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

1 Answer
Dec 24, 2016

Answer:

#hatu = < -sqrt6/6, -sqrt6/6, sqrt6/3 >#. See below.

Explanation:

To normalize a vector is to find a unit vector in the same direction as the given vector, where a unit vector, #hat u#, is a vector with a magnitude of #1#. In other words, the normalized vector #hat u# of a non-zero vector u is a unit vector in the direction of u.

We can find a unit vector in the same direction as the given vector by dividing the given vector by its magnitude.

#hat u = u /| u|#

Where the magnitude of a vector is given by:

#|u|=sqrt((u_x)^2+(u_y)^2+(u_z)^2)#

Given the vector, u #=< -1,-1,2 >#,

#|u|=sqrt((-1)^2+(-1)^2+(2)^2)#

#|u|=sqrt(1+1+4)#

#|u|=sqrt6#

We then have

#hat u= (< -1,-1,2 >) /sqrt6#

#=>hat u = < -1/sqrt6,-1/sqrt6,2/sqrt6 >#

Or equivalently, by rationalizing the denominator:

#=> hatu = < -sqrt6/6, -sqrt6/6, sqrt6/3 >#