The magnitude of scalar product of two vectors is `8` and of vector product is `8sqrt(3)`. The angle between them is:

Options

A. `30^(@)`
B. `60^(@)`
C. `120^(@)`
D. `150^(@)`

Explanation

To find the angle between two vectors given the magnitude of their scalar (dot) product and vector (cross) product, we can use the following relationships:

1. The scalar product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ A \cdot B = |A| |B| \cos \theta \]
where \( \theta \) is the angle between the two vectors.

2. The vector product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ A \times B = |A| |B| \sin \theta \]

Given:
- The magnitude of the scalar product \( A \cdot B = 8 \)
- The magnitude of the vector product \( A \times B = 8\sqrt{3} \)

### Step 1: Set up the equations
From the scalar product, we have:
\[ |A| |B| \cos \theta = 8 \quad \text{(1)} \]
From the vector product, we have:
\[ |A| |B| \sin \theta = 8\sqrt{3} \quad \text{(2)} \]

### Step 2: Divide the equations
To eliminate \( |A| |B| \), we can divide equation (2) by equation (1):
\[ \frac{|A| |B| \sin \theta}{|A| |B| \cos \theta} = \frac{8\sqrt{3}}{8} \]
This simplifies to:
\[ \tan \theta = \sqrt{3} \]

### Step 3: Find the angle \( \theta \)
The angle \( \theta \) for which \( \tan \theta = \sqrt{3} \) is:
\[ \theta = 60^\circ \quad \text{or} \quad \theta = 120^\circ \]

### Conclusion
Thus, the angle between the two vectors can be either \( 60^\circ \) or \( 120^\circ \).