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

1 Answer
Konstantinos Michailidis · Truong-Son N.
Dec 26, 2015

If you have a vector #veca# with coordinates (x,y,z) then the normalized vector #vecn# is

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

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

Hence for #x=2,y=-3,z=4# we have that

#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 >>)#

Related questions
See all questions in Introduction to Vectors
Impact of this question
2007 views around the world
You can reuse this answer
Creative Commons License