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

1 Answer
Jan 9, 2018

Answer:

#hatu=<2/sqrt(14),-3/sqrt(14),-1/sqrt(14)>#

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,-3,-1>#, we can calculate the magnitude of the vector:

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

#=>=sqrt((2)^2+(-3)^2+(-1)^2)#

#=>=sqrt(4+9+1)#

#=>=sqrt(14)#

We now have:

#hatu=(<2,-3,-1>)/sqrt(14)#

#=>hatu=<2/sqrt(14),-3/sqrt(14),-1/sqrt(14)>#

We can also rationalize the denominator for each of the components:

#=>hatu=<(sqrt(14))/7,-(3sqrt(14))/14,-(sqrt(14))/14>#

Which may also be written #((sqrt(14))/7i-(3sqrt(14))/14j-(sqrt(14))/14k)#

Hope that helps!