The magnitude of the vector product of two vectors is `sqrt(3)` times their scalar product. The angle between the two vectors is

Options

A. `pi//2`
B. `pi//6`
C. `pi//32`
D. `pi//4`

Explanation

To solve the problem, we need to find the angle between two vectors given that the magnitude of their vector product is \(\sqrt{3}\) times their scalar product.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that:
- The magnitude of the vector product (cross product) of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \]
- The scalar product (dot product) of the same vectors is given by:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]

2. **Setting Up the Equation**:
According to the problem, we have:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{3} (\mathbf{A} \cdot \mathbf{B}) \]
Substituting the expressions for the vector and scalar products, we get:
\[ |\mathbf{A}| |\mathbf{B}| \sin \theta = \sqrt{3} (|\mathbf{A}| |\mathbf{B}| \cos \theta) \]

3. **Cancelling Common Terms**:
Assuming \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are not zero, we can divide both sides by \(|\mathbf{A}| |\mathbf{B}|\):
\[ \sin \theta = \sqrt{3} \cos \theta \]

4. **Using the Tangent Function**:
Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ \frac{\sin \theta}{\cos \theta} = \sqrt{3} \]
This simplifies to:
\[ \tan \theta = \sqrt{3} \]

5. **Finding the Angle**:
The angle \(\theta\) for which \(\tan \theta = \sqrt{3}\) is:
\[ \theta = \frac{\pi}{3} \text{ or } 60^\circ \]

6. **Conclusion**:
Therefore, the angle between the two vectors is:
\[ \theta = \frac{\pi}{3} \]

### Final Answer:
The angle between the two vectors is \(\frac{\pi}{3}\).