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!
