In the cube abcd-a1b1c1d1 with edge length of 2, e is the midpoint of edge AB, and point P is in plane a1b1c1d1. If d1p ⊥ plane PCE, try to find the direction of line d1p My answer is 4 root 5 / 5 is correct?

Correct. If we make the vertical line of CE through D, we can prove that the vertical line is equal to the line segment d1p. By using the similarity or sine cosine, we can get that d1p is 4 root sign 5 / 5

Make a straight line through the midpoint of any two edges of parallel hexahedron abcd-a1b1c1d1, where the line parallel to plane dbb1d1 has ()
A. 4 B. 6 C. 8 d. 12

As shown in the figure, make a straight line through the midpoint of any two edges of parallel hexahedron abcd-a1b1c1d1. There are 12 lines parallel to plane dbb1d1, so D

Make a straight line through the midpoint of any two edges of the parallelepiped abcd-a1b1c1d1, in which there are several lines parallel to the plane dbb1d1
Why 12
Draw a picture
Brother Wu asks for advice

Draw your own picture. The plane of the midpoint of CD, BC, c1d1 and b1c1 is parallel to dbb1d1, so all the lines on the plane are parallel to dbb1d1. If the midpoint of the edge has four points on the plane, it can form six lines (upper 4, lower 2) = C. similarly, there is such a plane on the other side of dbb1d1, so there are 12 lines in total

It is known that the parallelepiped abcd-a'b'c'd ', AA' ⊥ plane ABCD, ab = 4, ad = 2, b'd ⊥ BC, the angle between the straight line b'd and the plane ABCD is 30 degrees,
Find the angle between db 'and CD' on the different plane
Find the volume of the parallelepiped
Don't use the system building method

First question:
Extend AB to e so that be = AB and connect B ′ E and de. it is obvious that AE = 2Ab = 8. Then take the midpoint of B ′ D as f
∵ ABCD-A ′ B ′ C ′ D ′ is a parallelepiped, ∵ ad ∥ BC, AA ∥ BB ′
From ad ∥ BC and B ′ D ⊥ BC, B ′ D ⊥ ad
From the AA ′⊥ plane ABCD and AA ′∥ BB ′, it is concluded that BB ′⊥ plane ABCD, ∥ BB ′⊥ BD, ∥ B ′ DB = 30 degrees,
∴B′D＝2BD、BB′＝√3BD.
According to Pythagorean theorem, there are: ab ′ ^ 2 = ad ^ 2 + B ′ ^ D ^ 2, and ab ′ ^ 2 = AB ^ 2 + BB ′ ^ 2,
∴AD^2＋B′D^2＝AB^2＋BB′^2,∴4＋4BD^2＝16＋3BD^2,∴BD^2＝12,∴BD＝2√3.
From ad = 2, ab = 4, BD = 2 √ 3, ad ^ 2 + BD ^ 2 = AB ^ 2,
Based on the inverse theorem of Pythagorean theorem, we get that ad ⊥ BD, cos ∠ bad = ad / AB = 2 / 4 = 1 / 2
According to the cosine theorem, de ^ 2 = ad ^ 2 + AE ^ 2-2ad × aecos ∠ bad = 4 + 64-2 × 2 × 8 × (1 / 2) = 52,
∴DE＝2√13.
∵B′D＝2BD、BB′＝√3BD,BD＝2√3,∴B′D＝4√3,BB′＝6,∴AA′＝6.
It is obvious that a ′ B ′ = AB = be, a ′ B ′‖ AE, a ′ B ′ EB are parallelograms,
∴B′E＝A′B＝√（AA′^2＋AB^2）＝√（36＋16）＝2√13.
From de = 2 √ 13, B ′ e = 2 √ 13, de = B ′ e, ｜ EF ⊥ B ′ F
∴cos∠DB′E＝B′F/B′E＝（B′D/2）/B′E＝2√3/（2√13）＝√39/13,
∴∠DB′E＝arccos（√39/13）.
∵ ABCD-A ′ B ′ C ′ D ′ is a parallelepiped, ∵ CD ′‖ a ′ B
∵ a ′ B ′ EB is a parallelogram, ∵ a ′ B ‖ B ′ E
From CD ′‖ a ′ B and a ′ B ‖ B ′ e, the angle between CD ′‖ B ′ E and CD ′ is obtained,
The angle between B ′ D and CD ′ is arccos (√ 39 / 13)
Second question:
Obviously, the area of △ abd = (1 / 2) ad × BD = (1 / 2) × 2 × 2 √ 3 = 2 √ 3,
The area of ABCD = 2 △ the area of abd = 4 √ 3
The volume of ABCD-A ′ B ′ C ′ D ′ = the area of ABCD × BB ′ = 4 √ 3 × 6 = 24 √ 3

In the cube abcd-a1b1c1d1 with edge length of 2, e is the midpoint of edge AB, and point P is in plane a1b1c1d1. If d1p ⊥ plane PCE, try to find the direction of line d1p My answer is 4 root 5 / 5 is correct?