Let `vecr` be a unit vector satisfying `vecr xx veca = vecb, " where " |veca|= sqrt3 and |vecb| = sqrt2`, then `(a)vecr= 2/3(veca+ veca xx vecb)` (b)`vecr= 1/3(veca+ veca xx vecb)` (c)`vecr= 2/3(veca- veca xx vecb)` (d)`vecr= 1/3(-veca+ veca xx vecb)`

Options

A. `vecr= 2/3(veca+ veca xx vecb)`
B. `vecr= 1/3(veca+ veca xx vecb)`
C. `vecr= 2/3(veca- veca xx vecb)`
D. `vecr= 1/3(-veca+ veca xx vecb)`

Explanation

To solve the problem, we will follow these steps systematically:

1. **Understanding the Given Information**:
We have a unit vector \(\vec{r}\) such that:
\[ \vec{r} \times \vec{a} = \vec{b} \]
with magnitudes:
\[ |\vec{a}| = \sqrt{3}, \quad |\vec{b}| = \sqrt{2}, \quad |\vec{r}| = 1 \]

2. **Taking Magnitudes**:
Taking the magnitude of both sides of the equation \(\vec{r} \times \vec{a} = \vec{b}\):
\[ |\vec{r}||\vec{a}|\sin\theta = |\vec{b}| \]
Substituting the known magnitudes:
\[ 1 \cdot \sqrt{3} \cdot \sin\theta = \sqrt{2} \]
This simplifies to:
\[ \sqrt{3} \sin\theta = \sqrt{2} \]
Thus:
\[ \sin\theta = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \]

3. **Finding \(\cos\theta\)**:
We can find \(\cos\theta\) using the Pythagorean identity:
\[ \cos^2\theta + \sin^2\theta = 1 \]
Substituting \(\sin^2\theta\):
\[ \cos^2\theta + \frac{2}{3} = 1 \]
Therefore:
\[ \cos^2\theta = 1 - \frac{2}{3} = \frac{1}{3} \]
Thus:
\[ \cos\theta = \pm \frac{1}{\sqrt{3}} \]

4. **Using the Triple Product**:
We can use the vector triple product identity:
\[ \vec{a} \times (\vec{r} \times \vec{a}) = (\vec{a} \cdot \vec{a}) \vec{r} - (\vec{a} \cdot \vec{r}) \vec{a} \]
Substituting \(\vec{r} \times \vec{a} = \vec{b}\):
\[ \vec{a} \times \vec{b} = 3\vec{r} - (\vec{a} \cdot \vec{r}) \vec{a} \]

5. **Calculating \(\vec{a} \cdot \vec{r}\)**:
We know:
\[ \vec{a} \cdot \vec{r} = |\vec{a}||\vec{r}|\cos\theta = \sqrt{3} \cdot 1 \cdot \left(\pm \frac{1}{\sqrt{3}}\right) = \pm 1 \]

6. **Substituting Back**:
Substituting back into the equation:
\[ \vec{a} \times \vec{b} = 3\vec{r} - (\pm 1) \vec{a} \]
Rearranging gives:
\[ 3\vec{r} = \vec{a} \times \vec{b} + \vec{a} \quad \text{or} \quad 3\vec{r} = \vec{a} \times \vec{b} - \vec{a} \]
Thus:
\[ \vec{r} = \frac{1}{3}(\vec{a} \times \vec{b} \pm \vec{a}) \]

7. **Final Expression**:
Therefore, the possible values for \(\vec{r}\) are:
\[ \vec{r} = \frac{1}{3}(\vec{a} + \vec{a} \times \vec{b}) \quad \text{or} \quad \vec{r} = \frac{1}{3}(-\vec{a} + \vec{a} \times \vec{b}) \]

8. **Checking Options**:
From the options provided, we see that:
- (b) \(\vec{r} = \frac{1}{3}(\vec{a} + \vec{a} \times \vec{b})\)
- (d) \(\vec{r} = \frac{1}{3}(-\vec{a} + \vec{a} \times \vec{b})\)

Thus, the correct answers are options (b) and (d).