If `vecaxx(vecbxxvecc)=vecbxx(veccxxveca)` and `[(vec, vecb, vecc)]!=0`
then `vecaxx(vecbxxvecc)` is equal to

Options

A. `vec0`
B. `vecaxxvecb`
C. `vecbxxvecc`
D. `veccxxveca`

Explanation

To solve the problem, we start with the given equation:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} \times (\vec{c} \times \vec{a}) \]

We need to simplify the left-hand side and the right-hand side using the vector triple product identity.

### Step 1: Use the Vector Triple Product Identity
The vector triple product identity states that:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

Applying this to the left-hand side:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

### Step 2: Simplify the Right-Hand Side
Now, we apply the vector triple product identity to the right-hand side:

\[ \vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]

### Step 3: Set the Two Sides Equal
Now we equate the two results from Step 1 and Step 2:

\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]

### Step 4: Rearranging the Equation
Rearranging gives us:

\[ (\vec{a} \cdot \vec{c}) \vec{b} + (\vec{b} \cdot \vec{c}) \vec{a} = (\vec{b} \cdot \vec{a}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{c} \]

### Step 5: Factor Out Common Terms
We can factor out common terms:

\[ (\vec{a} \cdot \vec{c}) \vec{b} + (\vec{b} \cdot \vec{c}) \vec{a} = 2(\vec{a} \cdot \vec{b}) \vec{c} \]

### Conclusion
Thus, we have shown that:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} \times (\vec{c} \times \vec{a}) \implies \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

Therefore, the expression \(\vec{a} \times (\vec{b} \times \vec{c})\) is equal to:

\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]