to ^on-Metrical Vector Algebra* \'M 



The plane section c = 0, and therefore any diametral section * 

 of Q, will be a conic whose equation will be of the form 



x 2 - +y 2 + 2a 12 xg = 1 . 



This will play, in the said plane, the part of our previous k. 

 Thus, defining ab as the corresponding coefficient of 2xy and 

 reasoning as before, we shall obtain the scalar product AB 

 with all its properties, provided that A, B are in a diametral 

 plane. But if A. B are not in a diametral plane we can 

 construct vectors equal to them in an appropriate plane, viz. 

 that passing through O and the straight Y'a'/b- Moreover, 

 we can draw these, and all other vectors, from as origin. 

 It is enough to declare, as in the case of two dimensions, 

 that A'B' = AB whenever A' = A and B=B, no matter where 

 A', B' are drawn. 



All fundamental questions are thus reduced to those treated 

 previously, and need not detain us here. 



It remains only to show how triads of mutually orthogonal 

 vectors, reducing the Q -ec l ua tio-n to ;e 2 +y 2 + z 2 = l, can be 

 constructed. And, by what has just been said, we can 

 without loss to generality, confine our attention to vectors 

 drawn from as orioin. 



Let, therefore, &=Oa be any given unit vector, and T a its 

 terminus, i. e. the cross of the a-line with the 7-plane. The 

 second vector b normal to a being required to be in a given 

 plane, lay that plane through and draw from T a a tangent 

 to /c a }„ the section of Q. The contact point b will be the 

 end-point of the required b. (The other tangent will give 

 b' = — b, as before.) It remains to construct a third vector c 

 such that ca = and cb = 0. Lay through T a T b a plane 

 tangent to Q. Then its contact point will be the end-point <* 

 of the required unit vector c. 



To verify this statement notice that if c is to be normal to a, b, 

 which are already normal to one another, the equation of Q with respect 

 to a, b, c as axes will be x--\-y 2 -\-z 2 = I, and therefore the equation of 

 the plane touching it at any point £, 17, £ will be easily found to be 



£v-rW+£2=l, 



so that the tangent plane at c (£=1, '/ = C=0) should simply be x= I. 



Now, this is precisely the plane through c\ T , T,. 



The second tangent plane through T a T h will touch Q in <•'. 

 the end-point of the vector — c. 



Thus any number of orthogonal triads a, b, c can be 



* I. e. by a plane passing through 0. 



