Vectors a and b makes an angle `theta = (2pi)/(3)`
If `|a| = 1, |b| = 2`, then `|(2a + b) xx (a + 2b)|^(2) =`

Options

A. 9
B. 18
C. 27
D. 81

Answer

To solve the problem, we need to find the square of the magnitude of the vector expression \(|(2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b})|^2\), given that \(|\mathbf{a}| = 1\), \(|\mathbf{b}| = 2\), and the angle \(\theta = \frac{2\pi}{3}\) between vectors \(\mathbf{a}\) and \(\mathbf{b}\).

### Step-by-step Solution:

1. **Identify the vectors and their magnitudes**:
- Let \(|\mathbf{a}| = 1\) and \(|\mathbf{b}| = 2\).
- The angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(\theta = \frac{2\pi}{3}\).

2. **Calculate the cross product**:
We need to find \((2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b})\).

3. **Use the distributive property of the cross product**:
\[ (2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b}) = 2\mathbf{a} \times \mathbf{a} + 4\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} + 2\mathbf{b} \times \mathbf{b} \]
Since \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\) and \(\mathbf{b} \times \mathbf{b} = \mathbf{0}\), we simplify this to:
\[ 4\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} = 4\mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{b} = 3\mathbf{a} \times \mathbf{b} \]

4. **Find the magnitude of the cross product**:
The magnitude of \(\mathbf{a} \times \mathbf{b}\) is given by:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin(\theta) \]
Substituting the values:
\[ |\mathbf{a} \times \mathbf{b}| = 1 \cdot 2 \cdot \sin\left(\frac{2\pi}{3}\right) = 2 \cdot \sin\left(\frac{2\pi}{3}\right) \]
We know that \(\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\). Therefore:
\[ |\mathbf{a} \times \mathbf{b}| = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

5. **Calculate the magnitude of the cross product of the combined vectors**:
\[ |(2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b})| = |3\mathbf{a} \times \mathbf{b}| = 3 |\mathbf{a} \times \mathbf{b}| = 3 \cdot \sqrt{3} \]

6. **Find the square of the magnitude**:
\[ |(2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b})|^2 = (3\sqrt{3})^2 = 9 \cdot 3 = 27 \]

### Final Answer:
\[ |(2\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b})|^2 = 27 \]