2.6. Calculate the direction cosines of the vector $\overrightarrow{A B}$, if $A(-3,2,0)$ and $B(3,-3,1)$.
The solution is as follows. The direction cosines of vector $\overrightarrow{A B}$ are determined by its coordinates using formulas (2.28). We find the coordinates of vector $\overrightarrow{A B}$ in the same way as it was done in problem (2.5). We have $\bar{A} B=\{3+3,-3+2,1-0\}=\{6,-1,1\}$. The modulus of vector $\overrightarrow{A B}$ is found using the formula

$$
\overrightarrow{A B}=\sqrt{36+1+1}=\sqrt{38}
$$

We now have, according to formulas (2.28):

$$
\begin{aligned}
& \cos \alpha=\frac{6}{\sqrt{38}} \\
& \cos \beta=\frac{-1}{\sqrt{38}} \\
& \cos \gamma=\frac{1}{\sqrt{38}}
\end{aligned}
$$