14. (1994 Japan 4th Mathematical Olympiad) Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ where the angle between the planes $A B_{1} D_{1}$ and $A_{1} B D$ is $\theta\left(0^{\circ} \leqslant \theta \leqslant 90^{\circ}\right)$, find the value of $\cos \theta$.
14. As shown in the figure, establish a spatial rectangular coordinate system $A-xyz$, with the edge length of the cube being 1. It is easy to get $\overrightarrow{A_{1} B}-(1,0,-1), \overrightarrow{A_{1} D}=(0,1,-1), \overrightarrow{A B_{1}}=(1,0,1), \overrightarrow{A D_{1}}=(0,1, 1)$. Let $\vec{m}=\left(x_{1}, y_{1}, z_{1}\right)$ and $\vec{n}=\left(x_{2}, y_{2}, z_{2}\right)$ be the normal vectors of the planes $A B_{1} D_{1}$ and $A_{1} B D$, respectively. From $\left\{\begin{array}{l}\vec{m} \cdot \overrightarrow{A_{1} B}=0 \\ \vec{m} \cdot \overrightarrow{A_{1} D}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}-z_{1}=0, \\ y_{1}-z_{1}=0 .\end{array}\right.\right.$
Let $z_{1}=1$, we get $\vec{m}=(1,1,1)$,
From $\left\{\begin{array}{l}\vec{n} \cdot \overrightarrow{A B_{1}}=0 \\ \vec{n} \cdot \overrightarrow{A D}_{1}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}+z_{2}=0, \\ y_{2}+z_{2}=0 .\end{array}\right.\right.$
Let $z_{2}=-1$, we get $\vec{n}=(1,1,-1)$,
Therefore, $\cos (\vec{m}, \vec{n})=|\vec{m} \cdot \vec{n}| /|\vec{m}||\vec{n}|=1 / 3$,
Thus, $\cos \theta=1 / 3$.