What is the area of the triangle with vertices `(0, 2,2), (2, 0, -1) and (3, 4, 0)` ?

Options

A. `(15)/(2)`sq unit
B. `15` sq unit
C. `(7)/(2)` sq unit
D. `7` sq unit

Explanation

To find the area of the triangle with vertices \( A(0, 2, 2) \), \( B(2, 0, -1) \), and \( C(3, 4, 0) \), we can use the formula for the area of a triangle formed by three points in 3D space. The area can be calculated using the cross product of two vectors formed by these points.

### Step-by-Step Solution:

1. **Define the vertices:**
Let the vertices be:
- \( A(0, 2, 2) \)
- \( B(2, 0, -1) \)
- \( C(3, 4, 0) \)

2. **Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):**
- The vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (2 - 0, 0 - 2, -1 - 2) = (2, -2, -3) \]
- The vector \( \overrightarrow{AC} \) is given by:
\[ \overrightarrow{AC} = C - A = (3 - 0, 4 - 2, 0 - 2) = (3, 2, -2) \]

3. **Calculate the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):**
The cross product can be calculated using the determinant of a matrix:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & -3 \\ 3 & 2 & -2 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} -2 & -3 \\ 2 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- For \( \hat{i} \):
\[ = \hat{i}((-2)(-2) - (-3)(2)) = \hat{i}(4 + 6) = 10\hat{i} \]
- For \( \hat{j} \):
\[ = -\hat{j}((2)(-2) - (-3)(3)) = -\hat{j}(-4 + 9) = -5\hat{j} \]
- For \( \hat{k} \):
\[ = \hat{k}((2)(2) - (-2)(3)) = \hat{k}(4 + 6) = 10\hat{k} \]

Thus, the cross product is:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = (10, -5, 10) \]

4. **Find the magnitude of the cross product:**
The magnitude of the vector \( (10, -5, 10) \) is given by:
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{10^2 + (-5)^2 + 10^2} = \sqrt{100 + 25 + 100} = \sqrt{225} = 15 \]

5. **Calculate the area of the triangle:**
The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \times 15 = \frac{15}{2} \]

### Final Answer:
The area of the triangle is \( \frac{15}{2} \) square units.

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