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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-resit/9/a/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 9 / a / Solution by

Codex  0 2026-10-25

Let

D = a \cdot (b \times c) \neq = 0

be the scalar triple product. The reciprocal basis satisfies

a \cdot a^{'} = 1

and all mismatched dot products vanish, with cyclic analogues. Apply the given vector triple product identity to obtain

b^{'} \times c^{'} = \frac{( c \times a ) \times ( a \times b )}{D ^{2}} = \frac{D a}{D ^{2}} = \frac{a}{D} .

(1)

Dotting with

a^{'}

gives

a^{'} \cdot (b^{'} \times c^{'}) = \frac{1}{D} .

(2)

The sign is retained: a negatively oriented original basis has a negatively oriented reciprocal basis. Equivalently, if the original vectors are columns of

B

, their reciprocals are columns of

B^{- T}

, whose determinant is

1/ det B

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