Question

Let ABC be a triangle, the position vectors of whose vertices are $7\hat{j}+10\hat{k},-\hat{i}+6\hat{j}+6\hat{k}$ and $-4\hat{i}+9\hat{j}+6\hat{k}$. Then $\Delta$ABC is

• A. isosceles
• B. equilateral
• C. right angled
• D. none of these

Answer 1

Answer: (A), (C)

isosceles
right angled

We have, $\overrightarrow{AB}=-\hat{i}-\hat{j}-4\hat{k},\overrightarrow{BC}=-3\hat{i}+3\hat{j}$ and $\overrightarrow{CA}=4\hat{i}-2\hat{j}+4\hat{k}$. Therefore,
$|\overrightarrow{AB}|=|\overrightarrow{BC}|=3\sqrt{2}$ and $|\overrightarrow{CA}|=6$
Clearly, $|\overrightarrow{AB}{|}^2+|\overrightarrow{BC}{|}^2=|\overrightarrow{AC}{|}^2$
Hence, the triangle is right-angled isosceles triangle.