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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/11/b/iii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 11 / b / iii / Solution by

Codex  0 2026-10-18

The plane equation

x + z = 0

has normal vector

(1, 0, 1)^{T} = 2 \hat{n}

. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis

e = (1, 0, - 1)^{T} / 2

and

f = (0, 1, 0)^{T}

of the plane, write

r = s e + t f

. The cylinder equation becomes

s^{2} + t^{2} = 4

. Equivalently, the intersection has the parametric equation

r (θ) = 2 cos θ e + 2 sin θ f, 0 \leq θ < 2 π .

(1)

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