Question

Prove that the triangle, whose position vectors of the vertices are $2\hat{i}+4\hat{j}-\hat{k},4\hat{i}+5\hat{j}+\hat{k}$ and $3\hat{i}+6\hat{j}-3\hat{k}$ respectively, is an isosceles right angled triangle.

Answer 1

Position vector of vertices are $A,B,C$
$\overrightarrow{OA}=2\hat{i}+4\hat{j}-\hat{k}$
$\overrightarrow{OB}=4\hat{i}+5\hat{j}+\hat{k}$
$\overrightarrow{OC}=3\hat{i}+6\hat{j}-3\hat{k}$
$\therefore \overrightarrow{AB}=$ Position vector of $B-$ Position vector of $A$
$\overrightarrow{AB}=(4\hat{i}+5\hat{j}+\hat{k})-(2\hat{i}+4\hat{j}-\hat{k})=2\hat{i}+\hat{j}+2\hat{k}$
$\left|\overrightarrow{AB}\right|=AB=\sqrt{2^2+1^2+2^2}=\mathrm{3....}\left(1\right)$
$\therefore \overrightarrow{BC}=$ Position vector of $C-$ Position vector of $B$
$\overrightarrow{BC}=(3\hat{i}+6\hat{j}-3\hat{k})-(4\hat{i}+5\hat{j}+\hat{k})=-\hat{i}+\hat{j}-4\hat{k}$
$\left|\overrightarrow{BC}\right|=BC=\sqrt{{(-1)}^2+{\left(1\right)}^2+{(-4)}^2}=\sqrt{18}=3$
$\therefore \overrightarrow{CA}=$ Position vector of $A-$ Position vector of $C$
$\overrightarrow{CA}=(2\hat{i}+4\hat{j}-\hat{k})-(3\hat{i}+6\hat{j}-3\hat{k})=-\hat{i}-2\hat{j}+2\hat{k}$
$\left|\overrightarrow{CA}\right|=CA=\sqrt{{(-1)}^2+{(-2)}^2+{\left(2\right)}^2}=3$
$\because$ In triangle $ABC,AB=BC$
$\therefore$ This is isosceles triangle
and ${\left(3\sqrt{2}\right)}^2=3^2+3^2$
${BC}^2={AB}^2+{AC}^2$
In triangle $ABC$, angle $BAC={90}^o$