Question

The position vectors of the points A, B, C are $2\hat{i}+\hat{j}-\hat{k},3\hat{i}-2\hat{j}+\hat{k}\mathrm{and}\hat{i}+4\hat{j}-3\hat{k}$ respectively. These points
(a) form an isosceles triangle
(b) form a right triangle
(c) are collinear
(d) form a scalene triangle

Answer 1

(a) form an isosceles triangle
Given: Position vectors of $A,B,C$ are $2\hat{i}+\hat{j}-\hat{k},3\hat{i}-2\hat{j}+\hat{k}$ and $\hat{i}+4\hat{j}-3\hat{k}$. Then,
$$\begin{gathered} \overrightarrow{AB}=\left(3\hat{i}-2\hat{j}+\hat{k}\right)-\left(2\hat{i}+\hat{j}-\hat{k}\right)=\hat{i}-3\hat{j}+2\hat{k} \\ \overrightarrow{BC}=\left(\hat{i}+4\hat{j}-3\hat{k}\right)-\left(3\hat{i}-2\hat{j}+\hat{k}\right)=-2\hat{i}+6\hat{j}-4\hat{k} \\ \overrightarrow{CA}=\left(2\hat{i}+\hat{j}-\hat{k}\right)-\left(\hat{i}+4\hat{j}-3\hat{k}\right)=\hat{i}-3\hat{j}+2\hat{k} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\overrightarrow{\left|AB\right|}=\sqrt{1^2+{\left(-3\right)}^2+2^2}=\sqrt{1+9+4}=\sqrt{14} \\ \left|\overrightarrow{CA}\right|=\sqrt{1^2+{\left(-3\right)}^2+2^2}=\sqrt{1+9+4}=\sqrt{14} \\ \left|\overrightarrow{BC}\right|=\sqrt{{\left(-2\right)}^2+6^2+\left(-4^2\right)}=\sqrt{4+36+16}=\sqrt{56} \\ \therefore \left|AB\right|=\left|\overrightarrow{CA}\right| \end{gathered}$$
Hence, the triangle is isosceles as two of its sides are equal.