Elementary Geometry to Crystallography. 347 



the straight lines XOX', YOY', ZOZ' its axes; the three lines 

 a, b, c, or any three lines in the same ratio, its jjararneiers ; and 

 let h, k, I, or any three whole numhers in the same ratio, and 

 having the same signs, be called the indices of any plane parallel 

 to HKL. Then 



-OH = ^OK=-OL. 

 aba 



When one of the indices h, k or I becomes 0^ the correspond- 

 ing point H, K or L will be indefinitely distant, and the plane Jikl 

 will be parallel to the corresponding axis. When two of the in- 

 dices become 0, the plane hkl will be parallel to the two corre- 

 sponding axes. 



When a numerical index is negative, or a literal index is taken 

 negatively, the negative sign will be placed over it. 



The planes hkl, hkl are obviously parallel, and on opposite 

 sides of the point 0. 



5. The line in which any two planes of the system intersect 

 will be called an edge. 



Let be the origin ; OX, OY, OZ the axes ; a, b, c the para- 

 meters of a system of planes. Let OB = Z>; and let the planes 

 hkl, pqr, passing through B, inter- 

 sect in the edge BM, meeting 

 the plane ZOX in M, and let them 

 meet OZ and OX in L, R and H, P. 

 Then 



-0H=|0B=-0L, 

 a c 



^0P = |0B=-0R. 

 a b c 



Therefore 



/.OL = ^e, h.OYi. = ka, r .0^ = qc, p .OV = qa, 

 Ir . \jYi={kr-lq)c, hp . V\{ = {hq-kp)a. 

 But by (2), 



HM.Lll.OP = ML,RO.PH. 

 Therefore if 



u = kr — lq, v=lp — hr, vf=hq — kp, 



w/.ML = u/i.HM, w;.LH = -v^.HM, uA.LH= -v/fc .ML. 

 Draw MD parallel to OZ meeting OX in D. 



LH . OD = ML . OH, and LH . MD = MH . OL. 



Hence v.OD=— ua, and v.MD = — wc. Take DE equal 

 and parallel to BO, EP equal and parallel to MD. Then 



