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

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

Use only the dot product to minimize the squared Euclidean distance from the origin:

D (t)^{2} = (a_{2} + t n_{2}) \cdot (a_{2} + t n_{2}) = ∥ a_{2} ∥^{2} + 2 t (a_{2} \cdot n_{2}) + t^{2} ∥ n_{2} ∥^{2} .

Completing the square gives

D (t)^{2} = ∥ n_{2} ∥^{2} (t + \frac{a _{2} \cdot n _{2}}{∥ n _{2} ∥ ^{2}})^{2} + ∥ a_{2} ∥^{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{∥ n _{2} ∥ ^{2}} .

The square is minimized at

t_{0} = - (a_{2} \cdot n_{2}) /∥ n_{2} ∥^{2}

, yielding

d_{0} = a_{2} \cdot a_{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{n _{2} \cdot n _{2}} .

Geometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point,

(a_{2} + t_{0} n_{2}) \cdot n_{2} = 0

, as required for perpendicular distance.

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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