Question

Let $ABC$ be a triangle. The position vectors of the vertices are $\hat{i}+2\hat{j}+4\hat{k},-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-3\hat{k}$ respectively. Then $\Delta$ABC is

• A. Isosceles
• B. Equilateral
• C. Right angled
• D. None of these

Answer 1

The correct option is C Right angled

We have

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=4\hat{i}+2\hat{j}-4\hat{k}$
Similarly, we have

$\overrightarrow{AB}=-3\hat{i}-3\hat{k}$,

$\overrightarrow{AC}=\hat{i}+2\hat{j}-7\hat{k}$

$|BC{|}^2=36,|AB{|}^2=18$, and $|AC{|}^2=54$

Clearly, $A{C}^2=B{C}^2+A{B}^2$
$\therefore \angle B={90}^o$
Hence, option $C$ is correct.