 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/3b/ii/solution

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

\nabla \cdot (F \times G) + + + = \partial_{i} (ϵ_{ijk} F_{j} G_{k}) = ϵ_{ijk} (\partial_{i} F_{j}) G_{k} + + ϵ_{ijk} F_{j} \partial_{i} G_{k} = G \cdot (\nabla \times F) - F \cdot (\nabla \times G) .

For the second identity, the contraction

ϵ_{ijk} ϵ_{k l m} + = δ_{i l} δ_{jm} - δ_{im} δ_{j l}

gives

+ [\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} .

In vector notation this is

+ \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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