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

Feb 4, 2016

Normalizing a vector involves dividing each of its elements by its length. In this case the normalized vector is $\left(\frac{15}{\sqrt{378}} i - \frac{3}{\sqrt{378}} j + \frac{12}{\sqrt{378}} k\right)$ or $\left(\frac{15}{19.4} i - \frac{3}{19.4} j + \frac{12}{19.4} k\right)$.

#### 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 $\left(a i + b j + c k\right)$ the formula is:

$l = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$

In this case:

$l = \sqrt{{15}^{2} + {\left(- 3\right)}^{2} + {12}^{2}} = \sqrt{225 + 9 + 144} = \sqrt{378}$ $\left(= 19.4\right)$

The final vector can be expressed as $\left(\frac{15}{\sqrt{378}} i - \frac{3}{\sqrt{378}} j + \frac{12}{\sqrt{378}} k\right)$ or $\left(\frac{15}{19.4} i - \frac{3}{19.4} j + \frac{12}{19.4} k\right)$.