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 that the modulus of their vector product is \(\frac{1}{\sqrt{3}}\) times their scalar product. Let's denote the two vectors as \(\vec{A}\) and \(\vec{B}\).

### Step-by-step Solution:

1. **Understand the given relationship**:
The problem states that:
\[ |\vec{A} \times \vec{B}| = \frac{1}{\sqrt{3}} |\vec{A} \cdot \vec{B}| \]

2. **Use the formulas for vector and scalar products**:
The magnitude of the vector product (cross product) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
The scalar product (dot product) is given by:
\[ |\vec{A} \cdot \vec{B}| = |\vec{A}| |\vec{B}| \cos \theta \]

3. **Substitute the formulas into the given relationship**:
Substitute the expressions for the vector and scalar products into the equation:
\[ |\vec{A}| |\vec{B}| \sin \theta = \frac{1}{\sqrt{3}} |\vec{A}| |\vec{B}| \cos \theta \]

4. **Cancel the common terms**:
Assuming \(|\vec{A}|\) and \(|\vec{B}|\) are not zero, we can divide both sides by \(|\vec{A}| |\vec{B}|\):
\[ \sin \theta = \frac{1}{\sqrt{3}} \cos \theta \]

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

6. **Find the angle**:
The angle \(\theta\) for which \(\tan \theta = \frac{1}{\sqrt{3}}\) is:
\[ \theta = 30^\circ \quad \text{or} \quad \theta = \frac{\pi}{6} \text{ radians} \]

### Final Answer:
The angle between the two vectors is \(30^\circ\) or \(\frac{\pi}{6}\) radians.