In the question number 20, a unit vector perpendicular to the direction of `vecA and vecB ` is

Options

A. `(-2hati-hatj-hatk)/(sqrt(6))`
B. `(2hati+hatj+hatk)/(sqrt(6))`
C. `(2hati-hatj-hatk)/(sqrt(6))`
D. `(2hati-hatj+hatk)/(sqrt(6))`

Explanation

To find a unit vector that is perpendicular to the vectors \(\vec{A}\) and \(\vec{B}\), we will follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{A} = -2 \hat{i} + 3 \hat{j} + 1 \hat{k} \]
\[ \vec{B} = 1 \hat{i} + 2 \hat{j} - 4 \hat{k} \]

### Step 2: Calculate the cross product \(\vec{A} \times \vec{B}\)
The cross product of two vectors \(\vec{A}\) and \(\vec{B}\) in determinant form is given by:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 1 \\ 1 & 2 & -4 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 1 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 1 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 3 & 1 \\ 2 & -4 \end{vmatrix} = (3)(-4) - (1)(2) = -12 - 2 = -14\)
2. \(\begin{vmatrix} -2 & 1 \\ 1 & -4 \end{vmatrix} = (-2)(-4) - (1)(1) = 8 - 1 = 7\)
3. \(\begin{vmatrix} -2 & 3 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (3)(1) = -4 - 3 = -7\)

Putting it all together:
\[ \vec{A} \times \vec{B} = -14 \hat{i} - 7 \hat{j} - 7 \hat{k} \]

### Step 3: Find the magnitude of \(\vec{A} \times \vec{B}\)
The magnitude of the cross product is given by:
\[ |\vec{A} \times \vec{B}| = \sqrt{(-14)^2 + (-7)^2 + (-7)^2} \]
Calculating:
\[ |\vec{A} \times \vec{B}| = \sqrt{196 + 49 + 49} = \sqrt{294} = 7\sqrt{6} \]

### Step 4: Find the unit vector
The unit vector \(\hat{n}\) 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:
\[ \hat{n} = \frac{-14 \hat{i} - 7 \hat{j} - 7 \hat{k}}{7\sqrt{6}} = -2 \hat{i} - \hat{j} - \hat{k} \cdot \frac{1}{\sqrt{6}} \]
Thus, the unit vector is:
\[ \hat{n} = -\frac{2}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \]

### Final Answer
The unit vector perpendicular to the directions of \(\vec{A}\) and \(\vec{B}\) is:
\[ \hat{n} = -\frac{2}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \]