How do you normalize # (-4i + 0j + 3k)#?

1 Answer
Jan 23, 2016

Answer:

To normalise a vector we divide each component by the length of the original vector, so the resultant vector is:

#(-4/5i+0/5j+3/5k)#

Explanation:

Normalising a vector means dividing each of its components by its length, to yield a unit vector (a vector 1 unit long) in the same direction.

The length of a 3D vector (a#i#+b#j#+c#k#) is given by:

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

In this case:

#l=sqrt((-4)^2+0^2+3^2) = sqrt(25) = 5# units

To normalise the vector, then, we divide each component by 5, so the resultant vector is:

#(-4/5i+0/5j+3/5k)#