Question

If the angle between the vectors $ \overrightarrow{A}$ and $ \overrightarrow{B}$ is $\theta, $ the value of the product $ ( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A}$ is equal to

• A. B $ A^2 \sin \theta$
• B. B $ A^2 \cos \theta$
• C. B $ A^2 \sin \theta \cos \theta$
• D. zero

Answer 1

The Correct Option is D

To solve this problem, we need to evaluate the expression $( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A}$. We begin by understanding the vector operations involved:

1. Cross Product: The cross product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is a vector perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$. The magnitude is given by $(|\overrightarrow{B}| |\overrightarrow{A}| \sin \theta)$, where $\theta$ is the angle between the two vectors.
2. Dot Product: The dot product of two vectors is a scalar and is calculated as the product of their magnitudes and the cosine of the angle between them: $(|\overrightarrow{A}| |\overrightarrow{B}| \cos \phi)$, where $\phi$ is the angle between the vectors.
3. The expression $( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A}$ involves taking the dot product of the vector $( \overrightarrow{B} \times \overrightarrow{A})$ with $\overrightarrow{A}$.

Understanding the geometrical implication:

• Since $( \overrightarrow{B} \times \overrightarrow{A})$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$, the angle between this resultant vector and $\overrightarrow{A}$ is $90^\circ$.
• Therefore, $( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A} = |\overrightarrow{B} \times \overrightarrow{A}| |\overrightarrow{A}| \cos 90^\circ$.
• Since $\cos 90^\circ = 0$, it follows that $( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A}$ evaluates to 0.

Conclusion: The value of the product $( \overrightarrow{B} \times \overrightarrow{A}) \cdot \overrightarrow{A}$ is zero. This is the correct answer, as the cross product of two vectors is perpendicular to the original vectors, making their dot product zero.