In the cube abcd-a1b1c1d1, e, F and G are the midpoint of Aa1, CD and d1d respectively

In the cube abcd-a1b1c1d1, e, F and G are the midpoint of Aa1, CD and d1d respectively

Question: plane a1bc is parallel to plane EFG answer: connecting CD1, there is CD1 parallel to A1B, because FG is the central defensive line of triangle, so FG / / A1B, eg / / BC, A1B intersects BC with B, eg intersects FG with G, eg belongs to plane EFG, FG belongs to EFG, A1B belongs to plane a1bc, BC belongs to plane a1bc, so plane EFG / / plane a1bc

The cube abcd-a1b1c1d1 takes D as the origin, Da DC dd1 as the X, y, Z axis to establish the space rectangular coordinate system, point P on the cube diagonal BD1 and point Q on CD
1. When point P is the key point of AB, Q moves on edge CD to seek the minimum value of PQ
2. When point P moves on AB and point Q moves on CD, the minimum value of PQ is found
I sincerely look forward to your reply

1, let the side length of the cube be 1, then the coordinates of each point are (d (0,0,0), a (1,0,0), B (1,1,0), C (0,1,0), D1 (0,0,1), A1 (1,0,1), B1 (1,1,1), C1 (0,1,1), respectively. Point P is the center of gravity of AB, then the coordinates are p (1 / 2,1 / 2,1 / 2). Let Q (0, y, 0), 0 ≤ y ≤ 1, then PQ & sup2; = (1 / 2-0) &