Show that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k},3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle.

Let $A=2\hat{i}-\hat{j}+\hat{k}B=\hat{i}-3\hat{j}-5\hat{k}C=3\hat{i}-4\hat{j}-4\hat{k}$

$|\overrightarrow{AB}|=\sqrt{41},|\overrightarrow{BC}|=\sqrt{6},|\overrightarrow{CA}|=\sqrt{35}$

${BC}^2+{CA}^2={AB}^2$

By pythogorean theorem $A,B,C$ form the vertices of a right angled triangle with right angle at$C$

Hence proved