If ` vec aa n d vec b` are two vectors such that `| vec axx vec b|=2,` then find the value of `[ vec a vec b vec axx vec b]dot`

Answer

To solve the problem, we need to find the value of the scalar triple product \([ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ]\) given that \(| \vec{a} \times \vec{b} | = 2\).

### Step-by-Step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product \([ \vec{a}, \vec{b}, \vec{c} ]\) can be expressed as \(\vec{a} \cdot (\vec{b} \times \vec{c})\). In our case, we need to evaluate \([ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ]\).

2. **Applying the Scalar Triple Product Property**:
We can rearrange the scalar triple product:
\[ [ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ] = \vec{a} \cdot (\vec{b} \times (\vec{a} \times \vec{b})) \]
By the vector triple product identity, we have:
\[ \vec{b} \times (\vec{a} \times \vec{b}) = (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \]

3. **Substituting Back**:
Now substituting this back into our expression:
\[ [ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ] = \vec{a} \cdot \left( (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \right) \]

4. **Calculating Each Term**:
- The first term is \(\vec{a} \cdot ((\vec{b} \cdot \vec{b}) \vec{a}) = (\vec{b} \cdot \vec{b}) (\vec{a} \cdot \vec{a})\).
- The second term is \(\vec{a} \cdot (-(\vec{b} \cdot \vec{a}) \vec{b}) = -(\vec{b} \cdot \vec{a}) (\vec{a} \cdot \vec{b})\).

5. **Combining the Results**:
Therefore, we have:
\[ [ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ] = (\vec{b} \cdot \vec{b}) (\vec{a} \cdot \vec{a}) - (\vec{b} \cdot \vec{a})^2 \]

6. **Using the Magnitude of the Cross Product**:
We know that \(| \vec{a} \times \vec{b} | = 2\). The magnitude of the cross product is given by:
\[ | \vec{a} \times \vec{b} | = |\vec{a}| |\vec{b}| \sin \theta = 2 \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

7. **Finding the Final Value**:
The scalar triple product can also be expressed as:
\[ [ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ] = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 \]
This is equivalent to the squared area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\), which is \(|\vec{a} \times \vec{b}|^2\).

Therefore:
\[ [ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ] = | \vec{a} \times \vec{b} |^2 = 2^2 = 4 \]

### Final Answer:
The value of \([ \vec{a}, \vec{b}, \vec{a} \times \vec{b} ]\) is **4**.