Given three vectors a,b,c such that b.c = 3 `a.c = (1)/(3)`. The vector r which satisfies `r xx a = b xxa` and `r.c =0` is

Options

A. `b + 9a`
B. `a + 9b`
C. `b -9a`
D. none of these

Answer

To solve the problem, we need to find the vector \( \mathbf{r} \) that satisfies the conditions \( \mathbf{r} \times \mathbf{a} = \mathbf{b} \times \mathbf{a} \) and \( \mathbf{r} \cdot \mathbf{c} = 0 \). We are given the dot products \( \mathbf{b} \cdot \mathbf{c} = 3 \) and \( \mathbf{a} \cdot \mathbf{c} = \frac{1}{3} \).

### Step-by-Step Solution:

1. **Start with the Cross Product Condition**:
\[ \mathbf{r} \times \mathbf{a} = \mathbf{b} \times \mathbf{a} \]
Rearranging gives:
\[ \mathbf{r} \times \mathbf{a} - \mathbf{b} \times \mathbf{a} = 0 \]
This implies that \( \mathbf{r} - \mathbf{b} \) is parallel to \( \mathbf{a} \). Therefore, we can express this as:
\[ \mathbf{r} - \mathbf{b} = \lambda \mathbf{a} \]
for some scalar \( \lambda \).

2. **Express \( \mathbf{r} \)**:
Rearranging the above equation gives:
\[ \mathbf{r} = \mathbf{b} + \lambda \mathbf{a} \]

3. **Use the Dot Product Condition**:
We know that \( \mathbf{r} \cdot \mathbf{c} = 0 \). Substituting for \( \mathbf{r} \):
\[ (\mathbf{b} + \lambda \mathbf{a}) \cdot \mathbf{c} = 0 \]
Expanding this gives:
\[ \mathbf{b} \cdot \mathbf{c} + \lambda (\mathbf{a} \cdot \mathbf{c}) = 0 \]

4. **Substituting Known Values**:
We substitute the known values \( \mathbf{b} \cdot \mathbf{c} = 3 \) and \( \mathbf{a} \cdot \mathbf{c} = \frac{1}{3} \):
\[ 3 + \lambda \left(\frac{1}{3}\right) = 0 \]

5. **Solve for \( \lambda \)**:
Rearranging gives:
\[ \lambda \left(\frac{1}{3}\right) = -3 \]
Multiplying both sides by 3:
\[ \lambda = -9 \]

6. **Substitute \( \lambda \) back into the equation for \( \mathbf{r} \)**:
\[ \mathbf{r} = \mathbf{b} - 9 \mathbf{a} \]

### Final Answer:
Thus, the vector \( \mathbf{r} \) is:
\[ \mathbf{r} = \mathbf{b} - 9 \mathbf{a} \]