 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/1/a/solution

[\nabla \times (A \times B)]_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} (A_{l} B_{m}) = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) \partial_{j} (A_{l} B_{m}) = \partial_{j} (A_{i} B_{j}) - \partial_{j} (A_{j} B_{i}) = B_{j} \partial_{j} A_{i} + A_{i} \partial_{j} B_{j} - B_{i} \partial_{j} A_{j} - A_{j} \partial_{j} B_{i} .

The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

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

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