 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-1/5b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Dotting

c = \sum_{j} γ_{j} a_{j}

with

e_{i}

gives

e_{i} \cdot c = γ_{i} λ_{i}

, proving

γ_{i} = (e_{i} \cdot c) / λ_{i}

. Also

a_{1} \cdot (a_{2} \times a_{3}) = \frac{λ _{1} λ _{2} λ _{3}}{Δ}, γ_{1} γ_{2} γ_{3} = - \frac{( c \cdot v + d ) ^{3}}{λ _{1} λ _{2} λ _{3}} .

Combining these with

V = ∣Δ∣/6

gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.

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