(1)证明:以D为原点,DA、DC、DD 1所在直线分别为x、y、z轴建立空间直角坐标系D-xyz
则D(0,0,0),A(2,0,0),B(2,2,0),C(0,2,0),A 1(2,0,4),B 1(2,2,4),C 1(0,2,4),D 1(0,0,4),E(0,2,1),
∴
BE =(-2,0,1).
∵
A 1 C =(-2,2,-4),
DB =(2,2,0),
∴
A 1 C •
BE =4+0-4=0且
A 1 C •
DB =-4+4+0=0,
∴
A 1 C ⊥
DB 且
A 1 C ⊥
BE ,
∵DB∩BE=B
∴A 1C⊥平面BDE;
(2)由(1)知
A 1 C =(-2,2,-4)是平面BDE的一个法向量,
∵
A 1 B =(0,2,-4),
∴cos<
A 1 C ,
A 1 B >=
A 1 C •
A 1 B
|
A 1 C ||
A 1 B | =
30
6 ,
∴A 1B与平面BDE所成角的正弦值为
30
6 .