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 \[\text{\vec{P}}\] and \[\text{\vec{Q}}\] whose resultant is \[\text{\vec{R}}\text{.}\] 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, \[\cos \theta =\cos {{60}^{o}}\] or \[\theta ={{60}^{o}}=\frac{\pi }{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
