 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014/10/a/iii/solution

The orthogonal projection of the origin onto the line occurs at

t_{*} = - p \cdot u /∣ u ∣^{2} = - 1/5

, giving closest point

(2, 4/5, - 2/5)

d = 4 + \frac{16 + 4}{25} = \frac{24}{5} = \frac{2 30}{5} .

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