2 Points $A, B, C, D$ in space satisfy $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=7$, $|\overrightarrow{C D}|=11,|\overrightarrow{D A}|=9$, then the value of $\overrightarrow{A C} \cdot \overrightarrow{B D}$ is ( ).
(A) only one
(B) two
(C) four
(D) infinitely many
Solve: Since $\overrightarrow{A B}^{2}+\overrightarrow{C D}^{2}=3^{2}+11^{2}=130=7^{2}+9^{2}=\overrightarrow{B C}^{2}+\overrightarrow{D A}^{2}$, from $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{0}$, we get $\overrightarrow{A B}+\overrightarrow{C D}=-(\overrightarrow{B C}+\overrightarrow{D A})$, squaring both sides gives $\overrightarrow{A B} \cdot \overrightarrow{C D}=\overrightarrow{B C} \cdot \overrightarrow{D A}$, hence $\overrightarrow{A B} \cdot \overrightarrow{C D}=-\overrightarrow{A D} \cdot \overrightarrow{B C}$, thus
$$
\begin{aligned}
\overrightarrow{A C} \cdot \overrightarrow{B D} & =(\overrightarrow{A B}+\overrightarrow{B C})(\overrightarrow{B C}+\overrightarrow{C D}) \\
& =(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}) \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D} \\
& =\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}=0 .
\end{aligned}
$$
$\overrightarrow{A C} \cdot \overrightarrow{B D}$ has only one value 0, so the answer is $\mathrm{A}$.