Question

Show that the points whose position vectors are $\overrightarrow{a}=4\hat{i}-3\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}-4\hat{j}+5\hat{k},\overrightarrow{c}=\hat{i}-\hat{j}$form a right triangle.

Answer 1

$$\begin{gathered} \text{Given that} \\ \overrightarrow{a}=\overrightarrow{OA}=4\hat{i}-3\hat{j}+\hat{k};\overrightarrow{b}=\overrightarrow{OB}=2\hat{i}-4\hat{j}+5\hat{k};\overrightarrow{c}=\overrightarrow{OC}=\hat{i}-\hat{j}+0\hat{k} \\ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\hat{i}-\hat{j}+4\hat{k} \\ \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=-\hat{i}+3\hat{j}-5\hat{k} \\ \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=3\hat{i}-2\hat{j}+\hat{k} \\ \overrightarrow{AB}.\overrightarrow{BC}=2-3-20=-21\ne 0 \\ \overrightarrow{BC}.\overrightarrow{CA}=-3-6-5=-14\ne 0 \\ \overrightarrow{AB}.\overrightarrow{CA}=-6+2+4=0 \\ \text{So,}\overrightarrow{AB}\text{is perpendicular to}\overrightarrow{CA}\text{.} \\ \text{So, ∆}\mathit{\text{ABC\hspace{0.17em}}}\text{is a right-angled triangle.} \end{gathered}$$