如图在直三棱柱ABC=A1B1C1中,AC=AB=4,B1B=4根号2，BM=根号2，∠BAC=90,点N在CA

Question

上，且CN=1/3NA1 1求证：MN//平面A1B1C1
2求点A1到平面AMC的距离
3求二面角C-A1M-B的正切值 两种解答方法

Answer 1

几何法:
(1)作ND⊥A1C1於D,由直三棱柱的性质得ND∥B1M∥CC1
∴ND/CC1=NA1/A1C=3/4,ND=3√2=B1M
连接B1D,则四边形MNDB1是平行四边形,∴MN∥B1D
∵B1D包含於面A1B1C1,∴MN∥面A1B1C1
(2)∵AA1⊥AC,AB⊥AC,∴AC⊥面AA1B1B
S△A1AM=1/2*AA1*AB=1/2*4√2*4=8√2
∵AM包含於面AA1B1B,∴AC⊥AM
勾股定理得AM=3√2,S△ACM=1/2*AC*AM=6√2
设A1到面ACM距离为h,体积法得h=S△A1AM*AC/S△ACM=16/3
(3)勾股定理得A1M=CM=√34,A1C=4√3,馀弦定理得cosCA1M=6/√102
∴sinCA1M=√11/√17
同理,cosBA1M=10/√102,sinBA1M=1/√51,cosCA1B=2/3
设二面角C-A1M-B为θ,则cosCA1B=cosBA1McosCA1M+sinBA1MsinCA1Mcosθ
解得cosθ=4/√33
∴tanθ=√17/4
向量法:
(1)以A为原点,AC,AB,AA1为轴建系,由题意得M(0,4,√2),N(3,0,√2)
MN→=(3,-4,0)
易证AA1→=(0,0,4√2)是面A1B1C1的法向量
∵MN→·AA1→=0,∴MN→⊥AA1→,∴MN∥面A1B1C1
(2)AM→=(0,4,√2),AC→=(4,0,0)
设面ACM法向量为p→=(a,b,1),则
4b+√2=0,4a=0,∴p→=(0,-√2/4,1)
设A1到面ACM距离为d,则d=|AA1→·p→|/|p→|=|0+0+4√2|/√(0+1+1/8)=16/3
(3)易证AC→=(4,0,0)是面A1BM的法向量
A1M→=(0,4,-3√2),A1C→=(4,0,-4√2)
设面A1CM法向量为n→=(x,y,1),则
4y-3√2=0,4x-4√2=0,∴n→=(√2,3√2/4,1)
cos<n→,AC→>=4√2/[4*√(2+9/8+1)]=4/√33
由图像得二面角C-A1M-B的馀弦为4/√33
∴正切为√17/4

Answer 2

给我看看你的 逼

Answer 3

第二题16/3