Given the edge length of cube abcd-a'b'c'd 'as a, find the angle a'B and b'c, a'B vertical AC'

Given the edge length of cube abcd-a'b'c'd 'as a, find the angle a'B and b'c, a'B vertical AC'

1. Link a'd
Because a'B '/ / CD and a'B' = CD
The quadrilateral a'b'cd is a parallelogram
So: a'd / / b'c
So, ba'd is the angle between a'B and b'c
Since a'B = a'd = BD, the triangle a'bd is an equilateral triangle
Then: ∠ Ba'd = 60 degree
The angle between a'B and b'c is 60 degrees
2. Proof: link ab '
In the square abb'a ', it is easy to know: a'B ⊥ ab'
B'c '⊥ plane abb'a' and a'B in plane abb'a '
So: b'c '⊥ a'B
That is to say, a'B is perpendicular to the two intersecting lines ab'b'c'in the plane ab'c '
Therefore, from the judgment theorem of perpendicularity of lines and planes, we can get the following results:
A'B ⊥ plane ab'c '
And AC 'is in the plane ab'c'
A'B⊥AC

It is known that the edge length of the cube ABCD-A "'b'c'd 'is m, so we can find: (1) the angle between a'B and b'c; (2) prove a'B ⊥ AC'

Connect d'c and d'b‘
Then: D 'C / / a' B and d 'C = a' B
And d'c, b'c, d'b 'are the diagonals of this square
The equilateral triangle, i.e. d'cb '= 60 degree
That is: the angle between a'B and b'c = 60 degrees
(2) Proof: connect DC 'and ab‘
∵ C 'B' ⊥ surface ab '∵ C' B '⊥ a' B
A'B and ab 'are the two diagonals of a square abb'a'; ⊥ a'B ⊥ ab‘
A'B ⊥ plane ab'c'd
∴A’B⊥AC‘