Question

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

Answer 1

Let $\overrightarrow{a}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}-5\overrightarrow{k}$
$\overrightarrow{c}=\overrightarrow{-3}i+\overrightarrow{4}j+\overrightarrow{4}k$
We see that
$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
$\therefore \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ forms a triangle
Further $\overrightarrow{a}.\overrightarrow{b}=(2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}).(\overrightarrow{i}-3\overrightarrow{j}-\overrightarrow{k})$
$=2+3-5$
$=0$
Therefore, $\overrightarrow{a}\perp \overrightarrow{b}$
$\therefore$ the vectors from the sides of a right angled triangle.