Let `vecu and vecv` be unit vectors such that `vecu xx vecv + vecu = vecw and vecw xx vecu = vecv` . Find the value of `[vecu vecv vecw ] `

Explanation

To solve the problem, we need to find the value of the scalar triple product \([ \vec{u}, \vec{v}, \vec{w} ]\) given the conditions involving the unit vectors \(\vec{u}\) and \(\vec{v}\).

### Step-by-step Solution:

1. **Understand the Given Information**:
We have two unit vectors \(\vec{u}\) and \(\vec{v}\) such that:
\[ \vec{u} \times \vec{v} + \vec{u} = \vec{w} \]
\[ \vec{w} \times \vec{u} = \vec{v} \]

2. **Substituting for \(\vec{w}\)**:
From the first equation, we can express \(\vec{w}\) as:
\[ \vec{w} = \vec{u} \times \vec{v} + \vec{u} \]

3. **Using the Second Equation**:
Substitute \(\vec{w}\) into the second equation:
\[ (\vec{u} \times \vec{v} + \vec{u}) \times \vec{u} = \vec{v} \]

4. **Applying the Vector Triple Product Identity**:
Using the identity \(\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\):
\[ (\vec{u} \times \vec{v}) \times \vec{u} + \vec{u} \times \vec{u} = \vec{v} \]
Since \(\vec{u} \times \vec{u} = \vec{0}\), we have:
\[ (\vec{u} \times \vec{v}) \times \vec{u} = \vec{v} \]

5. **Using the Vector Triple Product Again**:
Applying the vector triple product identity again:
\[ (\vec{u} \cdot \vec{u}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{u} = \vec{v} \]
Since \(\vec{u}\) is a unit vector, \(\vec{u} \cdot \vec{u} = 1\):
\[ \vec{v} - (\vec{u} \cdot \vec{v}) \vec{u} = \vec{v} \]

6. **Solving for \(\vec{u} \cdot \vec{v}\)**:
This implies:
\[ -(\vec{u} \cdot \vec{v}) \vec{u} = \vec{0} \]
Therefore, \(\vec{u} \cdot \vec{v} = 0\), meaning \(\vec{u}\) and \(\vec{v}\) are orthogonal.

7. **Finding \([ \vec{u}, \vec{v}, \vec{w} ]\)**:
The scalar triple product can be expressed as:
\[ [ \vec{u}, \vec{v}, \vec{w} ] = \vec{u} \cdot (\vec{v} \times \vec{w}) \]
We already have \(\vec{w} = \vec{u} \times \vec{v} + \vec{u}\). Thus:
\[ \vec{v} \times \vec{w} = \vec{v} \times (\vec{u} \times \vec{v} + \vec{u}) = \vec{v} \times (\vec{u} \times \vec{v}) + \vec{v} \times \vec{u} \]
Using the vector triple product identity again:
\[ \vec{v} \times (\vec{u} \times \vec{v}) = (\vec{v} \cdot \vec{v}) \vec{u} - (\vec{v} \cdot \vec{u}) \vec{v} = 1 \cdot \vec{u} - 0 \cdot \vec{v} = \vec{u} \]
Therefore:
\[ \vec{v} \times \vec{w} = \vec{u} + \vec{v} \times \vec{u} \]

8. **Final Calculation**:
Since \(\vec{u}\) and \(\vec{v}\) are orthogonal and both are unit vectors, we can conclude:
\[ [ \vec{u}, \vec{v}, \vec{w} ] = 1 \]

### Final Answer:
\[ [ \vec{u}, \vec{v}, \vec{w} ] = 1 \]