In the cube abcd-a1b1c1d1, the cosine of the dihedral angle formed by plane a1bd and plane c1bd is () A. 12B. 13C. 32D. 33

In the cube abcd-a1b1c1d1, the cosine of the dihedral angle formed by plane a1bd and plane c1bd is () A. 12B. 13C. 32D. 33

Let o, join, ob, OA1, a1c1, ∵ in BD be in the cube abcd-a1b1c1d1, let the edge length be 1, ∵ a1c1 = 2, OB = OA1 = 62. According to the geometric properties of the cube, it is concluded that BD ⊥ OA, BD ⊥ OC, BD ⊥ Aa1, BD ⊥ CC1, ≁ BD ⊥ plane oaa1, BD ⊥ plane occ1, OA1 ⊂ plane oaa1, OC1 ⊂ plane occ1, ∵ BD ⊥ OA1, BD ⊥ OC1, ∵ a1oc1 are the dihedral angle between plane a1bd and plane c1bd In △ a1oc1, cos ∠ a1oc1 = 32 + 32 − 22 × 62 × 62 = 13, so B is selected

In cube AC1 with edge length 1, the cosine of the angle between plane c1bd and plane cb1d1 is 0___ ．

Let the intersection of the diagonals C1d and CD1 of square cdd1c1 be m, and the intersection of the diagonals B1C and C1b of square cbb1c1 be n, then the intersection of plane bdc1 and plane b1d1c is Mn, and the edge length of cube AC1 is 1, then the diagonal of square C1d = 2, C1M = 22, c1n = 22, Mn is the median line of triangle c1db, Mn = BD2 = 22, triangle c1mn is an equilateral triangle, and the triangle CMN is also an equilateral triangle Point E, connecting CE and C1E, then CE ⊥ Mn, C1E ⊥ Mn, so ∠ c1ec is the plane angle of the dihedral angle formed by plane c1bd and plane cb1d1, C1E = CE = 32Mn = 64, in triangle c1ec, CC1 = 1, according to the cosine theorem, CC12 = c1e2 + ce2-2 · CE · c1ecos ⊥ c1ec, ｜ cos ⊥ c1ec = - 13, then the cosine value of the angle formed by plane c1bd and plane cb1d1 is - 13

As shown in the figure, in the cube abcd-a1b1c1d1, e and F are the midpoint of edge AB and CC1 respectively, and the straight line ()
A. There is no B. there is one C. There are two d. There are countless

Let's know that plane add1a1 and plane d1ef have a common point D1. From the axiom of the basic properties of plane, we know that there must be a common line L passing through this point. There are countless lines parallel to L in plane add1a1, and they are not in plane d1ef. From the judgment theorem of line plane parallel, we know that they are parallel to plane d1ef, so choose: D

As shown in the figure, in the cube abcd-a1b1c1d1, there are two points E and F on the side diagonal Ab1 and BC1 respectively, and b1e = C1F

Proof 1: make em ⊥ AB at point m through E and F, FN ⊥ BC at point n, connect Mn. ⊥ BB1 ⊥ plane ABCD, ⊥ BB1 ⊥ AB, BB1 ⊥ BC. ⊥ EM ∥ BB1, FN ⊥ EM ∥ FN. Then b1e = C1F, ⊥ EM = FN. So the quadrilateral MnFe is parallelogram. ⊥ EF ∥ Mn. And Mn is in plane ABCD, ⊥ EF ∥ plane ABCD. Proof 2: make eg ∥ AB through e, intersect BB1 at point G, connect GF, then b1eb1a = b1gb1 B. ∩ eg ∩ FG = g, ab ∩ BC = B, ∩ plane EFG ∩ plane ABCD. In plane EFG, EF ∩ plane ABCD