If the vertices A, B, C of a triangle ABC are (1, 1, 3) , (-1, 0, 0) , (0, 1, 2) respectively, then determine ABC . - Mathematics MCQ Mathematics Mock Test - 9 2026 | ExamTest.live

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If the vertices A, B, C of a triangle A BC are (1, 1, 3), (- 1, 0, 0), (0, 1, 2) respectively, then determine ∠ A BC .

Solution & Step-by-step Explanation

∠ A BC is the angle between the vectors B A and BC . B A = A - B = (1 - (- 1)) \hat{i} + (1 - 0) \hat{j} + (3 - 0) \hat{k} = 2 \hat{i} + \hat{j} + 3 \hat{k} BC = C - B = (0 - (- 1)) \hat{i} + (1 - 0) \hat{j} + (2 - 0) \hat{k} = \hat{i} + \hat{j} + 2 \hat{k} The angle θ is given by: cos θ = \frac{B A \cdot BC}{∣ B A ∣∣ BC ∣} B A \cdot BC = (2 \times 1) + (1 \times 1) + (3 \times 2) = 2 + 1 + 6 = 9 ∣ B A ∣ = 2^{2} + 1^{2} + 3^{2} = 4 + 1 + 9 = 14 ∣ BC ∣ = 1^{2} + 1^{2} + 2^{2} = 1 + 1 + 4 = 6 θ = cos^{- 1} (\frac{9}{14 6}) = cos^{- 1} (\frac{9}{84})

Try it yourself before checking the explanation above.

If the vertices A, B, C of a triangle A BC are (1, 1, 3), (- 1, 0, 0), (0, 1, 2) respectively, then determine ∠ A BC .

cos^{- 1} (\frac{10}{102})

cos^{- 1} (\frac{9}{14 \times 6})

cos^{- 1} (\frac{9}{14 \times 9})

cos^{- 1} (\frac{9}{84})

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