# How do you find the direction cosines of the vector u=2i -4j+4k and find the direction angles?

Sep 2, 2016

$\alpha = \arccos \left(\frac{1}{3}\right) = 1.231$
$\beta = \arccos \left(- \frac{2}{3}\right) = 2.3$
$\gamma = \arccos \left(\frac{2}{3}\right) = 0.841$

#### Explanation:

Given the unit base vectors $\hat{i} , \hat{j} , \hat{k}$ we have

$\left\langle\vec{u} , \hat{i}\right\rangle = \left\mid u \right\mid \cos \left(\alpha\right) = 2 \to \cos \left(\alpha\right) = \frac{2}{\sqrt{{2}^{2} + {4}^{2} + {4}^{2}}} = \frac{1}{3}$

$\left\langle\vec{u} , \hat{j}\right\rangle = \left\mid u \right\mid \cos \left(\beta\right) = - 4 \to \cos \left(\beta\right) = - \frac{4}{\sqrt{{2}^{2} + {4}^{2} + {4}^{2}}} = - \frac{2}{3}$

$\left\langle\vec{u} , \hat{k}\right\rangle = \left\mid u \right\mid \cos \left(\gamma\right) = 4 \to \cos \left(\gamma\right) = \frac{4}{\sqrt{{2}^{2} + {4}^{2} + {4}^{2}}} = \frac{2}{3}$

with

$\alpha = \arccos \left(\frac{1}{3}\right) = 1.231$
$\beta = \arccos \left(- \frac{2}{3}\right) = 2.3$
$\gamma = \arccos \left(\frac{2}{3}\right) = 0.841$