As shown in the figure, in the cube ABCD – a1b1c1d1, verify that plane aca1c1 is perpendicular to plane a1bd

As shown in the figure, in the cube ABCD – a1b1c1d1, verify that plane aca1c1 is perpendicular to plane a1bd

The intersection of plane aca1c1 and a1bd is a1o
A1d = A1B = BD do = Bo so a1o vertical BD
So plane aca1c1 is perpendicular to plane a1bd

If the edge length of cube AC1 is 1, the perpendicular line of plane a1bd is made through point a, and the perpendicular foot is point h, then the angle between line ah and BB1 is?
Why?

Δ a1bd is an equilateral triangle
Let the midpoint of BD be o, connected with a1o, then ah is perpendicular to a1o and H,
The angle between ah and BB1 is equal to the angle between ah and Aa1 ∠ a1ah. (A1A and B1B are parallel)
∠A1AH = ∠A1OA
A1A = 1 Ao = 2 / 2 root sign 2
Tan ∠ a1oa = Tan ∠ a1ah = 1 / (2 / 2 root sign 2)
So the angle between ah and BB1 satisfies Tan ∠ a1ah = root 2. The size is arctan √ 2