Example 2. In the cube $A C_{1}$, the edge length is $a$, find:
(1) The distance from point $A_{1}$ to the plane $B C_{1} D$; (2) The distance between the line $A B$ and $A_{1} C$; (3) The distance between the plane $A B_{1} D_{1}$ and the plane $B C_{1} D$.

保留源文本的换行和格式，直接输出翻译结果。
"Body only law". It is easy to prove that $A_{1} C \perp$ plane $B C_{1} D$, let the foot of the perpendicular be $S, A_{1} S$ is the desired length. $A_{1} C=\sqrt{3} a$, $C S=\frac{1}{\sqrt{3}} a$, so $A_{1} S=A_{1} C-C S$ $=\sqrt{3} a-\frac{1}{\sqrt{3}} a=\frac{2 \sqrt{3}}{3} a$.
(2) (Figure 3) This is to find the distance between skew lines. Draw $B_{1} C$, then $A B / /$ plane $A_{1} B_{1} C, A B$ to the distance of plane $A_{1} B_{1} C$ is the distance between $A B$ and $A_{1} C$, which can be converted to finding the distance from point $B$ to plane $A_{1} B_{1} C$. Connect $B C_{1}$, then $B C_{1} \perp C B_{1}$, and $2 C B_{1} \perp$ plane $A_{1} B_{1} C$, so $B O_{1} \perp$ plane $A_{1} B_{1} C, B O_{1}$ is
(3) (Figure 4) It is easy to prove that plane $A B_{1} D_{1} / /$ plane $D B C_{1}$, from (1) we know that $A_{1} C$ is perpendicular to plane $A B_{1} D_{1}$ and plane $D B C_{1}$, let the feet of the perpendiculars be $R$ and $S$ respectively, we can find that $A_{1} R=C S=\frac{\sqrt{3}}{3} a$, so $R S=\frac{\sqrt{3}}{3} a$, which is the desired length.

In a cube, many problems of finding angles and distances can be selected. This is not further elaborated here.