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 expression \( B \times A \cdot A \), given that the angle between vectors \( A \) and \( B \) is \( \theta \).

### Step-by-Step Solution:

1. **Understanding the Cross Product**:
The cross product \( B \times A \) results in a vector that is perpendicular to both \( B \) and \( A \). The magnitude of this vector can be calculated using the formula:
\[ |B \times A| = |B| |A| \sin(\theta) \]
where \( \theta \) is the angle between vectors \( A \) and \( B \).

2. **Direction of the Cross Product**:
The direction of the vector \( B \times A \) can be determined using the right-hand rule. It will be perpendicular to the plane formed by vectors \( A \) and \( B \).

3. **Dot Product with Vector A**:
Now, we need to calculate the dot product \( (B \times A) \cdot A \). The dot product of two vectors is given by:
\[ \text{If } \mathbf{C} = \mathbf{B} \times \mathbf{A}, \text{ then } \mathbf{C} \cdot \mathbf{A} = |C| |A| \cos(\phi) \]
where \( \phi \) is the angle between vector \( C \) (which is \( B \times A \)) and vector \( A \).

4. **Finding the Angle \( \phi \)**:
Since \( B \times A \) is perpendicular to both \( B \) and \( A \), the angle \( \phi \) between \( B \times A \) and \( A \) is \( 90^\circ \).

5. **Calculating the Dot Product**:
The cosine of \( 90^\circ \) is zero:
\[ \cos(90^\circ) = 0 \]
Therefore, the dot product becomes:
\[ (B \times A) \cdot A = |B \times A| |A| \cos(90^\circ) = |B \times A| |A| \cdot 0 = 0 \]

6. **Final Result**:
Thus, the value of \( B \times A \cdot A \) is:
\[ \boxed{0} \]