How do you normalize <2,0,-1>?

1 Answer
Sep 9, 2017

Answer:

#hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>#

Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

#hatu=vecv/(|vecv|)#

Given #vecv=<2,0,-1>#, we can calculate the magnitude of the vector:

#abs(vecv)=sqrt((v_x)^2+(v_y)^2+(v_z)^2)#

#=>=sqrt((2)^2+(0)^2+(-1)^2)#

#=>=sqrt(4+0+1)#

#=>=sqrt(5)#

We now have:

#hatu=(<2,0,-1>)/sqrt(5)#

#=>hatu=<2/sqrt(5),0,-1/sqrt(5)>#

We can also rationalize the denominator on the #hatx# #(hati)# and #hatz# #(hatk)# component:

#=>hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>#

Hope that helps!