537 
of Edinburgh, Session 1881-82. 
The equations (26) and (30) solve the question of finding the 
positions of the resultant force for any given point of space. When 
namely, p is given, the scalar-equation (26) will give theoretically 
four values of h , after the elimination of g by (25), and the vector- 
equation will give the four directions k' corresponding. 
Before attacking this question we will find it very convenient to 
have to deal with rectangular coordinates, and we make use of one 
of the solutions of Professor Tait’s of the equation of condition, 
(31) 0 = S«' = S/(^K = Srf'ftW. 
We consider, namely, <£>i ’ , <£/ by their expressions 
<K = SaS/&' = 2aSj3 0 i' 0 
4>f = = SctSftio , 
and designating in particular 
2aS (3 0 a> by <£ 0 CO 
we take for i ' 0 , j' 0 the solutions of the equation 
the third being kf . 
The function (</>' 0 <£ 0 ) is self-conjugate, and its principal axes are 
at right angles to one another ; we designate by if , jf perpendicular 
to kf , the unit vectors in the direction of these axes. The cubic, 
reduced to a quadric (because of (frfkf = 0), corresponding will be 
iv 2 - M 2 w + Mj = 0 . 
Calling a , b , the absolute values of g> 0 if , cj>f 0 , we have by (18) 
( M 9 = a 2 + b 2 , 
(32) ] 
(M , = 
because now V<£/ 0 </> 0 / 0 = ^o^o/o owing to (31). 
We have therefore 
(4* Mi o = ^ o a ‘ 2 > (4* Mj • 
We now call i, j , Jc, the system of trirectanglar unit vectors in 
the directions of <f>ff , 4>of oj an( ^ ^ o4of o = sM’^o/o • This 
gives us 
