 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-1/5b/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Put

τ = a \cdot (b \times c)

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

Cyclic invariance of the scalar triple product gives

(a \times b) \cdot c = τ

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

If an expansion entirely in pairwise dot products is desired, expand the Gram determinant of

a, b, c

∣ a ∣^{2} ∣ b ∣^{2} ∣ c ∣^{2} + 2 (a \cdot b) (b \cdot c) (c \cdot a) - ∣ a ∣^{2} (b \cdot c)^{2} - ∣ b ∣^{2} (c \cdot a)^{2} - ∣ c ∣^{2} (a \cdot b)^{2} .

Both expressions remain valid for degenerate, including linearly dependent, vectors.

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