Q86 / 150 · MHT CET 2022 (08 Aug Shift 1) · Mathematics

The position vectors of vertices of $\triangle A B C$ are $4 \hat{i}-2 \hat{j}$; $\hat{i}+4 \hat{j}-3 \hat{k}$ and $-\hat{i}+5 \hat{j}+\hat{k}$ respectively, then $m \angle A B C=$

  • A
    $\frac{\pi}{3}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • Correct answer
    D
    $\frac{\pi}{2}$

Solution

$\begin{aligned} & \angle A B C=\text { angle between } \overrightarrow{B A} \text { and } \overrightarrow{B C} \\ & =\cos ^{-1}\left(\frac{\overrightarrow{B A} \cdot \overrightarrow{B C}}{|\overrightarrow{B A}||\overrightarrow{B C}|}\right)=\cos ^{-1}\left(\frac{(3 \hat{i}-6 \hat{j}+3 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+4 \hat{k})}{\sqrt{3^2+(-6)^2+3^2} \sqrt{(-2)^2+1^2+4^2}}\right) \\ & =\cos ^{-1}\left(\frac{6-6+12}{\sqrt{54} \sqrt{21}}\right)=\cos ^{-1}(0)=\frac{\pi}{2}\end{aligned}$

Hence, the correct answer is option: 'D'.

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