The unit vector perpendicular to the vectors `hati-hatj and hati +hatj` forming a right handed system is

Options

A. `hatk`
B. `-hatk`
C. `(hati-hatj)/(sqrt(2))`
D. `(hati+hatj)/(sqrt(2))`

Answer

To find the unit vector perpendicular to the vectors \( \hat{i} - \hat{j} \) and \( \hat{i} + \hat{j} \), we can follow these steps:

### Step 1: Define the Vectors
Let:
- Vector \( \mathbf{A} = \hat{i} - \hat{j} \)
- Vector \( \mathbf{B} = \hat{i} + \hat{j} \)

### Step 2: Calculate the Cross Product
The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) gives us a vector that is perpendicular to both. We can calculate \( \mathbf{A} \times \mathbf{B} \) using the determinant method:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{vmatrix} \]

### Step 3: Calculate the Determinant
To compute the determinant, we expand it:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

- For \( \hat{i} \): \( (-1)(0) - (0)(1) = 0 \)
- For \( \hat{j} \): \( (1)(0) - (0)(1) = 0 \)
- For \( \hat{k} \): \( (1)(1) - (-1)(1) = 1 + 1 = 2 \)

So, we have:

\[ \mathbf{A} \times \mathbf{B} = 0\hat{i} - 0\hat{j} + 2\hat{k} = 2\hat{k} \]

### Step 4: Find the Magnitude of the Cross Product
The magnitude of \( \mathbf{A} \times \mathbf{B} \) is:

\[ |\mathbf{A} \times \mathbf{B}| = |2\hat{k}| = 2 \]

### Step 5: Find the Unit Vector
The unit vector \( \mathbf{C} \) in the direction of \( \mathbf{A} \times \mathbf{B} \) is given by:

\[ \mathbf{C} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{2\hat{k}}{2} = \hat{k} \]

### Final Answer
Thus, the unit vector perpendicular to the vectors \( \hat{i} - \hat{j} \) and \( \hat{i} + \hat{j} \) is:

\[ \hat{k} \]