 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-3/3c/i/solution

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

[\nabla \times (F \times G)]_{i} = ϵ_{ijk} \partial_{j} (ϵ_{k ℓ m} F_{ℓ} G_{m}) = \partial_{j} (F_{i} G_{j} - F_{j} G_{i}) = F_{i} \partial_{j} G_{j} - G_{i} \partial_{j} F_{j} + G_{j} \partial_{j} F_{i} - F_{j} \partial_{j} G_{i} .

\nabla \times (F \times G) = F (\nabla \cdot G) - G (\nabla \cdot F) + (G \cdot \nabla) F - (F \cdot \nabla) G .

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