Question

Prove that the points $\hat{i}-\hat{j},4\hat{i}+3\hat{j}+\hat{k}\mathrm{and}2\hat{i}-4\hat{j}+5\hat{k}$ are the vertices of a right-angled triangle.

Answer 1

Given the points $\hat{i}-\hat{j},4\hat{i}+3\hat{j}+\hat{k}$ and $2\hat{i}-4\hat{j}+5\hat{k}$ Are A, B and C respectively.
Then,
$$\begin{gathered} \overrightarrow{AB}=4\hat{i}+3\hat{j}+\hat{k}-\hat{i}+\hat{j}=3\hat{i}+4\hat{j}+\hat{k}. \\ \overrightarrow{BC}=2\hat{i}-4\hat{j}+5\hat{k}-4\hat{i}-3\hat{j}-\hat{k}=-2\hat{i}-7\hat{j}+4\hat{k}. \\ \overrightarrow{CA}=\hat{i}-\hat{j}-2\hat{i}+4\hat{j}-5\hat{k}=-\hat{i}+3\hat{j}-5\hat{k}. \end{gathered}$$
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=3\hat{i}+4\hat{j}+\hat{k}-2\hat{i}-7\hat{j}+4\hat{k}-\hat{i}+3\hat{j}-5\hat{k}=\overrightarrow{0}.$
The given points forms a vertices of a triangle.
Now,
$$\begin{gathered} \left|\overrightarrow{AB}\right|=\sqrt{9+16+1}=\sqrt{26}. \\ \left|\overrightarrow{BC}\right|=\sqrt{4+49+16}=\sqrt{69}. \\ \left|\overrightarrow{CA}\right|=\sqrt{1+9+25}=\sqrt{35}. \end{gathered}$$
${\left|\overrightarrow{AB}\right|}^2+{\left|\overrightarrow{CA}\right|}^2=26+35=61$≠ ${\left|\overrightarrow{BC}\right|}^2$
The given triangle is not right-angled.