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Theorem 4

THEOREM 4

Let A and B be two vectors in space which are not zero, and not parallel.

(i) A × B is perpendicular to both A and B.

(ii) Any vector perpendicular to both A and B is parallel to A × B.

PROOF

(i) We compute the inner products.

A · (A × B) = a1(a2b3 - a3b2) + a2(a3b1 - a1b3) + a3(a1b2 - a2b1) =

a1a2b3 - a1a3b2 + a2a3b1 - a1a2b3 + a1a3b2 - a2a3b1 = 0.

Similarly

B · (A × B) = 0.

It remains to prove that A × B0. At least one component of A, say a1, is nonzero. Let t = b1/a1 and let C = A × B. When we solve the equations

c3 = a1b2 - a2b1, c2 = a3b1 - a1b3 for b2 and b3,

we get

b1 = ta1, b2 = c3/a1 + ta2, b3 = c2/a1 + ta3.

Since B is not parallel to A, B ≠ tA. Therefore at least one of c2, c3 is nonzero, so C0.

(ii) Let C = A × B and let D be any other vector perpendicular to both A and B. Then

A · D = a1d1 + a2d2 + a3d3 = 0. B · D = b1d1 + b2d2 + b3d3 = 0.

At least one component of D, say d1, is nonzero. We may then solve the above equations for a1 and b1,

10_vectors-160.gif

Let s = a1/d1. Then c1 = sd1. Also,

10_vectors-161.gif

Similarly, c3 = sd3. Therefore C = sD, and C is parallel to D.

Warning: The Commutative Law and the Associative Law do not hold for the vector product. For example,

i × j = k,               j × i = -k

i × (j × j) = 0, (i × j) × j = -i.

However, vector products do satisfy the Distributive Laws

(sA + tB) × C = s(A × C) + t(B × C), C × (sA +tB) = s(C × A) + t(C × B).

The proof is left as an exercise (Problem 36 at the end of this section).


Last Update: 2006-11-25