$63 M$, $N$ are the centers of the faces $ABCD$, $BCC_{1}B_{1}$ of the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, respectively. Then the angle formed by the lines $D_{1}M$ and $A_{1}N$ is $\qquad$.
$6360^{\circ}$. As shown in Figure 8, let the midpoints of $A_{1} D_{1}$, $C C_{1}$, and $A B$ be $E$, $F$, and $G$ respectively, and the edge length of the cube be 2. Then $E F=A_{1} N=\sqrt{2^{2}+(\sqrt{2})^{2}}=\sqrt{6}, E G=D_{1} M=\sqrt{6}$, and $E F$ and $E G$ are parallel to $A_{1} N$ and $D_{1} M$ respectively.
Also, $G F=\sqrt{F C^{2}+C B^{2}+B G^{2}}=\sqrt{1+2^{2}+1}=\sqrt{6}$, so the angle formed by $A_{1} N$ and $D_{1} M$ $=\angle F E G=60^{\circ}$.