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

Toolbox:
• In a right angled triangle ${AB}^2+{BC}^2={AC}^2$
• If $\overrightarrow a$ and $\overrightarrow b$ are position vectors then $(\overrightarrow b-\overrightarrow a)=\overrightarrow{AB}$
• $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$
Step 1:
Let $\overrightarrow a=3\hat i-4\hat j-4\hat k$
$\quad\;\;\overrightarrow b=2\hat i-\hat j+\hat k$
$\quad\;\;\overrightarrow c=\hat i-3\hat j-5\hat k$
Step 2:
Let us find $\overrightarrow{AB}$
$\overrightarrow{AB}=\overrightarrow b-\overrightarrow a$
$\quad\;\;=(2\hat i-\hat j+\hat k)-(3\hat i-4\hat j-4\hat k)$
$\quad\;\;=-\hat i+3\hat j+5\hat k$
$|\overrightarrow{AB}|=\sqrt{(-1)^2+(3)^2+(5)^2}$
$\qquad=\sqrt{1+9+25}$
$\qquad=\sqrt{35}$
Therefore $|AB|^2=35$
Step 3:
Next let us find $\overrightarrow{BC}$
$\overrightarrow{BC}=\overrightarrow c-\overrightarrow b$
$\quad\;\;=(\hat i-3\hat j-5\hat k)-(2\hat i-\hat j+\hat k)$
$\quad\;\;=-\hat i-2\hat j-6\hat k$
$|\overrightarrow{BC}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}$
$\qquad=\sqrt{1+4+36}$
$\qquad=\sqrt{41}$
Therefore $|BC|^2=41.$
Step 4:
Next let us find $\overrightarrow{CA}$
$\overrightarrow{CA}=\overrightarrow a-\overrightarrow c$
$\quad\;\;=(3\hat i-4\hat j-4\hat k)-(\hat i-3\hat j-5\hat k)$
$\quad\;\;=2\hat i-\hat j+\hat k$
$|\overrightarrow{CA}|=\sqrt{(2)^2+(-1)^2+(1)^2}$
$\qquad=\sqrt{4+1+1}$
$\qquad=\sqrt{6}$
Therefore $|CA|^2=6.$
Step 5:
Hence $41=35+6$
$\Rightarrow |AB|^2+|CA|^2=|BC|^2$
Therefore ABC is a right angled triangle.