 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014-resit/12/i/solution

In components the scalar triple product is

A_{i} ϵ_{ijk} B_{j} C_{k}

ϵ_{ijk} = ϵ_{kij}

. Relabelling the summed indices therefore gives

A_{i} ϵ_{ijk} B_{j} C_{k} = ϵ_{kij} A_{i} B_{j} C_{k} = (A \times B)_{k} C_{k},

which proves

A \cdot (B \times C) = (A \times B) \cdot C

. Equivalently both expressions are the same three-column determinant.

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