 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015/10/a/solution

For the remaining distance calculation with

b = c

, every point on the line is

a + t b

. The nearest point to

p

b

t_{*} = \frac{( p - a ) \cdot b}{∣ b ∣ ^{2}}, d (p, S) = (p - a) - \frac{( p - a ) \cdot b}{∣ b ∣ ^{2}} b = \frac{∣ ( p - a ) \times b ∣}{∣ b ∣} .

Indeed, differentiating

∣ a + t b - p ∣^{2}

sets its component along

b

to zero. Its second derivative is

2∣ b ∣^{2} > 0

, proving this is the minimum.

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