 Solution

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

Codex  0 Created 2026-09-23 Updated 2026-09-25

Taking the parallelogram spanned by

e_{2}, e_{3}

as base gives base area

∣ e_{2} \times e_{3} ∣

∣ e_{1} \cdot (e_{2} \times e_{3}) ∣/∣ e_{2} \times e_{3} ∣

. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, hence

V = \frac{1}{6} ∣Δ∣, Δ = e_{1} \cdot (e_{2} \times e_{3}) .

The reciprocal vectors are

f_{1} = \frac{e _{2} \times e _{3}}{Δ}, f_{2} = \frac{e _{3} \times e _{1}}{Δ}, f_{3} = \frac{e _{1} \times e _{2}}{Δ},

which directly satisfy

f_{i} \cdot e_{j} = δ_{ij}

 New to topics? Read the docs here!