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} $
Correct Answer: B 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} $Show Answer
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