高考数学二轮复习 专题四 立体几何第3讲 立体几何中的向量方法配套课件

故).0 , 0 , 6(),22 , 2 , 2(FDAF, 0 AF, 0FD图图2 2|nnPMd,),0 , 4 , 2(),0 , 4 , 2(NCDENCDEEFFB1).0 , 2 , 2(),2, 2, 2(AFEF),4, 2 , 2(1FB,1EFFBAFFB1,1AFFBOSOCOB,. 0),26, 0 ,22(),0 , 0 ,22, 0(),0 ,22, 0(SDOCaaSDaOCaC则DS./.,1:2:.310).26),1 (22,22(,).26,22, 0(),26, 0 ,22(PACBEPACBEDSBEECSEtDSBEattaaCStBCCEBCBECStCEaaCSaaDS平面故内,不在平面而时即当而则设且PBCDCDPB,cosCDPB,PDAPAC,00ACmAPm则|sinmPDmPD),21, 1, 1 ( CM).0 ,21,21(SN),0 , 1 ,21(NC(2)(2)解解., 002121SNCMSNCM所以因为设a=(x,y,z)为平面CMN的一个法向量,则, 0, 0NCaCMa,22223211|,SNaSNaSNa.,. 0202, 0242).1 , 0 , 1 (),1 ,2, 1(),2,22 , 2(EFPCBFPCEFPCBFPCEFBFPCPC2,22AD22图图2 2).2, 1 , 0(),0 , 2 , 1(kBEBD,0022BEnBDnNEAM,101022521|,cosAMNEAMNEAMNEANANASEAASEAES, 0, 0ANESAMES21AS| AS).2, 0 , 2(),1, 1 , 1 (),0 , 1 , 0(PBFGEFFGyEFxPB,FGyEFxPBFG、EF、PB).0 , 2 , 2(),1, 2 , 1 (BDEG|cosBDEGBDEG. 0, 0EQnEFn|nnEACQ,/ ACAB.MByMAxMP ABBC解析解析.221052122|,cos),0 , 3 , 1(),0 , 4, 2(BCBABCBABCBABCBABCBA,DCBAEBFBFAE|,cosBFAEBEAEBFAEAEBF| , 1 AEBFAE|,cosBFAEBFAEBFAE答案答案 DMN11DC., 01111DCMNDCMN答案答案 B11DC2BA1AC|,cos111ACBAACBAACBACD1CB, 0, 01CBnCDn.54|,cos|sinnDAnDAnDADA3333AGAE, 023, 0zyxAEmzxAGmDG.111BAPAPNAMAMPN |,cos|nPNnPNnPN),21, 1 , 0(MMP, 0, 0MPmNPm1AA 返回返回