最新高中数学解题思路大全:空间向量与立体几何解答题精选优秀名师资料.doc
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1、2013年高中数学解题思路大全:空间向量与立体几何解答题精选(数学选修2-1)第三章 空间向量与立体几何解答题精选 1 已知四棱锥的底面为直角梯形,PABCD,ABDC/1,底面,且,ABCDPAADDC,,DAB,90,PA,2,是的中点 AB,1MPB(?)证明:面面; PCDPAD,(?)求与所成的角; ACPB(?)求面与面所成二面角的大小 AMCBMC证明:以为坐标原点长为单位长度,如图建立空间直角坐标系,则各点坐标为 AAD1 ABCDPM(0,0,0),(0,2,0),(1,1,0),(1,0,0),(0,0,1),(0,1,)2(?)证明:因 AP,(0,0,1),DC,(0,
2、1,0),故AP,DC,0,所以AP,DC.由题设知,且与是平面内的两条相交直线,由此得面ADDC,DC,APADPAD又在面上,故面?面 DCPCDPCDPADPAD(?)解:因 AC,(1,1,0),PB,(0,2,1),故|AC|,2,|PB|,5,AC,PB,2,所以AC,PB10cos,AC,PB,.5|AC|,|PB|MC(?)解:在上取一点,则存在使 NC,MC,Nxyz(,),R,11 NC,(1,x,1,y,z),MC,(1,0,),?x,1,y,1,z,.22,14要使,只需即解得 , ANMCANMCxz,00,.25412,可知当,时,N点坐标为(,1,),能使AN,M
3、C,0.555 1212此时,AN,(,1,),BN,(,1,),有BN,MC,05555由AN,MC,0,BN,MC,0得AN,MC,BN,MC.所以,ANB为 所求二面角的平面角 ,30304? |,|,.ANBNANBN,555,ANBN 2,?,cos(,).ANBN 3|ANBN,2故所求的二面角为arccos().,3高考学习网,中国最大高考学习网站G | 我们负责传递知识 2 如图,在四棱锥中,底面是正方形,侧面是正三角形, VABCD,ABCDVAD平面底面 VAD,ABCD(?)证明:平面; VADAB,(?)求面与面所成的二面角的大小 VADDB证明:以为坐标原点,建立如图
4、所示的坐标图系 D(?)证明:不防设作, A(1,0,0)13则, , V(,0,)B(1,1,0)2213 AB,(0,1,0),VA,(,0,)22由得,又,因而与平面内两条相交直线ABVA,VADAB,VA,0,ABAD,AB,都垂直 ?平面 VAVADADAB,13 (?)解:设为DV中点,则, E(,0,)E44333313 EA,(,0,),EB,(,1,),DV,(,0,).444422由 EB,DV,0,得EB,DV,又EA,DV.因此,是所求二面角的平面角, ,AEBEA,EB21 cos(EA,EB),7|EA|,|EB|21解得所求二面角的大小为 arccos.7PABC
5、D,ABCD3 如图,在四棱锥中,底面为矩形, VAB,3ABCDBC,1侧棱底面, PA,PA,2CD为的中点 EPDAC (?)求直线与所成角的余弦值; PBABNNE,PAC(?)在侧面内找一点,使面, PAB N并求出点到和的距离 ABAP解:(?)建立如图所示的空间直角坐标系, 则的坐标为、 ABCDPE,A(0,0,0)、 B(3,0,0)C(3,1,0)D(0,1,0)高考学习网,中国最大高考学习网站G | 我们负责传递知识 1、, E(0,1)P(0,0,2)2从而 AC,(3,1,0),PB,(3,0,2).设的夹角为,则 AC与PB,AC,PB337cos, 1427|AC
6、|,|PB|37?与所成角的余弦值为 ACPB14(?)由于点在侧面内,故可设点坐标为,则 NNPAB(,0,)xz1,由面可得, NE,PACNE,(,x,1,z)2,1,3z,1,0,(,x,1,z),(0,0,2),0,x,NE,AP,0,2 ? ,6即化简得,11,3x,,0.,NE,AC,0.,z,1,(,x,1,z),(3,1,0),0.,2,2,33NN即点的坐标为,从而点到和的距离分别为 (,0,1)ABAP1,66ABCDAECF4 如图所示的多面体是由底面为的长方体被截面所截面而得到的,其中 1ABBCCCBE,4,2,3,1 1(?)求的长; BFCAECF (?)求点到
7、平面的距离 1解:(I)建立如图所示的空间直角坐标系,则, D(0,0,0)B(2,4,0)ACEC(2,0,0),(0,4,0),(2,4,1),(0,4,3)设 Fz(0,0,)1AECF?为平行四边形, 1高考学习网,中国最大高考学习网站G | 我们负责传递知识 ?由AECF为平行四边形,1?由AF,EC得,(,2,0,z),(,2,0,2),1?z,2.?F(0,0,2).?EF,(,2,4,2).于是|BF|,26,即BF的长为26.(II)设为平面的法向量, AECFn11显然n不垂直于平面ADF,故可设n,(x,y,1)11,n,AE,0,0,x,4,y,1,0,1由得 ,xy,
8、2,0,2,0,n,AF,0,1,x,1,4y,1,0, 即?,1,2x,2,0,y,.,4,的夹角为,则 又CC,(0,0,3),设CC与n111CC,n343311cos,. 331|CC|,|n|113,1,116CAECF?到平面的距离为 1433433d,|CC|cos,3,,. 13311ABCDABCD,ADAAAB,1,25 如图,在长方体,中,点在棱上移EAD11111DEAD,动 (1)证明:; 11ACD (2)当为的中点时,求点到面的距离; EABE1,DECD, (3)等于何值时,二面角的大小为 AE14高考学习网,中国最大高考学习网站G | 我们负责传递知识 解:以
9、为坐标原点,直线分别为轴,建立空间直角坐标系,DADCDD,Dxyz,1设,则 AEx,ADExAC(1,0,1),(0,0,1),(1,0),(1,0,0),(0,2,0)11(1) 因为DA,DE,(1,0,1),(1,x,1),0,所以DA,DE.1111(2)因为为的中点,则,从而, DE,(1,1,1),AC,(,1,2,0)EABE(1,1,0)1,n,AC,0,,设平面的法向量为,则也即ACDAD,(,1,0,1)n,(a,b,c),11,n,AD,0,1,a,2b,a,2b,0,,得,从而,所以点到平面的距离为 ACDn,(2,1,2)E,1,a,c,0a,c,|DE,n|2,
10、1,211h,. 33|n|(3)设平面DEC的法向量,?n,(a,b,c)1CE,(1,x,2,0),DC,(0,2,1),DD,(0,0,1),11,n,DC,0,2b,c,0,1,由 令, bcax,?,1,2,2,abx,(,2),0.,n,CE,0,? n,(2,x,1,2).|n,DD|,2221cos,.依题意 2422|n|,|DD|(x,2),51?(不合,舍去), x,2,3x,2,312,AE,23DECD,?时,二面角的大小为 14ABCABC,BBCCCCCC,6 如图,在三棱柱中,侧面,为棱上异于的AB,E1111111,EAEB,ABBBBCBCC,,,2,2,1
11、,一点,已知,求: 1113EB (?)异面直线与的距离; AB1AEBA, (?)二面角的平面角的正切值 11高考学习网,中国最大高考学习网站G | 我们负责传递知识 解:(I)以为原点,、分别为轴建立空间直角坐标系 BBBAByz,1, 由于, ABBBBCBCC,,,2,2,1,113在三棱柱中有 ABCABC,1113133 , BAB(0,0,0),(0,0,2),(0,2,0)C(,0),C(,0)1122223 设 E(,a,0),由EA,EB,得EA,EB,0,即11233 0,(,a,2),(,2,a,0)22332 ,,a(a,2),a,2a,,44131331得(a,)(
12、a,),0,即a,或a,(舍去),故E(,0)222222 313333BE,EB,(,0),(,0),,,0,即BE,EB.11222244又BBCCABEB,侧面,故 因此是异面直线的公垂线, AB,ABBE,BE11131|BE|,,,1ABEB,则,故异面直线的距离为 1144,AEBA,(II)由已知有故二面角的平面角的大小为向EA,EB,BA,EB,111111量的夹角 BA与EA1131因BA,BA,(0,0,2),EA,(,2),1122EA,BA211,故cos, 3|EA|BA|112即tan,.2PABCD,ABCDABCD7 如图,在四棱锥中,底面为矩形,PD,底面,E
13、是AB上 1PFEC,PD,2,CD,2,AE,一点, 已知 2高考学习网,中国最大高考学习网站G | 我们负责传递知识 求(?)异面直线与的距离; ECPD(?)二面角的大小 EPCD,解:(?)以为原点,、分别为 DCDADPD轴建立空间直角坐标系 xyz,由已知可得 DPC(0,0,0),(0,0,2),(0,2,0)设 A(x,0,0)(x,0),则B(x,2,0),113PE,CE得PE,CE,0 由, E(x,0),PE,(x,2),CE,(x,0).2223331332即 由, x,0,故x,.DE,CE,(,0),(,0),0得DE,CE422222又,故是异面直线与的公垂线,
14、易得,故异面直线 CE|DE|,1PDDE,DEPD,的距离为 CEPD1DGPC,(?)作,可设 由DG,PC,0得 Gyz(0,)(0,y,z),(0,2,2),0EFPC,即作于,设, z,2y,故可取DG,(0,1,2),FFmn(0,),31则 EFmn,(,).22,31由, EFPCmnmn,0(,)(0,2,2)0,2120得即22,22312PC又由nmmnEF,,,2,1,(,).故在上得 F22222,EFDG与EPCD,因故的平面角的大小为向量的夹角 EFPCDGPC,DGEF,2,EPCD,.cos,故 即二面角的大小为 ,424|DGEF高考学习网,中国最大高考学习
15、网站G | 我们负责传递知识 ,英文版 , easily blame, to prevent the broken window effect. Supervise the leading cadres to play an exemplary role, take the lead in the strict implementation of the and , lead to safeguard the solemnity and authority of the party discipline, ensure that the party discipline and the law
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