Kinematics Question 4
Question: If the sum of two unit vectors its a unit vector, then magnitude of difference it’s [CPMT 1995; CBSE PMT 1989]
Options:
A) $ \sqrt{2} $
B) $ \sqrt{3} $
C) $ 1/\sqrt{2} $
D) $ \sqrt{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ {{\hat{n}}_1} $ and $ {{\hat{n}}_2} $ are the two unit vectors, then the suM is
$ {{\overrightarrow{n}} _{s}}={{\hat{n}}_1}+{{\hat{n}}_2} $ or $ n_s^{2}=n_1^{2}+n_2^{2}+2n_1n_2\cos \theta $
$ =1+1+2\cos \theta $
Since it’s given that $ n _{s} $ it’s also a unit vector,
therefore $ 1=1+1+2\cos \theta $
therefore $ \cos \theta =-\frac{1}{2} $ $ \theta =120{}^\circ $
Now the difference vector is $ {{\hat{n}} _{d}}={{\hat{n}}_1}-{{\hat{n}}_2} $ or $ n_d^{2}=n_1^{2}+n_2^{2}-2n_1n_2\cos \theta $
$ =1+1-2\cos (120{}^\circ ) $
$ n_d^{2}=2-2(-1/2)=2+1=3 $
$ \Rightarrow n _{d}=\sqrt{3} $
BETA
