# What is the norm of < 5 , -2, 3 >?

$\setminus \sqrt{38}$
Just using pithagoras theorem you know $h = \setminus \sqrt{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2} + \ldots}$ as you are in three dimensional vectorial space, you stop at the third component.
so $| V | = \setminus \sqrt{{5}^{2} + {\left(- 2\right)}^{2} + {3}^{2}} = \setminus \sqrt{38}$