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If the angle between the vectors $\vec{A}$ and $\vec{B}$ is $\theta$, the value of the product $(\vec{B} \times \vec{A})\cdot\vec{A}$ is equal to -
A
Zero
B
$BA^2\sin\theta\cos\theta$
C
$BA^2\cos\theta$
D
$BA^2\sin\theta$
Detailed Solution
$\vec{B} \times \vec{A} = BA\sin\theta\,\hat{n}$, where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$.
$(\vec{B} \times \vec{A})\cdot\vec{A} = (BA\sin\theta)(\hat{n}\cdot\vec{A})$
Since $\hat{n}$ is perpendicular to $\vec{A}$, $\hat{n}\cdot\vec{A} = A\cos 90^\circ = 0$.
$\Rightarrow (\vec{B} \times \vec{A})\cdot\vec{A} = 0$
In general, a scalar triple product with two identical vectors is always zero.
$(\vec{B} \times \vec{A})\cdot\vec{A} = (BA\sin\theta)(\hat{n}\cdot\vec{A})$
Since $\hat{n}$ is perpendicular to $\vec{A}$, $\hat{n}\cdot\vec{A} = A\cos 90^\circ = 0$.
$\Rightarrow (\vec{B} \times \vec{A})\cdot\vec{A} = 0$
In general, a scalar triple product with two identical vectors is always zero.