107 
the angles of crystals. 
Let a plane pqrh^ formed on the angle A ; then, since all 
the angles are symmetrical, we must have a coexistent plane 
at any other angle, as x. 
Let A/> = h, Aq^k, A rz=s 1; and let xP = h, x Q =: k, 
xR = I ; it is required to find the equation to the plane 
PQR. 
Draw X M andy K parallel to PQ and we have, if A<r = ^, 
xK = xy . AK = a(i —4)' 
% 
Also AM = A;-. ^ = — ^ 
* T 
Similarly if x N be parallel to PR, we shall find AN s= — ^ 
1 
Hence the equation of the plane N jt M is 
T+c-4)i+('-MT='’ 
" t + (t— tI^’ + It— t)^ = t- 
And the symbol of this plane will be 
And the plane PQR is parallel to N x M, and will have the 
same symbol. 
If Y ^ ~ symbol of the plane PQR 
will he[p\ p — q; p — r). 
In the same way we shall have at the angles y and «, planes 
g; q — r) and [p^r;q — r; r). 
But the edges Ax, Ay, Az are also similar, and therefore 
p, q, r may be permuted in any manner. Hence we have 
these co-existent planes 
{p,q,r),{p,p — q,p — r), (q-p,q,q-.r), {r-p,r~-q,r). 
