 Solution

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

Put

V = a \cdot (b \times c)

(b \times c) \times (c \times a) = c [(b \times c) \cdot a] - a [(b \times c) \cdot c] = V c .

Taking its dot product with

a \times b

and using cyclic invariance of the scalar triple product yields

(a \times b) \cdot [(b \times c) \times (c \times a)] = V^{2} .

There was no division by

V

, so this also proves the identity for coplanar vectors.

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