If `veca and vecb` are two unit vectors inclined at an angle `pi/3` then `{ veca xx (vecb+veca xx vecb)} .vecb` is equal to (a) `-3/4` (b) `1/4` (c) `3/4` (d) `1/2`

Options

A. `(-3)/4`
B. `1/4`
C. `3/4`
D. `1/2`

Explanation

To solve the problem, we need to evaluate the expression \(\vec{a} \times (\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{b}\) given that \(\vec{a}\) and \(\vec{b}\) are unit vectors inclined at an angle of \(\frac{\pi}{3}\).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Both \(\vec{a}\) and \(\vec{b}\) are unit vectors.
- The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\).
- Therefore, \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\).

2. **Finding the Dot Product \(\vec{a} \cdot \vec{b}\):**
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{3}\right) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]

3. **Expanding the Expression:**
We need to evaluate:
\[ \vec{a} \times (\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{b} \]
This can be rewritten as:
\[ \vec{a} \times \vec{b} + \vec{a} \times (\vec{a} \times \vec{b}) \]

4. **Using the Vector Triple Product Identity:**
The vector triple product identity states:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this identity to \(\vec{a} \times (\vec{a} \times \vec{b})\):
\[ \vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} \]
Since \(\vec{a} \cdot \vec{a} = 1\):
\[ \vec{a} \times (\vec{a} \times \vec{b}) = \left(\frac{1}{2}\right) \vec{a} - \vec{b} \]

5. **Substituting Back:**
Now substituting back into our expression:
\[ \vec{a} \times \vec{b} + \left(\frac{1}{2} \vec{a} - \vec{b}\right) \]
We can simplify this to:
\[ \vec{a} \times \vec{b} + \frac{1}{2} \vec{a} - \vec{b} \]

6. **Taking the Dot Product with \(\vec{b}\):**
Now we need to dot this entire expression with \(\vec{b}\):
\[ \left(\vec{a} \times \vec{b} + \frac{1}{2} \vec{a} - \vec{b}\right) \cdot \vec{b} \]
- The term \(\vec{a} \times \vec{b} \cdot \vec{b} = 0\) (since the cross product is perpendicular to both vectors).
- The term \(\frac{1}{2} \vec{a} \cdot \vec{b} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\).
- The term \(-\vec{b} \cdot \vec{b} = -1\).

So, combining these:
\[ 0 + \frac{1}{4} - 1 = \frac{1}{4} - 1 = -\frac{3}{4} \]

### Final Answer:
Thus, the value of \(\{\vec{a} \times (\vec{b} + \vec{a} \times \vec{b})\} \cdot \vec{b}\) is \(-\frac{3}{4}\).

### Conclusion:
The correct option is (a) \(-\frac{3}{4}\).