S2向量的共线与共面.11.证明：已知ABC三点不共线，故AB,AC线性无关.任意点M位于平面ABC上当且仅当AM,AB,AC共面，即：AM.AB.AC线性相关，当且仅当存在不全为0的实数m1m2.m3，使miAM+mAB+m3AC=0当且仅当对于定点O有：mi(OM-OA) +m2(OB-OA)+m3(OC-OA) = 0当且仅当miOM=(m1+m2+m3)OA-m2OB-mOC显然m1≠0,不然与AB,AC线性无关矛盾.因此若记1m(m+m2+m3),k2=ki=m1mim3k3 =mi则OM=kiOA+kOB+kaOC且 k1 + k2 + k3 = 1.11.设O是一个定点，证明：点M位于△ABC上（包括它的边）的充分必要条件是存在非负实数ki,k2,k3，使得OM=k,OA+k2OB+k3OC,且ki+k2+k3=1.证明：延长AM，必可交BC于D点.因此AM=IAD，其中0≤1≤1.由于D在线段BC上，根据例2.1，存在实数m1，m2，使得OD=mOB+m2OC,m1+m2=1,m1,m2≥0.于是OM=OA+AM = (1-I)OA+IOD=(1-I)OA+lmiOB+Im2OC令ki=1-l,k2=lm1,k3=lm2,即得OM =kOA+kzOB+kgOC, ki + k2 + k3 =1, k1, k2, kg ≥0
§ 2 þ¡ · 11 · y²: ® A B C n:Ø, −−→AB, −→AC 5Ã'. ?¿: M u² ¡ ABC þ = −−→AM, −−→AB, −→AC ¡, =: −−→AM, −−→AB, −→AC 5', = 3Ø 0 ¢êm1,m2,m3, ¦ m1 −−→AM + m2 −−→AB + m3 −→AC = 0, =éu½: O k: m1( −−→OM − −→OA) + m2( −−→OB − −→OA) + m3( −−→OC − −→OA) = 0, = m1 −−→OM = (m1 + m2 + m3) −→OA − m2 −−→OB − m3 −−→OC. w, m1 6= 0, Ø, −−→AB, −→AC 5Ã'gñ. ÏdeP: k1 = 1 m1 (m1 + m2 + m3), k2 = − m2 m1 , k3 = − m3 m1 , K −−→OM = k1 −→OA + k2 −−→OB + k3 −−→OC, k1 + k2 + k3 = 1. 11. O ´½:, y²: : M u △ABC þ ()§>) ¿© 7^´3K¢ê k1,k2,k3, ¦ −−→OM = k1 −→OA + k2 −−→OB + k3 −−→OC, k1 + k2 + k3 = 1. y²: ò AM, 7 BC u D :. Ïd −−→AM = l −−→AD, Ù¥ 0 ≤ l ≤ 1. du D 3ã BC þ, â~ 2.1, 3¢ê m1,m2, ¦ −−→OD = m1 −−→OB + m2 −−→OC, m1 + m2 = 1,m1,m2 ≥ 0. u´ −−→OM = −→OA + −−→AM = (1 − l) −→OA + l −−→OD = (1 − l) −→OA + lm1 −−→OB + lm2 −−→OC. - k1 = 1 − l, k2 = lm1, k3 = lm2, = −−→OM = k1 −→OA + k2 −−→OB + k3 −−→OC, k1 + k2 + k3 = 1, k1,k2,k3 ≥ 0.
.12 第一章向量代数反之，不妨设k1≠1,解方程组[=1-ki,(1 - = kik2可得人mimi = k21- kik3(1m2 = k3m2=1- ki则有0<1<1.mi +mz = 1,m,m2≥0,令OD = miOB +m2OC,则D点在线段BC上.由OM = (1-L)OA +IOD可以得出AM=IAD，因此M在线段AD上，从而在△ABC上.12.证明：任意不同的三点AB，C共线的充分必要条件是存在不全为零的实数ki,kz,k3，使得0=kiOA+kOB+kgOC,且ki+k2+kg=0.证明：ABC共线，当且仅当IAB+mAC=0(l,m都不为零)，当且仅当I(OB - OA) + m(OC- OA) = 0,当且仅当-(l+m)OA+1OB+mOC=0.令ki=-(l+m)k2=l,k3=m，显然它们不全为零，且：kOA+kOB+k3OC=0,k1+k2 + k3=0.13.证明：任意不同的四点A,B,C,D共面的充分必要条件是存在四个不全为零的实数，使得0 = kiOA+kOB+ kOC + k,OD,，且i+k2+k3+=0证明：ABCD共面当且仅当AB,AC,AD线性相关，当且仅当有不全为零的数l,m,n使：IAB+mAC+nAD=0
· 12 · 1Ù þê , Ø k1 6= 1, )§| 1 − l = k1 lm1 = k2 lm2 = k3 l = 1 − k1, m1 = k2 1 − k1 , m2 = k3 1 − k1 , Kk m1 + m2 = 1, m1,m2 ≥ 0, 0 < l ≤ 1. - −−→OD = m1 −−→OB + m2 −−→OC, K D :3ã BC þ. d −−→OM = (1 − l) −→OA + l −−→OD ±Ñ −−→AM = l −−→AD, Ïd M 3ã AD þ, l 3 △ABC þ. 12. y²: ?¿ØÓn: A,B,C ¿©7^´3Ø " ¢ê k1,k2,k3, ¦ 0 = k1 −→OA + k2 −−→OB + k3 −−→OC, k1 + k2 + k3 = 0. y²: A B C , = l −−→AB + m −→AC = 0 (l,m ÑØ"), = l( −−→OB − −→OA) + m( −−→OC − −→OA) = 0, = −(l + m) −→OA + l −−→OB + m −−→OC = 0. - k1 = −(l + m), k2 = l, k3 = m, w,§Ø ", : k1 −→OA + k2 −−→OB + k3 −−→OC = 0, k1 + k2 + k3 = 0. 13. y²: ?¿ØÓo: A,B,C,D ¡¿©7^´3oØ "¢ê, ¦ 0 = k1 −→OA + k2 −−→OB + k3 −−→OC + k4 −−→OD, k1 + k2 + k3 + k4 = 0. y²: A B C D ¡ = −−→AB, −→AC, −−→AD 5', =kØ "ê l,m,n ¦: l −−→AB + m −→AC + n −−→AD = 0,
.13 :82向量的共线与共面当且仅当I(OB- OA) + m(OC-OA) + n(OD -OA) = 0当且仅当-(I+m+n)OA+IOB+mOC+ nOD= 0记ki=-(l+m+n)，kz=l,k3=m，ka=n，显然它们不全为零，使得kiOA+k2OB+kgOC+k,OD=0,ki+k2+kg+k4=0.*14，用向量的方法证明契维定理：若△AABC的三条边AB.BC.CA依次被分割成AF：FB=k1:kz，BD:DC=k3：k1,CE：EA=kz:k3其中，ki,k2.k3均为正数.则△ABC的顶点与它对边的分点的连线交于一点M，且对于任意一点O有1OM =h ++k(kOA+kOB+ kaOC).AFB第11题图证明：根据分点D与E的定义可得k3k3BD:-BC.AE-AC-ki + k3k2 + k3于是k3k3AD=AB+BD=AB+BC= AB+(AC-AB)ki+ksk+kkik3_AB +-AC,ki +k3ki + k3k3_AC-AB.BE-AE-AB -k2 +k3设AD与BE交于M,则有AM=IAD.BM=mBE
§ 2 þ¡ · 13 · = l( −−→OB − −→OA) + m( −−→OC − −→OA) + n( −−→OD − −→OA) = 0, = −(l + m + n) −→OA + l −−→OB + m −−→OC + n −−→OD = 0. P k1 = −(l + m + n), k2 = l, k3 = m, k4 = n, w,§Ø ", ¦ k1 −→OA + k2 −−→OB + k3 −−→OC + k4 −−→OD = 0, k1 + k2 + k3 + k4 = 0. ∗14. ^þ{y²ê½n: e △ABC n^> AB, BC, CA g©¤ AF : FB = k1 : k2, BD : DC = k3 : k1, CE : EA = k2 : k3, Ù ¥, k1,k2.k3 þê. K △ABC º:§é>©:ëu: M, éu?¿: O k −−→OM = 1 k1 + k2 + k3 (k2 −→OA + k1 −−→OB + k3 −−→OC). A F B C D E M 1 11 Kã y²: â©: D E ½Â −−→BD = k3 k1 + k3 −−→BC, −→AE = k3 k2 + k3 −→AC. u´ −−→AD = −−→AB + −−→BD = −−→AB + k3 k1 + k3 −−→BC = −−→AB + k3 k1 + k3 ( −→AC − −−→AB) = k1 k1 + k3 −−→AB + k3 k1 + k3 −→AC, −−→BE = −→AE − −−→AB = k3 k2 + k3 −→AC − −−→AB. AD BE u M, Kk −−→AM = l −−→AD, −−→BM = m −−→BE.
:14 .第一章向量代数把前面得到的表达式代人以下等式：AM=AB+BM，得到k1ki3ks_AC-ABAB +-AB+mACka+kaki +k3ki +k3由于AB与AC线性无关，由上述等式得到方程组：Ik3mk3ki +k3店7kz + k3+ k3ki + k2kK解得ikiko+k:1-mm(ki+k3ki + k2 + k3即ki +k3AMADki +k2 + k3又设AD与CF相交于M.同理可得ki +k3ADAM'-k1 + k2 + k3即M与M'重合，因此AD,BE,CF交于同一点M.对任意点0.有ki + k3k2k3OM =OA+AM=OA+AB +ACk, +k2 + k3(ki + k3ki + k3kik3= OA +-(OC-OA)-(OB-OA) +ki + k2 + kski + kz + ks1-(k2OA+kiOB+k3OC)ki + k2 + k383用坐标表示向量1.设P,Q两点在标架[O:ei,e,el下的坐标分别是(2,2,1),（-1,-1,3)试画出P,Q点的位置解：见附图
· 14 · 1Ù þê rc¡Lª\±e ª: −−→AM = −−→AB + −−→BM, l k1 k1 + k3 −−→AB + k3 k1 + k3 −→AC = −−→AB + m k3 k2 + k3 −→AC − −−→AB . du −−→AB −→AC 5Ã', dþã ª§|: lk3 k1 + k3 = mk3 k2 + k3 lk1 k1 + k3 = 1 − m ) l = k1 + k3 k1 + k2 + k3 m = k2 + k3 k1 + k2 + k3 . = −−→AM = k1 + k3 k1 + k2 + k3 −−→AD. q AD CF u M′ , Ón −−→ AM′ = k1 + k3 k1 + k2 + k3 −−→AD, = M M′ Ü, Ïd AD,BE,CF uÓ: M. é?¿: O, k −−→OM = −→OA + −−→AM = −→OA + k1 + k3 k1 + k2 + k3 k2 k1 + k3 −−→AB + k3 k1 + k3 −→AC = −→OA + k1 k1 + k2 + k3 ( −−→OB − −→OA) + k3 k1 + k2 + k3 ( −−→OC − −→OA) = 1 k1 + k2 + k3 (k2 −→OA + k1 −−→OB + k3 −−→OC). § 3 ^IL«þ 1. P, Q ü:3Ie[O; −→e1 , −→e2 , −→e3 ]eI©O´(2, 2, 1), (−1, −1, 3). ÁxÑ P,Q : . ): Nã.
.15.S3用坐标表示向量Q(1, 1,3)e3P(2,2,1)ee1第2题图第1题图2. 对于平行四边形ABCD,求A,D,AD,DB在标架[C;AC,BDI下的坐标.解：CA=-AC=(-1)AC+0BD,点A坐标为(-1,0);CD-LCA+LBD=-LAC+LBD,221点D坐标为5121AD:AC+-BD20AD 坐标为DB=-BD =0AC +(-1)BDDB坐标为(0,-1).3.设，b，的坐标分别是(1,5,2)，(0,-3,4)，（-2,3,-1).求向量2a+，-3a+2b+4的坐标解：2a + = 2(1.5.2) +(-2.3.-1) = (2,10.4) +(-2,3,-1) (0, 13, 3)-3 +26 +4 =-3(1,5,2) +2(0,-3,4) +4(-2,3,-1)= (-3, -15, -6) + (0, -6, 8) + (-8, 12, -4) = (-11, -9, -2).4.已知A、B两点的坐标分别为(1,-2,3)，（4,1,2)(1）试确定点P的坐标，使点P分线段AB成定比3：2；(2)试确定点P的坐标，使点P分线段BA成定比-2：3
§ 3 ^IL«þ · 15 · b b O −→e1 −→e2 −→e3 P(2, 2, 1) Q(−1, −1, 3) 1 1 Kã A C B D O 1 2 Kã 2. éu²1o>/ABCD, ¦A,D, −−→AD, −−→DB3Ie[C; −→AC, −−→BD] e I. ): −→CA = − −→AC = (−1)−→AC + 0 −−→BD, : A I (−1, 0); −−→CD = 1 2 −→CA + 1 2 −−→BD = − 1 2 −→AC + 1 2 −−→BD, : D I − 1 2 , 1 2 ; −−→AD = 1 2 −→AC + 1 2 −−→BD, −−→AD I 1 2 , 1 2 ; −−→DB = − −−→BD = 0 −→AC + (−1)−−→BD, −−→DB I (0, −1). 3. −→a , −→b , −→c I©O´ (1, 5, 2), (0, −3, 4), (−2, 3, −1). ¦þ 2 −→a + −→c , −3 −→a + 2 −→b + 4−→c I. ): 2−→a + −→c = 2(1, 5, 2) + (−2, 3, −1) = (2, 10, 4) + (−2, 3, −1) = (0, 13, 3). − 3 −→a + 2 −→b + 4−→c = −3(1, 5, 2) + 2(0, −3, 4) + 4(−2, 3, −1) = (−3, −15, −6) + (0, −6, 8) + (−8, 12, −4) = (−11, −9, −2). 4. ® A!B ü:I©O (1, −2, 3), (4, 1, 2). (1) Á(½: P I, ¦: P ©ã AB ¤½' 3 : 2; (2) Á(½: P I, ¦: P ©ã BA ¤½' −2 : 3