176 Proceedings of Royal Society of Edinburgh. [sess. 
the resultant, it is OR' and not OR which is normal to the plane 
of OP and OQ. The expressions for the three vectors OR', OP, 
OQ are 
OR' = cosh c sinh b'/3 + cosh b sinh c*y — sinh b sinh c sin /3y’a 
OP = - sinh b~°—^f.p + sinh b . ^ - .y + cosh b'a 
sm /3y sin py ' 
OQ = - sinh c . ^ - sinh c y + cosh c'y 
sm py sm py 
from which it follows that S(OR')(OP) = 0 and S(OR')(OQ) = 0. 
Hence OR' is normal to the plane of POQ. How is the direction 
of OR related to that plane ? The plane of OA and OY (fig. 5) 
cuts the equilateral hyperboloid in an equilateral hyperbola ; and 
as it passes through the normal OR', it must cut the plane POQ 
orthogonally. 
Let OX he the line of intersection. Draw XM perpendicular to 
OA, draw XD a tangent to the equilateral hyperbola at X (fig. 5), 
and XA' parallel to OA. Let 0 denote the hyperbolic angle AOX. 
As OR' is normal to the plane POQ, it is perpendicular to OX ; 
but OY is perpendicular to OA, therefore the angle AOX is equal 
to the angle YOR'. How the angle AOR is the complement of 
ROY, and A'XD the complement of AOX; therefore the line OR 
is parallel to the tangent XD. Thus the direction of the directed 
sinh is that of the conjugate axis to the plane of OP and OQ. 
This idea of conjugate instead of normal also applies to the spherical 
case, from which it follows that ij=\J -Yk means that &is the 
axis conjugate to i and/ 
How sinh 6 = 
MX 
OA 
YR 
n/OY 2 - YR 2 
sinh b sinh c sin f3y 
J (cosh b cosh c + sinh b sinh c cos /3y) 2 - 1 
The above analysis shows that the product versor POQ may be 
specified by the following three elements : — First , c, a unit axis 
drawn perpendicular to 0 A in the plane of OA and the normal to 
the plane POQ ; second , 0, the hyperbolic angle determined by OA 
and OX, which is drawn at right angles to the normal in the plane 
of OA and the normal ; third , a , the angle of the hyperbolic sector 
