If `veca,vecb and vecc` are three non coplanar vectors and `vecr` is any vector in space, then `(vecaxxvecb)xx(vecrxxvecc)+(vecb xxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)=`

Options

A. `[veca vecb vecc]vecr`
B. `2 [veca vecb vecc]vecr`
C. `3 [veca vecb vecc]vecr`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression:

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

where \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-coplanar vectors, and \(\vec{r}\) is any vector in space.

### Step 1: Use the Vector Triple Product Identity

We can apply the vector triple product identity, which states that:

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

Using this identity, we can rewrite each term in the expression.

### Step 2: Rewrite Each Term

1. For the first term \((\vec{a} \times \vec{b}) \times (\vec{r} \times \vec{c})\):

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

2. For the second term \((\vec{b} \times \vec{c}) \times (\vec{r} \times \vec{a})\):

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

3. For the third term \((\vec{c} \times \vec{a}) \times (\vec{r} \times \vec{b})\):

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

### Step 3: Combine All Terms

Now we can combine all three rewritten terms:

\[ \begin{align*} & \left[(\vec{a} \times \vec{b}) \cdot \vec{c} \, \vec{r} - (\vec{a} \times \vec{b}) \cdot \vec{r} \, \vec{c}\right] + \left[(\vec{b} \times \vec{c}) \cdot \vec{a} \, \vec{r} - (\vec{b} \times \vec{c}) \cdot \vec{r} \, \vec{a}\right] + \left[(\vec{c} \times \vec{a}) \cdot \vec{b} \, \vec{r} - (\vec{c} \times \vec{a}) \cdot \vec{r} \, \vec{b}\right] \\ & = \left[(\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{a} + (\vec{c} \times \vec{a}) \cdot \vec{b}\right] \vec{r} \\ & - \left[(\vec{a} \times \vec{b}) \cdot \vec{r} \, \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{r} \, \vec{a} + (\vec{c} \times \vec{a}) \cdot \vec{r} \, \vec{b}\right] \end{align*} \]

### Step 4: Factor Out Common Terms

Let \( V = (\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{a} + (\vec{c} \times \vec{a}) \cdot \vec{b} \).

Thus, we can write:

\[ V \vec{r} - \left[(\vec{a} \times \vec{b}) \cdot \vec{r} \, \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{r} \, \vec{a} + (\vec{c} \times \vec{a}) \cdot \vec{r} \, \vec{b}\right] \]

### Step 5: Final Simplification

The expression simplifies to:

\[ V \vec{r} - \vec{r} \cdot V \]

This leads us to conclude that:

\[ = 3(\vec{a} \times \vec{b} \times \vec{c}) \cdot \vec{r} \]

### Final Answer

Thus, the final result is:

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