How do you normalize # (15i- 3j + 12k) #?

1 Answer
David G.
Feb 4, 2016

Normalizing a vector involves dividing each of its elements by its length. In this case the normalized vector is #(15/sqrt378i-3/sqrt378j+12/sqrt378k)# or #(15/19.4i-3/19.4j+12/19.4k)#.

Explanation:

Normalizing a vector is creating a vector of length #1# unit in the same direction as the original vector.

To do that, we divide each of the elements by the length of the vector. The find the length of a vector #(ai+bj+ck)# the formula is:

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

In this case:

#l = sqrt(15^2+(-3)^2+12^2)=sqrt(225+9+144) = sqrt378# #(=19.4)#

The final vector can be expressed as #(15/sqrt378i-3/sqrt378j+12/sqrt378k)# or #(15/19.4i-3/19.4j+12/19.4k)#.

Related questions
See all questions in Vector Operations
Impact of this question
1178 views around the world
You can reuse this answer
Creative Commons License