Kinematics Question 153
Question: The angle between the two vectors $ \vec{A}=3\hat{i}+4\hat{j}+5\hat{k} $ and $ \vec{B}=3\hat{i}+4\hat{j}-5\hat{k} $ will be [Pb. CET 2001]
Options:
A) $ 90{}^\circ $
B) $ 0{}^\circ $
C) $ 60{}^\circ $
D) $ 45{}^\circ $
Correct Answer: A $ \cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|A||B|}=\frac{(3\hat{i}+4\hat{j}+5\hat{k})(3\hat{i}+4\hat{j}-5\hat{k})}{\sqrt{9+16+25}\sqrt{9+16+25}} $ $ =\frac{9+16-25}{50}=0 $
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Answer:
Solution:
therefore $ \cos \theta =0 $ ,
$ \therefore $ $ \theta =90{}^\circ $
