If `vecaxx(vecaxxvecb)=vecbxx(vecbxxvecc)` and `veca.vecb!=0`, then `[(veca,vecb,vecc)]=`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

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

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

### Step 1: Use the vector triple product identity
We can use the vector triple product identity, which states that:

\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]

Applying this identity to both sides of the equation:

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

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

### Step 2: Set the results equal
Now we set the two results equal to each other:

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

### Step 3: Rearranging the equation
Rearranging gives us:

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

### Step 4: Grouping terms
We can group terms involving \(\vec{b}\) and \(\vec{c}\):

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

### Step 5: Dot product with \(\vec{b}\) and \(\vec{c}\)
Now we take the dot product of both sides with \(\vec{b} \times \vec{c}\):

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

### Step 6: Simplifying the dot products
Using the property that \(\vec{x} \cdot (\vec{y} \times \vec{z}) = 0\) if \(\vec{x}\) is in the plane formed by \(\vec{y}\) and \(\vec{z}\):

The left side simplifies to zero since \(\vec{b} \times \vec{c}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\).

### Step 7: Conclusion
Since \(\vec{a} \cdot \vec{b} \neq 0\) (as given), it implies that the scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = 0\).

Thus, we conclude:

\[ [\vec{a}, \vec{b}, \vec{c}] = 0 \]