# How do you normalize # (2i -3j + 4k)#?

##### 1 Answer

Dec 26, 2015

If you have a vector

#vecn = (veca)/(absa) = (<< x","y","z >>)/absa# where

#absa=sqrt(x^2+y^2+z^2)#

Hence for

#absa=sqrt(2^2+3^2+4^2)=sqrt(4+9+16)=sqrt29#

Thus the normalized vector is

#vecn = color(blue)(<< 2/sqrt29,-3/sqrt29,4/sqrt29 >>)#