438 



PROCEEDINGS OF THE AMERICAN ACADEMY. 



Consider next the product a'(bxc) of three 1-vectors. Here 



a'(bxc) = (a«c)b — (a«b)c. (34) 



Perhaps the simplest proof is obtained from the relation ^^ 



('b'C)c , C'(bxc) 



b = -^ '-, 



C'C c«c 



which states that a vector is equal to the sum of its components. 

 By clearing and transposing, and by permuting the letters, we have 



c«(bxc) = (c'c)b — (c'b)c, 

 b-(bxc) = (b-c)b — (b.b)c. 



If now d is any vector perpendicular to b and c, we have identically 



d.(bxc) = (d-c)b — (d-b)c = 0. 



If these equations be multiplied by x, y, z and added, we have 



(xc + 2/b + zd) • (bxc) = [(.TC + T/b + sd) • c]b — [(xc + yh-\- zd) • b]c, 



and any vector a may be represented in the form xc + ?/b + zd. 



From the rules (33), (34) combined with the rules (22)-(32) we may 

 obtain a number of other reduction formulas by simply taking comple- 

 ments of both sides of the equation. 



Thus 



(axb)-C = a-(b.C) = — b-(a.C). (35) 



28 With the aid of inner and outer products we may write down expressions 

 for the components of a 1-vector a along and perpendicular to another 1-vector 

 b or a 2-vector A. The components of a along b and perpendicular to b are 



b-b 



and 



(axbKb 

 b-b" 



The components of a along A and perpendicular to A are 



(a«A)«A , (axA)«A 



- "ATaT ^"^ A7A- 



The component of the plane A on the plane B is 



(A.B)B 

 B-B ' 



and a vector in the line of intersection of the two planes is 



A*'B or A.B*. 



