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 formed by the given position vectors, we will follow these steps:

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

### Step 2: Calculate the vectors AB and AC
To find the area of the triangle, we first need to find the vectors \( \vec{AB} \) and \( \vec{AC} \).

- The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + \hat{j} + \hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) \]
\[ = (2 - 1)\hat{i} + (1 + 1)\hat{j} + (1 - 2)\hat{k} = \hat{i} + 2\hat{j} - \hat{k} \]

- The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} = (3\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) \]
\[ = (3 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 - 2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \]

### Step 3: Calculate the cross product \( \vec{AB} \times \vec{AC} \)
Now, we will find the cross product of \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = \hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{AC} = 2\hat{i} \]

Using the determinant method:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 0 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -1 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} \]
\[ = \hat{i} (2 \cdot 0 - (-1) \cdot 0) - \hat{j} (1 \cdot 0 - (-1) \cdot 2) + \hat{k} (1 \cdot 0 - 2 \cdot 2) \]
\[ = 0\hat{i} - 2\hat{j} - 4\hat{k} \]
\[ = -2\hat{j} - 4\hat{k} \]

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

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

### Final Answer
The area of the triangle is \( \sqrt{5} \) square units.

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