100 Mr. Whewell on calculating 
secondary rhomboid becomes a regular hexagonal prism. 
(Example. Chaux Carbonatee Prismatique. Hauy.) 
In other cases the secondary form is an irregular hexago- 
nal prism, the angles being equal, three and three alternately. 
14. Prop. To find the symbol of a plane which truncates 
any edge of a given form. 
Let two faces {p;q; r) (/; q'; r') meet, and let (P ; Q ; R,) 
be a plane which truncates the edge formed by their inter- 
section : the plane must be parallel to this intersection ; and 
the equations to the intersection must be consistent with the 
equation V x + Qy + Rz=o. Now for the intersection we 
have p z q y r z =: o, p' X q^ y r'z z=: o : whence 
[pq' — p'q) X = [qr' — q'r)z, [p*r — p?'')x = (q r' — q'r)y. 
Multiply F X -[-Qy + R^!;=o by(^ r ' — q'r) and substitute, 
and we have 
P r'—cfr) + Q (/>' r—p / ) + R{pq'—p'q) = 0. 
And if P, Q, R fulfil this condition, (P ; Q ; R) will be a plane 
truncating the edge as required. 
15. Prop. To find the symbol of a plane which truncates 
an edge of any secondary rhomboid. 
This is a particular case of last Prop, when instead of (^ ; g ; r) 
[p'; q'; r'), the planes are {pi q \ q) {q p \ q)- Hence the 
equation of condition becomes 
^(q"—pq) + Q{q"-‘pq) + '^{p"—q")=o 
or Fq + Qq^R(p + q) = o 
Hence if R = P Q ~p .q. and with this condition, 
(P ’ Q 5 is the plane required. 
Ex. Required the planes which truncate the edges of the 
rhomboid produced by the law (3, 1 1). 
Here ^ ^ == 2 ; the values which may be given to 
