5.5. The vector $\overline{O A}$ forms angles with the axes $O x, O y, O z$ that are respectively equal to $\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{\pi}{4}$; the point $B$ has coordinates $(-2; -2; -2 \sqrt{2})$. Find the angle between the vectors $\overline{O A}$ and $\overline{O B}$.
5.5. Let $\bar{n}$ be a unit vector in the direction of $\overline{O A};$ then $\quad \bar{n}\left(\cos \frac{\pi}{3} ; \cos \frac{\pi}{3} ; \cos \frac{\pi}{4}\right)$, or $\quad \bar{n}\left(\frac{1}{2} ; \frac{1}{2} ; \frac{\sqrt{2}}{2}\right)$, and $\overline{O B}(-2 ;-2 ;-2 \sqrt{2})$. Denoting the angle between $\overline{O A}$ and $\overline{O B}$ by $\varphi$, we find

$$
\cos \varphi=\frac{\bar{n} \cdot \overline{O B}}{|\bar{n}||\overline{O B}|}=\frac{\frac{1}{2}(-2)+\frac{1}{2}(-2)+\frac{\sqrt{2}}{2}(-2 \sqrt{2})}{1 \cdot 4}=-1
$$

from which we obtain the answer: $\varphi=180^{\circ}$.

Answer: $180^{\circ}$.