# How do you normalize  (2i -3j + 4k)?

Dec 26, 2015

If you have a vector $\vec{a}$ with coordinates (x,y,z) then the normalized vector $\vec{n}$ is

$\vec{n} = \frac{\vec{a}}{\left\mid a \right\mid} = \frac{\left\langlex \text{,"y",} z\right\rangle}{\left\mid a \right\mid}$

where $\left\mid a \right\mid = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

Hence for $x = 2 , y = - 3 , z = 4$ we have that

$\left\mid a \right\mid = \sqrt{{2}^{2} + {3}^{2} + {4}^{2}} = \sqrt{4 + 9 + 16} = \sqrt{29}$

Thus the normalized vector is

$\vec{n} = \textcolor{b l u e}{\left\langle\frac{2}{\sqrt{29}} , - \frac{3}{\sqrt{29}} , \frac{4}{\sqrt{29}}\right\rangle}$