If `veca and vecb` are unequal unit vectors such that `(veca - vecb) xx[ (vecb + veca) xx (2 veca + vecb)] = veca+vecb` then angle `theta " between " veca and vecb` is

Options

A. 0
B. `pi//2`
C. `pi//4`
D. `pi`

Explanation

To solve the given problem, we need to analyze the vector equation provided and find the angle \( \theta \) between the unit vectors \( \vec{a} \) and \( \vec{b} \).

### Step-by-Step Solution:

1. **Understand the Given Equation**:
The equation given is:
\[ (\vec{a} - \vec{b}) \times [(\vec{b} + \vec{a}) \times (2\vec{a} + \vec{b})] = \vec{a} + \vec{b} \]

2. **Use Vector Identity**:
We can use the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Let \( \vec{x} = \vec{a} - \vec{b} \), \( \vec{y} = \vec{b} + \vec{a} \), and \( \vec{z} = 2\vec{a} + \vec{b} \).

3. **Calculate the Cross Product**:
We need to calculate \( (\vec{b} + \vec{a}) \times (2\vec{a} + \vec{b}) \):
\[ (\vec{b} + \vec{a}) \times (2\vec{a} + \vec{b}) = \vec{b} \times (2\vec{a} + \vec{b}) + \vec{a} \times (2\vec{a} + \vec{b}) \]
This simplifies to:
\[ 2(\vec{b} \times \vec{a}) + \vec{b} \times \vec{b} + 2(\vec{a} \times \vec{a}) + \vec{a} \times \vec{b} = 2(\vec{b} \times \vec{a}) + 0 + 0 + \vec{a} \times \vec{b} = 3(\vec{b} \times \vec{a}) \]

4. **Substituting Back**:
Now substitute back into the original equation:
\[ (\vec{a} - \vec{b}) \times [3(\vec{b} \times \vec{a})] = \vec{a} + \vec{b} \]
This simplifies to:
\[ 3(\vec{a} - \vec{b}) \times (\vec{b} \times \vec{a}) = \vec{a} + \vec{b} \]

5. **Using the Vector Identity Again**:
Apply the vector triple product identity again:
\[ (\vec{a} - \vec{b}) \times (\vec{b} \times \vec{a}) = (\vec{a} - \vec{b}) \cdot \vec{a} \vec{b} - (\vec{a} - \vec{b}) \cdot \vec{b} \vec{a} \]
This leads to:
\[ (\vec{a} - \vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} \]

6. **Dot Product**:
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
\[ |\vec{a}| = |\vec{b}| = 1 \quad \text{and} \quad \vec{a} \cdot \vec{b} = \cos \theta \]
Therefore, we can write:
\[ 1 - \cos \theta = \cos \theta \]
Rearranging gives:
\[ 1 = 2 \cos \theta \implies \cos \theta = \frac{1}{2} \]

7. **Finding the Angle**:
The angle \( \theta \) corresponding to \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = \frac{\pi}{3} \quad \text{or} \quad \theta = \frac{5\pi}{3} \]
However, since \( \vec{a} \) and \( \vec{b} \) are unequal unit vectors, we only consider \( \theta = \frac{\pi}{3} \).

### Final Answer:
The angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{3} \).