解析几何教程+(廖华奎王宝富)+课后习题

第一章向量代数习题1.11.试证向量加法的结合律，即对任意向量,,,,,,,,abcabcabcabc成立()().()().()().()().abcabcabcabcabcabcabcabc++=++++=++++=++++=++证明：作向量,,,,,,,,uuuruuuruuurABaBCbCDcABaBCbCDcABaBCbCDcABaBCbCDc============（如下图），AAAABBBBCCCCDDDDaaaabbbbccccabababab++++bcbcbcbc++++则()(),()(),()(),()(),uuuruuuruuuruuuruuuruuurabcABBCCDACCDADabcABBCCDACCDADabcABBCCDACCDADabcABBCCDACCDAD++=++=+=++=++=+=++=++=+=++=++=+=()(),()(),()(),()(),uuuruuuruuuruuuruuuruuurabcABBCCDABBDADabcABBCCDABBDADabcABBCCDABBDADabcABBCCDABBDAD++=++=+=++=++=+=++=++=+=++=++=+=故()().()().()().()().abcabcabcabcabcabcabcabc++=++++=++++=++++=++2.设,,,,,,,,abcabcabcabc两两不共线，试证顺次将它们的终点与始点相连而成一个三角形的充要条件是0.0.0.0.abcabcabcabc++=++=++=++=证明：必要性，设,,,,,,,,abcabcabcabc的终点与始点相连而成一个三角形ABCABCABCABC∆∆∆∆，AAAABBBBCCCCaaaabbbbcccc则0.0.0.0.uuuruuuruuuruuuruuuruuurabcABBCCAACCAAAabcABBCCAACCAAAabcABBCCAACCAAAabcABBCCAACCAAA++=++=+==++=++=+==++=++=+==++=++=+==充分性，作向量,,,,,,,,uuuruuuruuurABaBCbCDcABaBCbCDcABaBCbCDcABaBCbCDc============，由于0,0,0,0,uuuruuuruuuruuuruuuruuurabcABBCCDACCDADabcABBCCDACCDADabcABBCCDACCDADabcABBCCDACCDAD=++=++=+==++=++=+==++=++=+==++=++=+=所以点AAAA与DDDD重合，即三向量,,,,,,,,abcabcabcabc的终点与始点相连构成一个三角形。3.试证三角形的三中线可以构成一个三角形。证明：设三角形ABCABCABCABC∆∆∆∆三边,,,,,,,,ABBCCAABBCCAABBCCAABBCCA的中点分别是,,,,,,,,DEFDEFDEFDEF（如下图），并且记AAAABBBBaaaabbbbccccEEEEFFFFDDDD,,,,,,,,uuuruuuruuuraABbBCcCAaABbBCcCAaABbBCcCAaABbBCcCA============，则根据书中例1.1.1，三条中线表示的向量分别是111111111111(),(),(),(),(),(),(),(),(),(),(),(),222222222222uuuruuuruuurCDcbAEacBFbaCDcbAEacBFbaCDcbAEacBFbaCDcbAEacBFba=−=−=−=−=−=−=−=−=−=−=−=−所以，111111111111()()()0,()()()0,()()()0,()()()0,222222222222uuuruuuruuurCDAEBFcbacbaCDAEBFcbacbaCDAEBFcbacbaCDAEBFcbacba++=−+−+−=++=−+−+−=++=−+−+−=++=−+−+−=故由上题结论得三角形的三中线,,,,,,,,CDAEBFCDAEBFCDAEBFCDAEBF可以构成一个三角形。4.用向量法证明梯形两腰中点连线平行于上、下底且等于它们长度和的一半。证明：如下图，梯形ABCDABCDABCDABCD两腰,,,,BCADBCADBCADBCAD中点分别为,,,,EFEFEFEF，记向量,,,,uuuruuurABaFAbABaFAbABaFAbABaFAb========，AAAABBBBaaaabbbbEEEEFFFF则,,,,uuurDFbDFbDFbDFb====而向量uuurDCDCDCDC与uuurABABABAB共线且同向，所以存在实数0,0,0,0,λλλλ使得....uuuruuurDCABDCABDCABDCABλλλλ====现在,,,,uuurFBbaFBbaFBbaFBba=+=+=+=+,,,,uuurFCbaFCbaFCbaFCbaλλλλ=−+=−+=−+=−+由于EEEE是BCBCBCBC的中点，所以1111111111111111()()(1)(1).()()(1)(1).()()(1)(1).()()(1)(1).2222222222222222uuuruuuruuuruuurFEFBFCbaabaABFEFBFCbaabaABFEFBFCbaabaABFEFBFCbaabaABλλλλλλλλλλλλ=+=++−=+=+=+=++−=+=+=+=++−=+=+=+=++−=+=+且111111111111(1)()().(1)()().(1)()().(1)()().222222222222uuuruuuruuuruuuruuuruuurFEABABABABDCFEABABABABDCFEABABABABDCFEABABABABDCλλλλλλλλ=+=+=+=+=+=+=+=+=+=+=+=+故梯形两腰中点连线平行于上、下底且等于它们长度和的一半。5.试证命题1.1.2。CCCCDDDDCCCC证明：必要性，设,,,,,,,,abcabcabcabc共面，如果其中有两个是共线的，比如是,,,,abababab，则,,,,abababab线性相关，从而,,,,,,,,abcabcabcabc线性相关。现在设,,,,,,,,abcabcabcabc两两不共线，则向量cccc可以在两个向量,,,,abababab上的进行分解，即作以cccc为对角线，邻边平行于,,,,abababab的平行四边形，则存在实数,,,,λµλµλµλµ使得cabcabcabcabλµλµλµλµ=+=+=+=+，因而,,,,,,,,abcabcabcabc线性相关。充分性，设,,,,,,,,abcabcabcabc线性相关，则存在不全为零的数123123123123,,,,,,,,kkkkkkkkkkkk，使得1231231231230000kakbkckakbkckakbkckakbkc++=++=++=++=。不妨设33330000kkkk≠≠≠≠，则向量cccc可以表示为向量,,,,abababab的线性组合，因此由向量加法的平行四边形法则知道向量cccc平行于由向量,,,,abababab决定的平面，故,,,,,,,,abcabcabcabc共面。6.设,,,,,,,,ABCABCABCABC是不共线的三点，它们决定一平面ΠΠΠΠ，则点PPPP在ΠΠΠΠ上的充要条件是存在唯一的数组(,,)(,,)(,,)(,,)λµνλµνλµνλµν使得,,,,(*)(*)(*)(*)1,1,1,1,uuuruuuruuuruuurOPOAOBOCOPOAOBOCOPOAOBOCOPOAOBOCλµνλµνλµνλµνλµνλµνλµνλµν⎧⎧⎧⎧=++=++=++=++⎪⎪⎪⎪⎨⎨⎨⎨++=++=++=++=⎪⎪⎪⎪⎩⎩⎩⎩其中，OOOO是任意一点。PPPP在ABCABCABCABC∆∆∆∆内的充要条件是(*)与0,0,00,0,00,0,00,0,0λµνλµνλµνλµν≥≥≥≥≥≥≥≥≥≥≥≥同时成立。证明：必要性，作如下示意图，连接APAPAPAP并延长交直线BCBCBCBC于RRRR。AAAABBBBCCCCOOOORRRRPPPP则由三点,,,,,,,,BRCBRCBRCBRC共线，存在唯一的数组12121212,,,,kkkkkkkk使得12121212ORkOBkOCORkOBkOCORkOBkOCORkOBkOC=+=+=+=+uuuruuuruuur，并且121212121111kkkkkkkk+=+=+=+=。由三点,,,,,,,,APRAPRAPRAPR共线，存在唯一的数组12121212,,,,llllllll使得12121212OPlOAlOROPlOAlOROPlOAlOROPlOAlOR=+=+=+=+uuuruuuruuur，并且121212121111llllllll+=+=+=+=。于是1212122121212212121221212122OPlOAlORlOAlkOBlkOCOPlOAlORlOAlkOBlkOCOPlOAlORlOAlkOBlkOCOPlOAlORlOAlkOBlkOC=+=++=+=++=+=++=+=++uuuruuuruuuruuuruuuruuur，设12122121221212212122,,,,,,,,,,,,llklkllklkllklkllklkλµνλµνλµνλµν============由12121212,,,,kkkkkkkk，12121212,,,,llllllll的唯一性知道(,,)(,,)(,,)(,,)λµνλµνλµνλµν的唯一性，则,,,,OPOAOBOCOPOAOBOCOPOAOBOCOPOAOBOCλµνλµνλµνλµν=++=++=++=++uuuruuuruuuruuur且121221212212122121221111llklkllklkllklkllklkλµνλµνλµνλµν++=++=++=++=++=++=++=++=。充分性，由已知条件有(1)(1)(1)(1)OPOAOBOCOAOBOCOPOAOBOCOAOBOCOPOAOBOCOAOBOCOPOAOBOCOAOBOCλµνλµλµλµνλµλµλµνλµλµλµνλµλµ=++=++−−=++=++−−=++=++−−=++=++−−uuuruuuruuuruuuruuuruuuruuur()()()()()()()()OAOCOBOCOCOAOCOBOCOCOAOCOBOCOCOAOCOBOCOCλµλµλµλµ=−+−+=−+−+=−+−+=−+−+uuuruuuruuuruuuruuurCACBOCCACBOCCACBOCCACBOCλµλµλµλµ=++=++=++=++uuuruuuruuur,得到CPCACBCPCACBCPCACBCPCACBλµλµλµλµ=+=+=+=+uuuruuuruuur，因而向量,,,,,,,,CPCACBCPCACBCPCACBCPCACBuuuruuuruuur共面，即PPPP在,,,,,,,,ABCABCABCABC决定的平面上。如果PPPP在ABCABCABCABC∆∆∆∆内，则PPPP在线段ARARARAR内，RRRR在线段BCBCBCBC内，于是12121212121212120,,,10,,,10,,,10,,,1kkllkkllkkllkkll≤≤≤≤≤≤≤≤，则0,,10,,10,,10,,1λµνλµνλµνλµν≤≤≤≤≤≤≤≤。如果（*）成立且0,,10,,10,,10,,1λµνλµνλµνλµν≤≤≤≤≤≤≤≤，则有CPCACBCPCACBCPCACBCPCACBλµλµλµλµ=+=+=+=+uuuruuuruuur，这说明点PPPP在角ACBACBACBACB∠∠∠∠内。同样可得到APABACAPABAC