If the position vector of the vertices of a triangle are `hat(i)-hat(j)+2hat(k), 2hat(i)+hat(j)+hat(k)` & `3hat(i)-hat(j)+2hat(k)`, then find the area of the triangle.

Explanation

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

### Step 1: Identify the position vectors
The position vectors of the vertices of the triangle are given as:
- \( \mathbf{A} = \hat{i} - \hat{j} + 2\hat{k} \)
- \( \mathbf{B} = 2\hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{C} = 3\hat{i} - \hat{j} + 2\hat{k} \)

### Step 2: Find the vectors representing two sides of the triangle
We can find two vectors that represent the sides of the triangle:
- Vector \( \mathbf{AB} = \mathbf{B} - \mathbf{A} \)
- Vector \( \mathbf{AC} = \mathbf{C} - \mathbf{A} \)

Calculating \( \mathbf{AB} \):
\[ \mathbf{AB} = (2\hat{i} + \hat{j} + \hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) = (2 - 1)\hat{i} + (1 + 1)\hat{j} + (1 - 2)\hat{k} = \hat{i} + 2\hat{j} - \hat{k} \]

Calculating \( \mathbf{AC} \):
\[ \mathbf{AC} = (3\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) = (3 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 - 2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \]

### Step 3: Calculate the cross product of the two vectors
Now we need to find the cross product \( \mathbf{AB} \times \mathbf{AC} \):
\[ \mathbf{AB} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{AC} = 2\hat{i} \]
Using the determinant method:
\[ \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 0 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -1 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i} (2 \cdot 0 - (-1) \cdot 0) - \hat{j} (1 \cdot 0 - (-1) \cdot 2) + \hat{k} (1 \cdot 0 - 2 \cdot 2) \]
\[ = 0\hat{i} - 2\hat{j} - 4\hat{k} \]
Thus,
\[ \mathbf{AB} \times \mathbf{AC} = -2\hat{j} - 4\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we calculate the magnitude of the cross product:
\[ |\mathbf{AB} \times \mathbf{AC}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\mathbf{AB} \times \mathbf{AC}| = \frac{1}{2} (2\sqrt{5}) = \sqrt{5} \]

Thus, the area of the triangle is \( \sqrt{5} \) square units.