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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