How do you normalize #(- 3 i + 12j -5k)#?

1 Answer
Mar 5, 2017

Answer:

#u=<-3/sqrt(178),12/sqrt(178),-5/sqrt(178)>#

Explanation:

In normalizing the vector we are finding a unit vector (magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

#u=v/(|v|)#

Given #v=<-3,12,-5>#, we can calculate the magnitude of the vector:

#|v|=sqrt((v_x)^2+(v_y)^2+(v_z)^2)#

#|v|=sqrt((-3)^2+(12)^2+(-5)^2)#

#|v|=sqrt(9+144+25)#

#|v|=sqrt(178)#

We now have:

#u=(<-3,12,-5>)/sqrt(178)#

#=>u=<-3/sqrt(178),12/sqrt(178),-5/sqrt(178)>#

Hope that helps!