Answer
Direction cosines are: $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$
Direction angles are: $48 ^\circ,71 ^\circ, 48 ^\circ$
Work Step by Step
Let $v= \lt 2,1,2 \gt$
$|v|=\sqrt {2^2+1^2+2^2}=3$
Direction cosines are: $cos \alpha = \frac{2}{3}, cos \beta =\frac{1}{3}, cos \gamma=\frac{2}{3}$
Thus, the direction angles are:
$ \alpha =cos^{-1} \frac{2}{3}=48 ^\circ, \beta = cos^{-1} \frac{1}{3}=71 ^\circ, \gamma = cos^{-1} \frac{2}{3}=48^ \circ$
