 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/14/e/solution

With

b \cdot c = 0

, substitute

a = α b + β c

into the plane constraints. They become

α ∣ b ∣^{2} = λ

and

β ∣ c ∣^{2} = μ

α = \frac{λ}{∣ b ∣ ^{2}}, β = \frac{μ}{∣ c ∣ ^{2}} .

a

then gives the closest distance

d_{m i n} = ∣ a ∣ = \frac{λ ^{2}}{∣ b ∣ ^{2}} + \frac{μ ^{2}}{∣ c ∣ ^{2}} .

Zero values of either scalar are allowed, including zero distance if both planes contain the origin.

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