 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/1/b/solution

1 \cdot 0 + 2 (- 1) + (- 1) \cdot 1 = - 3

, while the Euclidean norms are

6

and

2

. Therefore the angle

θ \in [0, π]

satisfies

cos θ = \frac{- 3}{6 2} = - \frac{3}{2}, θ = \frac{5 π}{6} = 15 0^{\circ} .

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