126 Dr. L. Silberstein : Further Contributions 



normal to a, and lot the terminus of n be T n (this is also 

 obtained by simply drawing the tangent at a) . Draw T n b 

 crossing the a-line in N. Then the a-coordinate of N will 

 be ab or, vectorially, 



ON= (ab)a. ...... (8) 



To see this, take the orthogonal pair a, n as axes of #, y, 

 so that the equation of k will be 



Now, if a, v are the coordinates of b with these axes, we 

 have 



b = aa + vn, u 2 -\-v 2 = l. 



Thus, if x ' , y be the coordinates of any point (x, y) with 

 respect to a, b as axes, we have 



,,>a + ^n = a?'a + /b = (&•' + a/)a + vy'n, 

 wdience 



x =x' +«y ; y=^', 



and the conic equation with a, b as axes becomes 



Therefore, by the original definition of ab, as given in (1), 



which is the «- coordinate of b, i. e. that of N. This proves 

 the statement made in (8). 



Thus ab is obtained by the orthogonal projection of b 

 upon a or, similarly, by that of a upon b, exactly as in 

 ordinary vector algebra, notwithstanding the different, more 

 general meaning of our concepts. 



Now, such being the case, the distributivity of the scalar 

 product follows at once. In fact, if A, B, C be any vectors 

 and B + C = S, we have, first of all, 



A(B + C)= A . aS. 



Now, whatever B, C, we can always reduce them to a 

 chain, so that S will be the vector drawn from the origin 

 of B to the end-point of C. But the orthogonal projection 

 of S upon a is instantly seen to be equal to the algebraic 

 sum of the orthogonal projections of B and of C. Cf. fig. 5, 

 in which T n is the T-point of the vector n normal to a. 

 Notice that the intercepts of the three perpendiculars will 



