Question

The points $A(2\hat{i}-\hat{j}+\hat{k}),B(\hat{i}-3\hat{j}-5\hat{k})$ and $C(3\hat{i}-4\hat{j}-4\hat{k})$ represents the vertices of a $△ABC$. Then $△ABC$ is

• A. Obtuse
• B. Isoceles
• C. Right angled
• D. Equilateral

Answer 1

The correct option is C Right angled

The position vectors of $A,B$ and $C$ are $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$ respectively.
$\therefore \overrightarrow{AB}=-\hat{i}-2\hat{j}-6\hat{k},\overrightarrow{BC}=2\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{CA}=-\hat{i}+3\hat{j}+5\hat{k}$

Now, $|\overrightarrow{AB}|=\sqrt{1+4+36}=\sqrt{41},|\overrightarrow{BC}|=\sqrt{4+1+1}=\sqrt{6},|\overrightarrow{CA}|=\sqrt{1+9+25}=\sqrt{35}$
Clearly, $|\overrightarrow{AB}{|}^2=|\overrightarrow{BC}{|}^2+|\overrightarrow{CA}{|}^2$
$\Rightarrow A{B}^2=B{C}^2+C{A}^2$
Hence, $△ABC$ is a right triangle, right angled at $C$