123 
the angles of crystals. 
we suppose it to be determined by angular distances like the 
longitude and latitude on a globe, assuming as the axis of the 
ellipsoid that about which the figure is symmetrical. 
43. (1) In the rhomboid^ Here = 
suppose lA be taken as the axis ; and a plane API being 
drawn, let the angle between this plane and lAx be called 
the longitude (x) of the point P ; and let the complement Of 
AI P be called the latitude {jf) of P. 
Let the co-ordinates of P be called X, Y, Z. Then the 
plane API has a point A, of which the co-ordinates are 0,0,0; 
a point I, of which the co-ordinates are 1,1,1; a point P, of 
which the co-ordinates are X, Y, Z. Hence its equation is 
(Y — Z) X -f (Z — X)y -}- (X — Y ) ^ = o. And the equation 
to lAx is y — z = o. Therefore by the formula for the angle 
of two planes. Art. 8, 
— 2X + Y-fZ — (zX — Y— Z) cos. a 
cos. X = — r~7 ^ ^ 
v/ J 2((Y-Zy + (Z-X)‘+(X-Y)^42(X* + Y* + Z*-XY~XZ-YZ) cos. a 
If the symbol of the plane be (p; q; r) its equation is 
‘ P^r + qy + r z = m; and hence 
— X 
cos. X 
— similarly for Y and Z. Hence 
(2 p — g — r) (1 cos, g) 
2 I ( r'"— P(I — pr — qr) (i + cos. a) | 
2p — q — r , , X 
2 sj — p q — p r — q r) ' ^ ^ ' COS. aj. 
To find jtc ; if we draw PM perpendicular in AI, and call 
IP, r, we shall have IM = r sin. (^, and [x will be greater as 
IM is greater. Now if IM, NO, OP to the co-ordinates of 
P measured from I, and if we draw perpendiculars from 
N and O on I A, w'e shall see that IM = (a — X) cos. ^ -}- 
{a — Y) cos. ^ — Z) cos. ^ where ^is the angle wdiich AI 
makes with Ax, Ay or A%. 
