7. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $P$ is the midpoint of $D D_{1}$, $O$, $M$, and $N$ are the centers of the faces $A_{1} B_{1} C_{1} D_{1}$, $B B_{1} C_{1} C$, and $A B C D$ respectively. Find the angle formed by the skew lines $P N$ and $O M$.
7. As shown in Figure $20-19$, take $D$ as the origin, $D A, D C, D D_{1}$ as the $x, y, z$ axes to establish a rectangular coordinate system, and take the edge length of the cube as 2, then $P(0,0,1), N(1,1,0), M(1,2,1), o(1,1,2)$.
Since $\overrightarrow{P N} \cdot \overrightarrow{O M}=(1,1,-1) \cdot(0,1,-1)=2$,
$|\overrightarrow{P N}| \cdot|\overrightarrow{O M}|=\sqrt{3} \sqrt{2}=\sqrt{6}$,
thus the angle $\theta$ formed by $P N$ and $O M$ satisfies $\cos \theta=\frac{|\overrightarrow{P N}| \cdot|\overrightarrow{O M}|}{|\overrightarrow{P N}||\overrightarrow{O M}|}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}$,
$$
\theta=\arccos \frac{\sqrt{6}}{3} \text {. }
$$