5. In the spatial quadrilateral $ABCD$, $AB=2, BC=$ $3, CD=4, DA=5$. Then $\overrightarrow{AC} \cdot \overrightarrow{BD}=$ $\qquad$
$$
\begin{array}{l}
\text { Then } \overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D} \\
\Rightarrow \overrightarrow{A D}^{2}=(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D})^{2} \\
\Rightarrow \overrightarrow{A D}^{2}=\overrightarrow{A B}^{2}+\overrightarrow{B C}^{2}+\overrightarrow{C D}^{2}+2 \overrightarrow{A B} \cdot \overrightarrow{B C}+ \\
2 \overrightarrow{A B} \cdot \overrightarrow{C D}+2 \overrightarrow{B C} \cdot \overrightarrow{C D} \\
\Rightarrow 25=4+9+16+ \\
2(\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}+\overrightarrow{B C} \cdot \overrightarrow{C D}) \\
\Rightarrow \overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}+\overrightarrow{B C} \cdot \overrightarrow{C D}=-2 \text {. } \\
\text { Therefore } \overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C}) \cdot(\overrightarrow{B C}+\overrightarrow{C D}) \\
=\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}+\overrightarrow{B C} \cdot \overrightarrow{C D}+\overrightarrow{B C} \cdot \overrightarrow{B C} \\
=-2+9=7 \text {. } \\
\end{array}
$$