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

1 Answer
Mar 5, 2017

Answer:

#u=<-2/sqrt(17),-3/sqrt(17),2/sqrt(17)>#

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=<-2,-3,2>#, we can calculate the magnitude of the vector:

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

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

#|v|=sqrt(4+9+4)#

#|v|=sqrt(17)#

We now have:

#u=(<-2,-3,2>)/sqrt(17)#

#=>u=<-2/sqrt(17),-3/sqrt(17),2/sqrt(17)>#

Hope that helps!