# What is the norm of < -3, -1 , -3 >?

The norm is otherwise described as the normalized version of a vector: a unit vector in the same direction as the original vector. In this case: $< - \frac{3}{\sqrt{19}} , - \frac{1}{\sqrt{19}} , - \frac{3}{\sqrt{19}} >$.
We need to find the length of the initial vector. For a vector $< a , b , c >$
$l = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {\left(- 1\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{9 + 1 + 9} = \sqrt{18}$
Then we simply divide each of the 3 elements by this length to yield a unit vector: $< - \frac{3}{\sqrt{19}} , - \frac{1}{\sqrt{19}} , - \frac{3}{\sqrt{19}} >$.