The unit vector perpendicular to vector `i -j and i + j` forming a right handed system is

Options

A. `k`
B. `-k`
C. `(1)/(sqrt(2)) (i-j)`
D. `(1)/(2) (i + j`)

Answer

To find the unit vector that is perpendicular to the vectors \( \mathbf{A} = \mathbf{i} - \mathbf{j} \) and \( \mathbf{B} = \mathbf{i} + \mathbf{j} \) and forms a right-handed system, we will follow these steps:

### Step 1: Identify the vectors
We have two vectors:
- \( \mathbf{A} = \mathbf{i} - \mathbf{j} \)
- \( \mathbf{B} = \mathbf{i} + \mathbf{j} \)

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of the vectors \( \mathbf{A} \) and \( \mathbf{B} \).

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

### Step 3: Calculate the determinant
Expanding this determinant, we get:

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} = (-1)(0) - (0)(1) = 0 \)

2. \( \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} = (1)(0) - (0)(1) = 0 \)

3. \( \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \)

Putting it all together, we have:

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

### Step 4: Find the magnitude of the cross product
The magnitude of \( \mathbf{A} \times \mathbf{B} \) is:

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

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

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

### Final Answer
The unit vector perpendicular to the vectors \( \mathbf{i} - \mathbf{j} \) and \( \mathbf{i} + \mathbf{j} \) forming a right-handed system is:

\[ \mathbf{k} \]

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