 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/11/c/ii/solution

Here

a_{2} \cdot a_{2} = 1

a_{2} \cdot n_{2} = 1

and

n_{2} \cdot n_{2} = 3

. Therefore

d_{0} = 1 - \frac{1}{3} = \frac{2}{3} .

The closest point has parameter

t_{0} = - 1/3

(- 1/3, - 1/3, 2/3)

, whose norm gives the same result.

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