41 L 
FfiftF of (fig. 36*) drawn on the plane of the paper. (Fig. 
38*) the plane YAHH \G Y also drawn on the plane of the paper. 
Join FF 1 and GG V In plane FF^Gfi draw KF parallel to 
GG V and therefore perpendicular F^G X ; also in plane GG.RH 
draw GL parallel to Then KFGG 1 and HGLH 1 are rect- 
angular parallelograms and their opposite sides are equal. 
Then (fig. 37*) FF 1 2 =F 1 K 2 +KF 2 = (F 1 G 1 --KG 1 ¥+G,G*. 
But ( % 88.) 
G 1 G~GL 2 +G 1 L 2 =HE 2 + (G^-LE.) 2 . 
= (AE 1 —AE) 2 + (G 1 E 1 —GE) 2 = (tCj— *)*+ (y t — y) 2 . 
And FF 2 = (*!-*)*+ ( yi -y)*+ («!-«)*. 
We Lave seen (§105) that R*=x 2 +y 2 +z'>, and that 
R 2 R 2 , R 2 
*=— > y — ^r, and z=— 
a b c 
Similarly R 1 *=x 1 * + y 1 *+z*, and * 1= ^L, y 1= 5 L, and z x =?X. 
In triangle FF'A, fig. 39, if we put 0 for the ano-le FAF° or 
the angle between the normals AF, AF V or R and R, at the 
point A ; we have 
FF 2 = AF 2 + AF 2 —2AF 1 . AF cos 9=Ry + R 2 -2RR 1 cos 9; 
but FF*= (p 1 -x) 2 + (y 1 -y) 2 +(z 1 -z) 2 . 
Hence ( i r 1 -*) 2 + &-*)* + (z 1 -z) 2 =R 2 +R 2 -2RR 1 cos 9-, 
or £Bx 2 — 2iB* 1 + x 2 + y 1 2 —2y 1 y + y 2 +z 1 2 —2z 1 z + z 2 =R. 2 + R 2 — 
2 BB X cos 6. 
B u t B 2 = Xj 2 + y 2 -J- z 2 and B~ = x 2 + y 2 A z 2 . 
Hence x x xAy x y + zz 1 =BB 1 cos 6 , 
B 2 B 2 ^B 2 B 2 B 2 B 12 
01 
ClCl- 
b\ 
CC-, 
■ BB 1 cos 0. 
cos 
9=RR 1 f-L+J_ + i\ 
W x b\ ccj 
but R 2 -x 2 + y 2 + z 2 =^ +~ + ~ and J? 2 = 1 
a~ b 2 6 l 1 
also B 2 — 
-+1+1 
a 2 b^c* 
X+l + 1 
Therefore cos 0 = 
A+JLa 1 
__ b\ cc 1 
