Vector Algebra Question 131
Question: If $ \overset{\to }{\mathop{p}}, $ and $ \overset{\to }{\mathop{q}}, $ are two unit vectors inclined at an angle $ \alpha $ to each other than $ |\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},|<1 $ if
Options:
A) $ \frac{2\pi }{3}<\alpha <\frac{4\pi }{3} $
B) $ \frac{4\pi }{3}<\alpha <2\pi $
C) $ 0<\alpha <\frac{\pi }{3} $
D) $ \alpha =\frac{\pi }{2} $
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Answer:
Correct Answer: A
Solution:
- [a] $ |\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},|=(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},).(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},) $ $ =|\overset{\to }{\mathop{p}},{{|}^{2}}+|\overset{\to }{\mathop{q}},{{|}^{2}}+2\overset{\to }{\mathop{p}},\cdot \overset{\to }{\mathop{q}},=2+2\cos \alpha , $ Where $ \alpha $ is the angle between $ \overset{\to }{\mathop{p}}, $ and $ \overset{\to }{\mathop{q}}, $ $ =2(1+cos,\alpha )=4,cos^{2}( \frac{\alpha }{2} ) $ $ |\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},{{|}^{2}}<1\Rightarrow ( 4,{{\cos }^{2}}\frac{\alpha }{2}-1 )<0 $ $ ( 2,\cos \frac{\alpha }{2}-1 )( 2,\cos \frac{\alpha }{2}+1 )<0,-\frac{1}{2}<\cos \frac{\alpha }{2}<\frac{1}{2} $
$ \Rightarrow \frac{\pi }{3}<\frac{\alpha }{2}<\frac{2\pi }{3}\Rightarrow \frac{2\pi }{3}<\alpha <\frac{4\pi }{3} $
BETA
