Question

The position vectors 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

The correct option is A Form an Isosceles Triangle

Given that $\overrightarrow{OA}=2\hat{i}+\hat{j}-\hat{k}$,

$\overrightarrow{OB}=3\hat{i}-2\hat{j}+\hat{k}$

$\overrightarrow{OC}=\hat{i}+4\hat{j}-3\hat{k}$

We have

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(3-2)\hat{i}+(-2-1)\hat{j}+(1+1)\hat{k}$

$=\hat{i}-3\hat{j}+2\hat{k}$

$\left|\overrightarrow{AB}\right|=\sqrt{1^2+{(-3)}^2+2^2}=\sqrt{1+9+4}=\sqrt{14}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(1-3)\hat{i}+(4+2)\hat{j}+(-3-1)\hat{k}$

$=-2\hat{i}+6\hat{j}-4\hat{k}$

$\left|\overrightarrow{BC}\right|=\sqrt{{(-2)}^2+6^2+{(-4)}^2}=\sqrt{4+36+16}=\sqrt{56}$

$\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=(2-1)\hat{i}+(1-4)\hat{j}+(-1+3)\hat{k}$

$=\hat{i}-3\hat{j}+2\hat{k}$

$\left|\overrightarrow{CA}\right|=\sqrt{1^2+{(-3)}^2+{\left(2\right)}^2}=\sqrt{1+9+4}=\sqrt{14}$

It is clear that side $CA$ and $AB$ of the triangle are equal.

$\therefore$ Points $A,B,C$ from an isosceles triangle.