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

Options

A. `hatk`
B. `-hatk`
C. `(hati-hatj)/(sqrt2)`
D. `(hati+hatj)/(sqrt2)`

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
To find a vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \), we compute the cross product \( \mathbf{A} \times \mathbf{B} \).

Using the determinant form for the cross product:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Calculating the determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} = 0 \)
3. \( \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \)

Thus, 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
Now, we find the magnitude of \( \mathbf{A} \times \mathbf{B} \):
\[ |\mathbf{A} \times \mathbf{B}| = |2 \hat{k}| = 2 \]

### Step 5: Find the Unit Vector
The unit vector \( \mathbf{C} \) in the direction of \( \mathbf{A} \times \mathbf{B} \) is given by:
\[ \mathbf{C} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{2 \hat{k}}{2} = \hat{k} \]

### Conclusion
The unit vector perpendicular to the vectors \( \hat{i} - \hat{j} \) and \( \hat{i} + \hat{j} \) is:
\[ \hat{k} \]

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