What is the norm of #< -3, -1 , 8 >#?

1 Answer
Jan 18, 2016

Answer:

#sqrt74#

Explanation:

For any vector #A=(a_1,a_2,....,a_n)# in any finite n-dimensional vector space, the norm is defined as follows :

#||A||=sqrt(a_1^2+a_2^2+....+a_n^2)#.

So in this particular case we work in #RR^3# and get :

#||((-3,-1,8))||=sqrt(3^2+1^2+8^2)=sqrt74#.