121 
the angles of crystals. 
p qr he p X + (iy + rz = m, we shall have the co-ordinates 
of the point O by combining these equations. Hence we 
have px-{-qx-{-rx = m, or x = 
^ But if the co-ordinates x,y,z be projected upon AO, we 
shall have AO = A x cos. x AO + Ay cos. y AO A % cos. 
z AO. And since cos. x AO = cos. y AO = cos. z AO = -y, 
AO = l±l±i = x. ,.AO = — 
3 p+q+r 
Now let p'x + c(y + r's: = m be the equation to a plane 
which cuts Ax, Ay, Aw in the same manner in which 
{p\q\r) cuts Ax, Ay,' A 2;. Therefore the portion cut off 
from ry A produced will be y^Y~r Also the portions from 
Ax and Ay, are y, . 
Hence ^ = 
V 
m <m 
?'+ »•' ~ 
P'=P, 9'=?./+ 9'+ r'= — r . ; r'=— /> + </ + r. 
Hence if (/> ; g ; ;*) be a plane {p \q \ — ^ is a co- 
existent plane. 
Arso the axes of x,y, 2: being symmetrical, {p \q',r) has 
co-existent planes [p, (/, r). And making — p + g + r=Sy 
w'e have the planes 
(/>. 9 . '■) iP> 9 . iP> ^ ( 9 > ^ *)• 
Each of these symbols gives six permutations, so that we 
have in all 24 co-existent planes. 
§ 8. On the arrangement of secondary faces. 
42. When crystals have faces determined by the laws con- 
sidered in the preceding pages, they will have the form of 
polyhedrons bounded by polygons ; and in order to deter- 
mine the dihedral angles, &c. it will be necessary to know 
MDCCCXXV. R 
