Question

Show that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle.

Answer 1

Let vectors $\overrightarrow{OA}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{OB}=\hat{i}-3\hat{j}-5\hat{k},\overrightarrow{OC}=3\hat{i}-4\hat{j}-4\hat{k}$ be the position vectors of points $A,B$ and $C$ respectively.

Now,$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$=\hat{i}-3\hat{j}-5\hat{k}-2\hat{i}+\hat{j}-\hat{k}$

$=-\hat{i}-2\hat{j}-6\hat{k}$

$\left|\overrightarrow{AB}\right|=\sqrt{{(-1)}^2+{(-2)}^2+{(-6)}^2}$

$=\sqrt{1+4+36}=\sqrt{41}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$

$=3\hat{i}-4\hat{j}-4\hat{k}-\hat{i}+3\hat{j}+5\hat{k}$

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

$\left|\overrightarrow{BC}\right|=\sqrt{{\left(2\right)}^2+{(-1)}^2+{\left(1\right)}^2}$

$=\sqrt{4+1+1}=\sqrt{6}$

$\overrightarrow{CA}=\overrightarrow{OC}-\overrightarrow{OA}$

$=3\hat{i}-4\hat{j}-4\hat{k}-2\hat{i}+\hat{j}-\hat{k}$

$=\hat{i}-3\hat{j}-5\hat{k}$

$\left|\overrightarrow{CA}\right|=\sqrt{{\left(1\right)}^2+{(-3)}^2+{(-5)}^2}$

$=\sqrt{1+9+25}=\sqrt{35}$

$\therefore \left|\overrightarrow{AB}\right|=\sqrt{41},\left|\overrightarrow{BC}\right|=\sqrt{6}$ and $\left|\overrightarrow{CA}\right|=\sqrt{35}$

$\therefore {AB}^2={BC}^2+{CA}^2$

Hence $△ABC$ is right-angled at $C$ where $\angle C={90}^{\circ }$