If the angle between the vectors A and B is `theta`, the value of the product(BxA)). A is equal to

Options

A. `BA^2 cos theta`
B. `BA^2 sin theta`
C. `BA^2 sin theta cos theta`
D. zero

Explanation

To solve the problem, we need to find the value of the product \( \mathbf{B} \times \mathbf{A} \cdot \mathbf{A} \) given that the angle between the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \theta \).

### Step-by-Step Solution:

1. **Understand the Cross Product**:
The cross product \( \mathbf{B} \times \mathbf{A} \) gives a vector that is perpendicular to both \( \mathbf{B} \) and \( \mathbf{A} \). The magnitude of the cross product can be expressed as:
\[ |\mathbf{B} \times \mathbf{A}| = |\mathbf{B}| |\mathbf{A}| \sin(\theta) \]
where \( \theta \) is the angle between the two vectors.

2. **Introduce the Unit Normal Vector**:
Let \( \hat{n} \) be the unit vector in the direction of \( \mathbf{B} \times \mathbf{A} \). Thus, we can write:
\[ \mathbf{B} \times \mathbf{A} = |\mathbf{B} \times \mathbf{A}| \hat{n} = |\mathbf{B}| |\mathbf{A}| \sin(\theta) \hat{n} \]

3. **Dot Product with \( \mathbf{A} \)**:
Now, we need to compute the dot product \( (\mathbf{B} \times \mathbf{A}) \cdot \mathbf{A} \):
\[ (\mathbf{B} \times \mathbf{A}) \cdot \mathbf{A} = (|\mathbf{B}| |\mathbf{A}| \sin(\theta) \hat{n}) \cdot \mathbf{A} \]

4. **Evaluate the Dot Product**:
Since \( \hat{n} \) is perpendicular to both \( \mathbf{B} \) and \( \mathbf{A} \), the dot product \( \hat{n} \cdot \mathbf{A} \) is zero:
\[ \hat{n} \cdot \mathbf{A} = 0 \]

5. **Final Result**:
Therefore, we conclude that:
\[ (\mathbf{B} \times \mathbf{A}) \cdot \mathbf{A} = |\mathbf{B}| |\mathbf{A}| \sin(\theta) \cdot 0 = 0 \]

Thus, the value of the product \( \mathbf{B} \times \mathbf{A} \cdot \mathbf{A} \) is \( 0 \).