99 
the angles of crystals. 
Let q; r)be the plane, and (o, i, o) (o, o, i) the two pri- 
mary planes ; d and 9' the given angles 
cos. 9: 9-(P + r) cos. « 
V* p*+ 9*+ c * — z (p g+ pr~\- g r) cos. 
COS. 9 ' = 
r — ( p cos. a 
r *— 2 (pq-\-pr+qr) cos. a | 
whence q and r must be found in terms of p, as in last pro- 
position. 
Or we may find them directly thus. Since one of the three 
p, q, r is indeterminate, assume^^-h 9 ®+ r^^^2(^p q~[- pr-\^qr) 
cos. a == 1 . 
cos. 6=q — rcos. a — p cos. a ; cos. 6 '=;’— ^ cos. a — p cos. u. 
Eliminating, we have 
q sin."* a = cos. 9 + cos. at cos. ^ P cos. u (l + COS. a) ; 
r sin.^ a = COS. 9' + cos. a COS. 9+p COS. a (l + COS. cl). 
If we substitute these values in the assumed equation multi- 
plied by sin.'^ii;, viz. 
\p^+ q '+^^ — +P^ + q^) cos. a I sin.'^ar^ sin.'^a 
we shall have a quadratic equation in p ; and hence p, q, r are 
found. 
13 . Prop. To find what laws will give prisms parallel to 
the axis of the primary rhomboid. 
For this purpose the planes must be parallel to the axis ; 
and the equation of a plane must be consistent with the equa- 
tions of the axis, which are 
y = JT, z = x. 
Let {p\ q \ r) be the plane ; p oc q y + r 2 = o is the 
equation to it, supposing it to pass through the origin; and 
since y == j:, 2 ; == x ; we have p x-f q x-|-r x=o p = 
If rz=q, p = — 2 q; the planes are ( — 2 , 1 , 1 ) and the 
