(1) Verification: Plane AEC⊥ Plane PDB (Normal Vector Method)
(2) When PD=√ (root number) 2AB and E is the midpoint of PB, find the angle formed by AE and the plane PDB (set vector).
(1) Analysis: ∵ The bottom of the quadrilateral P-ABCD is a square, and the bottom of PD⊥ ABCD.
Establish a spatial rectangular coordinate system D-xyz with D as the origin, DC as the X axis, DA as the Y axis and DP as the positive direction of the Z axis.
Let AB= 1.
Then point coordinates:
D(0,0,0),A(0, 1,0),B( 1, 1,0),C( 1,0,0)
P(0,0,z 1),E(x,y,z)
The vector PD = (0,0, -z 1) and the vector PB=( 1, 1, -z 1).
Let the vector m be the normal vector of the surface PDB:
Vector m= vector PD× vector PB=(z 1, -z 1, 0)
The vector EA=(-x, 1-y, -z) and the vector EC=( 1-x, -y, -z).
Let vector n be the normal vector of surface EAC:
Vector n= vector EA× vector EC=(-z, -z, x-y- 1)
Vector m* vector n=-zz 1+zz 1+0=0.
∴ Vector m⊥ Vector n, ∴ Plane AEC⊥ Plane PDB
(2) Analysis: ∫PD =√2, and e is the midpoint of PB.
Then point coordinates:
D(0,0,0),A(0, 1,0),B( 1, 1,0),C( 1,0,0)
P(0,0,√2),E( 1/2, 1/2,√2/2)
Vector EA = (- 1/2, 1/2, -√2/2)= > | vector EA|= 1
Vector m=(√2,-√ 2,0) = = > Vector m|=2
Vector EA* vector m=-√2
Cos< vector EA, vector m>= (vector EA* vector m)/(| vector EA|*| vector m|)=-√2/2.
∴AE makes an angle of 45 with the plane PDB.