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

##### 1 Answer

Jan 9, 2018

#### 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

#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)#

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

Hope that helps!