If `vecu` and `vecv` be unit vectors. If `vecw` is a vector such that `vecw+(vecwxxvecu)=vecv` then `vecu.(vecvxxvecw)` will be equal to

Options

A. `1-vecv.vecw`
B. `1-|vecv|^(2)`
C. `|vecv|^(2)- (vecu.vecv)^(2)`
D. None of these

Answer

To solve the problem, we need to find the value of \( \vec{u} \cdot (\vec{v} \times \vec{w}) \) given the conditions that \( \vec{u} \) and \( \vec{v} \) are unit vectors and that \( \vec{w} + (\vec{w} \times \vec{u}) = \vec{v} \).

### Step-by-Step Solution:

1. **Understand the given conditions**:
- \( \vec{u} \) and \( \vec{v} \) are unit vectors, which means \( |\vec{u}| = 1 \) and \( |\vec{v}| = 1 \).
- The equation \( \vec{w} + (\vec{w} \times \vec{u}) = \vec{v} \) is given.

2. **Rearranging the equation**:
- We can rearrange the equation to isolate \( \vec{w} \):
\[ \vec{w} = \vec{v} - (\vec{w} \times \vec{u}) \]

3. **Taking the dot product with \( \vec{u} \)**:
- Now, we take the dot product of both sides of the rearranged equation with \( \vec{u} \):
\[ \vec{u} \cdot \vec{w} = \vec{u} \cdot \vec{v} - \vec{u} \cdot (\vec{w} \times \vec{u}) \]

4. **Using properties of the dot and cross products**:
- The term \( \vec{u} \cdot (\vec{w} \times \vec{u}) \) is zero because the dot product of a vector with a vector that is perpendicular to it (like \( \vec{w} \times \vec{u} \)) is zero:
\[ \vec{u} \cdot (\vec{w} \times \vec{u}) = 0 \]

5. **Substituting back**:
- Therefore, we have:
\[ \vec{u} \cdot \vec{w} = \vec{u} \cdot \vec{v} \]

6. **Using the property of unit vectors**:
- Since \( \vec{u} \) and \( \vec{v} \) are unit vectors, we can express their dot product as:
\[ \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
- Thus, we have:
\[ \vec{u} \cdot \vec{w} = \cos \theta \]

7. **Finding \( \vec{u} \cdot (\vec{v} \times \vec{w}) \)**:
- We know that \( \vec{u} \cdot (\vec{v} \times \vec{w}) \) can be expressed using the scalar triple product:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = \vec{v} \cdot (\vec{w} \times \vec{u}) \]
- By the properties of the scalar triple product, we can write:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = 1 - \vec{v} \cdot \vec{w} \]

8. **Final Result**:
- Therefore, the value of \( \vec{u} \cdot (\vec{v} \times \vec{w}) \) is:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = 1 - \vec{v} \cdot \vec{w} \]