2. Four points in space $A, B, C, D$ 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 ( ).
(National High School Competition 2005)
A. Only one
B. Two
C. Four
D. Infinitely many
2. Choose A. Reason: 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 $\overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C})(\overrightarrow{B C}+\overrightarrow{C D})=(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}) \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}=0$.
Therefore, $\overrightarrow{A C} \cdot \overrightarrow{B D}$ has only one value 0.