A unit vector n perpendicular to the plane determined by the points A (0, -2, 1), B (1, -2, -2) and C (-1, 1, 0)

Options

A. `1/3(2i+j+2k)`
B. `1//4sqrt(6)(8i+4j+4k)`
C. `1/sqrt(3)(i-j+k)`
D. `1/sqrt(14)(3i+j+2k)`

Answer

To find a unit vector \( \mathbf{n} \) that is perpendicular to the plane determined by the points \( A(0, -2, 1) \), \( B(1, -2, -2) \), and \( C(-1, 1, 0) \), we can follow these steps:

### Step 1: Find the vectors \( \mathbf{AB} \) and \( \mathbf{AC} \)

First, we need to find the vectors \( \mathbf{AB} \) and \( \mathbf{AC} \).

\[ \mathbf{AB} = B - A = (1 - 0, -2 - (-2), -2 - 1) = (1, 0, -3) \]

\[ \mathbf{AC} = C - A = (-1 - 0, 1 - (-2), 0 - 1) = (-1, 3, -1) \]

### Step 2: Compute the cross product \( \mathbf{AB} \times \mathbf{AC} \)

Next, we compute the cross product of \( \mathbf{AB} \) and \( \mathbf{AC} \) to find a vector that is perpendicular to the plane.

\[ \mathbf{n} = \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ -1 & 3 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 0 & -3 \\ 3 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ -1 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 0 \\ -1 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 0 & -3 \\ 3 & -1 \end{vmatrix} = (0)(-1) - (-3)(3) = 9 \)
2. \( \begin{vmatrix} 1 & -3 \\ -1 & -1 \end{vmatrix} = (1)(-1) - (-3)(-1) = -1 - 3 = -4 \)
3. \( \begin{vmatrix} 1 & 0 \\ -1 & 3 \end{vmatrix} = (1)(3) - (0)(-1) = 3 \)

Putting it all together:

\[ \mathbf{n} = 9\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \]

### Step 3: Find the magnitude of \( \mathbf{n} \)

Now, we find the magnitude of the vector \( \mathbf{n} \):

\[ |\mathbf{n}| = \sqrt{9^2 + 4^2 + 3^2} = \sqrt{81 + 16 + 9} = \sqrt{106} \]

### Step 4: Find the unit vector \( \mathbf{n} \)

Finally, we find the unit vector \( \mathbf{n} \):

\[ \mathbf{n}_{unit} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{9\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}}{\sqrt{106}} = \left( \frac{9}{\sqrt{106}}, \frac{4}{\sqrt{106}}, \frac{3}{\sqrt{106}} \right) \]

### Final Answer

The unit vector \( \mathbf{n} \) perpendicular to the plane determined by the points \( A \), \( B \), and \( C \) is:

\[ \mathbf{n} = \left( \frac{9}{\sqrt{106}}, \frac{4}{\sqrt{106}}, \frac{3}{\sqrt{106}} \right) \]