A) \[\pi \]
B) \[\pi /3\]
C) \[\pi /2\]
D) \[\pi /4\]
Correct Answer: A
Solution :
\[(\overrightarrow{A}\times \overrightarrow{B})=(\overrightarrow{B}\times \overrightarrow{A})\] (Given) \[\Rightarrow \] \[(\overrightarrow{A}\times \overrightarrow{B})-(\overrightarrow{B}\times \overrightarrow{A})=\overrightarrow{0}\] \[(\overrightarrow{A}\times \overrightarrow{B})+(\overrightarrow{A}\times \overrightarrow{B})=\overrightarrow{0}\] \[[\because (\overrightarrow{B}\times \overrightarrow{A})=-(\overrightarrow{A}\times \overrightarrow{B})]\] \[\Rightarrow \] \[2(\overrightarrow{A}\times \overrightarrow{B})=\overrightarrow{0}\] \[\Rightarrow \] \[2AB\sin \theta =0\] \[\Rightarrow \] \[sin\theta =0\] \[\because \] \[[|\overrightarrow{A}|]=A\ne 0;|\overrightarrow{B}|=B\ne 0\] \[\therefore \] \[\theta =0\]or \[\pi \]
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