 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/5a/b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Put

L = ∣ x_{n} ∣

e, f

such that

x_{n} = L e

and

x_{n + 1} = λ L f

. Direction vectors for the two lines are

x_{n + 2} - x_{n + 1} = - λ L (λ e + f)

and

x_{n + 3} - x_{n} = - L (e + λ^{3} f) .

Their scalar product and norms give

cos θ = \frac{λ ^{2} L ^{2} ( 1 + λ ^{2} )}{λ L ^{2} 1 + λ ^{2} 1 + λ ^{6}} .

Therefore the acute angle between the straight lines satisfies

cos θ = \frac{λ 1 + λ ^{2}}{1 + λ ^{6}} .

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