4. Given that line segment $A D / /$ plane $\alpha$, and the distance to plane $\alpha$ is 8, point $B$ is a moving point on plane $\alpha$, and satisfies $A B=10$. If $A D=21$, then the minimum distance between point $D$ and point $B$ is $\qquad$ .
4.17.

As shown in Figure 4, let the projections of points $A$ and $D$ on plane $\alpha$ be $O$ and $C$, respectively, then we have $A O = C D = 8$.
$$
\begin{array}{l}
\because A B = 10, \\
\therefore O B = 6 \text{ (constant). }
\end{array}
$$

Therefore, the trajectory of point $B$ in $\alpha$ is a circle with $O$ as the center and 6 as the radius.
$$
\begin{array}{l}
\because C D \perp \alpha, \text{ then } \\
B D^{2} = B C^{2} + C D^{2} \\
= B C^{2} + 8^{2},
\end{array}
$$
$\therefore$ The minimum value of $B D$ is to find the minimum value of $B C$.
Since point $C$ is outside the circle $\odot O$, connect $O C$ intersecting $\odot O$ at $B_{0}$. By plane geometry knowledge, we easily get
$$
\begin{array}{l}
(B C)_{\min} = B_{0} C = C O - O B_{0} = 21 - 6 = 15. \\
\text{ Hence } (B D)_{\min} = \sqrt{(B C)_{\min}^{2} + 8^{2}} = 17.
\end{array}
$$