If the position vectors of three points A, B, C are respectively `i+j+k, 2i + 3j -4k and 7i+4j+9k`, then the unit vector perpendicular to the plane of triangle ABCis

Options

A. `(31i-38j-9k)`
B. `(31i-38j-9k)/(sqrt(2486))`
C. `(31i + 38j +9k)/(sqrt(2486))`
D. none of these

Answer

To find the unit vector perpendicular to the plane of triangle ABC given the position vectors of points A, B, and C, we can follow these steps:

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k} \)
- \( \vec{C} = 7\hat{i} + 4\hat{j} + 9\hat{k} \)

### Step 2: Find the vectors AB and BC
To find the vectors \( \vec{AB} \) and \( \vec{BC} \), we can use the formula:
- \( \vec{AB} = \vec{B} - \vec{A} \)
- \( \vec{BC} = \vec{C} - \vec{B} \)

Calculating \( \vec{AB} \):
\[ \vec{AB} = (2\hat{i} + 3\hat{j} - 4\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = (2-1)\hat{i} + (3-1)\hat{j} + (-4-1)\hat{k} = \hat{i} + 2\hat{j} - 5\hat{k} \]

Calculating \( \vec{BC} \):
\[ \vec{BC} = (7\hat{i} + 4\hat{j} + 9\hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k}) = (7-2)\hat{i} + (4-3)\hat{j} + (9+4)\hat{k} = 5\hat{i} + 1\hat{j} + 13\hat{k} \]

### Step 3: Find the cross product \( \vec{AB} \times \vec{BC} \)
The cross product will give us a vector perpendicular to the plane formed by points A, B, and C. We can calculate it using the determinant of a matrix:

\[ \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -5 \\ 5 & 1 & 13 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -5 \\ 1 & 13 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -5 \\ 5 & 13 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & -5 \\ 1 & 13 \end{vmatrix} = (2 \cdot 13) - (-5 \cdot 1) = 26 + 5 = 31 \)
2. \( \begin{vmatrix} 1 & -5 \\ 5 & 13 \end{vmatrix} = (1 \cdot 13) - (-5 \cdot 5) = 13 + 25 = 38 \)
3. \( \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} = (1 \cdot 1) - (2 \cdot 5) = 1 - 10 = -9 \)

Putting it all together:
\[ \vec{AB} \times \vec{BC} = 31\hat{i} - 38\hat{j} - 9\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of the vector \( \vec{P} = 31\hat{i} - 38\hat{j} - 9\hat{k} \) is given by:
\[ |\vec{P}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486} \]

### Step 5: Find the unit vector
The unit vector \( \hat{n} \) perpendicular to the plane of triangle ABC is given by:
\[ \hat{n} = \frac{\vec{P}}{|\vec{P}|} = \frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}} \]

### Final Answer
Thus, the unit vector perpendicular to the plane of triangle ABC is:
\[ \hat{n} = \frac{31}{\sqrt{2486}} \hat{i} - \frac{38}{\sqrt{2486}} \hat{j} - \frac{9}{\sqrt{2486}} \hat{k} \]