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 equal to

9 sq units

The correct answer is Option (1) → 9 sq units ##

Given points are $A(0,4,1), B(2,3,-1), C(4,5,0)$ and $D(2,6,2)$.

We have, $\vec{AB} = (2-0)\hat{i} + (3-4)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k}$

and $\vec{BC} = (4-2)\hat{i} + (5-3)\hat{j} + (0+1)\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$

$∴\text{Area of quadrilateral } ABCD = |\vec{AB} \times \vec{BC}| = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix}$

$= |\hat{i}(-1+4) - \hat{j}(2+4) + \hat{k}(4+2)|$

$= |3\hat{i} - 6\hat{j} + 6\hat{k}| = \sqrt{9 + 36 + 36} = 9 \text{ sq. units}$