Given, two vectors are `hati - hatj` and `hati + 2hatj`, the unit vector coplanar with the two vectors and perpendicular to first is:

Options

A. `1/sqrt(2)(hati +hatj)`
B. `1/sqrt(2)(2hati + hatj)`
C. `+-1/sqrt(2)(hati +hatj)`
D. None of these

Answer

To solve the problem, we need to find a unit vector \( \mathbf{C} \) that is coplanar with the vectors \( \mathbf{A} = \hat{i} - \hat{j} \) and \( \mathbf{B} = \hat{i} + 2\hat{j} \), and is also perpendicular to the vector \( \mathbf{A} \).

### Step-by-Step Solution:

1. **Define the vectors**:
Let \( \mathbf{A} = \hat{i} - \hat{j} \) and \( \mathbf{B} = \hat{i} + 2\hat{j} \).

2. **Set up the vector \( \mathbf{C} \)**:
Let \( \mathbf{C} = x\hat{i} + y\hat{j} + z\hat{k} \). Since we are looking for a vector coplanar with \( \mathbf{A} \) and \( \mathbf{B} \), we can assume \( z = 0 \) (as the vectors lie in the \( xy \)-plane). Thus, \( \mathbf{C} = x\hat{i} + y\hat{j} \).

3. **Condition for coplanarity**:
The vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are coplanar if their scalar triple product is zero. This can be calculated using the determinant:
\[ \begin{vmatrix} 1 & -1 & 0 \\ 1 & 2 & 0 \\ x & y & 0 \end{vmatrix} = 0 \]
Expanding this determinant:
\[ 0 = 1 \cdot (2 \cdot 0 - (-1) \cdot 0) - (-1) \cdot (1 \cdot 0 - 1 \cdot 0) + 0 = 0 \]
The determinant simplifies to:
\[ 0 = 0 \]
This condition is satisfied for any \( x \) and \( y \).

4. **Condition for perpendicularity**:
The vector \( \mathbf{C} \) must be perpendicular to \( \mathbf{A} \). This means:
\[ \mathbf{C} \cdot \mathbf{A} = 0 \]
Substituting the values:
\[ (x\hat{i} + y\hat{j}) \cdot (\hat{i} - \hat{j}) = x \cdot 1 + y \cdot (-1) = 0 \]
This gives us:
\[ x - y = 0 \implies x = y \]

5. **Condition for unit vector**:
The vector \( \mathbf{C} \) must also be a unit vector, which means:
\[ |\mathbf{C}| = 1 \]
The magnitude of \( \mathbf{C} \) is:
\[ |\mathbf{C}| = \sqrt{x^2 + y^2} = 1 \]
Since \( x = y \), we can substitute \( y \) with \( x \):
\[ \sqrt{x^2 + x^2} = 1 \implies \sqrt{2x^2} = 1 \implies \sqrt{2} |x| = 1 \implies |x| = \frac{1}{\sqrt{2}} \]
Therefore, \( x = \frac{1}{\sqrt{2}} \) or \( x = -\frac{1}{\sqrt{2}} \).

6. **Final expression for \( \mathbf{C} \)**:
Thus, we have:
\[ \mathbf{C} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \quad \text{or} \quad \mathbf{C} = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]
Since we are looking for a unit vector, we can take:
\[ \mathbf{C} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]

### Conclusion:
The required unit vector \( \mathbf{C} \) that is coplanar with the two given vectors and perpendicular to the first vector is:
\[ \mathbf{C} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]