# Show that the points A, B and C with position vectors, $$\overrightarrow a = 3\hat i − 4\hat j − 4\hat k, \overrightarrow b = 2\hat i − \hat j + \hat k$$ and $$\overrightarrow c = \hat i − 3\hat j − 5\hat k$$ , respectively form the vertices of a right angled triangle.

Toolbox:
• In a $\Delta$ABC, if ${AB}^2+{BC}^2={AC}^2$,then it obeys Pythagoras rule and hence it is a right angled triangle.
Step 1:
Let $\overrightarrow a=3\hat i-4\hat j-4\hat k,\overrightarrow{b}=2\hat i-\hat j+\hat k$ and $\overrightarrow c=\hat i-3\hat j-5\hat k$
Let us check if the three sides of the triangle obeys Pythagoras rule.
To check this,let us find the magnitudes of the three vectors.
Step 2:
$\mid\overrightarrow{a}\mid=\sqrt{3^2+(-4)^2+(-4)^2}$
$\qquad=\sqrt{9+16+16}$
$\qquad=\sqrt{41}$
$\mid\overrightarrow{b}\mid=\sqrt{2^2+(-1)^2+(1)^2}$
$\qquad=\sqrt{4+1+1}$
$\qquad=\sqrt{6}$
$\mid\overrightarrow{c}\mid=\sqrt{1^2+(-3)^2+(5)^2}$
$\qquad=\sqrt{1+9+25}$
$\qquad=\sqrt{35}$
Step 3:
It is clear that $b^2+c^2=a^2$
(i.e)$\qquad\quad(\sqrt 6)^2+(\sqrt{35})^2=(\sqrt{41})^2$
$\Rightarrow 6+35=41.$
Hence the vectors form the sides of a right angled triangle.