Find the area of the triangle, the position vectors of whose vertices are
`a=i-2j+3k, b=j-k" and "c=2i+j`

Answer

To find the area of the triangle whose vertices are given by the position vectors \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \), \( \mathbf{b} = \mathbf{j} - \mathbf{k} \), and \( \mathbf{c} = 2\mathbf{i} + \mathbf{j} \), we can follow these steps:

### Step 1: Find the position vectors of the sides of the triangle
The position vectors of the sides of the triangle can be found by taking the difference of the position vectors of the vertices.

Let:
- \( \mathbf{A} = \mathbf{a} \)
- \( \mathbf{B} = \mathbf{b} \)
- \( \mathbf{C} = \mathbf{c} \)

We can define two vectors that represent two sides of the triangle:
\[ \mathbf{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{j} - \mathbf{k}) - (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = -\mathbf{i} + 3\mathbf{j} - 4\mathbf{k} \]
\[ \mathbf{AC} = \mathbf{c} - \mathbf{a} = (2\mathbf{i} + \mathbf{j}) - (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + 3\mathbf{j} - 3\mathbf{k} \]

### Step 2: Calculate the cross product of the two vectors
The area of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \|\mathbf{AB} \times \mathbf{AC}\| \]

Now, we need to calculate the cross product \( \mathbf{AB} \times \mathbf{AC} \).

Using the determinant formula for the cross product:
\[ \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & -4 \\ 1 & 3 & -3 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 3 & -4 \\ 3 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -1 & -4 \\ 1 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & 3 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 3 & -4 \\ 3 & -3 \end{vmatrix} = (3)(-3) - (3)(-4) = -9 + 12 = 3 \)
2. \( \begin{vmatrix} -1 & -4 \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (1)(-4) = 3 + 4 = 7 \)
3. \( \begin{vmatrix} -1 & 3 \\ 1 & 3 \end{vmatrix} = (-1)(3) - (1)(3) = -3 - 3 = -6 \)

Putting it all together:
\[ \mathbf{AB} \times \mathbf{AC} = 3\mathbf{i} - 7\mathbf{j} - 6\mathbf{k} \]

### Step 3: Find the magnitude of the cross product
Now we find the magnitude of the vector \( \mathbf{AB} \times \mathbf{AC} \):
\[ \|\mathbf{AB} \times \mathbf{AC}\| = \sqrt{(3)^2 + (-7)^2 + (-6)^2} = \sqrt{9 + 49 + 36} = \sqrt{94} \]

### Step 4: Calculate the area of the triangle
Finally, the area of the triangle is:
\[ \text{Area} = \frac{1}{2} \|\mathbf{AB} \times \mathbf{AC}\| = \frac{1}{2} \sqrt{94} \]

Thus, the area of the triangle is:
\[ \text{Area} = \frac{\sqrt{94}}{2} \]