A) \[{{A}^{2}}B\]
B) Zero
C) \[{{A}^{2}}B\sin \theta \]
D) \[{{A}^{2}}B\cos \theta \]
Correct Answer: B
Solution :
Let \[\overrightarrow{A}\,.(\overrightarrow{B}\,\times \overrightarrow{A})=\overrightarrow{A}\,.\,\overrightarrow{C}\,\] Here \[\overrightarrow{C}=\overrightarrow{B}\times \overrightarrow{A}\] Which is perpendicular to both vector \[\overrightarrow{A}\] and \[\overrightarrow{B}\] \ \[\,\overrightarrow{A}\,.\overrightarrow{\,C}=0\]You need to login to perform this action.
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