Question

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

Answer 1

`\sqrt 38`

Explanation:

Just using pithagoras theorem you know `h=\sqrt(x_1^2+x_2^2+x_3^2+x_4^2+...)` as you are in three dimensional vectorial space, you stop at the third component.

so `|V|=\sqrt(5^2+(-2)^2+3^2)=\sqrt(38)`