Example 11 In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the degree of the dihedral angle $A-B D_{1}-A_{1}$ is $\qquad$ .
(2004 National High School League Question)
Let the cube be a unit cube, and establish a spatial rectangular coordinate system as shown in Figure 20-10. Then, $D(0,0,0), A(1,0,0), D_{1}(0,0,1), B(1,1,0), A_{1}(1,0,1)$.

Thus, $\overrightarrow{A D_{1}}=(-1,0,1), \overrightarrow{A B}=(0,1,0), \overrightarrow{A_{1} B}=(0,1,-1), \overrightarrow{A_{1} D_{1}}=(-1,0,0)$.

Let the normal vectors of plane $A B D_{1}$ and plane $A_{1} B D_{1}$ be $\boldsymbol{n}_{1}=\left(x_{1}, y_{1}, z_{1}\right), \boldsymbol{n}_{2}=\left(x_{2}, y_{2}, z_{2}\right)$. From $\boldsymbol{n}_{1} \perp \overrightarrow{A B}, \boldsymbol{n}_{1} \perp \overrightarrow{A D_{1}}, \boldsymbol{n}_{2} \perp \overrightarrow{A_{1} B}, \boldsymbol{n}_{2} \perp \overrightarrow{A_{1} D_{1}}$, we get
$$
y_{1}=0,-x_{1}+z_{1}=0, y_{2}-z_{2}=0,-x_{2}=0 \text {. }
$$

Let $x_{1}=a$, then $z_{1}=a$; let $y_{2}=b$, then $z_{2}=b$.
Thus, $\boldsymbol{n}_{1}=(a, 0, a), \boldsymbol{n}_{2}=(0, b, b)$, and let the angle between $\boldsymbol{n}_{1}$ and $\boldsymbol{n}_{2}$ be $\alpha$.
Therefore, $\cos \alpha=\frac{\boldsymbol{n}_{1} \cdot \boldsymbol{n}_{2}}{\left|\boldsymbol{n}_{1}\right| \cdot\left|\boldsymbol{n}_{2}\right|}=\frac{(a, 0, a) \cdot(0, b, b)}{\sqrt{2} a \cdot \sqrt{2} b}=\frac{1}{2}$.
So, $\alpha=60^{\circ}$. It is also easy to see that the dihedral angle $A-B D_{1}-A_{1}$ has a plane angle $\theta=\alpha$.
Therefore, the dihedral angle $A-B D_{1}-A_{1}$ is $60^{\circ}$.