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

Here, $\overrightarrow {AB} = \vec b - \vec a$
$= (2\hat i - \hat j + \hat k) - (3\hat i - 4\hat j - 4\hat k) = - \hat i + 3\hat j + 5\hat k$

$\overrightarrow {BC} = \vec c - \vec b = (\hat i - 3\hat j - 5\hat k) - (2\hat i - \hat j + \hat k) = - \hat i - 2\hat j - 6\hat k$

$\overrightarrow {CA} = \vec a - \vec c = (3\hat i - 4\hat j - 4\hat k) - (\hat i - 3\hat j - 5\hat k) = 2\hat i - \hat j + \hat k$

$|\overrightarrow {AB} {|^2} = {( - 1)^2} + {3^2} + {5^2} = 1 + 9 + 25 = 35$
$|\overrightarrow {BC} {|^2} = {( - 1)^2} + {( - 2)^2} + {( - 6)^2} = 1 + 4 + 36 = 41$

and $|\overrightarrow {CA} {|^2} = {2^2} + {( - 1)^2} + {1^2} = 4 + 1 + 1 = 6$

So, $|\overrightarrow {BC} {|^2} = |\overrightarrow {AB} {|^2} + |\overrightarrow {CA} {|^2}$

Hence, the triangle is a right angled triangle