Find with the help of vectors, the area of the triangle with vertices `A(3,-1,2), B(1,-1,-3)` and `C(4,-3,1)`

Answer

To find the area of the triangle with vertices \( A(3, -1, 2) \), \( B(1, -1, -3) \), and \( C(4, -3, 1) \) using vectors, we can follow these steps:

### Step 1: Define the position vectors of the points
Let the position vectors of points A, B, and C be defined as:
\[ \vec{A} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} \]

### Step 2: Find the vectors \( \vec{CA} \) and \( \vec{CB} \)
The vectors from point C to points A and B can be calculated as follows:
\[ \vec{CA} = \vec{A} - \vec{C} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 - 4 \\ -1 + 3 \\ 2 - 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \]
\[ \vec{CB} = \vec{B} - \vec{C} = \begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 - 4 \\ -1 + 3 \\ -3 - 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ -4 \end{pmatrix} \]

### Step 3: Calculate the cross product \( \vec{CA} \times \vec{CB} \)
To find the area of the triangle, we need to compute the cross product \( \vec{CA} \times \vec{CB} \):
\[ \vec{CA} \times \vec{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ -3 & 2 & -4 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 1 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ -3 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ -3 & 2 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i} (2 \cdot -4 - 1 \cdot 2) - \hat{j} (-1 \cdot -4 - 1 \cdot -3) + \hat{k} (-1 \cdot 2 - 2 \cdot -3) \]
\[ = \hat{i} (-8 - 2) - \hat{j} (4 - 3) + \hat{k} (-2 + 6) \]
\[ = -10\hat{i} - 1\hat{j} + 4\hat{k} \]
Thus,
\[ \vec{CA} \times \vec{CB} = \begin{pmatrix} -10 \\ -1 \\ 4 \end{pmatrix} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of the cross product gives us twice the area of the triangle:
\[ |\vec{CA} \times \vec{CB}| = \sqrt{(-10)^2 + (-1)^2 + 4^2} = \sqrt{100 + 1 + 16} = \sqrt{117} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\vec{CA} \times \vec{CB}| = \frac{1}{2} \sqrt{117} \]

### Final Answer
The area of the triangle with vertices \( A(3, -1, 2) \), \( B(1, -1, -3) \), and \( C(4, -3, 1) \) is:
\[ A = \frac{\sqrt{117}}{2} \text{ square units} \]