It is known that the three edges of the parallelepiped abcd-a1b1c1d1 with the same vertex a as the end point are all equal to 1 and the angle between them is 60 degrees, and the length of the ball AC1

It is known that the three edges of the parallelepiped abcd-a1b1c1d1 with the same vertex a as the end point are all equal to 1 and the angle between them is 60 degrees, and the length of the ball AC1

In Δ ABC, ab = BC = 1, ∠ ABC = 120 & ordm;, get 3 under AC = root. In Δ acc1, AC = 3 under root, CC1 = 1, the rest is to find ∠ acc1. ∠ acc1 = 180 & ordm; - caa1

In the parallel hexahedron abcd-a1b1c1d1, the lengths of three edges with vertex a as the end point are all 1 and the included angles are all 60 ° to find the distance between AD1 and BC1
The answer is root 6 / 3.1, how to do it with geometric method; 2, if you use vector method, the origin should be selected in the center of the bottom surface, then how to find the coordinates of A1, B1 and other points

As shown in the figure, make be ⊥ ad, connect a1e and A1B, make BF ⊥ a1e and eg ⊥ A1B
∵∠A1AE=∠BAE=60°、A1A=AB
∴⊿A1AE≌⊿BAE
A1e ⊥ ad
Ad ‖ BC
So plane a1be ⊥ plane AD1, BC1
The BF length is the distance between AD1 and BC1
In ⊿ a1be, a1e = be = √ 3 / 2, A1B = 1
In isosceles ⊿ a1be, eg = √ 2 / 2
∴S⊿A1BE=（1/2）A1E×BF=（1/2）EG×A1B
So BF = √ 6 / 3
In the vector method, it is suggested to take a as the origin and plane AC as the plane of XY coordinate system,
Connect AC to be and a1h
It is easy to know from the above solution that a1h ⊥ plane AC, and a1h = BF = √ 6 / 3,
In the case of RT ⊿ a1ah, ah = √ 3 / 3
∵∠ DAC = ∠ BAC, so ∠ HAB = 30 °
The H coordinate of the point is (1 / 2, √ 3 / 6)
The coordinates of point A1 are (1 / 2, √ 3 / 6, √ 6 / 3)
The coordinates of A1, B1, C1 and D1 need only add the corresponding values of A1 on the basis of a, B, C and D coordinates