Question

The position vector of the points $A,B,C$ are $(2\hat{i}+\hat{j}-\hat{k}),(3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

• A. Form an isosceles triangle
• B. Form a right angled triangle
• C. Are collinear
• D. Form a scalene triangle

Answer 1

Answer: (C)

Are collinear

$\overrightarrow{AB}=(3-2)\hat{i}+(-2-1)\hat{j}+(1+1)\hat{k}$
$=\hat{i}-3\hat{j}+2\hat{k}$
$\Rightarrow \left|\overrightarrow{AB}\right|=\sqrt{1+9+4}=\sqrt{14}$
$\overrightarrow{BC}=(1-3)\hat{i}+(4+2)\hat{j}+(-3-1)\hat{k}$
$=-2\hat{i}+6\hat{j}-4\hat{k}$
$\Rightarrow \left|\overrightarrow{BC}\right|=\sqrt{4+36+16}$
$=\sqrt{56}=2\sqrt{14}$
$\overrightarrow{CA}=(2-1)\hat{i}+(1-4)\hat{j}+(-1+3)\hat{k}$
$=\hat{i}-3\hat{j}+2\hat{k}$
$\Rightarrow \left|\overrightarrow{CA}\right|=\sqrt{1+9+4}=\sqrt{14}$
So, $\left|\overrightarrow{AB}\right|+\left|\overrightarrow{AC}\right|=\left|\overrightarrow{BC}\right|$ and angle between $AB$ and $BC$ is $180$.
$\therefore$ Points $A,B,C$ cannot form an isosceles triangle.
Hence, $A,B,C$ are collinear.