A unit vector perpendicular to each of the vectors `2hat(i) - hat(j) + hat(k ) and 3hat(i) - 4hat(i) - hat(k)` is

Options

A. `(1)/(sqrt3) hat(i) + (1)/(sqrt3) hat(j) - (1)/(sqrt3) hat(k)`
B. `(1)/(sqrt2) hat(i) + (1)/(2) hat(j) + (1)/(2) hat(k)`
C. `(1)/(sqrt3) hat(i) - (1)/(sqrt3) hat(j) -(1)/(sqrt3) hat(k)`
D. `(1)/(sqrt3) hat(i) + (1)/(3) hat(j) + (1)/(sqrt3) hat(k)`

Answer

To find a unit vector that is perpendicular to the given vectors \( \mathbf{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = 3\hat{i} - 4\hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Define the vectors
Let:
\[ \mathbf{a} = 2\hat{i} - \hat{j} + \hat{k} \]
\[ \mathbf{b} = 3\hat{i} - 4\hat{j} - \hat{k} \]

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

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -4 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -1 & 1 \\ -4 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 3 & -4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -1 & 1 \\ -4 & -1 \end{vmatrix} = (-1)(-1) - (1)(-4) = 1 + 4 = 5 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} = (2)(-1) - (1)(3) = -2 - 3 = -5 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & -1 \\ 3 & -4 \end{vmatrix} = (2)(-4) - (-1)(3) = -8 + 3 = -5 \]

Putting it all together:
\[ \mathbf{a} \times \mathbf{b} = 5\hat{i} + 5\hat{j} - 5\hat{k} \]

### Step 4: Simplify the cross product
We can factor out 5:
\[ \mathbf{a} \times \mathbf{b} = 5(\hat{i} + \hat{j} - \hat{k}) \]

### Step 5: Calculate the magnitude of the cross product
The magnitude of \( \mathbf{a} \times \mathbf{b} \) is:
\[ \|\mathbf{a} \times \mathbf{b}\| = \sqrt{(5)^2 + (5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 6: Find the unit vector
The unit vector \( \hat{n} \) that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is given by:
\[ \hat{n} = \frac{\mathbf{a} \times \mathbf{b}}{\|\mathbf{a} \times \mathbf{b}\|} = \frac{5(\hat{i} + \hat{j} - \hat{k})}{5\sqrt{3}} = \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} \]

### Final Answer:
Thus, the required unit vector is:
\[ \hat{n} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k} \]