 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv/3/b/solution

L : x = r (θ, ϕ) + λ (2 cos θ, sin θ cos ϕ, sin θ sin ϕ), λ \in R .

For

0 < θ < π

, the

y

- and

z

-coordinates vanish together exactly when

λ = - 1

. At that point, the

x

\frac{c o s θ}{2} - 2 cos θ = - \frac{c o s θ}{2} .

Thus

a = - \frac{1}{2} .

θ = 0

θ = π

, both the position vector and normal vector lie along the

x

-axis, so $L$ is the whole $x$ -axis in either case.

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