Find a unit vector normal to the plane through the points `(1, 1, 1), (-1, 2, 3) and (2, -1, 3)`.

Answer

To find a unit vector normal to the plane defined by the points \( (1, 1, 1) \), \( (-1, 2, 3) \), and \( (2, -1, 3) \), we can follow these steps:

### Step 1: Find two vectors in the plane
We can find two vectors that lie in the plane by subtracting the coordinates of the points.

Let:
- \( A = (1, 1, 1) \)
- \( B = (-1, 2, 3) \)
- \( C = (2, -1, 3) \)

Now, we can find vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = B - A = (-1 - 1, 2 - 1, 3 - 1) = (-2, 1, 2) \]
\[ \vec{AC} = C - A = (2 - 1, -1 - 1, 3 - 1) = (1, -2, 2) \]

### Step 2: Find the normal vector using the cross product
The normal vector \( \vec{n} \) to the plane can be found by taking the cross product of \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{n} = \vec{AB} \times \vec{AC} \]

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 2 \\ -2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 2 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} \]

Calculating each of these determinants:
1. \( \begin{vmatrix} 1 & 2 \\ -2 & 2 \end{vmatrix} = (1)(2) - (2)(-2) = 2 + 4 = 6 \)
2. \( \begin{vmatrix} -2 & 2 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (2)(1) = -4 - 2 = -6 \)
3. \( \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = (-2)(-2) - (1)(1) = 4 - 1 = 3 \)

Putting these together:
\[ \vec{n} = 6\hat{i} + 6\hat{j} + 3\hat{k} = (6, 6, 3) \]

### Step 3: Find the unit normal vector
To find the unit vector, we need to divide the normal vector by its magnitude.

First, we calculate the magnitude of \( \vec{n} \):
\[ |\vec{n}| = \sqrt{6^2 + 6^2 + 3^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9 \]

Now, we can find the unit normal vector \( \hat{n} \):
\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \left(\frac{6}{9}, \frac{6}{9}, \frac{3}{9}\right) = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) \]

### Final Answer
The unit vector normal to the plane is:
\[ \hat{n} = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) \]