Question

Let $ABC$ be a triangle, the position vector of whose vertices are respectively $-10\hat{i}+10\hat{k},-\hat{i}+6\hat{j}+6\hat{k}$ and $-4\hat{i}+9\hat{j}+6\hat{k}$ then $△ABC$ is

• A. Isosceles
• B. Scalene
• C. Equilateral
• D. Right angled

Answer 1

The correct option is A Isosceles

Position vector of vertex $A=-10\hat{i}+10\hat{k}$

Position vector of vertex $B=-\hat{i}+6\hat{j}+6\hat{k}$

Position vector of vertex $C=-4\hat{i}+9\hat{j}+6\hat{k}$

$\overrightarrow{AB}=-\hat{i}+6\hat{j}+6\hat{k}-(-10\hat{i}+10\hat{k})=9\hat{i}+6\hat{j}-4\hat{k}$

$\overrightarrow{AC}=-4\overrightarrow{i}+9\overrightarrow{j}+6\overrightarrow{k}-(-10\overrightarrow{i}+10\overrightarrow{k})=6\overrightarrow{i}+9\overrightarrow{j}-4\overrightarrow{k}\overrightarrow{BC}=-4\hat{i}+9\hat{j}+6\hat{k}-(-\hat{i}+6\hat{j}+6\hat{k})=-3\hat{i}+3\hat{j}$

Now,

$\left|\overrightarrow{AB}\right|=\sqrt{9^2+6^2+4^2}=\sqrt{81+36+16}=\sqrt{133}\left|\overrightarrow{AC}\right|=\sqrt{6^2+9^2+4^2}=\sqrt{81+36+16}=\sqrt{133}\left|\overrightarrow{BC}\right|=\sqrt{9+9}=9\sqrt{2}$

Since $\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|$ we have two equal sides equal .

Hence $△ABC$ is isosceles.