Question

Prove that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$ are sides of a right angled triangle.

Answer 1

2 things need to be checked which are the satisfaction of Pythagoras theorem and that the angle between the shorter 2 sides is ${90}^{\circ }$.

$\therefore |\overrightarrow{A}{|}^2=|2\hat{i}-\hat{j}+\hat{k}{|}^2=4+1+1=6$

$\therefore |\overrightarrow{B}{|}^2=|\hat{i}-3\hat{j}-5\hat{k}{|}^2=1+9+25=35$

$\therefore |\overrightarrow{C}{|}^2=|3\hat{i}-4\hat{j}+4\hat{k}{|}^2=9+16+16=41$

Now as $|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2=|\overrightarrow{C}{|}^2$, the Pythagoras theorem is satisfied. Also, $\overrightarrow{A}$ and $\overrightarrow{B}$ are the smaller sides.

$\therefore \overrightarrow{A}\cdot \overrightarrow{B}=2+3-5=0⟹$ angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is ${90}^{\circ }$.

Hence, proved.