Physics Two Dimensional Motion Question 132
Question: The resultant of two vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is perpendicular to the vector $ \overrightarrow{A} $ and its magnitude is equal to half the magnitude of vector $ \overrightarrow{B} $ . The angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is
Options:
A) $ 120{}^\circ $ B)$ 150{}^\circ $ C) $ 135{}^\circ $ D)$ 180{}^\circ $
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Answer:
Correct Answer: B
Solution:
[b] $ \frac{\text{B}}{2}=\sqrt{{{A^{2}}\text{+}{{B^{2}}\text{+2AB cos}\theta } $ …… (i) $ \
Therefore \tan 90{}^\circ =\frac{\text{B}\sin \theta }{\text{A+B cos}\theta }\Rightarrow \text{A+B cos}\theta =0 $ $ \
Therefore \cos \theta =-\frac{\text{A}}{\text{B}} $ Hence, from (i) $ \frac{{{B^{2}}}{\text{A}}={{A^{2}}\text{+}{{B^{2}}-\text{2}{{A^{2}} $ $ \Rightarrow \text{A=}\sqrt{3}\frac{\text{B}}{2}\Rightarrow \cos \theta =-\frac{\text{A}}{\text{B}}=-\frac{\sqrt{3}}{2} $ $ \
Therefore \theta =150{}^\circ $
BETA
