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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-1/1a/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 1A / ii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention

(A \times B)_{i} = ϵ_{ijk} A_{j} B_{k}

. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identity

ϵ_{ijk} ϵ_{k pq} = δ_{i p} δ_{j q} - δ_{i q} δ_{j p},

where

δ

is the Kronecker delta. Thus

((A \times B) \times (A \times C))_{i} = ϵ_{ijk} ϵ_{j ℓ m} ϵ_{k pq} A_{ℓ} B_{m} A_{p} C_{q} = (δ_{i p} δ_{j q} - δ_{i q} δ_{j p}) ϵ_{j ℓ m} A_{ℓ} B_{m} A_{p} C_{q} = A_{i} ϵ_{q ℓ m} C_{q} A_{ℓ} B_{m} - C_{i} ϵ_{p ℓ m} A_{p} A_{ℓ} B_{m} = A_{i} ϵ_{ℓ m q} A_{ℓ} B_{m} C_{q} .

The second term is zero because

A_{p} A_{ℓ}

is symmetric in

p, ℓ

and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product

A \cdot (B \times C)

. Hence

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

This cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

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