Question

高三数学 正确答案 立体几何

Answer 1

Answer 2

几何法(1)由长方体性质得A1D1∥BC,∴A1D1和O1B所成角为∠O1BC
取底面中心O,连接OO1,则OO1=AA1=3,OO1⊥面ABCD
易证OB=OC=√2,勾股定理得O1B=O1C=√11

余弦定理得cos∠O1BC=√11/11,∴异面直线所成角为arccos(√11/11)
(2)连接O1D,作DE⊥O1B于E
∵O1B在面ABCD的射影是OB,E∈O1B,∴E在面ABCD的射影在OB上
∴DE在面ABCD的射影是直线BD
∵AC⊥BD,∴AC⊥DE
∵A1C1∥AC,∴DE⊥A1C1
∵A1C1∩O1B=O1,∴DE⊥面A1BC1,即DE是D到面A1BC1的距离
面积法得DE=OO1*BD/O1B=6√2/√11=6√22/11

向量法(1)以A为原点,AB,AD,AA1为轴建系,则
A1(0,0,3),D1(0,2,3),O1(1,1,3),B(2,0,0)
A1D1→=(0,2,0),O1B→=(1,-1,-3)

设异面直线所成角为Θ,则cosΘ=|cos<A1D1,O1B>|=|-2|/[√(0+4+0)*√(1+1+9)=√11/11
∴Θ=arccos(√11/11)
(2)D(0,2,0),A1B→=(2,0,-3),A1C1→=(2,2,0)
设面A1BC1法向量为n→=(x,y,2),则
2x-6=0,x=3.2x+2y=0,y=-3,∴n→=(3,-3,2)
DB→=(2,-2,0),设D到面A1BC1距离为d,则d=|DB→·n→|/|n→|=|6+6+0|/√(9+9+4)=6√22/11

Answer 3