A) \[90{}^\circ \]
B) \[60{}^\circ \]
C) \[120{}^\circ \]
D) \[0{}^\circ \]
Correct Answer: C
Solution :
The resultant of the two vectors P and Q is \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ}\cos \theta \] Here, \[R=P=Q\] So, \[P=\sqrt{2{{P}^{2}}+2{{P}^{2}}\cos \theta }\] \[\therefore \] \[{{P}^{2}}=2{{P}^{2}}+2{{P}^{2}}\cos \theta \] \[\Rightarrow \] \[\cos \theta =-\frac{1}{2}\] \[=\cos {{120}^{o}}\] Hence, \[\theta ={{120}^{o}}\]
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