Vectors `veca` and `vec b` are inclined at an angle `theta = 120^@` . If `|veca| = |vecb| = 2` , then `[(veca + 3vecb) xx (3veca + vecb)]^2` is equal to

Options

A. 190
B. 275
C. 300
D. 768

Answer

To solve the problem, we need to find the value of \([( \vec{a} + 3 \vec{b} ) \times ( 3 \vec{a} + \vec{b} )]^2\) given that \(|\vec{a}| = |\vec{b}| = 2\) and the angle \(\theta = 120^\circ\) between them.

### Step-by-Step Solution:

1. **Identify the vectors and their magnitudes:**
\[ |\vec{a}| = 2, \quad |\vec{b}| = 2, \quad \theta = 120^\circ \]

2. **Calculate the cross product:**
We need to compute:
\[ \vec{u} = \vec{a} + 3\vec{b}, \quad \vec{v} = 3\vec{a} + \vec{b} \]
Then, we find \(\vec{u} \times \vec{v}\):
\[ \vec{u} \times \vec{v} = (\vec{a} + 3\vec{b}) \times (3\vec{a} + \vec{b}) \]

3. **Expand the cross product using distributive property:**
\[ \vec{u} \times \vec{v} = \vec{a} \times (3\vec{a}) + \vec{a} \times \vec{b} + 3\vec{b} \times (3\vec{a}) + 3\vec{b} \times \vec{b} \]
Since \(\vec{a} \times \vec{a} = 0\) and \(\vec{b} \times \vec{b} = 0\), we simplify:
\[ \vec{u} \times \vec{v} = 0 + \vec{a} \times \vec{b} + 9(\vec{b} \times \vec{a}) + 0 \]
\[ = \vec{a} \times \vec{b} - 9(\vec{a} \times \vec{b}) = -8(\vec{a} \times \vec{b}) \]

4. **Find the magnitude of the cross product:**
The magnitude of \(\vec{a} \times \vec{b}\) is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Substituting the values:
\[ |\vec{a} \times \vec{b}| = 2 \cdot 2 \cdot \sin(120^\circ) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

5. **Calculate \(|\vec{u} \times \vec{v}|\):**
\[ |\vec{u} \times \vec{v}| = |-8(\vec{a} \times \vec{b})| = 8 |\vec{a} \times \vec{b}| = 8 \cdot 2\sqrt{3} = 16\sqrt{3} \]

6. **Square the magnitude:**
Now we need to square this value:
\[ [|\vec{u} \times \vec{v}|]^2 = (16\sqrt{3})^2 = 256 \cdot 3 = 768 \]

### Final Answer:
Thus, the value of \([( \vec{a} + 3 \vec{b} ) \times ( 3 \vec{a} + \vec{b} )]^2\) is \(768\).