 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/3b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

[\nabla \times (ψ A)]_{i} = ϵ_{ijk} \partial_{j} (ψ A_{k}) = ψ ϵ_{ijk} \partial_{j} A_{k} + ϵ_{ijk} (\partial_{j} ψ) A_{k} .

This is

\nabla \times (ψ A) = ψ \nabla \times A + \nabla ψ \times A .

Similarly,

\nabla \cdot (A \times B) = \partial_{i} (ϵ_{ijk} A_{j} B_{k}) = ϵ_{ijk} (\partial_{i} A_{j}) B_{k} + ϵ_{ijk} A_{j} (\partial_{i} B_{k}) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B),

after cyclically relabelling the first term and swapping two indices in the second. Hence

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

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