`veca xx [(vecr-vecb)xxveca]+vecb xx [(vecr-vecc)xxvecb]+vecc xx [(vecr-veca)xxvecc]=0 and veca , vecb and vecc` are unit vector mutually perpendicular to each other then the `vecr` is :

Options

A. `(veca-vecb-vecc)/2`
B. `(veca-vecb+vecc)/2`
C. `(veca+vecb+vecc)/2`
D. `(veca+vecb-vecc)/2`

Answer

To solve the given vector equation, we start with the expression:

\[ \vec{a} \times [(\vec{r} - \vec{b}) \times \vec{a}] + \vec{b} \times [(\vec{r} - \vec{c}) \times \vec{b}] + \vec{c} \times [(\vec{r} - \vec{a}) \times \vec{c}] = 0 \]

where \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors that are mutually perpendicular.

### Step 1: Apply the Vector Triple Product Identity
Using the vector triple product identity, we know that:

\[ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C} \]

We apply this identity to each term in the equation.

1. For the first term:
\[ \vec{a} \times [(\vec{r} - \vec{b}) \times \vec{a}] = (\vec{a} \cdot \vec{a})(\vec{r} - \vec{b}) - (\vec{a} \cdot (\vec{r} - \vec{b}))\vec{a} \]
Since \(\vec{a}\) is a unit vector, \(\vec{a} \cdot \vec{a} = 1\):
\[ = (\vec{r} - \vec{b}) - (\vec{a} \cdot \vec{r} - \vec{a} \cdot \vec{b})\vec{a} \]

2. For the second term:
\[ \vec{b} \times [(\vec{r} - \vec{c}) \times \vec{b}] = (\vec{b} \cdot \vec{b})(\vec{r} - \vec{c}) - (\vec{b} \cdot (\vec{r} - \vec{c}))\vec{b} \]
Again, since \(\vec{b}\) is a unit vector:
\[ = (\vec{r} - \vec{c}) - (\vec{b} \cdot \vec{r} - \vec{b} \cdot \vec{c})\vec{b} \]

3. For the third term:
\[ \vec{c} \times [(\vec{r} - \vec{a}) \times \vec{c}] = (\vec{c} \cdot \vec{c})(\vec{r} - \vec{a}) - (\vec{c} \cdot (\vec{r} - \vec{a}))\vec{c} \]
And since \(\vec{c}\) is a unit vector:
\[ = (\vec{r} - \vec{a}) - (\vec{c} \cdot \vec{r} - \vec{c} \cdot \vec{a})\vec{c} \]

### Step 2: Combine All Terms
Now we combine all three results:

\[ [(\vec{r} - \vec{b}) - (\vec{a} \cdot \vec{r} - \vec{a} \cdot \vec{b})\vec{a}] + [(\vec{r} - \vec{c}) - (\vec{b} \cdot \vec{r} - \vec{b} \cdot \vec{c})\vec{b}] + [(\vec{r} - \vec{a}) - (\vec{c} \cdot \vec{r} - \vec{c} \cdot \vec{a})\vec{c}] = 0 \]

### Step 3: Simplify the Equation
Grouping the terms gives:

\[ 3\vec{r} - (\vec{b} + \vec{c} + \vec{a}) - [(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}] = 0 \]

### Step 4: Solve for \(\vec{r}\)
Rearranging the equation leads to:

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

### Step 5: Final Expression for \(\vec{r}\)
Assuming \(\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}\), we can express:

\[ \vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c}) \]

### Conclusion
Thus, the value of \(\vec{r}\) is:

\[ \vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c}) \]