.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 óê/. d. w,k: −−→OA1 + · · · + −−→OAn = 0. 2 w n Ûê/: ·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 < π n < π, −→q −→p ز1, −→p + −→q = 0 = −→p = −→q = 0. ∗14. O õ>/ A1A2 · · · An ¥%, P ´?¿:. y²: −−→PA1 + −−→PA2 + · · · + −−→PAn = n −−→PO. y²: Ï −−→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, Kþ −→c = 3−→a + −→b −→d = 2−→a − −→b ´Ä5 '? ): 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 þÑUüþ −→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 CþêL§ê, ½k|") 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 5', l ¡. 3. y²nþ k1 −→a − k2 −→b , k2 −→b − k3 −→c , k3 −→c − k1 −→a ¡. y²: d ª (k1 −→a − k2 −→b ) + (k2 −→b − k3 −→c ) + (k3 −→c − k1 −→a ) = 0, ù 3 þ5', ¤±¡. 4. −→a1 = 2 −→b1 + 3 −→b2 − −→b3 , −→a2 = −→b2 − −→b3 , −→a3 = −→b2 + −→b3 . y²þ −→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 Ø " = 2k1, 3k1 + k2 + k3, −k1 − k2 + k3 Ø ". = −→a1, −→a2, −→a3 5' = −→b1 , −→b2 , −→b3 5 '. 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 ): Ï −−→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©5, | −−→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¡þkn/ △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) ¦' 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 .
$2向量的共线与共面.9.2.(2)设OE=kOA=k,则由OE=OD+mDC可知k：6+35→4b2a解得k=，因此0E：OA=4：5.m5!38.在△ABC中，点M分线段AB为2：1.点N分线段AC为3：2.设CM与BN的交点为P,直线AP与边BC交于点Q.试用AB,AC来表示AP 和AQ.2AB, AN = 2AC,所以 BN = AN - AB-2A解：因为AM=AC3.2AB-AC.设CP=kCM,BP= mBN.则AB, CM = AM - AC -3CP-CB+BP得(AB-AC))-A-AC+m(AC-AB)?(32，所以解出k=32(2AB-AC)AP=AC+CP-AC+ 3=AB+IAC:又点Q在BC及AP的延长线上，所以AQ=IAP=AB+sBC.即(AB+IAC) =AB +s(AC-AB),394AB+AC.，即有AQ=解出1=LQM第8题图第9题图9.设ABCD是平行四边形，P,Q分别是边BC,CD的中点．证明AP,AQ与对角线BD相交于E,F,而将BD三等分证明：设AB=α,AD=b,则BD=AD-AB=b -a
§ 2 þ¡ · 9 · (2) −−→OE = k −→OA = k −→a , Kd −−→OE = −−→OD + m −−→DC k −→a = 2 3 −→b + m 2 −→a − 5 3 −→b . ) k = 4 5 , Ïd OE : OA = 4 : 5. 8. 3 △ABC ¥, : M ©ã AB 2 : 1, : N ©ã AC 3 : 2. CM BN : P, AP > BC u: Q. Á^ −−→AB, −→AC 5L« −→AP Ú −→AQ. ): Ï −−→AM = 2 3 −−→AB, −−→AN = 3 5 −→AC, ¤± −−→BN = −−→AN − −−→AB = 3 5 −→AC − −−→AB, −−→CM = −−→AM − −→AC = 2 3 −−→AB − −→AC. −−→CP = k −−→CM, −−→BP = m −−→BN. K −−→CP = −−→CB + −−→BP k 2 3 −−→AB − −→AC = −−→AB − −→AC + m 3 5 −→AC − −−→AB . )Ñ k = 2 3 . ¤± −→AP = −→AC + −−→CP = −→AC + 2 3 2 3 −−→AB − −→AC = 4 9 −−→AB + 1 3 −→AC. q: Q 3 BC 9 AP òþ, ¤± −→AQ = l −→AP = −−→AB + s −−→BC. = l 4 9 −−→AB + 1 3 −→AC = −−→AB + s( −→AC − −−→AB). )Ñ l = 9 7 , =k −→AQ = 4 7 −−→AB + 3 7 −→AC. B C M A Q N P 1 8 Kã A C B D Q P F E 1 9 Kã 9. ABCD ´²1o>/, P,Q ©O´> BC,CD ¥:. y² AP,AQ é BD u E,F, ò BD n ©. y²: −−→AB = −→a , −−→AD = −→b , K −−→BD = −−→AD − −−→AB = −→b − −→a ,
.10 .第一章向量代数126BC=a+AP-AB+2211AQ=AD+DC=6+a22又设AE=kAPAF=mAQ3(k>0)，3(m>0)则kb,AF=mb+a.AE=ka+!22但是AE=AB+tBD=a+t(6 -a)=(1-t)a+tb (t>0),所以=(1-)+6,ka+2即：(k+t-1)a =由于与6不平行所以2k=k+t-1=0131即k?-0t=3:2.同理，由AF=AB+$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), : 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: éuØ3þ 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.