A unit vector normal to the plane through the point i,2j,3k is :

Options

A. `6i + 3j + 2k`
B. `i + 2j + 3k`
C. `(6i + 3j + 2k)/(7)`
D. `(6i + 3j + 2k)/(9)`

Answer

To find a unit vector normal to the plane through the point \( \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \), we can follow these steps:

### Step 1: Identify the Points
We have a point \( A = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \). Let's denote this point as \( A(1, 2, 3) \).

### Step 2: Define Vectors
We need two other points to define the plane. Let's denote them as:
- \( B = \mathbf{i} \) (Point B at \( (1, 0, 0) \))
- \( C = 2\mathbf{j} \) (Point C at \( (0, 2, 0) \))

### Step 3: Calculate Vectors
Now, we can find the vectors \( \mathbf{AB} \) and \( \mathbf{AC} \):
- \( \mathbf{AB} = B - A = (1, 0, 0) - (1, 2, 3) = (0, -2, -3) \)
- \( \mathbf{AC} = C - A = (0, 2, 0) - (1, 2, 3) = (-1, 0, -3) \)

### Step 4: Cross Product
To find the normal vector to the plane, we take the cross product of \( \mathbf{AB} \) and \( \mathbf{AC} \):
\[ \mathbf{n} = \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -2 & -3 \\ -1 & 0 & -3 \end{vmatrix} \]

### Step 5: Calculate the Determinant
Calculating the determinant:
\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} -2 & -3 \\ 0 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & -3 \\ -1 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & -2 \\ -1 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -2 & -3 \\ 0 & -3 \end{vmatrix} = (-2)(-3) - (0)(-3) = 6 \)
2. \( \begin{vmatrix} 0 & -3 \\ -1 & -3 \end{vmatrix} = (0)(-3) - (-1)(-3) = -3 \)
3. \( \begin{vmatrix} 0 & -2 \\ -1 & 0 \end{vmatrix} = (0)(0) - (-1)(-2) = -2 \)

Putting it all together:
\[ \mathbf{n} = 6\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} \]

### Step 6: Find the Magnitude of the Normal Vector
Now, we calculate the magnitude of \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]

### Step 7: Find the Unit Normal Vector
The unit normal vector \( \mathbf{u} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{6\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}}{7} = \frac{6}{7}\mathbf{i} + \frac{3}{7}\mathbf{j} - \frac{2}{7}\mathbf{k} \]

### Final Answer
The unit vector normal to the plane through the point \( \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) is:
\[ \mathbf{u} = \frac{6}{7}\mathbf{i} + \frac{3}{7}\mathbf{j} - \frac{2}{7}\mathbf{k} \]