Question

The area of the triangle formed by the point whose position vectors are $-3\hat{i}+\hat{j},5\hat{i}+2\hat{j}+\hat{k}$ and $\hat{i}-2\hat{j}+3\hat{k}$ is

• A. $\sqrt{23}$ sq. units
• B. $\sqrt{21}$ sq. units
• C. $\sqrt{305}$ sq. units
• D. $\sqrt{33}$ sq. units

Answer 1

Answer: (C)

$\sqrt{305}$ sq. units

$\overrightarrow{OA}=-3\hat{i}+\hat{j}$, $\overrightarrow{OB}=5\hat{i}+2\hat{j}+\hat{k}$, $\overrightarrow{OC}=\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=5\hat{i}+2\hat{j}+\hat{k}+3\hat{i}-\hat{j}=8\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\hat{i}-2\hat{j}+3\hat{k}-5\hat{i}-2\hat{j}-\hat{k}=-4\hat{i}-4\hat{j}+2\hat{k}$

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& 1& 1\\ -4& -4& 2\end{array}\right|$

$=\hat{i}(2+4)-\hat{j}(16+4)+\hat{k}(-32+4)$

$=6\hat{i}-20\hat{j}-28\hat{k}$

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\sqrt{36+400+784}=\sqrt{1220}$

$=2\sqrt{305}$

Area of $\Delta ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{BC}|=\frac{1}{2}\times 2\sqrt{305}$

$=\sqrt{305}$ sq. units.