What is the area of the triangle with vertices `(1, 2, 3), (2, 5, -1) and (-1, 1, 2)` ?

Options

A. `(sqrt155)/(2)`
B. `(sqrt175)/(2)`
C. `(sqrt155)/(4)`
D. `(sqrt175)/(4)`

Answer

To find the area of the triangle with vertices \( A(1, 2, 3) \), \( B(2, 5, -1) \), and \( C(-1, 1, 2) \), we can use the formula for the area of a triangle formed by three points in 3D space. The area \( A \) is given by:

\[ A = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \]

where \( \vec{AB} \) and \( \vec{AC} \) are the vectors from point \( A \) to points \( B \) and \( C \) respectively, and \( \times \) denotes the cross product.

### Step 1: Find the vectors \( \vec{AB} \) and \( \vec{AC} \)

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (2 - 1, 5 - 2, -1 - 3) = (1, 3, -4) \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = C - A = (-1 - 1, 1 - 2, 2 - 3) = (-2, -1, -1) \]

### Step 2: Compute the cross product \( \vec{AB} \times \vec{AC} \)

To compute the cross product, we can use the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{AB} \) and \( \vec{AC} \):

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -4 \\ -2 & -1 & -1 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 3 & -4 \\ -1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -4 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ -2 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ 3 \cdot (-1) - (-4) \cdot (-1) = -3 - 4 = -7 \]

2. For \( \hat{j} \):
\[ 1 \cdot (-1) - (-4) \cdot (-2) = -1 - 8 = -9 \]

3. For \( \hat{k} \):
\[ 1 \cdot (-1) - 3 \cdot (-2) = -1 + 6 = 5 \]

Thus, the cross product is:
\[ \vec{AB} \times \vec{AC} = -7\hat{i} + 9\hat{j} + 5\hat{k} \]

### Step 3: Find the magnitude of the cross product

The magnitude of the vector \( \vec{AB} \times \vec{AC} \) is given by:
\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{(-7)^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155} \]

### Step 4: Calculate the area of the triangle

Finally, we can find the area of the triangle:
\[ A = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \sqrt{155} \]

### Final Answer
The area of the triangle is:
\[ \frac{\sqrt{155}}{2} \]