Which one of the following is the unit vector perpendicular to both `vec(a)=-hat(i)+hat(j)+hat(k) and vec(b)=hat(i)-hat(j)+hat(k)` ?

Options

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

Explanation

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

### Step 1: Write down the vectors
Given:
\[ \vec{a} = -\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors.

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (1)(1) = -1 - 1 = -2 \quad \text{(so it becomes } +2\hat{j}\text{)} \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = (-1)(-1) - (1)(1) = 1 - 1 = 0 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = 2\hat{i} + 2\hat{j} + 0\hat{k} = 2\hat{i} + 2\hat{j} \]

### Step 4: Find the magnitude of \(\vec{a} \times \vec{b}\)
The magnitude of the vector \(2\hat{i} + 2\hat{j}\) is calculated as follows:
\[ |\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 5: Find the unit vector
The unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]
Substituting the values:
\[ \hat{n} = \frac{2\hat{i} + 2\hat{j}}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} \hat{i} + \frac{2}{2\sqrt{2}} \hat{j} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]

### Final Result
Thus, the unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is:
\[ \hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]