409 
105. To find the normal to a plane from the centre of the 
cubical axes in terms of the indices of that plane. 
Let BGD (fig. 33*, Plate IV.*) be a plane cutting the three 
cubical axes AB, AG, and AD, in the points B, C, and D. Let 
AB=ct, AG—b, and AD = c, be the three indices of this plane. 
Through A draw AE perpendicular to BG in triangle ABG. 
Join ED. 
Through A draw AF perpendicular to DE in triangle ADE. 
Then AF is perpendicular to the plane ABG. Let AF=R , 
then B is the normal drawn through A to the plane whose 
indices are a, b, c. 
Through F in triangle ADE draw FG perpendicular to AE, 
and in triangle ABG draw GH perpendicular to AB. 
Let AH=x, GH=y, and FG=z, are called the rectangular 
co-ordinates of the point F, referred to the rectangular axes 
AB, AG, AD, or AX, AY, AZ (fig. 33*, Plate IV.*), is drawn in 
perspective. (Fig. 35*) is the triangle AGB of (fig. 33*), drawn 
on the plane of the paper ; (fig. 34) the triangle DAE of the 
same figure, also on the plane of the paper. 
Let angle AEF=(3. Then by construction AFG=l 3, 
DAF=j 3, ADF= 90°-j3, and FAE=9 0°-/3. 
z=FG=AF sin FAG=B cos j3. 
Also B^AF—AD sin ADF—c cos (3. 
Hence z— — . 
c 
Again, in triangle AGF, AG—AF sin AFG—R sin (3. 
Also in triangle ABG, let a = angle ABG, then by construc- 
tion GAE=a, AGII—a, EGA— 99 — a, and EAB=90—a „ 
In triangle AG1I, x=AH=AG sin AGE=AG sin a = 
B sin (3 sin a. 
Also in triangle AEB, AE=a sin a and sin a~ — 
a 
In triangle AFE, B=AF=AE sin (3 and sin S = — 
H AE 
But x=R sin j3 sin a = B . 2^- . ^2 — 21 
AE a a 
Again in triangle AGH, y=GE=AG cos a = R sin ft cos «. 
In triangle AGE, 
AE—AG cos GAE=b cos a ; and cos a 
But sin (3 = 2L 
' AE 
Hence y = A sin [3 cos ci= R 
AE 
b 
AE 
AE 
b : 
m 
b 
