Example 7 In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, find the size of the acute dihedral angle formed by the planes $A B_{1} D_{1}$ and $A_{1} B D$.
As shown in Figure 15-11, let the edge length of the cube be 1.
Connect $A_{1} B$ intersecting $A B_{1}$ at point $E$, and connect $A_{1} D$ intersecting $A D_{1}$ at point $F$,
It is easy to know that $A F=A E=E F=\frac{\sqrt{2}}{2}$, thus $\angle A E F=60^{\circ}$
In the equilateral triangle $A_{1} B D$, $\angle D E F=30^{\circ}, D E=\frac{\sqrt{3}}{2} A_{1} B=\frac{\sqrt{6}}{2}$
Therefore, $\cos \angle A E D=\frac{A E^{2}+D E^{2}-A D^{2}}{2 A E \cdot D E}=\frac{\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\frac{\sqrt{6}}{2}\right)^{2}-1}{2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{6}}{2}}=\frac{\sqrt{3}}{3}$
Let the acute dihedral angle formed by plane $A B_{1} D_{1}$ and plane $A_{1} B D$ be $\theta$, then by the first cosine theorem of trihedral angles, we have
$$
\cos \theta=\frac{\cos \angle A E D-\cos \angle A E F \cos \angle D E F}{\sin \angle A E F \sin \angle D E F}=\frac{\frac{\sqrt{3}}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}}=\frac{1}{3}
$$

Thus, the acute dihedral angle formed by plane $\Lambda B_{1} D_{1}$ and plane $A_{1} B D$ is $\arccos \frac{1}{3}$.