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

1 Answer
Feb 3, 2016

\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)