM y, PM = z be the coordinates of P, and 
 AM = a the radius of the circular base of the cylinder in the plane xy. 
 
 . . z = na cos" 1 , and j 2 + x^ <r (i) 
 
 are the required equations of the helix. 
 
 Or we may represent the curve by the two equations 
 
 z na cos" 1 . z na sin" 1 (2): 
 a a 
 
 or the same in the forms 
 
 z z 
 
 x = a cos , y = a sin , n) and 
 na na 
 
 z / z \ z / z 
 
 since cos cos 2m 7t -\ and sin - = sin 2m n-\ 
 
 na \ na J na \ na 
 
 the same values of x and y correspond to an infinite number of 
 values of z. The equations (i) (2) and (3) show that the projec 
 tions of the helix on the planes xz, and yz give the curve of sines, 
 and the projection on xy is the circle. 
 
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