Answer
Direction cosines are: $\frac{1}{\sqrt {14}},\frac{-2}{\sqrt {14}}, \frac{-3}{\sqrt {14}}$
Direction angles are: $74 ^\circ, 122 ^\circ, 143 ^\circ$
Work Step by Step
Let $v= i-2j-3k=\lt 1,-2,-3 \gt$
$|v|=\sqrt {1^2+(-2)^2+(-3)^2}=\sqrt {14}$
Direction cosines are: $cos \alpha = \frac{1}{\sqrt {14}}, cos \beta =\frac{-2}{\sqrt {14}}, cos \gamma=\frac{-3}{\sqrt {14}}$
Thus, the direction angles are:
$ \alpha =cos^{-1} \frac{1}{\sqrt {14}}=74 ^\circ, \beta = cos^{-1} \frac{-2}{\sqrt {14}}=122 ^\circ, \gamma = cos^{-1} \frac{-3}{\sqrt {14}}=143^ \circ$
