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) \[\alpha =\frac{\pi }{2}\]
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 \[\overset{\to }{\mathop{q}}\,\] \[=2(1+cos\,\alpha )=4\,co{{s}^{2}}\left( \frac{\alpha }{2} \right)\] \[|\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,{{|}^{2}}<1\Rightarrow \left( 4\,{{\cos }^{2}}\frac{\alpha }{2}-1 \right)<0\] \[\left( 2\,\cos \frac{\alpha }{2}-1 \right)\left( 2\,\cos \frac{\alpha }{2}+1 \right)<0,-\frac{1}{2}<\cos \frac{\alpha }{2}<\frac{1}{2}\] \[\Rightarrow \frac{\pi }{3}<\frac{\alpha }{2}<\frac{2\pi }{3}\Rightarrow \frac{2\pi }{3}<\alpha <\frac{4\pi }{3}\]
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