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 can follow these steps:

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

### Step 2: Compute the Cross Product
The unit vector perpendicular to both vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using the cross product:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]

### Step 3: Evaluate the Determinant
Calculating the determinant:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
- For \( \hat{i} \): \( 1 \cdot 1 - 0 \cdot 1 = 1 \)
- For \( \hat{j} \): \( 1 \cdot 1 - 0 \cdot 0 = 1 \)
- For \( \hat{k} \): \( 1 \cdot 1 - 1 \cdot 0 = 1 \)

Thus, we have:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \]

### Step 4: Calculate the Magnitude of the Cross Product
Now, we find the magnitude of \( \mathbf{a} \times \mathbf{b} \):
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 5: Find the Unit Vector
The unit vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]

Thus, the unit vector perpendicular to both vectors is:
\[ \mathbf{u} = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

### Final Answer
The unit vector perpendicular to both \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) is:
\[ \mathbf{u} = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

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