A unit vector perpendicular to the plane passing through the points whose position vectors are `hati-hatj+2hatk, 2hati-hatk` and `2hati+hatk` is

Options

A. `2hati+hatj+hatk`
B. `(1)/(sqrt(6))(hati+2hatj+hatk)`
C. `+-((hati-hatj)/(sqrt(2)))`
D. `(1)/(sqrt(6))(3hati+4hatj-hatk)`

Answer

To find a unit vector perpendicular to the plane defined by the points with position vectors \( \mathbf{a} = \hat{i} - \hat{j} + 2\hat{k} \), \( \mathbf{b} = 2\hat{i} - \hat{k} \), and \( \mathbf{c} = 2\hat{i} + \hat{k} \), we will follow these steps:

### Step 1: Determine the position vectors
We have the position vectors:
- \( \mathbf{a} = \hat{i} - \hat{j} + 2\hat{k} \)
- \( \mathbf{b} = 2\hat{i} - \hat{k} \)
- \( \mathbf{c} = 2\hat{i} + \hat{k} \)

### Step 2: Find the vectors \( \mathbf{AB} \) and \( \mathbf{BC} \)
The vector \( \mathbf{AB} \) is given by:
\[ \mathbf{AB} = \mathbf{b} - \mathbf{a} = (2\hat{i} - \hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) \]
Calculating this, we get:
\[ \mathbf{AB} = (2 - 1)\hat{i} + (0 + 1)\hat{j} + (-1 - 2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k} \]

Next, we find the vector \( \mathbf{BC} \):
\[ \mathbf{BC} = \mathbf{c} - \mathbf{b} = (2\hat{i} + \hat{k}) - (2\hat{i} - \hat{k}) \]
Calculating this, we have:
\[ \mathbf{BC} = (2 - 2)\hat{i} + (0 + 1)\hat{j} + (1 + 1)\hat{k} = 0\hat{i} + 0\hat{j} + 2\hat{k} = 2\hat{k} \]

### Step 3: Find the cross product \( \mathbf{AB} \times \mathbf{BC} \)
To find the unit vector perpendicular to the plane, we calculate the cross product \( \mathbf{AB} \times \mathbf{BC} \):
\[ \mathbf{AB} = \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}, \quad \mathbf{BC} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \]

The cross product is given by the determinant:
\[ \mathbf{n} = \mathbf{AB} \times \mathbf{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ 0 & 0 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{n} = \hat{i} \begin{vmatrix} 1 & -3 \\ 0 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -3 \\ 0 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} \]
\[ = \hat{i}(1 \cdot 2 - 0 \cdot -3) - \hat{j}(1 \cdot 2 - 0 \cdot -3) + \hat{k}(1 \cdot 0 - 1 \cdot 0) \]
\[ = 2\hat{i} - 2\hat{j} + 0\hat{k} = 2\hat{i} - 2\hat{j} \]

### Step 4: Normalize the vector \( \mathbf{n} \)
To find the unit vector, we need to normalize \( \mathbf{n} \):
\[ \|\mathbf{n}\| = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
Thus, the unit vector \( \hat{n} \) is:
\[ \hat{n} = \frac{\mathbf{n}}{\|\mathbf{n}\|} = \frac{2\hat{i} - 2\hat{j}}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]

### Step 5: Include the opposite direction
Since the question asks for a unit vector perpendicular to the plane, we can also include the opposite direction:
\[ \hat{n} = \pm \left( \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \right) \]

### Final Answer
The unit vector perpendicular to the plane is:
\[ \hat{n} = \pm \frac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \]