The modulus of the vector product of two vector is `(1)/(sqrt(3)` times their scalar product . The angle between vectors is

Options

A. `(pi)/(6)`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(3)`

Explanation

To solve the problem, we need to find the angle between two vectors given the relationship between their vector product and scalar product.

### Step-by-Step Solution:

1. **Understand the Given Information**:
We are given that the modulus of the vector product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \frac{1}{\sqrt{3}} \) times their scalar product.
\[ |\mathbf{A} \times \mathbf{B}| = \frac{1}{\sqrt{3}} |\mathbf{A} \cdot \mathbf{B}| \]

2. **Use the Formulas for Vector and Scalar Products**:
The magnitude of the vector product is given by:
\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \]
The scalar product is given by:
\[ |\mathbf{A} \cdot \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \cos \theta \]

3. **Substitute These Formulas into the Given Equation**:
Substituting the expressions for the vector and scalar products into the given equation:
\[ |\mathbf{A}| |\mathbf{B}| \sin \theta = \frac{1}{\sqrt{3}} |\mathbf{A}| |\mathbf{B}| \cos \theta \]

4. **Cancel Common Terms**:
Assuming \( |\mathbf{A}| \) and \( |\mathbf{B}| \) are not zero, we can cancel them from both sides:
\[ \sin \theta = \frac{1}{\sqrt{3}} \cos \theta \]

5. **Rearranging the Equation**:
We can rearrange the equation to express it in terms of \( \tan \theta \):
\[ \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \]
This implies:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

6. **Finding the Angle**:
The angle \( \theta \) whose tangent is \( \frac{1}{\sqrt{3}} \) is:
\[ \theta = 30^\circ \quad \text{(or } \frac{\pi}{6} \text{ radians)} \]

### Conclusion:
The angle between the two vectors is \( 30^\circ \).

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