3.( 12 points) Solution 1: (1) Connect BD, take DC midpoint G, and connect BG.
So DG = GC = BG = 1, that is, △DBC is a right triangle, so BC⊥BD.
And SD⊥ plane ABCD, so BC⊥SD,
So BC⊥ aircraft BDS, BC⊥DE.
For BK⊥EC, k is vertical feet. Because the plane EDC⊥, the plane SBC, the plane EDC, BK⊥DE.DE are perpendicular to the two intersecting lines BK and BC in the plane SBC.
De⊥sb. de⊥ecde⊥ Aircraft Company
SB= =,
DE= =,
EB= =,SE=SB-EB=,
So se = 2eb.
(2) from SA = =, AB = 1, SE = 2eb, AB⊥SA, known AE = = 1, AD = 1
So △ADE is an isosceles triangle.
Take the midpoint f of ED, connect AF, and then AF⊥DE, af = =.
If FG is connected, then fg∨EC, FG⊥DE.
So ∠AFG is the plane angle of dihedral angle A-DE-C. 。
Add AG, ag =, fg = =,
cos∠AFG= =-。
So the dihedral angle A-DE-C is 120.
Solution 2: Take D as the coordinate origin and ray DA as the positive semi-axis of the X axis, and establish the rectangular coordinate system Dxyz as shown in the figure.
Let A( 1, 0,0), then B( 1, 1, 0), C (0,2,0), S (0 0,0,2).
( 1) =(0,2,-2),
=(- 1, 1,0).
Let the normal vector of plane SBC be
n=(a,b,c),
Take n from n⊥n⊥? =0,n? =0.
So 2b-2c = 0, -a+b = 0.
Let A = 1, then B = 1, C = 1, n = (1, 1).
Let = λ (λ > 0),
Then e (,,).
=( , ,), =(0,2,0).
Let the normal vector m = (x, y, z) of the plane CDE,
From m⊥, m⊥, get
m? =0,m? =0.
Therefore ++= 0, 2y = 0.
Let x = 2, then m = (2 2,0,-λ).
M⊥n,m? n=0,2-λ=0,λ=2。
So se = 2eb.
(2) We know E (,,) from (1), and if the midpoint of DE is F, then F (,) = (,-,-),
So what? = 0, from which FA⊥DE is obtained.
= (-,-) Here we go again, so? = 0, thus obtaining EC⊥DE,
The angle between vector and is equal to the plane angle of dihedral angle ADEC.
So cos = =-,
So the dihedral angle A-DE-C is 120.