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

Jan 23, 2016

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

$\left(- \frac{4}{5} i + \frac{0}{5} j + \frac{3}{5} k\right)$

#### 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:

$l = \sqrt{{\left(- 4\right)}^{2} + {0}^{2} + {3}^{2}} = \sqrt{25} = 5$ units

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

$\left(- \frac{4}{5} i + \frac{0}{5} j + \frac{3}{5} k\right)$