But there is something wrong with this process:

1. Establish the system first.

Take D as the coordinate origin, and the directions of vector DA, vector DC and vector DD 1 are the positive directions of X-axis, Y-axis and Z-axis, respectively, to establish a spatial rectangular coordinate system (as shown in the figure).

2. Write 4a+5c=0 4b-5c=0. Write a=5 here, then b = -5 and c = -4, so the vector n 1=(5, -5, -4).

3. By observing the direction of the normal vector, judge whether the angle formed by the normal vector and the plane angle of the dihedral angle are equal or complementary.

Finally, the dihedral angle A 1-BD-A is acute, so the cosine of dihedral angle A 1-BD-A is two 66/33, so θ = arccos (two 66)/33.

PS: This problem is simple by direct method. Taking the midpoint m of BD, it is easy to prove that ∠AMA 1 is the plane angle of dihedral angle a1-BD-a.