the area of triangle whose vertices are (1,2,3),(2,5-1) and (-1,1,2) is

Options

A. 150 sq. unit
B. 145 sq unit
C. `(sqrt(155))/(2)` sq unit
D. `(155)/(2)` sq unit

Explanation

To find the area of the triangle whose vertices are given by the points \( A(1, 2, 3) \), \( B(2, 5, -1) \), and \( C(-1, 1, 2) \), we can use the formula for the area of a triangle in 3D space. The area \( A \) of the triangle can be calculated using the formula:

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

where \( \vec{AB} \) and \( \vec{AC} \) are the vectors formed by the points A, B, and C.

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

First, we calculate the vectors \( \vec{AB} \) and \( \vec{AC} \):

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

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

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

Next, we calculate the cross product \( \vec{AB} \times \vec{AC} \):

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

Calculating the 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. \( \begin{vmatrix} 3 & -4 \\ -1 & -1 \end{vmatrix} = (3 \cdot -1) - (-4 \cdot -1) = -3 - 4 = -7 \)

2. \( \begin{vmatrix} 1 & -4 \\ -2 & -1 \end{vmatrix} = (1 \cdot -1) - (-4 \cdot -2) = -1 - 8 = -9 \)

3. \( \begin{vmatrix} 1 & 3 \\ -2 & -1 \end{vmatrix} = (1 \cdot -1) - (3 \cdot -2) = -1 + 6 = 5 \)

Putting it all together:

\[ \vec{AB} \times \vec{AC} = -7\hat{i} + 9\hat{j} + 5\hat{k} = (-7, 9, 5) \]

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

Now, we find the magnitude of the cross product:

\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{(-7)^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155} \]

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

Finally, we calculate the area of the triangle:

\[ A = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \sqrt{155} \]

Thus, the area of the triangle is:

\[ \boxed{\frac{\sqrt{155}}{2}} \]