What is the unit vector perpendicular to the following Vector `2hat(i)+2hat(j)-hat(k)` and `6hat(i)-3hat(j)+2hat(k)?`

Options

A. `(hati +10hat j-18 hatk)/(5sqrt(17))`
B. `(hati +10hat j+18 hatk)/(5sqrt(17))`
C. `(hati -10hat j-18 hatk)/(5sqrt(17))`
D. `(hati +10hat j+18 hatk)/(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 will follow these steps:

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

### Step 2: Compute 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.

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

### Step 3: Calculate the determinant
Calculating the determinant, we expand it as follows:

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

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ 2 \cdot 2 - (-1) \cdot (-3) = 4 - 3 = 1 \]

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

3. For \( \hat{k} \):
\[ 2 \cdot (-3) - 2 \cdot 6 = -6 - 12 = -18 \]

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: Find the magnitude of the cross product
Now, we need to find the magnitude of \( \mathbf{A} \times \mathbf{B} \):
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(1)^2 + (-10)^2 + (-18)^2} \]
\[ = \sqrt{1 + 100 + 324} = \sqrt{425} \]

### Step 5: Calculate the unit vector
The unit vector \( \mathbf{U} \) in the direction of \( \mathbf{A} \times \mathbf{B} \) is given by:
\[ \mathbf{U} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{\hat{i} - 10\hat{j} - 18\hat{k}}{\sqrt{425}} \]

### Final Result
Thus, the unit vector perpendicular to both vectors is:
\[ \mathbf{U} = \frac{1}{\sqrt{425}} \hat{i} - \frac{10}{\sqrt{425}} \hat{j} - \frac{18}{\sqrt{425}} \hat{k} \]