The angle between vectors `(vecA xx vecB) and (vecB xx vecA)` is :

Options

A. `pi` rad
B. `(pi)/(2)` rad
C. `(pi)/(4)` rad
D. zero

Explanation

To find the angle between the vectors \( \vec{A} \times \vec{B} \) and \( \vec{B} \times \vec{A} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Cross Product Properties**:
- The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \]
- This means that \( \vec{B} \times \vec{A} \) is equal to the negative of \( \vec{A} \times \vec{B} \).

2. **Set Up the Dot Product**:
- To find the angle \( \theta \) between the two vectors \( \vec{A} \times \vec{B} \) and \( \vec{B} \times \vec{A} \), we can use the dot product formula:
\[ \vec{A} \times \vec{B} \cdot \vec{B} \times \vec{A} = |\vec{A} \times \vec{B}| |\vec{B} \times \vec{A}| \cos(\theta) \]

3. **Substitute the Cross Product Relation**:
- Since \( \vec{B} \times \vec{A} = -(\vec{A} \times \vec{B}) \), we can substitute this into our equation:
\[ \vec{A} \times \vec{B} \cdot (-\vec{A} \times \vec{B}) = |\vec{A} \times \vec{B}| |\vec{A} \times \vec{B}| \cos(\theta) \]

4. **Simplify the Dot Product**:
- The left-hand side simplifies to:
\[ -|\vec{A} \times \vec{B}|^2 \]
- The right-hand side simplifies to:
\[ |\vec{A} \times \vec{B}|^2 \cos(\theta) \]

5. **Equate and Solve for Cosine**:
- Setting the two sides equal gives:
\[ -|\vec{A} \times \vec{B}|^2 = |\vec{A} \times \vec{B}|^2 \cos(\theta) \]
- Dividing both sides by \( |\vec{A} \times \vec{B}|^2 \) (assuming it is not zero):
\[ -1 = \cos(\theta) \]

6. **Find the Angle**:
- The cosine of \( \theta \) being -1 implies:
\[ \theta = \pi \text{ radians} \]

### Final Answer:
The angle between the vectors \( \vec{A} \times \vec{B} \) and \( \vec{B} \times \vec{A} \) is \( \pi \) radians.