《高等代数与解析几何》课程教学资源（书籍教材）高等代数与解析几何习题解答（共十四章）

第一章向量代数81向量的线性运算1.如图，已知平行六面体ABCD-A,B,CD1，E、F分别是棱BCC,D,的中点设AB=,AD=,AA=.试用，,表示下列向量：(4) EF.(1) AC;(3) AF;(2) BDI;D1C2第1题图解：(1)因为BC-AD,CC=AA,ACI=AB+BC+CC所以AC=a+b+c.(2)因为BD=BD +DDI,而BD=AD-AB=-6-d, DD, =AA所以BD= b-a+.(3) AF=AD +DDI+D,F,而DD,=AAI, DIF=DIC =AB2

1Ù •þê § 1 •þ‚5$Ž 1. Xã, ®²18¡N ABCD−A1B1C1D1, E!F ©O´c BC! C1D1 ¥:.  −−→AB = −→a , −−→AD = −→b , −−→AA1 = −→c . Á^ −→a , −→b , −→c L«e •þ: (1) −−→AC1; (2) −−→BD1; (3) −→AF; (4) −−→EF. A B D C A1 B1 D1 C1 F E −→a −→c −→b 1 1 Kã ): (1) Ϗ −−→BC = −−→AD, −−→CC1 = −−→AA1, −−→AC1 = −−→AB + −−→BC + −−→CC1, ¤± −−→AC1 = −→a + −→b + −→c . (2) Ϗ −−→BD1 = −−→BD + −−→DD1, −−→BD = −−→AD − −−→AB = −→b − −→a , −−→DD1 = −−→AA1. ¤± −−→BD1 = −→b − −→a + −→c . (3) −→AF = −−→AD + −−→DD1 + −−→D1F, −−→DD1 = −−→AA1, −−→D1F = 1 2 −−−→ D1C1 = 1 2 −−→AB,

.2 .第一章向量代数所以1AF=a+b+z2(4) EF = AF - AE = AF - (AB + BE) = AF - (AB + BE) =111.(AB+BC)AFa++++++222.已知平行四边形ABCD的对角线为AC和BD.设AC=a.BD6.求AB,CD.DA解：如图，AG+B-AB-AO+OB- 1(α- b)CD--AB- --(a-),DA = -AD = -(AO + OD) = -(α + b).DCMBN第3题图第2题图3.在△ABC中，点M.N是AB边上的三等分点.设CA=a.CB6. 求CM,CN解：如图，因为12AB,ANABAM:-33所以21+3IAB-CA+CM-CA+AM-CA+--CA) :CR-a33.12(CB-CA) =CN=CA+AN-CA+AB-CA+2+a3334.设L,M.N分别是△ABC的三边BC.CA.AB的中点.证明三中线向量AL，BMCN可以构成一个三角形

· 2 · 1Ù •þê ¤± −→AF = 1 2 −→a + −→b + −→c . (4) −−→EF = −→AF − −→AE = −→AF − ( −−→AB + −−→BE) = −→AF − ( −−→AB + −−→BE) = −→AF −  −−→AB + 1 2 −−→BC = 1 2 −→a + −→b + −→c − −→a − 1 2 −→b = − 1 2 −→a + 1 2 −→b + −→c . 2. ®²1o>/ ABCD 邏 AC Ú BD.  −→AC = −→a , −−→BD = −→b . ¦ −−→AB, −−→CD, −−→DA. ): Xã, −−→AB = −→AO + −−→OB = 1 2 −→AC + 1 2 −−→DB = 1 2 ( −→a − −→b ), −−→CD = − −−→AB = − 1 2 ( −→a − −→b ), −−→DA = − −−→AD = −( −→AO + −−→OD) = − 1 2 ( −→a + −→b ). A C B D O 1 2 Kã A B M N C 1 3 Kã 3. 3 △ABC ¥, : M,N ´ AB >þn©:.  −→CA = −→a , −−→CB = −→b . ¦ −−→CM, −−→CN. ): Xã, Ϗ −−→AM = 1 3 −−→AB, −−→AN = 2 3 −−→AB, ¤± −−→CM = −→CA + −−→AM = −→CA + 1 3 −−→AB = −→CA + 1 3 ( −−→CB − −→CA) = 1 3 −→b + 2 3 −→a , −−→CN = −→CA + −−→AN = −→CA + 2 3 −−→AB = −→CA + 2 3 ( −−→CB − −→CA) = 2 3 −→b + 1 3 −→a . 4.  L,M,N ©O´ △ABC n> BC,CA,AB ¥:. y²n¥‚ •þ −→AL, −−→BM, −−→CN Œ±¤‡n/.

S1向量的线性运算:3.证明：因为AL,BM,CN可以构成一个三角形，当且仅当将这三个向量之和为零向量.由AL =(AB + AC),BM = I(BA+ BC),CN = (CA +CB),21可得：AL+BM+CN=0.LC6CBL第4题图第5题图5.设O是△ABC的重心，证明：OA+OB+OC= 0解：如图，设AL,BM,CN是3条中线，O是三角形的重心：则OA：S22LA=6BM,OC=-32cN，因此由第4题，AL.OB-32OA+OB +OC =-2(AL + BM +CN) =0.6.在四面体O-ABC中，设点G是△ABC的重心.用OA,OB,OC来表示向量OG解：因为 GO=GA+AO,GO=GB+BO.GO=GC+CO.而由第5题知GA+GB+GC=0.因此3GO= AO+BO+COOG = I(OA + OB +OC)7.设ABCDEF为正六边形，求AB+AC+AD+AE+AF解：因为AD=AC+AF=AE+AB,所以AB+AC+AD+AE+AF3AD

§ 1 •þ‚5$Ž · 3 · y²: Ϗ −→AL, −−→BM, −−→CN Œ±¤‡n/, …=òùn‡•þƒ ڏ"•þ. d −→AL = 1 2 ( −−→AB + −→AC), −−→BM = 1 2 ( −−→BA + −−→BC), −−→CN = 1 2 ( −→CA + −−→CB), Œ: −→AL + −−→BM + −−→CN = 0. B L C N M A 1 4 Kã B L C N M A O 1 5 Kã 5.  O ´ △ABC ­%, y²: −→OA + −−→OB + −−→OC = 0. ): Xã,  AL,BM,CN ´ 3 ^¥‚, O ´n/­%. K −→OA = 2 3 −→LA = − 2 3 −→AL, −−→OB = − 2 3 −−→BM, −−→OC = − 2 3 −−→CN, Ïdd1 4 K, −→OA + −−→OB + −−→OC = − 2 3 ( −→AL + −−→BM + −−→CN) = 0. 6. 3o¡N O−ABC ¥, : G ´ △ABC ­%. ^ −→OA, −−→OB, −−→OC 5 L«•þ −−→OG. ): Ϗ −−→GO = −→GA + −→AO, −−→GO = −−→GB + −−→BO, −−→GO = −−→GC + −−→CO. d1 5 K −→GA + −−→GB + −−→GC = 0. Ïd 3 −−→GO = −→AO + −−→BO + −−→CO. −−→OG = 1 3 ( −→OA + −−→OB + −−→OC). 7.  ABCDEF 8>/, ¦ −−→AB + −→AC + −−→AD + −→AE + −→AF. ): Ϗ −−→AD = −→AC+ −→AF = −→AE+ −−→AB, ¤± −−→AB+ −→AC+ −−→AD+ −→AE+ −→AF = 3 −−→AD.

第一章向量代数:4:YFC第6题图第7题图8.在四边形ABCD中，AB=a+26.BC=-4a-.D-5α-3b（，b是不共线的非零向量）.证明ABCD为梯形证明：因为AD=AB+BC+CD=-8α-26=2BC,所以AD/BC但|AD|=2|BCI，所以ABCD是梯形.9.设A,B,C.D是一个四面体的四个顶点，M,N分别是边AB.CD的中点.证明：MN - I(AD + BC).2证明：如图，I(CA+CB),CN=CD,CM-所以MN-CN-CM-ICD-I(CA + CB) = (AD + BC),第10题图第9题图10.设M是平行四边形ABCD的中心，O是任意一点证明：OA+OB+OC+OD=4OM

· 4 · 1Ù •þê B C G O A 1 6 Kã A D F B E C 1 7 Kã 8. 3o>/ ABCD ¥, −−→AB = −→a + 2 −→b , −−→BC = −4 −→a − −→b , −−→CD = −5 −→a − 3 −→b ( −→a , −→b ´Ø
‚š"•þ). y² ABCD F/. y²: Ϗ −−→AD = −−→AB+ −−→BC+ −−→CD = −8 −→a −2 −→b = 2 −−→BC, ¤± −−→AD//−−→BC. | −−→AD| = 2| −−→BC|, ¤± ABCD ´F/. 9.  A,B,C,D ´‡o¡No‡º:, M,N ©O´> AB, CD  ¥:. y²: −−→MN = 1 2 ( −−→AD + −−→BC). y²: Xã, −−→CM = 1 2 ( −→CA + −−→CB), −−→CN = 1 2 −−→CD, ¤± −−→MN = −−→CN − −−→CM = 1 2 −−→CD − 1 2 ( −→CA + −−→CB) = 1 2 ( −−→AD + −−→BC). b b B D C A M N 1 9 Kã b A C B D M O 1 10 Kã 10.  M ´²1o>/ ABCD ¥%, O ´?¿:. y²: −→OA + −−→OB + −−→OC + −−→OD = 4 −−→OM.

.581向量的线性运算证明：如图，因为OM =I(OA +OC), OM = I(OB+OD),所以OA+OB+OC+OD=4OM11.要使下列各式成立，向量，6应满足什么条件？(1) /a + 6/= [a/+[6l;(2) [ + 6/= [[-[6];(3) /a - bl=[a|-[bl;(4) [a - b/= [al+[b/解：（1）利用“三角形两边之和大于第三边”可知：a/b：且要使a+=+必须：与同向，或，中至少有一为0.(2)令=,则=-,原式化为-=所以/且反向.由此可得：//6，反向，且a≥61或=0.(3)令=-,则=+,原式化为:1+/=+由(1)知：//且同向.所以a//6且同向.又因-b≥0,所以a|≥b1或6=0.(4)令=,则=,原式化为:=由(2)知：/6且反向，或6=0,同时，≥6.所以//6且反向，或b=0或可=012.证明下列不等式，并说明等号什么时候成立.(1)16 -a[≥[a]-[bl:(2) [a+6 +c/≤/a/+/b/+/cl证明：（1）如图，利用“三角形两边之差小于第三边”可得欲证的不等式.等式成立的条件可参见习题11(3)：a/b同向，且a|≥b1,或6=0b-aa第12(1)题图(2)令=+. 则:++=++=[a+6l≤a+b+l等号成立当且仅当(i)a，b，互相平行且同向，或(i)α，b，中至少两个为0（也可看成(i)的特例).*13.O为正多边形A1A2.*A的中心.证明：OA +OA2+..+OA, = 0

§ 1 •þ‚5$Ž · 5 · y²: Xã, Ϗ −−→OM = 1 2 ( −→OA + −−→OC), −−→OM = 1 2 ( −−→OB + −−→OD), ¤± −→OA + −−→OB + −−→OC + −−→OD = 4 −−→OM. 11. ‡¦eˆª¤á, •þ −→a , −→b A÷vŸo^‡? (1) | −→a + −→b | = | −→a | + | −→b |; (2) | −→a + −→b | = | −→a | − |−→b |; (3) | −→a − −→b | = | −→a | − |−→b |; (4) | −→a − −→b | = | −→a | + | −→b |. ): (1) |^“n/ü>ƒÚŒu1n>”Œ: −→a //−→b . …‡¦ | −→a + −→b | = | −→a | + | −→b | 7L: −→a † −→b ӕ, ½ −→a , −→b ¥k 0. (2) - −→c = −→a + −→b , K −→a = −→c − −→b , ªz: | −→c − −→b | = | −→c | + | −→b |. ¤± −→b //−→c …‡•. ddŒ: −→a //−→b , ‡•, … | −→a | ≥ |−→b |, ½ −→b = 0. (3) - −→c = −→a − −→b , K −→a = −→b + −→c , ªz: | −→b |+| −→c | = | −→b + −→c |. d (1) : −→b //−→c …Ó•. ¤± −→a //−→b …Ó•. qÏ| −→a − −→b | ≥ 0, ¤±| −→a | ≥ |−→b |, ½ −→b = 0. (4) - −→c = −→a − −→b , K −→a = −→b + −→c , ªz: | −→b + −→c | = | −→c | − |−→b |. d (2) : −→c //−→b …‡•, ½ −→b = 0, Ӟ, | −→c | ≥ |−→b |. ¤±−→a //−→b …‡•, ½ −→b = 0 ½ −→a = 0. 12. y²eت, ¿`²ҟožÿ¤á. (1) | −→b − −→a | > | −→a | − |−→b |; (2) | −→a + −→b + −→c | ≤ |−→a |+| −→b |+| −→c |. y²: (1) Xã, |^“n/ü>ƒu1n>”Œyت.  ª¤á^‡Œë„SK 11(3): −→a //−→b , ӕ, … | −→a | ≥ |−→b |, ½ −→b = 0. −→b − −→a −→b −→a 1 12(1) Kã (2) - −→d = −→b + −→c . K: | −→a + −→b + −→c | = | −→a + −→d | ≤ |−→a | + | −→d | = | −→a | + | −→b + −→c | ≤ |−→a | + | −→b | + | −→c |. Ò¤á…=(i) −→a , −→b , −→c pƒ² 1…Ó•, ½(ii) −→a , −→b , −→c ¥ü‡ 0 (Œw¤ (i) A~). ∗13. O õ>/ A1A2 · · · An ¥%. y²: −−→OA1 + −−→OA2 + · · · + −−→OAn = 0.

.6.第一章向量代数证明：先考虑n为偶数的情形.此时.显然有：OA+..+OA，=0.再看n为奇数的情形：我们增加一倍顶点B..B.使原来正n边形A，.A.成为：ABAzB2An-1Bn-1A,B，这是—个2n边形所以OA+OB+OA+OB,+..+OA,+OB,=0注意到OB,是由OA,旋转一个定角=而得到，若记：P -OA +...+0A.q=OBi+...+OBn.那么是由旋转角而得到。由于，与不平行，故n=0当且仅当==0.*14.O为正多边形A1A2·.·An的中心，P是任意一点.证明：PA +PA,+...+PA, =nPO证明：因为PO-PA,+A.O (i=1,2,.,n),所以nPO- PAI +...+PA, +(AIO+...+A,O) =PAI +...+PA.(利用第13题的结论)92向量的共线与共面1.己知,不共线，则向量=3a+与=2a-是否线性相关？解：设有km使：k+md=0.即3ka+kb+2ma-mb=0整理后为(3k + 2m)a +(k-m)6 = 0.由于，不共线，故，线性无关，所以3k+2m=0解得k=m=0,k-m=0

· 6 · 1Ù •þê y²: kÄ n óêœ/. dž. w,k: −−→OA1 + · · · + −−→OAn = 0. 2 w n Ûêœ/: ·‚O\º: B1, · · · ,Bn ¦5 n >/ A1 · · · An ¤: A1B1A2B2 · · · An−1Bn−1AnBn, ù´‡ 2n >/. ¤± −−→OA1 + −−→OB1 + −−→OA2 + −−→OB2 + · · · + −−→OAn + −−→OBn = 0. 5¿ −−→OBi ´d −−→OAi ^=‡½ π n , eP: −→p = −−→OA1 + · · · + −−→OAn, −→q = −−→OB1 + · · · + −−→OBn, @o −→q ´d −→p ^= π n  . du 0 / A1A2 · · · An ¥%, P ´?¿:. y²: −−→PA1 + −−→PA2 + · · · + −−→PAn = n −−→PO. y²: Ϗ −−→PO = −−→PAi + −−→AiO (i = 1, 2, · · · ,n), ¤± n −−→PO = −−→PA1 + · · · + −−→PAn + (−−→A1O + · · · + −−→AnO) = −−→PA1 + · · · + −−→PAn (|^1 13 K(Ø). § 2 •þ
‚†
¡ 1. ® −→a , −→b Ø
‚, K•þ −→c = 3−→a + −→b † −→d = 2−→a − −→b ´Ä‚5 ƒ'? ): k k m ¦: k −→c + m −→d = 0, = 3k −→a + k −→b + 2m−→a − m −→b = 0, n￾ (3k + 2m) −→a + (k − m) −→b = 0. du −→a , −→b Ø
‚,  −→a , −→b ‚5Ã', ¤±    3k + 2m = 0 k − m = 0 ) k = m = 0,

82向量的共线与共面.7.即,线性无关2.如果3个向量都能被两个向量a，6线性表示，那么这3个向量一定共面证明：设=Ca+C126,=C21a+C226,=C31a+ C326.则+22+=(1C11+2C21+3C31)+(1C12+22C22+3C32)方程组C111+C212+C3123=0C121+C2272+C323=0的变量个数超过方程个数，一定有一组非零解21=k1，2=k2，23=k3，使得ki+h2+ka=(kiC11+k2C21+kgC31)a+(kiC12+k2C22+kacC32)=0.因此P，可，下线性相关，从而共面3.证明三个向量ki-kzb,kzb-k3,k3-kia共面.证明：由等式(i-)+(-k)+(-)=0,可知这3个向量线性相关，所以共面。4.设ai=26+3bz-ba,az=b2-b3,a=b2+b3.证明向量ai,a2,a3共面的充分必要条件是b1,b2,b3共面.证明：kiai +k2a2 + k3as = 2kib1 + (3ki +k2 + k3)b2 +(-k1 - k2 + k3)b3从方程组(2k) = 03k1+k2+k3=0(-ki-k2+k3= 0解得ki=kz=ks=0.也就是说，ki,k2,ks不全为零当且仅当2k1,3k1+k2+k3，-k1-k2+k3不全为零即ai,a2,as线性相关当且仅当bi,b2,b3线性相关.从而ai,a2,a3共面当且仅当b1,b2b3共面5.设D是△ABC的边BC上的点，满足BD=kDC.试用AB,AC来表示 AD

§ 2 •þ
‚†
¡ · 7 · = −→c , −→d ‚5Ã'. 2. XJ 3 ‡•þÑUü‡•þ −→a , −→b ‚5L«, @où 3 ‡•þ½
¡. y²:  −→p = c11 −→a + c12 −→b , −→q = c21 −→a + c22 −→b , −→r = c31 −→a + c32 −→b . K x1 −→p + x2 −→q + x3 −→r = (x1c11 + x2c21 + x3c31) −→a + (x1c12 + x2c22 + x3c32) −→b . §|    c11x1 + c21x2 + c31x3 = 0 c12x1 + c22x2 + c32x3 = 0 Cþ‡ê‡L§‡ê, ½k|š") x1 = k1, x2 = k2, x3 = k3, ¦ k1 −→p +k2 −→q +k3 −→r = (k1c11 +k2c21 +k3c31) −→a + (k1c12 +k2c22 +k3c32) −→b = 0. Ïd −→p , −→q , −→r ‚5ƒ', l
¡. 3. y²n‡•þ k1 −→a − k2 −→b , k2 −→b − k3 −→c , k3 −→c − k1 −→a
¡. y²: dª (k1 −→a − k2 −→b ) + (k2 −→b − k3 −→c ) + (k3 −→c − k1 −→a ) = 0, Œù 3 ‡•þ‚5ƒ', ¤±
¡. 4.  −→a1 = 2 −→b1 + 3 −→b2 − −→b3 , −→a2 = −→b2 − −→b3 , −→a3 = −→b2 + −→b3 . y²•þ −→a1, −→a2, −→a3
¡¿©7‡^‡´ −→b1 , −→b2 , −→b3
¡. y²: k1 −→a1 + k2 −→a2 + k3 −→a3 = 2k1 −→b1 + (3k1 + k2 + k3) −→b2 + (−k1 − k2 + k3) −→b3 . l§|    2k1 = 0 3k1 + k2 + k3 = 0 −k1 − k2 + k3 = 0 ) k1 = k2 = k3 = 0. Ò´`, k1,k2,k3 ؏"…= 2k1, 3k1 + k2 + k3, −k1 − k2 + k3 ؏". = −→a1, −→a2, −→a3 ‚5ƒ'…= −→b1 , −→b2 , −→b3 ‚5ƒ '. l −→a1, −→a2, −→a3
¡…= −→b1 , −→b2 , −→b3
¡. 5.  D ´ △ABC > BC þ:, ÷v −−→BD = k −−→DC. Á^ −−→AB, −→AC 5 L« −−→AD.

第一章向量代数:8.解：因为BD=AD-AB,DC=AC-AD.代人BD=kDC.得F1AC.AD-AB= kAC-kAD,解得 AD =-AB+1+k1+k0DTL第5题图第6题图6.设AT是△ABC中ZA的平分线（与BC交于T点)，将AT用AB,AC来表示.解：设BT=kBC,则TC=(1-k)BC.由角平分线的性质可知,IAB[ABI:IAC= k : (1 -k),因此k=于是[AB| + [AC]AT=AB+BT=AB+kBC=(1-k)AB+kAC1(IACIAB+ABAC)[AB| +|AC]7平面上有一个三角形△OAB，点B和C关于中心A对称，点D把线段OB分成2：1,DC和OA交于点E.设OA=a，OB=6(1)试用，来表示C和D:(2)求比值OE：OA.EAH第7题图解：(1)因B和 C关于中心A对称,BC=2BA=2(α_6).又因2OB-2,DB-1OB-1D,得 DC-DB+BC-2a -56OD-a3333OC=OB+BC=2a - b

· 8 · 1Ù •þê ): Ϗ −−→BD = −−→AD − −−→AB, −−→DC = −→AC − −−→AD. \ −−→BD = k −−→DC,  −−→AD − −−→AB = k −→AC − k −−→AD, ) −−→AD = 1 1 + k −−→AB + k 1 + k −→AC. B C D A 1 5 Kã B C T A 1 6 Kã 6.  AT ´ △ABC ¥ ∠A ²©‚ († BC u T :), ò −→AT ^ −−→AB, −→AC 5L«. ):  −→BT = k −−→BC, K −→TC = (1 − k) −−→BC. d²©‚5ŸŒ, | −−→AB| : | −→AC| = k : (1 − k), Ïd k = | −−→AB| | −−→AB| + | −→AC| . u´ −→AT = −−→AB + −→BT = −−→AB + k −−→BC = (1 − k) −−→AB + k −→AC = 1 | −−→AB| + | −→AC| (| −→AC| −−→AB + | −−→AB| −→AC). 7 ²¡þk‡n/ △OAB, : B Ú C 'u¥% A é¡, : D r‚ ã OB ©¤ 2 : 1, DC Ú OA u: E.  −→OA = −→a , −−→OB = −→b . (1) Á^ −→a , −→b 5L« −−→OC Ú −−→DC; (2) ¦'Š OE : OA. C A B O D E 1 7 Kã ): (1) Ï B Ú C 'u¥% A é¡, −−→BC = 2 −−→BA = 2(−→a − −→b ). qÏ −−→OD = 2 3 −−→OB = 2 3 −→b , −−→DB = 1 3 −−→OB = 1 3 −→b ,  −−→DC = −−→DB + −−→BC = 2−→a − 5 3 −→b . −−→OC = −−→OB + −−→BC = 2−→a − −→b .

$2向量的共线与共面.9.2.(2)设OE=kOA=k,则由OE=OD+mDC可知k：6+35→4b2a解得k=，因此0E：OA=4：5.m5!38.在△ABC中，点M分线段AB为2：1.点N分线段AC为3：2.设CM与BN的交点为P,直线AP与边BC交于点Q.试用AB,AC来表示AP 和AQ.2AB, AN = 2AC,所以 BN = AN - AB-2A解：因为AM=AC3.2AB-AC.设CP=kCM,BP= mBN.则AB, CM = AM - AC -3CP-CB+BP得(AB-AC))-A-AC+m(AC-AB)?(32，所以解出k=32(2AB-AC)AP=AC+CP-AC+ 3=AB+IAC:又点Q在BC及AP的延长线上，所以AQ=IAP=AB+sBC.即(AB+IAC) =AB +s(AC-AB),394AB+AC.，即有AQ=解出1=LQM第8题图第9题图9.设ABCD是平行四边形，P,Q分别是边BC,CD的中点．证明AP,AQ与对角线BD相交于E,F,而将BD三等分证明：设AB=α,AD=b,则BD=AD-AB=b -a

§ 2 •þ
‚†
¡ · 9 · (2)  −−→OE = k −→OA = k −→a , Kd −−→OE = −−→OD + m −−→DC Œ k −→a = 2 3 −→b + m  2 −→a − 5 3 −→b  . ) k = 4 5 , Ïd OE : OA = 4 : 5. 8. 3 △ABC ¥, : M ©‚ã AB  2 : 1, : N ©‚ã AC  3 : 2.  CM † BN : P, †‚ AP †> BC u: Q. Á^ −−→AB, −→AC 5L« −→AP Ú −→AQ. ): Ϗ −−→AM = 2 3 −−→AB, −−→AN = 3 5 −→AC, ¤± −−→BN = −−→AN − −−→AB = 3 5 −→AC − −−→AB, −−→CM = −−→AM − −→AC = 2 3 −−→AB − −→AC.  −−→CP = k −−→CM, −−→BP = m −−→BN. K −−→CP = −−→CB + −−→BP  k  2 3 −−→AB − −→AC = −−→AB − −→AC + m  3 5 −→AC − −−→AB . )Ñ k = 2 3 . ¤± −→AP = −→AC + −−→CP = −→AC + 2 3  2 3 −−→AB − −→AC = 4 9 −−→AB + 1 3 −→AC. q: Q 3 BC 9 AP ò‚þ, ¤± −→AQ = l −→AP = −−→AB + s −−→BC. = l  4 9 −−→AB + 1 3 −→AC = −−→AB + s( −→AC − −−→AB). )Ñ l = 9 7 , =k −→AQ = 4 7 −−→AB + 3 7 −→AC. B C M A Q N P 1 8 Kã A C B D Q P F E 1 9 Kã 9.  ABCD ´²1o>/, P,Q ©O´> BC,CD ¥:. y² AP,AQ †é‚ BD ƒu E,F, ò BD n©. y²:  −−→AB = −→a , −−→AD = −→b , K −−→BD = −−→AD − −−→AB = −→b − −→a ,

.10 .第一章向量代数126BC=a+AP-AB+2211AQ=AD+DC=6+a22又设AE=kAPAF=mAQ3(k>0)，3(m>0)则kb,AF=mb+a.AE=ka+!22但是AE=AB+tBD=a+t(6 -a)=(1-t)a+tb (t>0),所以=(1-)+6,ka+2即：(k+t-1)a =由于与6不平行所以2k=k+t-1=0131即k?-0t=3:2.同理，由AF=AB+$BD=(1-$)6+$6(s >0):可得：2TIEm=+8-1=01322即:-m=0.s=3最后得到：211BF_3BD,BD.BE133说明E，F是线段BD的三等分点10.设O是一个定点.证明：对于不在一直线上的3个点A.B.C.点M位于平面ABC上的充分必要条件是存在实数ki,k2,k3，使得OM = k,OA+ kOB + kaOC,，且ki+kz+k3=1

· 10 · 1Ù •þê −→AP = −−→AB + 1 2 −−→BC = −→a + 1 2 −→b , −→AQ = −−→AD + 1 2 −−→DC = −→b + 1 2 −→a . q −→AE = k −→AP (k > 0), −→AF = m −→AQ (m > 0), K −→AE = k −→a + k 2 −→b , −→AF = m −→b + m 2 −→a . ´ −→AE = −−→AB + t −−→BD = −→a + t( −→b − −→a ) = (1 − t) −→a + t −→b (t > 0). ¤± k −→a + k 2 −→b = (1 − t) −→a + t −→b , =: (k + t − 1)−→a =  t − k 2  −→b , du −→a † −→b ز1, ¤±    k + t − 1 = 0 t − k 2 = 0 =    k = 2 3 t = 1 3 . Ón, d −→AF = −−→AB + s −−→BD = (1 − s) −→b + s −→b (s > 0), Œ:    m 2 + s − 1 = 0 s − m = 0, =:    m = 2 3 s = 2 3 . ￾: −−→BF = 2 3 −−→BD, −−→BE = 1 3 −−→BD, `² E,F ´‚ã BD n©:. 10.  O ´‡½:, y²: éuØ3†‚þ 3 ‡: A,B,C, : M  u²¡ ABC þ¿©7‡^‡´3¢ê k1,k2,k3, ¦ −−→OM = k1 −→OA + k2 −−→OB + k3 −−→OC, … k1 + k2 + k3 = 1.