 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-resit/8/a/solution

\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) .

The first contraction uses the cyclic identity

ϵ_{ijk} = ϵ_{kij}

ϵ_{ijk} = - ϵ_{jik}

. This fixes the relative minus sign without guessing the orientation of the cross products.

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