The area of the quadrilateral ABCD where `A (0,4,1) , B(2,3,-1), c (4,5,0)and D(2,6,2)` is to :

Options

A. 9 sq units
B. 18 sq units
C. 27 sq units
D. 81 sq units

Answer

To find the area of the quadrilateral ABCD with the given vertices A(0, 4, 1), B(2, 3, -1), C(4, 5, 0), and D(2, 6, 2), we can follow these steps:

### Step 1: Write down the position vectors
The position vectors of the points are:
- \( \vec{A} = 0\hat{i} + 4\hat{j} + 1\hat{k} \)
- \( \vec{B} = 2\hat{i} + 3\hat{j} - 1\hat{k} \)
- \( \vec{C} = 4\hat{i} + 5\hat{j} + 0\hat{k} \)
- \( \vec{D} = 2\hat{i} + 6\hat{j} + 2\hat{k} \)

### Step 2: Find the vectors AB and BC
To find vector \( \vec{AB} \):
\[ \vec{AB} = \vec{B} - \vec{A} = (2 - 0)\hat{i} + (3 - 4)\hat{j} + (-1 - 1)\hat{k} = 2\hat{i} - 1\hat{j} - 2\hat{k} \]

To find vector \( \vec{BC} \):
\[ \vec{BC} = \vec{C} - \vec{B} = (4 - 2)\hat{i} + (5 - 3)\hat{j} + (0 - (-1))\hat{k} = 2\hat{i} + 2\hat{j} + 1\hat{k} \]

### Step 3: Calculate the cross product \( \vec{AB} \times \vec{BC} \)
We will set up the determinant for the cross product:
\[ \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = (-1)(1) - (-2)(2) = -1 + 4 = 3 \)
2. \( \begin{vmatrix} 2 & -2 \\ 2 & 1 \end{vmatrix} = (2)(1) - (-2)(2) = 2 + 4 = 6 \)
3. \( \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} = (2)(2) - (-1)(2) = 4 + 2 = 6 \)

Putting it all together:
\[ \vec{AB} \times \vec{BC} = 3\hat{i} - 6\hat{j} + 6\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of the vector \( \vec{AB} \times \vec{BC} \) is:
\[ |\vec{AB} \times \vec{BC}| = \sqrt{3^2 + (-6)^2 + 6^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9 \]

### Step 5: Find the area of the quadrilateral
The area of the quadrilateral ABCD is given by:
\[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{BC}| = \frac{1}{2} \times 9 = 4.5 \]

### Final Answer
The area of the quadrilateral ABCD is \( 4.5 \) square units.