To find the sum of two vectors, draw them emanating from the origin. Complete the parallelogram. The diagonal, from the origin, is the sum of these two vectors.
A scalar multiplier may be factored out of either the inner (dot) product or the cross product. It is obvious from the respective definitions of each of the products that
(s a) . b = s (a . b)
(s a) x b = s (a x b).
Likewise for the right-scalar multiplier.
(a + b) . c = a . c + b . c.
Proof: Let the angle from c to a be alpha, to b be beta, and to (a + b) be gamma.
Substitute into the foregoing, to obtain
|a + b| |c| cos(gamma) =? (|a| cos(alpha) + |b| cos(beta)) |c|.
Divide through by |c|, to obtain
|a + b| cos(gamma) =? (|a| cos(alpha) + |b| cos(beta)),
which is obvious. QED.
Corollary: The right-distribution holds, as well.
(a + b) x c = a x c + b x c.
Proof is obvious from the isomorphism of the parallelogram under the mapping induced by the cross-multiplication by c.
Corollary: The right-distribution holds, as well.
a x (b x c) = (a . c) b - (a . b) c.
Proof: Employ the Gramm-Schmidt theorem, to obtain unit vectors i, j, and k, in the directions of a, b, and c, respectively. Then, check
the equation for each of the possible six permutations of i, j, and k. Only six; because, in the a-direction, only i is possible, and in the
b-direction, only either i or j is possible. Three of the products are zero: ix(ixi), ix(jxj), and ix(jxk).
The remaining three products are: ix(ixj) = -j, ix(ixk) = -k, and ix(jxi) = j. Finally, employ the foregoing two theorems. QED.
Copyright (c) 2000 R. I. 'Scibor-Marchocki ;ast modified Wednesday 24-th May 2000.