A unit vector normal to the plane through the points
`hati, 2hatj and 3hatk ` is

Options

A. `6hati + 3hatj + 2hatk`
B. `hati + 2hati + 3hatk`
C. `(6hati + 3hatj + 2hatk)/7`
D. `(hati + 2hatj + 3hatk)/7`

Answer

To find a unit vector normal to the plane through the points represented by the vectors \( \hat{i} \), \( 2\hat{j} \), and \( 3\hat{k} \), we can follow these steps:

### Step 1: Identify the points
Let:
- Point A = \( \hat{i} = (1, 0, 0) \)
- Point B = \( 2\hat{j} = (0, 2, 0) \)
- Point C = \( 3\hat{k} = (0, 0, 3) \)

### Step 2: Find vectors in the plane
We can find two vectors that lie in the plane formed by these points:
- Vector \( \vec{AB} = \vec{B} - \vec{A} = (0, 2, 0) - (1, 0, 0) = (-1, 2, 0) \)
- Vector \( \vec{AC} = \vec{C} - \vec{A} = (0, 0, 3) - (1, 0, 0) = (-1, 0, 3) \)

### Step 3: Find the normal vector using the cross product
The normal vector \( \vec{N} \) to the plane can be found using the cross product of vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{N} = \vec{AB} \times \vec{AC} \]

Calculating the cross product:
\[ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{N} = \hat{i} \begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ -1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ -1 & 0 \end{vmatrix} \]

Calculating each of the determinants:
1. \( \begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} = 2 \cdot 3 - 0 \cdot 0 = 6 \)
2. \( \begin{vmatrix} -1 & 0 \\ -1 & 3 \end{vmatrix} = -1 \cdot 3 - 0 \cdot -1 = -3 \)
3. \( \begin{vmatrix} -1 & 2 \\ -1 & 0 \end{vmatrix} = -1 \cdot 0 - 2 \cdot -1 = 2 \)

Putting it all together:
\[ \vec{N} = 6\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 4: Find the magnitude of the normal vector
The magnitude of \( \vec{N} \) is given by:
\[ |\vec{N}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]

### Step 5: Find the unit normal vector
The unit normal vector \( \hat{n} \) is given by:
\[ \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{6\hat{i} + 3\hat{j} + 2\hat{k}}{7} = \frac{6}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{2}{7}\hat{k} \]

### Conclusion
Thus, the unit vector normal to the plane through the points \( \hat{i} \), \( 2\hat{j} \), and \( 3\hat{k} \) is:
\[ \hat{n} = \frac{6\hat{i} + 3\hat{j} + 2\hat{k}}{7} \]