The area ( in sq. units ) of the triangle having vertices with position vectors `hat(i) - 2 hat(j) +3 hat(k), - 2 hat(i) + 3 hat(j) - hat(k)` and `4 hat(i)- 7 hat(j) + 7 hat(k)` is

Options

A. 36
B. 0
C. 39
D. 11

Answer

To find the area of the triangle formed by the given position vectors, we can follow these steps:

### Step 1: Identify the position vectors
Let the position vectors of the vertices of the triangle be:
- \( \mathbf{A} = \hat{i} - 2\hat{j} + 3\hat{k} \)
- \( \mathbf{B} = -2\hat{i} + 3\hat{j} - \hat{k} \)
- \( \mathbf{C} = 4\hat{i} - 7\hat{j} + 7\hat{k} \)

### Step 2: Calculate the vectors \( \mathbf{AB} \) and \( \mathbf{BC} \)
The vector \( \mathbf{AB} \) is given by:
\[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (-2\hat{i} + 3\hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) \]
Calculating this:
\[ \mathbf{AB} = (-2 - 1)\hat{i} + (3 + 2)\hat{j} + (-1 - 3)\hat{k} = -3\hat{i} + 5\hat{j} - 4\hat{k} \]

The vector \( \mathbf{BC} \) is given by:
\[ \mathbf{BC} = \mathbf{C} - \mathbf{B} = (4\hat{i} - 7\hat{j} + 7\hat{k}) - (-2\hat{i} + 3\hat{j} - \hat{k}) \]
Calculating this:
\[ \mathbf{BC} = (4 + 2)\hat{i} + (-7 - 3)\hat{j} + (7 + 1)\hat{k} = 6\hat{i} - 10\hat{j} + 8\hat{k} \]

### Step 3: Calculate the cross product \( \mathbf{AB} \times \mathbf{BC} \)
To find the area of the triangle, we need to compute the cross product \( \mathbf{AB} \times \mathbf{BC} \):
\[ \mathbf{AB} \times \mathbf{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 5 & -4 \\ 6 & -10 & 8 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 5 & -4 \\ -10 & 8 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & -4 \\ 6 & 8 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & 5 \\ 6 & -10 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 5 & -4 \\ -10 & 8 \end{vmatrix} = (5)(8) - (-4)(-10) = 40 - 40 = 0 \)

2. \( \begin{vmatrix} -3 & -4 \\ 6 & 8 \end{vmatrix} = (-3)(8) - (-4)(6) = -24 + 24 = 0 \)

3. \( \begin{vmatrix} -3 & 5 \\ 6 & -10 \end{vmatrix} = (-3)(-10) - (5)(6) = 30 - 30 = 0 \)

Thus,
\[ \mathbf{AB} \times \mathbf{BC} = 0\hat{i} - 0\hat{j} + 0\hat{k} = \mathbf{0} \]

### Step 4: Calculate the area of the triangle
The area \( A \) of triangle \( ABC \) is given by:
\[ A = \frac{1}{2} |\mathbf{AB} \times \mathbf{BC}| \]
Since \( \mathbf{AB} \times \mathbf{BC} = \mathbf{0} \):
\[ A = \frac{1}{2} \cdot 0 = 0 \]

### Conclusion
The area of the triangle is \( 0 \) square units, which implies that the points are collinear.

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