If `|veca|=2, |vecb|=5 and veca*vecb=0`, then `veca xx (vecaxx(vecaxx(vecaxx(vecaxx(veca xx vecb)))))` is equal to :

Options

A. `64veca `
B. `64vecb`
C. `-64veca`
D. `-64vecb`

Answer

To solve the problem, we need to evaluate the expression \( \vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b})))) \) given that \( |\vec{a}| = 2 \), \( |\vec{b}| = 5 \), and \( \vec{a} \cdot \vec{b} = 0 \).

### Step-by-step Solution:

1. **Understanding the Cross Product**:
Since \( \vec{a} \cdot \vec{b} = 0 \), it implies that \( \vec{a} \) and \( \vec{b} \) are orthogonal (perpendicular) vectors.

2. **First Cross Product**:
Let's denote \( \vec{c} = \vec{a} \times \vec{b} \).
The magnitude of \( \vec{c} \) can be calculated using the formula:
\[ |\vec{c}| = |\vec{a}| |\vec{b}| \sin(\theta) \]
where \( \theta = 90^\circ \) (since they are perpendicular), thus:
\[ |\vec{c}| = 2 \cdot 5 \cdot 1 = 10 \]

3. **Second Cross Product**:
Now we need to evaluate \( \vec{a} \times \vec{c} \):
Using the vector triple product identity:
\[ \vec{a} \times \vec{c} = \vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} \]
Since \( \vec{a} \cdot \vec{b} = 0 \), we have:
\[ \vec{a} \times \vec{c} = 0 \cdot \vec{a} - (|\vec{a}|^2) \vec{b} = -|\vec{a}|^2 \vec{b} = -4 \vec{b} \]

4. **Third Cross Product**:
Now we need to evaluate \( \vec{a} \times (-4 \vec{b}) \):
\[ \vec{a} \times (-4 \vec{b}) = -4 (\vec{a} \times \vec{b}) = -4 \vec{c} \]
We already found \( |\vec{c}| = 10 \), thus:
\[ \vec{a} \times (-4 \vec{b}) = -4 \vec{c} \]

5. **Fourth Cross Product**:
Next, we evaluate \( \vec{a} \times (-4 \vec{c}) \):
\[ \vec{a} \times (-4 \vec{c}) = -4 (\vec{a} \times \vec{c}) = -4 (-4 \vec{b}) = 16 \vec{b} \]

6. **Fifth Cross Product**:
Finally, we evaluate \( \vec{a} \times (16 \vec{b}) \):
\[ \vec{a} \times (16 \vec{b}) = 16 (\vec{a} \times \vec{b}) = 16 \vec{c} \]
Since \( \vec{c} = \vec{a} \times \vec{b} \) and we already found \( |\vec{c}| = 10 \), we have:
\[ \vec{a} \times (16 \vec{b}) = 16 \vec{c} \]

7. **Final Result**:
Putting it all together, we find that:
\[ \vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b})))) = -64 \vec{b} \]

Thus, the final answer is:
\[ \boxed{-64 \vec{b}} \]