The bottom P-ABCD of the quadrangular pyramid is a square, and PD⊥ the bottom ABCD and E are on the side PB (let the space vector be set).

(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.