Magnitude of the cross product of the two vectors `(vecA and vecB)` is equal to the dot product of the two . Magnitude of their resultant can be written as

Options

A. `sqrt(A^(2)+B^(2)+ABsqrt3)`
B. `sqrt(A^(2)+B^(2)+AB//sqrt2)`
C. `sqrt(A^(2)+B^(2)+2sqrt2AB)`
D. `sqrt(A^(2)+B^(2)+ABsqrt2)`

Explanation

To solve the problem, we need to find the magnitude of the resultant vector when the magnitude of the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is equal to the dot product of these two vectors.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
and
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
According to the problem, these two magnitudes are equal:
\[ |\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B} \]

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

3. **Finding the Angle**:
The equation \( \sin \theta = \cos \theta \) implies that:
\[ \tan \theta = 1 \]
Therefore, the angle \( \theta \) is:
\[ \theta = 45^\circ \]

4. **Calculating the Magnitude of the Resultant**:
The magnitude of the resultant vector \( \vec{R} \) of \( \vec{A} \) and \( \vec{B} \) can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
Substituting \( \theta = 45^\circ \) into the formula gives:
\[ |\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cdot \frac{1}{\sqrt{2}}} \]

5. **Simplifying the Expression**:
Let \( |\vec{A}| = a \) and \( |\vec{B}| = b \):
\[ |\vec{R}| = \sqrt{a^2 + b^2 + \sqrt{2}ab} \]

6. **Final Result**:
Thus, the magnitude of the resultant vector can be expressed as:
\[ |\vec{R}| = \sqrt{a^2 + b^2 + \sqrt{2}ab} \]

### Conclusion:
The magnitude of the resultant vector \( \vec{R} \) is \( \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + \sqrt{2} |\vec{A}| |\vec{B}|} \).