For

A = (A_{1}, A_{2}, A_{3})

and

B = (B_{1}, B_{2}, B_{3})

, the scalar product is the inner product

A \cdot B = A_{1} B_{1} + A_{2} B_{2} + A_{3} B_{3} .

A \times B = (A_{2} B_{3} - A_{3} B_{2},, A_{3} B_{1} - A_{1} B_{3},, A_{1} B_{2} - A_{2} B_{1}) .

It is perpendicular to both vectors and has magnitude

∣ A ∣∣ B ∣ sin θ

ii

Solution

A \times (B \times C) = B (A \cdot C) - C (A \cdot B) .

Applying the same identity after reversing the outer cross product gives

(A \times B) \times C = B (A \cdot C) - A (B \cdot C) .

The formulas differ because the cross product is not associative.

Let

Δ = A \cdot (B \times C) .

De lt a

is nonzero. Applying the vector triple-product formulas to

(A \times B) \times (C \times D)

recovers the coefficients of

D

in the basis

A, B, C

. Equivalently, the reciprocal basis to

A, B, C

\frac{B \times C}{Δ}, \frac{C \times A}{Δ}, \frac{A \times B}{Δ} .

Taking scalar products with

D

therefore yields

D = \frac{D \cdot ( B \times C )}{Δ} A + \frac{D \cdot ( C \times A )}{Δ} B + \frac{D \cdot ( A \times B )}{Δ} C .

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