5. In trapezoid $A B C D$, $\overrightarrow{A B}=2 \overrightarrow{D C},|\overrightarrow{B C}|=6, P$ is a point on the plane of trapezoid $A B C D$, and satisfies $\overrightarrow{A P}+\overrightarrow{B P}+4 \overrightarrow{D P}=0, \overrightarrow{D A} \cdot \overrightarrow{C B}=|\overrightarrow{D A}| \cdot|\overrightarrow{D P}|, Q$ is a moving point on side $A D$, then the minimum value of $|\overrightarrow{P Q}|$ is $\qquad$.
5. $\frac{4 \sqrt{2}}{3}$.

As shown in Figure 4, take the midpoint $M$ of $A B$. From $\overrightarrow{A P}+\overrightarrow{B P}+4 \overrightarrow{D P}=0$, we get $\overrightarrow{P M}=2 \overrightarrow{D P}$, so $P$ is the point on segment $D M$ closer to point $D$ and divides it into a ratio of 1:2. According to the problem, $\overrightarrow{D A} \cdot \overrightarrow{C B}=\overrightarrow{D A} \cdot \overrightarrow{D M}=|\overrightarrow{D A}| \cdot|\overrightarrow{D M}| \cos \angle A D M=$ $|\overrightarrow{D A}| \cdot|\overrightarrow{D P}|$, so $\cos \angle A D M=\frac{1}{3}$, then $\sin \angle A D M=\frac{2 \sqrt{2}}{3}$, thus the minimum value of $|\overrightarrow{P Q}|$ is $2 \sin \angle A D M=\frac{4 \sqrt{2}}{3}$.