If `vecb` is a unit vector, then `(veca. vecb)vecb+vecbxx(vecaxxvecb)` is a equal to

Options

A. `|veca|^(2)vecb`
B. `(veca.vecb)veca`
C. `veca`
D. `(veca. vecb)vecb`

Explanation

To solve the problem, we need to simplify the expression \((\vec{a} \cdot \vec{b}) \vec{b} + \vec{b} \times (\vec{a} \times \vec{b})\), given that \(\vec{b}\) is a unit vector.

### Step-by-Step Solution:

1. **Understanding the Components**:
- We know that \(\vec{b}\) is a unit vector, which means \(|\vec{b}| = 1\).
- The expression consists of two parts: the dot product and the cross product.

2. **Evaluate the Dot Product**:
- The first part of the expression is \((\vec{a} \cdot \vec{b}) \vec{b}\).
- This represents a vector in the direction of \(\vec{b}\) scaled by the scalar \(\vec{a} \cdot \vec{b}\).

3. **Evaluate the Cross Product**:
- The second part is \(\vec{b} \times (\vec{a} \times \vec{b})\).
- We can use the vector triple product identity: \(\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z}\).
- Here, let \(\vec{x} = \vec{b}\), \(\vec{y} = \vec{a}\), and \(\vec{z} = \vec{b}\).
- Applying the identity, we get:
\[ \vec{b} \times (\vec{a} \times \vec{b}) = (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \]
- Since \(\vec{b}\) is a unit vector, \(\vec{b} \cdot \vec{b} = 1\).
- Thus, the expression simplifies to:
\[ \vec{b} \times (\vec{a} \times \vec{b}) = \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \]

4. **Combine the Results**:
- Now, substituting back into the original expression:
\[ (\vec{a} \cdot \vec{b}) \vec{b} + \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \]
- Notice that \((\vec{a} \cdot \vec{b}) \vec{b}\) and \(-(\vec{b} \cdot \vec{a}) \vec{b}\) are equal (since the dot product is commutative).
- Therefore, they cancel each other out:
\[ (\vec{a} \cdot \vec{b}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{b} + \vec{a} = \vec{a} \]

5. **Final Result**:
- The final result of the expression is:
\[ \vec{a} \]

### Conclusion:
The expression \((\vec{a} \cdot \vec{b}) \vec{b} + \vec{b} \times (\vec{a} \times \vec{b})\) simplifies to \(\vec{a}\).