Question

If $ABCD$ is a quadrilateral such that $\overrightarrow{AB}=\hat{i}+2\hat{j},\overrightarrow{AD}=\hat{j}+2\hat{k}$ and $\overrightarrow{AC}=2(\hat{i}+2\hat{j})+3(\hat{j}+2\hat{k})$. Then area of the quadrilateral $ABCD$ is

• A. $\frac{5\sqrt{21}}{2}$
• B. $\frac{3\sqrt{21}}{2}$
• C. $\frac{\sqrt{21}}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

$\frac{\sqrt{21}}{2}$

$\overrightarrow{AC}=2(\overrightarrow{i}+2\overrightarrow{j})+3(\overrightarrow{j}+2\overrightarrow{k})$

$=2\overrightarrow{AB}+3\overrightarrow{AD}$

Area of quadrilateral ABCD = Area of $△ABC$ + Area of $△ADC$

$=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|+\frac{1}{2}|\overrightarrow{AD}\times \overrightarrow{AC}|.$

$\overrightarrow{AB}\times \overrightarrow{AC}=\overrightarrow{AB}\times (2\overrightarrow{AB}+3\overrightarrow{AD})$

$=3(\overrightarrow{AB}+3\overrightarrow{AD})$

$\overrightarrow{AD}\times \overrightarrow{AC}=\overrightarrow{AD}\times (2\overrightarrow{AB}+3\overrightarrow{AD})$

$=2.\overrightarrow{AD}\times \overrightarrow{AB}+0.$

Quad. ABCD $=\frac{1}{2}\left|3\right(\overrightarrow{AB}\times \overrightarrow{AD})-2(\overrightarrow{AB}\times \overrightarrow{AD}\left)\right|$

$=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AD}|$

$\overrightarrow{AB}=\overrightarrow{i}+\overrightarrow{2j};\overrightarrow{AD}=\overrightarrow{j}+2\overrightarrow{K}$

$\overrightarrow{AB}\times \overrightarrow{AD}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 2& 0\\ 0& 1& 2\end{array}\right|=0$

$=4\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k}$

$|\overrightarrow{AB}\times \overrightarrow{AD}|=\sqrt{4^2+2^2+1}=\sqrt{21}$

Area $=\frac{1}{2}\sqrt{21}.$