3. In rhombus $A B C D$, $\angle D A B=60^{\circ}$, fold $\triangle A B D$ along $B D$ to get $\triangle A_{1} B D$. If the dihedral angle $A_{1}-B D-C$ is $60^{\circ}$, then the cosine value of the angle formed by the skew lines $D A_{1}$ and $B C$ is $\qquad$ .
$E$ is the midpoint of $B D$, and $A_{1} F \perp$ plane $B C D$. Establish a rectangular coordinate system as shown in the figure. Let $A B=4$, then $A_{1} E=C E=2 \sqrt{3}, \angle A_{1} E C=60^{\circ}$ $\Rightarrow \triangle A_{1} E C$ is an equilateral triangle,
thus $E F=\sqrt{3}, A_{1} F=3 \Rightarrow A_{1}(0,0,3)$, $D(\sqrt{3},-2,0), B(\sqrt{3}, 2,0), C(-\sqrt{3}, 0,0)$ $\Rightarrow \overrightarrow{D A_{1}}=(-\sqrt{3}, 2,3), \overrightarrow{B C}=(-2 \sqrt{3},-2,0)$. Therefore, $\cos \left\langle\overrightarrow{D A_{1}}, \overrightarrow{B C}\right\rangle=\frac{6-4}{\sqrt{16} \cdot \sqrt{16}}=\frac{1}{8}$.