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 þ5$ 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) Ï −−→BC = −−→AD, −−→CC1 = −−→AA1, −−→AC1 = −−→AB + −−→BC + −−→CC1, ¤± −−→AC1 = −→a + −→b + −→c . (2) Ï −−→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 é 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, Ï −−→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 ±¤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 þ5$ · 3 · y²: Ï −→AL, −−→BM, −−→CN ±¤n/, =òùnþ Ú"þ. d −→AL = 1 2 ( −−→AB + −→AC), −−→BM = 1 2 ( −−→BA + −−→BC), −−→CN = 1 2 ( −→CA + −−→CB), : −→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. ): Ï −−→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. ): Ï −−→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"þ). y² ABCD F/. y²: Ï −−→AD = −−→AB+ −−→BC+ −−→CD = −8 −→a −2 −→b = 2 −−→BC, ¤± −−→AD//−−→BC. | −−→AD| = 2| −−→BC|, ¤± ABCD ´F/. 9. A,B,C,D ´o¡Noº:, 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 þ5$ · 5 · y²: Xã, Ï −−→OM = 1 2 ( −→OA + −−→OC), −−→OM = 1 2 ( −−→OB + −−→OD), ¤± −→OA + −−→OB + −−→OC + −−→OD = 4 −−→OM. 11. ¦eª¤á, þ −→a , −→b A÷vo^? (1) | −→a + −→b | = | −→a | + | −→b |; (2) | −→a + −→b | = | −→a | − |−→b |; (3) | −→a − −→b | = | −→a | − |−→b |; (4) | −→a − −→b | = | −→a | + | −→b |. ): (1) |^“n/ü>Úu1n>”: −→a //−→b . ¦ | −→a + −→b | = | −→a | + | −→b | 7L: −→a −→b Ó, ½ −→a , −→b ¥k 0. (2) - −→c = −→a + −→b , K −→a = −→c − −→b , ªz: | −→c − −→b | = | −→c | + | −→b |. ¤± −→b //−→c . dd: −→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, Ó, | −→c | ≥ |−→b |. ¤±−→a //−→b , ½ −→b = 0 ½ −→a = 0. 12. y²eØ ª, ¿`² Òoÿ¤á. (1) | −→b − −→a | > | −→a | − |−→b |; (2) | −→a + −→b + −→c | ≤ |−→a |+| −→b |+| −→c |. y²: (1) Xã, |^“n/ü> u1n>”yØ ª. ª¤á^ëSK 11(3): −→a //−→b , Ó, | −→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 p² 1 Ó, ½(ii) −→a , −→b , −→c ¥ü 0 (w¤ (i) A~). ∗13. O õ>/ A1A2 · · · An ¥%. y²: −−→OA1 + −−→OA2 + · · · + −−→OAn = 0.