Q113 / 160 · JIPMER 2007 · Physics
If the angle between the vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ is $\theta$, the value of the product $(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{A}}$ is equal to
- A$B A^2 \cos \theta$
- B$B A^2 \sin \theta$
- C$B A^2 \sin \theta \cos \theta$
- Correct answerDzero
Solution
$(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{A}}$
$\begin{aligned} & =B A \cos \theta \overrightarrow{\mathbf{n}} \cdot \overrightarrow{\mathbf{A}} \\ & =0\end{aligned}$
Here $\hat{\mathbf{n}}$ is perpendicular to both $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$.
Alternative : $(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{A}}$
Interchange the cross and dot, we have, $(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{A}}=\overrightarrow{\mathbf{B}} \cdot(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}})=0 \quad(\because \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}}=0)$
NOTE: The volume of a parallelopiped bounded by vectors $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{A}}$ can be obtained by giving formula $(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{A}}$
Hence, the correct answer is option: 'D'.