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MHT CET · Maths · Vector Algebra

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 \(m \angle A B C=\)

  1. A \(\frac{\pi}{3}\)
  2. B \(\frac{\pi}{4}\)
  3. C \(\frac{\pi}{6}\)
  4. D \(\frac{\pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\pi}{2}\)

Step-by-step Solution

Detailed explanation

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