Find the unit vector perpendicular to both `2hat(i) + 3hat(j)+ hat(k) and hat(i)-hat(j)+ 4hat(k)`

Answer

To find the unit vector perpendicular to both vectors \( \mathbf{A} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \mathbf{B} = \hat{i} - \hat{j} + 4\hat{k} \), we will follow these steps:

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

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
To find a vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \), we compute the cross product \( \mathbf{A} \times \mathbf{B} \) using the determinant of a matrix:

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

### Step 3: Expand the determinant
Calculating the determinant, we have:

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

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 3 & 1 \\ -1 & 4 \end{vmatrix} = (3 \cdot 4) - (1 \cdot -1) = 12 + 1 = 13 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = (2 \cdot 4) - (1 \cdot 1) = 8 - 1 = 7 \]

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

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 13\hat{i} - 7\hat{j} - 5\hat{k} \]

### Step 4: Find the magnitude of \( \mathbf{A} \times \mathbf{B} \)
The magnitude \( |\mathbf{A} \times \mathbf{B}| \) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(13)^2 + (-7)^2 + (-5)^2} = \sqrt{169 + 49 + 25} = \sqrt{243} \]

### Step 5: Calculate the unit vector
The unit vector \( \mathbf{n} \) in the direction of \( \mathbf{A} \times \mathbf{B} \) is given by:
\[ \mathbf{n} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{13\hat{i} - 7\hat{j} - 5\hat{k}}{\sqrt{243}} \]

### Step 6: Simplify the unit vector
We can simplify \( \sqrt{243} \):
\[ \sqrt{243} = \sqrt{81 \cdot 3} = 9\sqrt{3} \]
Thus, the unit vector is:
\[ \mathbf{n} = \frac{13}{9\sqrt{3}}\hat{i} - \frac{7}{9\sqrt{3}}\hat{j} - \frac{5}{9\sqrt{3}}\hat{k} \]

### Final Answer
The unit vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \) is:
\[ \mathbf{n} = \frac{13}{9\sqrt{3}}\hat{i} - \frac{7}{9\sqrt{3}}\hat{j} - \frac{5}{9\sqrt{3}}\hat{k} \]