Elementary Calculus: 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 × B ≠ 0. 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 - a₁b₃for 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 C ≠ 0.

(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,

Let s = a₁/d₁. Then c₁ = sd₁. Also,

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).

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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 × B ≠ 0. 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 - a₁b₃for 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 C ≠ 0. (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, Let s = a₁/d₁. Then c₁ = sd₁. Also, 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).
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