In the cube ABCD ABCD with edge length 1, m and N are the midpoint of the line segments BB and BC respectively, and the distance between the line Mn and ACD is calculated

3√2 / 4
BAC is parallel to ACD. The distance is √ 2 / 2
The distance from Mn to BAC is √ 2 / 4

In the cube abcd-a'b'c'd 'with edge length a, M is the midpoint of BB'. The cross-sectional area of a.m.d 'is obtained by calculating the plane section cube of a.m.d'

In fact, the cross section is the area of a.m.d '
Because am is perpendicular to ad '
therefore
Ad = (radical 2) a
Am = (radical 5) a / 2
In the right triangle m.a.d ', area = 1 / 2 * ad * am = (radical 10) a ^ 2 / 4

The cube abcd-a'b'c'd'has an edge length of 1, e is the midpoint of BB 'and F is the midpoint of b'c'
(1) To prove d'f ‖ plane a'de (2) to find the cosine value of dihedral angle a-de-a '

(1) Take the midpoint g of BC to connect eg and dg. because eg / / a1d, A1, D, G and E are coplanar
So, DG is in plane a1de. And d1f is not in plane a1de, and d1f / / DG. So, d1f / / plane a1de
(2) Join AE. Triangle a1de congruent triangle ade (three sides equal)
Make a1h vertical De, perpendicular foot h, connect ah, then ah vertical de. so angle a1ha is the plane angle of dihedral angle a-de-a1
In triangle ade, ad = 1, de = 3 / 2, AE = √ 5 / 2
AD^2-DH^2=AE^2-(3/2-DH)^2,DH=2/3.
AH^2=AD^2-DH^2=1-4/9=5/9,A1H=AH=√5/3.
In the triangle aha1, cosaha1 = (ah ^ 2 + a1h ^ 2-aa1 ^ 2) / (2ah * a1h) = 1 / 10
Therefore, the cosine value of dihedral angle a-de-a1 is 1 / 10

In the cube abcd-a1b1c1d1 with edge length a, M is the midpoint of AB, then the distance from point C to plane a1dm is (3 / 3 root 6a)

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In the cube ABCD ABCD with edge length 1, m and N are the midpoint of the line segments BB and BC respectively, and the distance between the line Mn and ACD is calculated