 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/d/iii/solution

d (t)^{2} = (t + 1)^{2} + 1/2

is minimized when its first term vanishes. Therefore

t_{m i n} = - 1, d_{m i n} = \frac{1}{2} = s .

This agrees with the distance between skew lines found in part (b). The corresponding parameter on the first line is its orthogonal projection value

λ_{m i n} = \frac{( r _{2} ( - 1 ) - a _{1} ) \cdot n _{1}}{∥ n _{1} ∥ ^{2}} = - \frac{3}{2} .

The closest pair is

(- 1/2, - 3/2, 0)

and

(- 1, - 1, 0)

; their connecting vector

(- 1/2, 1/2, 0)

is perpendicular to both direction vectors.

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