If `|vecaxxvecb|=2,|veca.vecb|=2`, then `|veca|^(2)|vecb|^(2)` is equal to

Options

A. 6
B. 2
C. 8
D. 20

Answer

To solve the problem, we need to find the value of \(|\vec{a}|^2 |\vec{b}|^2\) given that \(|\vec{a} \times \vec{b}| = 2\) and \(|\vec{a} \cdot \vec{b}| = 2\).

### Step-by-Step Solution:

1. **Use the formula for the magnitude of the cross product**:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Given that \(|\vec{a} \times \vec{b}| = 2\), we can write:
\[ |\vec{a}| |\vec{b}| \sin \theta = 2 \tag{1} \]

2. **Use the formula for the magnitude of the dot product**:
\[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos \theta \]
Given that \(|\vec{a} \cdot \vec{b}| = 2\), we can write:
\[ |\vec{a}| |\vec{b}| \cos \theta = 2 \tag{2} \]

3. **Square both equations**:
- Squaring equation (1):
\[ (|\vec{a}| |\vec{b}| \sin \theta)^2 = 2^2 \]
This gives:
\[ |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta = 4 \tag{3} \]

- Squaring equation (2):
\[ (|\vec{a}| |\vec{b}| \cos \theta)^2 = 2^2 \]
This gives:
\[ |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 4 \tag{4} \]

4. **Add equations (3) and (4)**:
\[ |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 4 + 4 \]
Factoring out \(|\vec{a}|^2 |\vec{b}|^2\):
\[ |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 8 \]

5. **Use the Pythagorean identity**:
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:
\[ |\vec{a}|^2 |\vec{b}|^2 (1) = 8 \]
Therefore:
\[ |\vec{a}|^2 |\vec{b}|^2 = 8 \]

### Final Answer:
\[ |\vec{a}|^2 |\vec{b}|^2 = 8 \]