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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iv/1/d/ii/solution

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Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 1 / d / ii / Solution by

Codex  0 2026-10-25

The cross product

b \times c

is perpendicular to both factors. Thus

a^{'} \cdot b = \frac{( b \times c ) \cdot b}{V} = 0, a^{'} \cdot c = \frac{( b \times c ) \cdot c}{V} = 0.

Cyclic permutations give all the off-diagonal dot products between the original basis and its reciprocal basis.

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