If P = i + 3j - 2k and Q = 3i + j - 2k , then what is the position vector of the bisector of POQ ? - Mathematics MCQ Mathematics Mock Test - 3 2026 | ExamTest.live

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If P = \hat{i} + 3 \hat{j} - 2 \hat{k} and Q = 3 \hat{i} + \hat{j} - 2 \hat{k}, then what is the position vector of the bisector of ∠ POQ ?

Solution & Step-by-step Explanation

Magnitudes: ∣ OP ∣ = 1^{2} + 3^{2} + (- 2)^{2} = 14 ∣ OQ ∣ = 3^{2} + 1^{2} + (- 2)^{2} = 14 Since ∣ OP ∣ = ∣ OQ ∣, the vector along the angle bisector is simply OP + OQ : R = (1 + 3) \hat{i} + (3 + 1) \hat{j} + (- 2 - 2) \hat{k} = 4 \hat{i} + 4 \hat{j} - 4 \hat{k} Any scalar multiple of this vector acts as the bisector. Dividing by 2 : 2 \hat{i} + 2 \hat{j} - 2 \hat{k}

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If P = \hat{i} + 3 \hat{j} - 2 \hat{k} and Q = 3 \hat{i} + \hat{j} - 2 \hat{k}, then what is the position vector of the bisector of ∠ POQ ?

\hat{i} - \hat{j} - \hat{k}

2 \hat{i} + 2 \hat{j} - 2 \hat{k}

\hat{i} + \hat{j} + \hat{k}

None of the above

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