Question

Prove that $\overrightarrow{a}=2\hat{j}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{j}-3\hat{j}-5\hat{k}$ and $\overrightarrow{c}=3\hat{j}-4\hat{j}-4\overrightarrow{k}$ are the vector sides of right angle triangle

Answer 1

$\overrightarrow{a}=\sqrt{2^2+(-1{)}^2+1^2}=\sqrt{6}$ unit

$\overrightarrow{b}=\sqrt{1^2+(-3{)}^2+(-5{)}^2}=\sqrt{35}$ unit

$\overrightarrow{c}=\sqrt{3^2+(-4{)}^2+(-4{)}^2}=\sqrt{41}$ unit

If above sides form a right angle then

${\overrightarrow{c}}^2={\overrightarrow{a}}^2+{\overrightarrow{b}}^2$

$(\sqrt{41}{)}^2=(\sqrt{35}{)}^2+(\sqrt{6}{)}^2$

$41=41$

Hence, proved.