A) \[B{{A}^{2}}\cos \theta \]
B) \[B{{A}^{2}}\sin \theta \]
C) \[B{{A}^{2}}\sin \theta \cos \theta \]
D) zero
Correct Answer: D
Solution :
\[(\vec{B}\times \vec{A}).\vec{A}\] \[=B\,A\,\cos \,\theta \,\hat{n}\,.\,\vec{A}\] = 0 Here \[\hat{n}\]is perpendicular to both \[\vec{A}\,and\,\vec{B}\]. Alternative: \[(\vec{B}\times \vec{A})\cdot \vec{A}\] Interchange the cross and dot, we have, \[(\vec{B}\times \vec{A})\cdot \vec{A}=\vec{B}\cdot (\vec{A}\times \vec{A})=0\] \[(\because \vec{A}\times \vec{A}=0)\] Note: The volume of a parallelepiped bounded by vectors \[\vec{A},\,\vec{B}\] and \[\vec{C}\] can be obtained by giving formula \[(\vec{A}\times \vec{B})\cdot \vec{C}\].
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