How do you normalize # (-5i + 12j+ 2 k)#?

1 Answer
Jan 24, 2016

Answer:

Normalizing a vector means making a unit vector in the same direction by dividing each component of the vector by the length of the original vector. In this case the normalized vector is # (−5/sqrt(173)i+12/sqrt(173)j+2/sqrt(173)k)#

Explanation:

First find the length of the vector. If a vector is #(ai+bk+cj)# its length is given by:

#l=sqrt(a^2+b^2+c^2#

The length of the given vector, # (−5i+12j+2k)#, is:

#l=sqrt((-5)^2+12^2+2^2) =sqrt(173) =13.2#

We can express the normalized vector in two equivalent ways:

# (−5/sqrt(173)i+12/sqrt(173)j+2/sqrt(173)k)#

OR

# (−5/13.2i+12/13.2j+2/13.2k)#