The unit vector perpendicular to `vec A = 2 hat i + 3 hat j + hat k` and ` vec B = hat i - hat j + hat k` is

Options

A. `(4hati-hatj-5hatk)/(sqrt(42))`
B. `(4hati-hatj+5hatk)/(sqrt(42))`
C. `(4hati+hatj+5hatk)/(sqrt(42))`
D. `(4hati+hatj-5hatk)/(sqrt(42))`

Explanation

To find the unit vector that is perpendicular to the given vectors \( \vec{A} = 2 \hat{i} + 3 \hat{j} + \hat{k} \) and \( \vec{B} = \hat{i} - \hat{j} + \hat{k} \), we will follow these steps:

### Step 1: Calculate the Cross Product of Vectors A and B
The cross product \( \vec{A} \times \vec{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of vectors \( \vec{A} \) and \( \vec{B} \).

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

### Step 2: Calculate the Determinant
Now we will calculate the determinant using the cofactor expansion:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

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

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

Putting it all together, we have:

\[ \vec{A} \times \vec{B} = 4 \hat{i} - 1 \hat{j} - 5 \hat{k} \]

### Step 3: Find the Magnitude of the Cross Product
Now we will find the magnitude of the vector \( \vec{A} \times \vec{B} \):

\[ |\vec{A} \times \vec{B}| = \sqrt{(4)^2 + (-1)^2 + (-5)^2} = \sqrt{16 + 1 + 25} = \sqrt{42} \]

### Step 4: Calculate the Unit Vector
The unit vector \( \hat{n} \) in the direction of \( \vec{A} \times \vec{B} \) is given by:

\[ \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{4 \hat{i} - 1 \hat{j} - 5 \hat{k}}{\sqrt{42}} \]

### Final Answer
Thus, the unit vector perpendicular to \( \vec{A} \) and \( \vec{B} \) is:

\[ \hat{n} = \frac{4}{\sqrt{42}} \hat{i} - \frac{1}{\sqrt{42}} \hat{j} - \frac{5}{\sqrt{42}} \hat{k} \]