If the scalar and vector products of two vectors `vecA and vecB` are equal in magnitude, then the angle between the two vectors is

Options

A. `45^(@)`
B. `90^(@)`
C. `180^(@)`
D. `120^(@)`

Explanation

To solve the problem, we need to find the angle between two vectors \(\vec{A}\) and \(\vec{B}\) given that their scalar product and vector product are equal in magnitude.

### Step-by-Step Solution:

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

2. **Understand the Vector Product**:
The vector (cross) product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin \theta \]
The magnitude of the vector product is given by \(|\vec{A} \times \vec{B}|\).

3. **Set the Scalar and Vector Products Equal**:
According to the problem, the magnitudes of the scalar and vector products are equal:
\[ |\vec{A} \cdot \vec{B}| = |\vec{A} \times \vec{B}| \]
Substituting the expressions for the scalar and vector products, we have:
\[ |\vec{A}| |\vec{B}| |\cos \theta| = |\vec{A}| |\vec{B}| |\sin \theta| \]

4. **Simplify the Equation**:
Assuming \(|\vec{A}|\) and \(|\vec{B}|\) are not zero, we can divide both sides by \(|\vec{A}| |\vec{B}|\):
\[ |\cos \theta| = |\sin \theta| \]

5. **Solve for \(\theta\)**:
The equation \(|\cos \theta| = |\sin \theta|\) implies that:
\[ \tan \theta = 1 \]
The angles that satisfy this condition are:
\[ \theta = 45^\circ \quad \text{or} \quad \theta = 225^\circ \]
However, since we are looking for the angle between two vectors, we consider only the acute angle:
\[ \theta = 45^\circ \]

### Final Answer:
The angle between the two vectors \(\vec{A}\) and \(\vec{B}\) is \(45^\circ\).