164 Mr Brill, A new Geometrical Interpretation [Nov. 28, 
a {ee (Y,Z) “ cos (Z, X) 4 cos (X, =| a. Ls x asa “ih 
124s V EWE VIE P, Po Ps 
3 = (B, C) i cos (C. A) a cos (A, =} 
P2Ps PsP1 P:P2 
15. Asa final example of formule involving products of pairs 
of planes, we will take the equation 
S.s,(s,+s,')+8.s,(s, +s,)+S.s,(s,/+5,) 
=8S.s/(s,+8,)+8.s,(s,+5,) +8.5s, (8, +5,). 
Suppose that we have six given planes (4), (B), (C), (D), (£), P). 
Let (L), (Mf), LV), (X), (Y), (Z) be the ee polar planes of 
the origin with respect to (B) and (C), (C) and (A), (A) and (B), 
(D) and (£), (£) and (Ff), (7) and (D). Then the above equation 
becomes 
S.(A)(X) +8. (B)(VY)4+8.(C)(Z) 
=S.D)(H+4+8.(4L)M)+8.(F)). 
Hence, if we use p, to denote the perpendicular from the origin 
on (A), and adopt a similar notation for the other perpendiculars, 
we have 
cos (A, X) 4 £08 (Gai 1%) 4, 098 (C, Z) 
PaPa PoPy PcP- 
_ cos (D, L) A GORE) 08 (F, NV) 
DaPr P-Pm (os 
16. We will new proceed to the discussion of the scalar portion 
of the product of three planes, making use of the construction and 
notation that we adopted in Articles 10 and 11; premising at the 
same time that PA, PB, PC are to be taken-as the edges of that 
particular solid angle, contained by the three planes, within which 
the origin lies. 
We have the equations 
IS 2S,S)SA— ISIS S352 1S) Sh V/S1S) 11S ese Visrsee 
12°93 
Now the perpendicular from the origin on V s,s, is parallel to 
AP, and T'V (s,s,)=sin A/p,p,- Thus 
S.s,Vs,s, =— cos (7 — 8) sin A/p,p,p,5 
where @ is the angle between PA and the normal to s,, 2.e 
S.s,Vs,s,= cos @ sin A/p,p,p,=sin p, sin A/p, p, ps. 
Thus we have 
snp,.sinA_sinp,.sinB_ sinp,.sinC 
S.s.s.s,= 
ita PPPs P:P2Ps P:P2Ps 
