`[a " " b " " axx b]` is equal to

Options

A. `a^(2)b^(2)`
B. `(a.b)^(2)`
C. `(a xxb)^(2)`
D. `|axxb|^(2)`

Answer

To solve the problem `[a " " b " " axx b]`, we need to evaluate the scalar triple product of the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{a} \times \mathbf{b} \).

### Step-by-Step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product of three vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) is defined as:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \]
In our case, we have \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \). Therefore, we need to evaluate:
\[ \mathbf{a} \cdot (\mathbf{b} \times (\mathbf{a} \times \mathbf{b})) \]

2. **Using the Vector Triple Product Identity**:
We can simplify \( \mathbf{b} \times (\mathbf{a} \times \mathbf{b}) \) using the vector triple product identity:
\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]
Here, let \( \mathbf{x} = \mathbf{b} \), \( \mathbf{y} = \mathbf{a} \), and \( \mathbf{z} = \mathbf{b} \). Thus:
\[ \mathbf{b} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{b} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \]
Since \( \mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 \), we can rewrite this as:
\[ |\mathbf{b}|^2 \mathbf{a} - (\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \]

3. **Substituting Back**:
Now we substitute this back into our scalar triple product:
\[ \mathbf{a} \cdot \left( |\mathbf{b}|^2 \mathbf{a} - (\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \right) \]
This expands to:
\[ |\mathbf{b}|^2 (\mathbf{a} \cdot \mathbf{a}) - (\mathbf{b} \cdot \mathbf{a})(\mathbf{a} \cdot \mathbf{b}) \]

4. **Using the Dot Product**:
We know that \( \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 \) and \( \mathbf{b} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b} \). Thus, we can simplify our expression to:
\[ |\mathbf{b}|^2 |\mathbf{a}|^2 - (\mathbf{a} \cdot \mathbf{b})^2 \]

5. **Final Result**:
Therefore, the final result of the scalar triple product \( [\mathbf{a}, \mathbf{b}, \mathbf{a} \times \mathbf{b}] \) is:
\[ |\mathbf{a}|^2 |\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2 \]