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)/(3)`
D. `(pi)/(4)`

Answer

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.

Let's denote the two vectors as \(\vec{A}\) and \(\vec{B}\).

1. **Understanding the Vector and Scalar Products**:
- The scalar product (dot product) of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
- The vector product (cross product) of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]

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

3. **Canceling Common Terms**:
- We can cancel \(|\vec{A}| |\vec{B}|\) from both sides (assuming neither vector is zero):
\[ \sin \theta = \sqrt{3} \cos \theta \]

4. **Dividing Both Sides**:
- Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ \tan \theta = \sqrt{3} \]

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

Thus, the angle between the two vectors is \(60^\circ\) or \(\frac{\pi}{3}\) radians.