 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/5/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The face

PQR

q - p = (- 1, 1, 1)

and

r - p = (- 2, - 1, 2)

(3, 0, 3)

, so choose the outward normal vector

n = (1, 0, 1)

cos θ = \frac{p \cdot n}{∣ p ∣∣ n ∣} = \frac{3}{6 2} = \frac{3}{2}, θ = \frac{π}{6} = 3 0^{\circ} .

Reversing the chosen normal gives the supplementary angle

15 0^{\circ}

; the acute angle with the normal line is

3 0^{\circ}

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