What is the unit vector perpendicular to the following vectors `2hati+2hatj-hatk` and `6hati-3hatj+2hatk`

Options

A. `(hati+10hatj-18hatk)/(5sqrt(17))`
B. `(hati-10hatj+18hatk)/(5sqrt(17))`
C. `(hati-10hatj-18hatk)/(5sqrt(17))`
D. `(hati+10hatj+18hatk)/(5sqrt(17))`

Explanation

To find the unit vector that is perpendicular to the vectors \( \mathbf{A} = 2\hat{i} + 2\hat{j} - \hat{k} \) and \( \mathbf{B} = 6\hat{i} - 3\hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Write down the vectors
We have:
\[ \mathbf{A} = 2\hat{i} + 2\hat{j} - \hat{k} \]
\[ \mathbf{B} = 6\hat{i} - 3\hat{j} + 2\hat{k} \]

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

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 6 & -3 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant gives:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 2 & -1 \\ -3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 6 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 2 \\ 6 & -3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} 2 & -1 \\ -3 & 2 \end{vmatrix} = (2)(2) - (-1)(-3) = 4 - 3 = 1 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & -1 \\ 6 & 2 \end{vmatrix} = (2)(2) - (-1)(6) = 4 + 6 = 10 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & 2 \\ 6 & -3 \end{vmatrix} = (2)(-3) - (2)(6) = -6 - 12 = -18 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(10) + \hat{k}(-18) = \hat{i} - 10\hat{j} - 18\hat{k} \]

### Step 4: Resultant vector
Thus, the resultant vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \) is:
\[ \mathbf{R} = \hat{i} - 10\hat{j} - 18\hat{k} \]

### Step 5: Calculate the magnitude of the resultant vector
The magnitude of \( \mathbf{R} \) is given by:
\[ |\mathbf{R}| = \sqrt{(1)^2 + (-10)^2 + (-18)^2} = \sqrt{1 + 100 + 324} = \sqrt{425} = 5\sqrt{17} \]

### Step 6: Find the unit vector
The unit vector \( \hat{r} \) in the direction of \( \mathbf{R} \) is given by:
\[ \hat{r} = \frac{\mathbf{R}}{|\mathbf{R}|} = \frac{\hat{i} - 10\hat{j} - 18\hat{k}}{5\sqrt{17}} \]

### Final Answer
Thus, the unit vector perpendicular to the given vectors is:
\[ \hat{r} = \frac{1}{5\sqrt{17}} \hat{i} - \frac{10}{5\sqrt{17}} \hat{j} - \frac{18}{5\sqrt{17}} \hat{k} \]