Question

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

• A. $\sqrt{23}$
• B. $\sqrt{21}$
• C. $\sqrt{29}$
• D. $\sqrt{33}$

Answer 1

Answer: (C)

$\sqrt{29}$

$\overrightarrow{v}=3\hat{i}+\hat{j}$
$\overrightarrow{u}=5\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{w}=\hat{i}-2\hat{j}+3\hat{k}$
$\overrightarrow{U}-\overrightarrow{V}=2\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{w}-\overrightarrow{v}=-2\hat{i}-3\hat{j}+3\hat{k}$
Area $=\frac{1}{2}|(\overrightarrow{u}-\overrightarrow{v})\times (\overrightarrow{w}-\overrightarrow{v})|$
$(\overrightarrow{u}-\overrightarrow{v})\times (\overrightarrow{w}-\overrightarrow{v})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 1\\ -2& -3& 3\end{array}\right|=6\hat{i}-8\hat{j}-4\hat{k}$
$area=\frac{1}{2}\sqrt{36+64+16}=\sqrt{29}$