If $\alpha ,\beta$ and $\gamma$ be the angles made by the vector $3\overline{i}-6\overline{j}+2\overline{k}$ with the positive directions of the coordinate axes, then find $\cos \alpha ,\cos \beta$ and $\cos \gamma .$ 

Given vector is $3\overline{i}-6\overline{j}+2\overline{k}$

$\therefore \cos \alpha =\frac{3}{\sqrt{(3{)}^2+(-6{)}^2+(2{)}^2}}$

$=\frac{3}{\sqrt{9+36+4}}=\frac{3}{\sqrt{49}}\boxed{\therefore \cos \alpha =\frac{3}{7}}$

$\cos \beta =\frac{-6}{\sqrt{(3{)}^2+(-6{)}^2+(2{)}^2}}$

$=\frac{-6}{\sqrt{9+36+4}}=\frac{-6}{\sqrt{49}}=-\frac{6}{7}\boxed{\therefore \cos \beta =-\frac{6}{7}}$

$\cos \gamma =\frac{2}{\sqrt{(3{)}^2+(-6{)}^2+(2{)}^2}}=\frac{2}{\sqrt{9+36+4}}=\frac{2}{\sqrt{49}}=\frac{2}{7}$

$\boxed{\therefore \cos \gamma =\frac{2}{7}}$