Example 8 As shown in Figure 20-8, in the parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$, the base $A B C D$ is a square with side length 1, the side edge $A A_{1}=2$, and the angle between $A A_{1}$ and $A B, A D$ is $120^{\circ}$.
(1) Find the length of the line segment $A C_{1}$;
(2) If $K$ is the midpoint of $B C_{1}$, find the length of $A C$;
(3) Find the size of the angle formed by the skew lines $B D_{1}$ and $A C$;
(4) Prove: $A A_{1} \perp B D$.
(1) Since $\overrightarrow{A C_{1}}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C C_{1}}$, then
$$
\begin{aligned}
\left|\overrightarrow{A C_{1}}\right|^{2} & =\left(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C C_{1}}\right)^{2} \\
& =|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}+\left|\overrightarrow{C C_{1}}\right|^{2}+2 \overrightarrow{A B} \cdot \overrightarrow{B C}+2 \overrightarrow{A B} \cdot \overrightarrow{C C_{1}}+2 \overrightarrow{B C} \cdot \overrightarrow{C C}_{1} \\
& =1+1+4+4+2 \cdot \cos 120^{\circ}+4 \cdot \cos 120^{\circ}=2 .
\end{aligned}
$$

Thus, $A C_{1}=\sqrt{2}$.
(2) Since $\overrightarrow{A K}=\frac{1}{2}\left(\overrightarrow{A C_{1}}+\overrightarrow{A B}\right)=\frac{1}{2}\left(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A A_{1}}+\overrightarrow{A B}\right)=\frac{1}{2}\left(2 \overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A A_{1}}\right)$, then $|\overrightarrow{A K}|^{2}=\frac{1}{4}\left(2 \overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A A_{1}}\right)^{2}=\frac{3}{4}$, hence $A K=\frac{\sqrt{3}}{2}$.
(3) Since $\overrightarrow{B D_{1}}=\overrightarrow{B A}+\overrightarrow{A D}+\overrightarrow{D D_{1}}$, then $\left|\overrightarrow{B D_{1}}\right|^{2}=\left(\overrightarrow{B A}+\overrightarrow{A D}+\overrightarrow{D D_{1}}\right)^{2}=6$, i.e., $\left|\overrightarrow{B D_{1}}\right|=\sqrt{6}$. Let the angle between $B D_{1}$ and $A C$ be $\theta$, then
$$
\begin{aligned}
\cos \theta & =\left|\frac{\overrightarrow{B D_{1}} \cdot \overrightarrow{A C}}{\left|\overrightarrow{B D_{1}}\right| \cdot|\overrightarrow{A C}|}\right|=\left|\frac{\left(\overrightarrow{B D}+\overrightarrow{D D_{1}}\right) \cdot \overrightarrow{A C}}{\left|\overrightarrow{B D_{1}}\right| \cdot|\overrightarrow{A C}|}\right| \\
& =\frac{\left|\overrightarrow{D D_{1}} \cdot \overrightarrow{A C}\right|}{\sqrt{2}}=\frac{\left|\overrightarrow{D D_{1}} \cdot(\overrightarrow{A B}+\overrightarrow{B C})\right|}{\sqrt{12}}=\frac{\left|2 \cdot \cos 120^{\circ}+2 \cdot \cos 120^{\circ}\right|}{\sqrt{12}}=\frac{\sqrt{3}}{3},
\end{aligned}
$$

Thus, $\theta=\arccos \frac{\sqrt{3}}{3}$.
(4) Since $\overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B}$, then $\overrightarrow{A A_{1}} \cdot \overrightarrow{B D}=\overrightarrow{A A_{1}} \cdot(\overrightarrow{A D}-\overrightarrow{A B})=2 \cdot \cos 120^{\circ}-2 \cdot \cos 120^{\circ}=0$, hence $A A_{1} \perp A D$.