Kinematics Question 19
Question: The resultant of two vectors A and b is perpendicular to the vector A and it’s magnitude is equal to half the magnitude of vector B. The angle between A and b is
Options:
A) $ 120{}^\circ $
B) $ 150{}^\circ $
C) $ 135{}^\circ $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{B}{2}=\sqrt{A^{2}+B^{2}+2AB\ \cos \theta } $ -(i)
$ \tan 90{}^\circ =\frac{B\sin \theta }{A+B\cos \theta }\Rightarrow A+B\cos \theta =0 $ $ \cos \theta =-\frac{A}{B} $
Hence, from (i) $ \frac{B^{2}}{4}=A^{2}+B^{2}-2A^{2}\Rightarrow A=\sqrt{3}\frac{B}{2} $
therefore $ \cos \theta =-\frac{A}{B}=-\frac{\sqrt{3}}{2} $ $ \theta =150{}^\circ $
BETA
