A unit vector perpendicular to both ` hati + hatj and hatj + hatk` is

Options

A. `hati -hatj +hatk`
B. ` hati +hatj + hatk`
C. ` ( hati +hatj +hatk)/(sqrt3)`
D. ` (hati -hatj +hatk)/( sqrt3)`

Explanation

To find a unit vector that is perpendicular to both vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \), we will follow these steps:

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

### Step 2: Calculate the Cross Product
The cross product \( \mathbf{A} \times \mathbf{B} \) will give us a vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).

To calculate the cross product, we set up the determinant:

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

### Step 3: Compute the Determinant
Calculating the determinant, we have:

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

Calculating each of the 2x2 determinants:

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

Thus, we have:

\[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \hat{i} - \hat{j} + \hat{k} \]

### Step 4: Find the Magnitude of the Resulting Vector
The magnitude of the vector \( \hat{i} - \hat{j} + \hat{k} \) is calculated as follows:

\[ |\mathbf{C}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 5: Calculate the Unit Vector
To find the unit vector, we divide the vector \( \hat{i} - \hat{j} + \hat{k} \) by its magnitude \( \sqrt{3} \):

\[ \mathbf{u} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

### Final Answer
The unit vector perpendicular to both \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) is:

\[ \mathbf{u} = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]