If `|vec(a)|=2 and |vec(b)|=3`, then `|vec(a) xx vec(b)|^(2)+ |vec(a).vec(b)|^(2)` is equal to

Options

A. 72
B. 64
C. 48
D. 36

Answer

To solve the problem, we need to find the value of \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 \) given that \( |\vec{a}| = 2 \) and \( |\vec{b}| = 3 \).

### Step-by-step Solution:

1. **Understand the Magnitudes**:
We know that:
- \( |\vec{a}| = 2 \)
- \( |\vec{b}| = 3 \)

2. **Calculate \( |\vec{a} \times \vec{b}|^2 \)**:
The magnitude of the cross product can be calculated using the formula:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) \]
where \( \theta \) is the angle between the vectors \( \vec{a} \) and \( \vec{b} \). Therefore,
\[ |\vec{a} \times \vec{b}|^2 = (|\vec{a}| |\vec{b}| \sin(\theta))^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2(\theta) \]
Substituting the values:
\[ |\vec{a} \times \vec{b}|^2 = (2^2)(3^2) \sin^2(\theta) = 4 \cdot 9 \sin^2(\theta) = 36 \sin^2(\theta) \]

3. **Calculate \( |\vec{a} \cdot \vec{b}|^2 \)**:
The magnitude of the dot product can be calculated using the formula:
\[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos(\theta) \]
Therefore,
\[ |\vec{a} \cdot \vec{b}|^2 = (|\vec{a}| |\vec{b}| \cos(\theta))^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2(\theta) \]
Substituting the values:
\[ |\vec{a} \cdot \vec{b}|^2 = (2^2)(3^2) \cos^2(\theta) = 4 \cdot 9 \cos^2(\theta) = 36 \cos^2(\theta) \]

4. **Combine the Results**:
Now we can combine both results:
\[ |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 36 \sin^2(\theta) + 36 \cos^2(\theta) \]
Factoring out the common term:
\[ = 36 (\sin^2(\theta) + \cos^2(\theta)) \]
Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ = 36 \cdot 1 = 36 \]

### Final Answer:
Thus, the value of \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 \) is **36**.