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

1 Answer
Jan 9, 2018

hatu=<2/sqrt(14),-3/sqrt(14),-1/sqrt(14)>

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 vecv=<2,-3,-1>, we can calculate the magnitude of the vector:

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)

=>hatu=<2/sqrt(14),-3/sqrt(14),-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 ((sqrt(14))/7i-(3sqrt(14))/14j-(sqrt(14))/14k)

Hope that helps!