Find the ate area of the triangle whose vertices are :
A (1,2,3) , B (2,-1,4) and C (4,5,-1).

Answer

To find the area of the triangle with vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1), we can use the vector approach. Here’s a step-by-step solution:

### Step 1: Find the position vectors of points A, B, and C.
The position vectors are:
- \( \vec{A} = (1, 2, 3) \)
- \( \vec{B} = (2, -1, 4) \)
- \( \vec{C} = (4, 5, -1) \)

### Step 2: Calculate the vectors \( \vec{BA} \) and \( \vec{BC} \).
- \( \vec{BA} = \vec{A} - \vec{B} = (1 - 2, 2 - (-1), 3 - 4) = (-1, 3, -1) \)
- \( \vec{BC} = \vec{C} - \vec{B} = (4 - 2, 5 - (-1), -1 - 4) = (2, 6, -5) \)

### Step 3: Compute the cross product \( \vec{BA} \times \vec{BC} \).
The cross product can be calculated using the determinant of a matrix:
\[ \vec{BA} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -1 \\ 2 & 6 & -5 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 3 & -1 \\ 6 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -1 \\ 2 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 3 \\ 2 & 6 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 3 & -1 \\ 6 & -5 \end{vmatrix} = (3)(-5) - (-1)(6) = -15 + 6 = -9 \)
2. \( \begin{vmatrix} -1 & -1 \\ 2 & -5 \end{vmatrix} = (-1)(-5) - (-1)(2) = 5 + 2 = 7 \)
3. \( \begin{vmatrix} -1 & 3 \\ 2 & 6 \end{vmatrix} = (-1)(6) - (3)(2) = -6 - 6 = -12 \)

Putting it all together:
\[ \vec{BA} \times \vec{BC} = -9\hat{i} - 7\hat{j} - 12\hat{k} = (-9, -7, -12) \]

### Step 4: Find the magnitude of the cross product.
The magnitude is given by:
\[ |\vec{BA} \times \vec{BC}| = \sqrt{(-9)^2 + (-7)^2 + (-12)^2} = \sqrt{81 + 49 + 144} = \sqrt{274} \]

### Step 5: Calculate the area of the triangle.
The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\vec{BA} \times \vec{BC}| = \frac{1}{2} \sqrt{274} \]

### Final Answer:
The area of the triangle is \( \frac{\sqrt{274}}{2} \) square units.

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