If the vectors `veca` and `vecb` are mutually perpendicular, then `vecaxx{vecaxx{vecaxx(vecaxxvecb)}}` is equal to

Options

A. `|veca|^(2)vecb`
B. `|veca|^(3)vecb`
C. `|veca|^(4)vecb`
D. none of these

Explanation

To solve the problem, we need to find the value of \( \vec{A} \times (\vec{A} \times (\vec{A} \times \vec{B})) \) given that the vectors \( \vec{A} \) and \( \vec{B} \) are mutually perpendicular.

### Step-by-Step Solution:

1. **Understand the Given Information:**
Since \( \vec{A} \) and \( \vec{B} \) are mutually perpendicular, we have:
\[ \vec{A} \cdot \vec{B} = 0 \]

2. **Use the Vector Triple Product Identity:**
We will use the vector triple product identity:
\[ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C} \]
In our case, we can apply this identity to simplify \( \vec{A} \times (\vec{A} \times \vec{B}) \).

3. **Calculate \( \vec{A} \times (\vec{A} \times \vec{B}) \):**
Let \( \vec{C} = \vec{B} \). Then:
\[ \vec{A} \times (\vec{A} \times \vec{B}) = (\vec{A} \cdot \vec{B}) \vec{A} - (\vec{A} \cdot \vec{A}) \vec{B} \]
Since \( \vec{A} \cdot \vec{B} = 0 \):
\[ \vec{A} \times (\vec{A} \times \vec{B}) = 0 \cdot \vec{A} - (\vec{A} \cdot \vec{A}) \vec{B} = -|\vec{A}|^2 \vec{B} \]

4. **Substitute Back into the Original Expression:**
Now we need to find \( \vec{A} \times (-|\vec{A}|^2 \vec{B}) \):
\[ \vec{A} \times (-|\vec{A}|^2 \vec{B}) = -|\vec{A}|^2 (\vec{A} \times \vec{B}) \]

5. **Final Result:**
Since \( \vec{A} \) and \( \vec{B} \) are perpendicular, \( \vec{A} \times \vec{B} \) is a vector perpendicular to both \( \vec{A} \) and \( \vec{B} \). Thus, we can conclude:
\[ \vec{A} \times (\vec{A} \times (\vec{A} \times \vec{B})) = -|\vec{A}|^2 (\vec{A} \times \vec{B}) \]

### Conclusion:
The final expression for \( \vec{A} \times (\vec{A} \times (\vec{A} \times \vec{B})) \) is:
\[ -|\vec{A}|^2 (\vec{A} \times \vec{B}) \]