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