Question

If the position vectors ofthe vertices A, B, C of a triangle ABC are $7\hat{j}+10\hat{k},-\hat{i}+6\hat{j}+6\hat{k}$ and $-4\hat{i}+9\hat{j}+6\hat{k}$ respectively, the triangle is

• A. equilateral
• B. isosceles
• C. scalene
• D. right angled and isosceles also

Answer 1

The correct option is D right angled and isosceles also

Given, $\overrightarrow{OA}=7\hat{j}+10\hat{k},\overrightarrow{OB}=-\hat{i}+6\hat{j}+6\hat{k},\overrightarrow{OC}=-4\hat{i}+9\hat{j}+6\hat{k}$
$\Rightarrow \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-\hat{i}-\hat{j}-4\hat{k}$
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=-3\hat{i}+3\hat{j}$
$\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=4\hat{i}-2\hat{j}+4\hat{k}$
$\Rightarrow |\overrightarrow{AB}|=\sqrt{1^2+1^2+4^2}=\sqrt{18}=3\sqrt{2}$
$|\overrightarrow{BC}|=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$
$|\overrightarrow{CA}|=\sqrt{4^2+2^2+4^2}=\sqrt{36}=6$
$3\sqrt{2},3\sqrt{2}$ & 6 are sides of a right angled $\Delta$.
$\because (3\sqrt{2}{)}^2+(3\sqrt{2}{)}^2=36=6^2$
Hence, the $\Delta$ ABC is a right-angled and isosceles also