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

For a point

p = r_{2} (t)

, the magnitude of

(p - a_{1}) \times n_{1}

is the area of a parallelogram. Dividing by its base length

∥ n_{1} ∥

p - a_{1} = (t - 1, t, t + 1), (p - a_{1}) \times n_{1} = (- (t + 1), t + 1, - 1) .

Hence

d (t) = \frac{2 ( t + 1 ) ^{2} + 1}{2} = (t + 1)^{2} + \frac{1}{2} .

This use of the cross product removes any component parallel to the first line automatically.

ii

The graph is the upper branch of the hyperbola

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

. It is symmetric about

t = - 1

, positive everywhere, and has its lowest point at

(- 1, 1/ 2)

d^{'} (t) = \frac{t + 1}{d ( t )}, d^{''} (t) = \frac{1}{2 d ( t ) ^{3}} > 0,

so it is strictly convex. Since

d (t) - ∣ t + 1∣ = \frac{1/2}{( t + 1 ) ^{2} + 1/2 + ∣ t + 1∣} ⟶ 0,

its two oblique asymptotes are

d = t + 1 (t \to + \infty), d = - t - 1 (t \to - \infty) .

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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