A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
Let\[\theta \]be the angle between vectors\[\overrightarrow{P}\]and\[\overrightarrow{Q}\] whose resultant is \[\overrightarrow{R}\]. Here,\[P=Q\]and \[{{R}^{2}}=3PQ=3{{P}^{2}}\] As \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \] \[\therefore \] \[3{{P}^{2}}={{P}^{2}}+{{P}^{2}}+2{{P}^{2}}\cos \theta \] Or \[3{{P}^{2}}-2{{P}^{2}}=2{{P}^{2}}\cos \theta \] Or \[{{P}^{2}}=2{{P}^{2}}\cos \theta \] Or \[1=2\cos \theta \] \[\therefore \] \[\cos \theta =\frac{1}{2},\]thus, \[\cos \theta =\cos 60{}^\circ \] Or \[\theta =60{}^\circ =\frac{\pi }{3}\]
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