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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-3/6e/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 6E / b / Solution by

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

Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by

a, b, c

. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by

a, c - b

and

b, a - c

. Their orthogonality gives

a \cdot c = a \cdot b, b \cdot a = b \cdot c .

Hence all three pairwise dot products have the same value

s

. The third pair satisfies

c \cdot (b - a) = c \cdot b - c \cdot a = s - s = 0,

so the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product gives

∣ a ∣^{2} + ∣ c - b ∣^{2} = ∣ b ∣^{2} + ∣ a - c ∣^{2} = ∣ c ∣^{2} + ∣ b - a ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} - 2 s .

Thus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.

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