How do you normalize # <3,-6,4>#?

1 Answer
Mar 4, 2017

Answer:

#u=<3/sqrt(61),-6/sqrt(61),4/sqrt(61)>#

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,-6,4>#, we can calculate the magnitude of the vector:

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

#|v|=sqrt((3)^2+(-6)^2+(4)^2)#

#|v|=sqrt(9+36+16)#

#|v|=sqrt(61)#

We now have:

#u=(<3,-6,4>)/sqrt(61)#

#=>u=<3/sqrt(61),-6/sqrt(61),4/sqrt(61)>#

Hope that helps!