Kinematics Question 584
Question: The resultant of two vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is perpendicular to the vector $ \overrightarrow{A} $ and it’s magnitude is equal to half the magnitude of vector $ \overrightarrow{B} $ . The angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ it’s
Options:
A) $ 120{}^\circ $
B)$ 150{}^\circ $
C) $ 135{}^\circ $
D)$ 180{}^\circ $
Show Answer
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 $
