《高等代数与解析几何》课程教学资源（习题解答）第一章 向量代数

习题解答 第一章向量代数 习题1-1 1.如图,已知平行六面体ABCD-A1B1C1D1,E、F分别是棱BC、C1D1的中点.设AB=a, AD=b,AA1=.试用,b,表示下列向量 解:(1)因为 BC=AD. CC= AA. AC=AB+ BC+CC 所以 AC=0+b+c (2)因为BD=B+DD1,而 BD=AD-AB=b-C DD 所以 BD,= 6 (3)=A+DD+D1F,而 DD,=AA. D,=-D,C 所以 AF=a+b (4)EF=AF-AE-AF-(AB+BE)=AF-(AB+BE)=AF-(AB+Bc) 第1题图 第3(1)题图 2.要使下列各式成立向量a,b应满足什么条件? a+b|=|+|b (2)+b|=|d-|b; (3)|a-b|=|-|b| (4)|d-b|=|+|b1 解:(1)利用“三角形两边之和大于第三边可知7∥b且要使+b=7+b必须: 与b同向或云,b中至少有一为0

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1–1 1.  ,   ABCD−A1B1C1D1, E￾ F  BC￾ C1D1 .  −→AB = −→a , −→AD = −→b , −−→AA1 = −→c .  −→a , −→b , −→c  : (1) −−→AC1; (2) −−→BD1; (3) −→AF; (4) −→EF. : (1) !" −→BC = −→AD, −−→CC1 = −−→AA1, −−→AC1 = −→AB + −→BC + −−→CC1, #$ −−→AC1 = −→a + −→b + −→c . (2) !" −−→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, #$ −→AF = 1 2 −→a + −→b + −→c . (4) −→EF = −→AF − −→AE = −→AF − ( −→AB + −→BE) = −→AF − ( −→AB + −→BE) = −→AF − ³−→AB + 1 2 −→BC´ = 1 2 −→a + −→b + −→c − −→a − 1 2 −→b = − 1 2 −→a + 1 2 −→b + −→c . uuu       uuu      
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2. &'()*+,  −→a , −→b ,-./012? (1) | −→a + −→b | = | −→a | + | −→b |; (2) | −→a + −→b | = | −→a | − |−→b |; (3) | −→a − −→b | = | −→a | − |−→b |; (4) | −→a − −→b | = | −→a | + | −→b |. : (1) 3“456789:;: −→a //−→b . ?&' | −→a + −→b | = | −→a | + | −→b | @A: −→a B −→b C, D −→a , −→b EFGH" 0. · 1 ·

2)令可=可+b,则可=可-b,原式化为|可-b=|+b1.所以b∥且反向由 此可得:a∥b,反向,且d≥|b1,或b=0 所以可∥b且同向又因-b1≥0,所以可≥b,或b=a+1.由()知b/且同向 (3)令己=可-b,则7=b+,原式化为:b+=|b (4)令乙=a-b,则a=b+,原式化为:|b+=|c1-1b1由(2)知:c∥b且反向 或b=0.,同时,|C≥|b,.所以a∥b且反向,或b=0或a=0. 3.证明下列不等式,并说明等号什么时候成立 )|b-|≥|可|-|b (2)|a+b+≤|+|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23):a∥b,同向,且≥|b,或b=0 (2)令d=b+己.则+b+=可+d≤同a+|=|a+|b+|≤|+|b'+|7 等号成立当且仅当()可,b,互相平行且同向或(G1)z,b,中至少两个为0(也可看成()的特 例) 4.在四边形ABCD中,AB=可+2b,BC=-4-b,CD=-57-3b(可,b都是非零 向量).证明ABCD为梯形 证明∵A=A方+BC+CD=-87-2b=2BC,:AD/BC.但国b=2B,所以 ABCD是梯形 5.设 ABCDEF为正六边形,求AB+AC+AD+AE+A 解:∵AD=AC+AF=AE+AB,∴AB+AC+AD+AE+AF=3AD 第6题图 第5题图 6.设L,M,N分别是三角形ABC的三边BC,CA,AB的中点证明三中线向量五,BM,CN可 以构成一个三角形 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 A 2(4B+北 (BA+BC), CN=(CA+ CB) 可得A+BM+CN=0 7.在三角形ABC中求一点O,使

(2) I −→c = −→a + −→b , J −→a = −→c − −→b , K)L": | −→c − −→b | = | −→c | + | −→b |. #$ −→b //−→c ?M. N O>P: −→a //−→b , M, ? | −→a | > | −→b |, D −→b = 0. (3) I −→c = −→a − −→b , J −→a = −→b + −→c , K)L": | −→b | + | −→c | = | −→b + −→c |. N (1) : −→b //−→c ?C. #$ −→a //−→b ?C. Q! | −→a − −→b | > 0, #$ | −→a | > | −→b |, D −→b = 0. (4) I −→c = −→a − −→b , J −→a = −→b + −→c , K)L": | −→b + −→c | = | −→c | − |−→b |. N (2) : −→c //−→b ?M, D −→b = 0, CR, | −→c | > | −→b |. #$ −→a //−→b ?M, D −→b = 0 D −→a = 0. 3. STUV), WXTVY/0RZ*+. (1) | −→b − −→a | > | −→a | − |−→b |; (2) | −→a + −→b + −→c | 6 | −→a | + | −→b | + | −→c |. : (1)  , 3“456789[\P]SUV). V)*+12>^_` a 2(3): −→a //−→b , C, ? | −→a | > | −→b |, D −→b = 0. (2) I −→d = −→b + −→c . J: | −→a + −→b + −→c | = | −→a + −→d | 6 | −→a |+| −→d | = | −→a |+| −→b + −→c | 6 | −→a |+| −→b |+| −→c |. VY*+b?cb (i) −→a , −→b , −→c de ?C, D (ii) −→a , −→b , −→c EF7f" 0 (g>h* (i) i j). 4. kl86 ABCD , −→AB = −→a + 2 −→b , −→BC = −4 −→a − −→b , −→CD = −5 −→a − 3 −→b ( −→a , −→b mno  ). ST ABCD "p6. :  −→AD = −→AB + −→BC + −→CD = −8 −→a − 2 −→b = 2 −→BC,  −→AD//−→BC. q | −→AD| = 2| −→BC|, #$ ABCD p6. 5.  ABCDEF "r86, s −→AB + −→AC + −→AD + −→AE + −→AF. :  −→AD = −→AC + −→AF = −→AE + −→AB,  −→AB + −→AC + −→AD + −→AE + −→AF = 3 −→AD.                 uuuu@P n m n m n m n m n m n m n m n +                 P@uuuu  +n m n m n m n m n m n m n m n uuuuuuuur F. @P - . - . - . . - . - . - . - . - . - . - . - . - . RSRSRSRSRSRSRSRSRSRRSRSRSRZ   =  ce u t 5C%4 KL ][ ;: EF A D F B E F ￾ 5
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 :  / GUH  $u*Hf456. : !" −→AL, −−→BM, −−→CN >$u*Hf456, b?cbvw4f 9:"o . N −→AL = 1 2 ( −→AB + −→AC), −−→BM = 1 2 ( −→BA + −→BC), −−→CN = 1 2 ( −→CA + −→CB), >P: −→AL + −−→BM + −−→CN = 0. 7. k456 ABC sH O, ' −→OA + −→OB + −→OC = 0. · 2 ·

解设L是BC的中点则五=是(AB+C在线段AL上取O点使0=3互=喜(B+C, 则O点在△ABC的内部.验证: +OB+ B)+(OA+AC)=30A+ AB+ -(AB+AC)+AB+AC=0 8.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 N D+ Bcy 证明如图, CA+C 所以 MN=CN-CM=C CA+CB)=t(AD+BO 第9题图 9.设M是平行四边形ABCD的中心,O是.,一点.证明 +Oc+OD= 40M 证明:如图,因为 OM ),OM=( A+OB+∂C+OD 10.设ABCD是平行四边形,P,Q分别是边BC,CD的中点.证明AP,AQ与.角线BD相于 E,F,而将BD三等分 证明设A=,AD=b,则 AB= b b 又设 AE=kAP (k>0), AF: m> AE=ka+k 6. AF=m6+a 但是 =AB+tB=a+t(b-a)=(1-1)+tb(t>0)

: LBC , J −→AL = 1 2 ( −→AB+ −→AC). ktxALyzO, ' −→AO = 2 3 −→AL = 1 3 ( −→AB+ −→AC), J O k 4ABC {|. }S: −→OA + −→OB + −→OC = −→OA + (−→OA + −→AB) + (−→OA + −→AC) = 3−→OA + −→AB + −→AC = −( −→AB + −→AC) + −→AB + −→AC = 0. 8.  A, B, C, D Hfllf~, M, N 8 AB, CD . ST: −−→MN = 1 2 ( −→AD + −→BC). :  , −−→CM = 1 2 ( −→CA + −→CB), −−→CN = 1 2 −→CD, #$ −−→MN = −−→CN − −−→CM = 1 2 −→CD − 1 2 ( −→CA + −→CB) = 1 2 ( −→AD + −→BC). ( ( ( ( ( ( ( ( ( ( ( ( ( ( s s s s s `                ?  XQQRQQQQQQQ
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H. ST: −→OA + −→OB + −→OC + −→OD = 4−−→OM. :  , !" −−→OM = 1 2 ( −→OA + −→OC), −−→OM = 1 2 ( −→OB + −→OD), #$ −→OA + −→OB + −→OC + −→OD = 4−−→OM. 10.  ABCD  l86, P, Q 8 BC, CD . ST AP, AQ B5t BD e 0), −→AF = m −→AQ (m > 0), J −→AE = k −→a + k 2 −→b , −→AF = m −→b + m 2 −→a . q −→AE = −→AB + t −→BD = −→a + t( −→b − −→a ) = (1 − t) −→a + t −→b (t > 0). · 3 ·

所以 (1-t)a+ (k+t-1)7=(t-5)b 由于矿与b不平行,所以 即 AF=AB+sBD=(l-sb+sb(s>0). 可得 +s-1=0 即: 最后得到 BD 说明E,F是线段BD的三等分点 11.O为正多边形A1A2…An的中心证明 O41+OA2+……+OAn=0. 证明先考虑n为偶数的情形此.显然有OA1+…+OA=0.再看n为奇数的情形:我们增 加一倍顶点B1,…,Bn使原来正n边形A1…An成为:A1B1A2B2……An-1Bn-1AnBn,这是一个2n边 形.所以 41+B+O4+DB2+…+0An+OBn=0 注意到OB1是由OA旋转一个定角6而得到(0<6<丌)若记 p=OA1+……+O4 q=OB1+…+OBn 那可是由旋转θ角而得到由于0<6<丌,可与不平行,故+了=0当且仅当p=可=0 12.O为正多边形A1A2…An的中心P是任意一点证明 +PA2+……+PA 证明因为 A+A2O(i=1,2,……,n), 所以 nO=PA1+…+PAn+(A1+…+An6)=PA1+…+PAn (利用第11题的结论) 习题1-2 1.已知平行四边形ABCD的对角线为AC和BD.设AC=可,BD=b求AB,CD,DA :如图 有B=6+B=16+1bB

#$ k −→a + k 2 −→b = (1 − t) −→a + t −→b , : (k + t − 1)−→a = µ t − k 2 ¶ −→b , N 0), >P: ( m 2 + s − 1 = 0 s − m = 0, : ( m = 2 3 s = 2 3 . P: −→BF = 2 3 −→BD, −→BE = 1 3 −→BD, XT E, F tx BD 4V. 11. O "r 86 A1A2 · · · An . ST: −−→OA1 + −−→OA2 + · · · + −−→OAn = 0. :  n " 6. OR. G: −−→OA1 + · · · + −−→OAn = 0. h n "6:  H~ B1, · · · , Bn 'Kr n 86 A1 · · · An *": A1B1A2B2 · · · An−1Bn−1AnBn, wHf 2n 8 6. #$ −−→OA1 + −−→OB1 + −−→OA2 + −−→OB2 + · · · + −−→OAn + −−→OBn = 0. 
 −−→OBi N −−→OAi Hf5 θ %P (0 < θ < π), : −→p = −−→OA1 + · · · + −−→OAn, −→q = −−→OB1 + · · · + −−→OBn, 0 −→q N −→p  θ 5%P. N< 0 < θ < π, −→q B −→p U , ! −→p + −→q = 0 b?cb −→p = −→q = 0. 12. O "r 86 A1A2 · · · An , P ￾
H. ST: −−→P A1 + −−→P A2 + · · · + −−→P An = n −→P O. : !" −→P O = −−→P Ai + −−→AiO (i = 1, 2, · · · , n), #$ n −→P O = −−→P A1 + · · · + −−→P An + (−−→A1O + · · · + −−→AnO) = −−→P A1 + · · · + −−→P An (3= 11 a"#).
1–2 1.  l86 ABCD 5t" AC : BD.  −→AC = −→a , −→BD = −→b . s −→AB, −→CD, −→DA. :  , −→AB = −→AO + −→OB = 1 2 −→AC + 1 2 −→DB = 1 2 ( −→a − −→b ), · 4 ·

CD B (a-b), A=-A=-(A0+D 了∠个 第1题图 第2题图 2.在三角形ABC中,点M,N是AB边上的三等分点设CA=a,CB=b.求CM,CN 解:如图,因为 AM 1B. AN=LAB 所以 CM=CA+AM=CA+AB=CA+(CB-CA)=3+3 不N=可+不N=+2=可A+2(CB-A=2万+1 3.在四面体O-ABC中,设点G是三角形ABC的重心.用OA,OB,OC来表示向量O 解:如图,由 4D、1 (AB + AC) 可得G=+而菇B=DB-可C=0-,所以 =1(B+0-20,6=1 B+O 4题图 第3题图 4.设AT是三角形ABC中∠A的平分线(与BC交于T点),将7用ABC来表 解:设B7=kBC,则7C=(1-)BC.由角平分线的性质可知函B:A(6=k:(1-k,因此 JAB+ACI AT=AB+BT=AB+kBC=(1-k)AB+kAC aBL+ -(AC AB +ABAC) 5.已知,b不线,则向量己=3+b与d=2a-b是否线性相关?

−→CD = − −→AB = − 1 2 ( −→a − −→b ), −→DA = − −→AD = −( −→AO + −→OD) = − 1 2 ( −→a + −→b ). 010010010100100100 100 10 10 010 0100 7                 Puuuuuu               7 uuuuuuP  '^\ ] \ \ \ \ \ ] \ \ \ \ \ ] \ \ \ SZ 7D)7 % f c A C B D O ￾ 1
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2. k456 ABC ,  M, N  AB 8y4V.  −→CA = −→a , −→CB = −→b . s −−→CM, −−→CN. :  , !" −−→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 . 3. kl O − ABC ,  G 456 ABC $.  −→OA, −→OB, −→OC  −→OG. :  , N −→OG = −→OA + −→AG, −→AG = 2 3 −→AD, −→AD = 1 2 ( −→AB + −→AC), >P −→AG = 1 3 ( −→AB + −→AC). % −→AB = −→OB − −→OA, −→AC = −→OC − −→OA, #$ −→AG = 1 3 ( −→OB + −→OC − 2 −→OA), −→OG = 1 3 ( −→OA + −→OB + −→OC).
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o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  565565565565565565565565565 , | −→AB| : | −→AC| = k : (1 − k), !O k = | −→AB| | −→AB| + | −→AC| . < −→AT = −→AB + −→BT = −→AB + k −→BC = (1 − k) −→AB + k −→AC = 1 | −→AB| + | −→AC| (| −→AC| −→AB + | −→AB| −→AC). 5.  −→a , −→b U(t, J −→c = 3−→a + −→b B −→d = 2−→a − −→b )t&e*? · 5 ·

质:kC+md=0,即 b=0 整理后为 )a+(k-m)b=0. 可,b、共,故可,b、性无关所重 3k+2m=0 解得k 即z,d、性无关 6.、明。个向量k1-k2b,k2b-kE,k7-k回共面 交明:、等否 (k17-k2b)+(k2b-k37)+(k37-k1a)=0 可这3个向量、性关所重共面 7.O/、个定点、明:对、在直上的3个点A,B,C,点M位1面ABC上的充分-性 条件/存在实数k1,k2,k3,质得 OM=k1OA+k2OB+k3OC, E k1+k2+k3=1 交明:已/A、B、C点共,故AB,AC、性无关.任意点M位-′面ABC上当且仅当 AM,AB,AC共面即:AM,AB,AC、性关,当且仅当存在、全为0的实数m,m2m3,质 C=0, 当且仅当对定点O m1(OM-OA)+m2(0B-OA)+m3(OC-OA)=0, 当且仅当 显然m1≠0,、然与BAC、性无关矛盾因此若记 k (m1+m2+m3),k2 OM=k10A+k20B+koC 且k1+k2+k3 8.O/、个定点、明:点M位一△ABC上(包括它的边)的充分-性条件/存在负实数 k1,k2,k3,质得 OM=k10A+koB+k3OC, Ek1+k+k3=1 交明:延长AM,-可交BC-D点因此AM=1AD,其中0≤1≤1、-D在段BC上根据 例2.1,存在实数m1,m2,质得 +m 1 AM

: G k￾ m ': k −→c + m −→d = 0,  3k −→a + k −→b + 2m−→a − m −→b = 0, +" (3k + 2m) −→a + (k − m) −→b = 0. Nw 3 f t&e*, #$(. 7. O Hf, ST: BC j 2.1, 1k2 m1, m2, 'P −→OD = m1 −→OB + m2 −→OC, m1 + m2 = 1, m1, m2 > 0. < −−→OM = −→OA + −−→AM = (1 − l) −→OA + l −→OD = (1 − l) −→OA + lm1 −→OB + lm2 −→OC. · 6 ·

令k1=1-l,k2=lm1,k3=lm2,即得 OM=k,0A+k20B+k30C, k1+k2+k3=1, k,k2, k3>0 反之,不妨设k1≠1,解方程组 1-1=k1 l=1-k1 m1=k2可得{m1=1-k Im2= k3 n. 则有 m1+m2=1,m1,m2≥0,0<l≤1. 令 OD=m,oB +m,O 则D点在线段BC上.由 DM=(1-1)0A+10D 可以得出AM=1AD,因此M在线段AD上,从而在△ABC上 9.证明:任意不同的三点A,B,C共线的充分必要条件是存在不全为零的实数k1,k2,k3,使得 0=k1OA+k20B+k3OC, Ek1+k2+k3=0 证明:A、B、C共线,当且仅当LAB+mAC=0(,m都不为零),当且仅当 OA)+m(OC-O 当且仅当 -(L+ m)OA+lOB +mOC=0. 令k1=-(+m),k2=l,k3=m,显然它们不全为零,且: k,0A+k20B+k30c=0, k1+k2+k3=0 10.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=k0A+k0B+kaoC+kOD. k+k+katk=o 证明:A、B、C、D共面当且仅当AB,ACD线性相关,当且仅当有不全为零的数1,m,n使 LAB+mAC+nAD=0 当且仅当 l(OB-OA)+m(OC-0A)+n(OD-OA)=0 当且仅当 (L+m+n)Oa+loB+mOC +nOD=0 记k1=-(+m+m),k2=l,k3=m,k4=n,显然它们不全为零,使得 610A+k20B +k30C+kOD=0, k1+k2+k3+k=0

I k1 = 1 − l, k2 = lm1, k3 = lm2, P −−→OM = k1 −→OA + k2 −→OB + k3 −→OC, k1 + k2 + k3 = 1, k1, k2, k3 > 0. M9, U? k1 6= 1, -@AB    1 − l = k1 lm1 = k2 lm2 = k3 >P    l = 1 − k1, m1 = k2 1 − k1 , m2 = k3 1 − k1 , JG m1 + m2 = 1, m1, m2 > 0, 0 $P% −−→AM = l −→AD, !O M ktx AD y, C%k 4ABC y. 9. ST: ￾
UC4 A, B, C (t0@&121kU3"o2 k1, k2, k3, 'P 0 = k1 −→OA + k2 −→OB + k3 −→OC, ? k1 + k2 + k3 = 0. : A￾ B￾ C (t, b?cb l −→AB + m −→AC = 0 (l, m mU"o), b?cb l( −→OB − −→OA) + m( −→OC − −→OA) = 0, b?cb −(l + m) −→OA + l −→OB + m −→OC = 0. I k1 = −(l + m), k2 = l, k3 = m, 8U3"o, ?: k1 −→OA + k2 −→OB + k3 −→OC = 0, k1 + k2 + k3 = 0. 10. ST: ￾
UCl A, B, C, D (0@&121klfU3"o2, 'P 0 = k1 −→OA + k2 −→OB + k3 −→OC + k4 −→OD, ? k1 + k2 + k3 + k4 = 0. : A￾ B￾ C￾ D (b?cb −→AB, −→AC, −→AD t&e*, b?cbGU3"o l, m, n ': l −→AB + m −→AC + n −→AD = 0, b?cb l( −→OB − −→OA) + m( −→OC − −→OA) + n( −→OD − −→OA) = 0, b?cb −(l + m + n) −→OA + l −→OB + m −→OC + n −→OD = 0.  k1 = −(l + m + n), k2 = l, k3 = m, k4 = n, 8U3"o, 'P k1 −→OA + k2 −→OB + k3 −→OC + k4 −→OD = 0, k1 + k2 + k3 + k4 = 0. · 7 ·

第11题图 *11.注记那们方法证明契维定理:若三角形ABC们三条边AB,BC,CA依次被然割成AF:FB= k1:k2,BD:DC=k3:k1,CE:EA=k2:k3,其增,k1,k2.k3均结正数.则三角形ABC们顶加与它对 边们然加们连线交于一加M,且对于任意一加O有 OM 1+k2+k3 (k2OA +k10B+k3OC) 证即:根据然加D与E们定义可得 BD BO k1+k3 于奇 AD=AB+ BD=AB (AC-AB k1 AB+ k1+k3 E=AE-AB AC-A 倍AD与BE交于M,则有 AM D BM BE 把前情得到们旋达式代入以定等式AM=AB+BM,得到 k1_AB+ AB +m AC-AB k1+k3 k1+k3 k2+ k 由于AB与AC线性无关,由上述等式得到方程组 mka k1+k3 k2+k I=ki+k 解得 k1+k2+k3 k k2+k 即 k1+k2+k3 又倍AD与CF相交于M,同理可得 AM k1+k3 Ad k1+k2+k3 即M与M重合,故此AD,BE,CF交于同一加M 对任意加O,有 OM=0A+AZ k1+k3 k1+k2+k3(k1+k3 ky (OB-OA)+ k1+k2+k3 OA+hoB+koC) k1+k2+k3

                 o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  54555455545554555455545554K VKLKLKKLKLKLKLKKLKLKLK A F B C D E M ￾ 11
∗11.  @DSTEF: 456ABC 418AB, BC, CAGHIJ*AF : F B = k1 : k2, BD : DC = k3 : k1, CE : EA = k2 : k3,  D B E M>P −→BD = k3 k1 + k3 −→BC, −→AE = k3 k2 + k3 −→AC. P −−→ AM0 = k1 + k3 k1 + k2 + k3 −→AD,  M B M0 $T, !O AD, BE, CF <CH M. ￾
 O, G −−→OM = −→OA + −−→AM = −→OA + k1 + k3 k1 + k2 + k3 µ 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). · 8 ·

1.设P,Q两点在标架O;可1,,]下的坐标分别是(2,2,1),(-1,-1,3)试画出PQ点的位置 解:见附图. 第1题图 2.对于平行四边形ABCD,求A,D,AD,DB在标架C;AC,BD下的坐标 解:A=-AC=(-1)AC+0BD,点A坐标为(-1,0) Cd=2CA+ 2BD BD, 点D坐标为(-2 万D坐标为(号 B=-BD=0AC+(-1BD DB坐标为(0,-1) 3.设a,b,乙的坐标分别是(1,5,2),(0,-3,4,(-2,3,-1).求向量2a+c,-3a+2b+4 的坐标 解:2a+=2(1,5,2)+(-2,3,-1)=(2,10,4)+(-2,3,-1)=(0,13,3) 3a+2b+4=-3(1,5,2)+2(0,-3,4)+4(-2,3,-1) (-3-15,-6)+(0,-6,8)+(_8,12,-4 4.证明三角形的三条中线交于一点(重心) 证明设D,E,F分别是边BC,CA,AB上的中点AD与BE交于G,AD与CF交于G.则 AG=kAD=kl-AB+-Ac AB+-AC 若建立仿射标架A(小则点G坐标为(,)又 BG=mBE=m(BA+BC)=m-AB+3(AC-AB mAC-mAB 所以BG坐标为(-m2)但xG=万B+BG所以 (会)=00+(-m)

1–3 1.  P, Q 7kUV [O; −→e1 , −→e2 , −→e3 ] WU (2, 2, 1), (−1, −1, 3). X% P, Q /Y. : _Z . . . / . . . u@ . l x x x x x   & C  55 ;: EF r r ~ ~  . . . / . . uur . l x x x x x G  G  G  G    P P P . l . l . l . l . l . l . l . l . l . l . l O −→e1 −→e2 −→e3 P(2, 2, 1) Q(−1, −1, 3) ￾ 1
0100 100 10 10 010 010010010100100100 7                 Puuuuuu               7 uuuuuuP  '^\ ] \ \ \ \ \ ] \ \ \ \ \ ] \ \ \ A C B D O ￾ 2
2. < l86ABCD, sA, D, −→AD, −→DBkUV[C; −→AC, −→BD] WU. : −→CA = − −→AC = (−1)−→AC + 0 −→BD,  A WU" (−1, 0); −→CD = 1 2 −→CA + 1 2 −→BD = − 1 2 −→AC + 1 2 −→BD,  D WU" ³ − 1 2 , 1 2 ´ ; −→AD = 1 2 −→AC + 1 2 −→BD, −→AD WU" ³ 1 2 , 1 2 ´ ; −→DB = − −→BD = 0 −→AC + (−1)−→BD, −→DB WU" (0, −1). 3.  −→a , −→b , −→c WU (1, 5, 2), (0, −3, 4), (−2, 3, −1). s 2 −→a + −→c , −3 −→a + 2 −→b + 4−→c WU. : 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. ST45641t<H ($). :  D, E, F 8 BC, CA, AB y. AD B BE < G, AD B CF < G0 . J −→AG = k −→AD = k µ 1 2 −→AB + 1 2 −→AC¶ = k 2 −→AB + k 2 −→AC. [+\]UV [A; −→AB, −→AC], J G WU" ³ k 2 , k 2 ´ . Q −→BG = m −→BE = m µ 1 2 −→BA + 1 2 −→BC¶ = m · − 1 2 −→AB + 1 2 ( −→AC − −→AB) ¸ = m 2 −→AC − m −→AB, #$ −→BG WU" ³ −m, m 2 ´ . q −→AG = −→AB + −→BG, #$, µ k 2 , k 2 ¶ = (1, 0) + ³ −m, m 2 ´ , · 9 ·

解方程与 得k=m=3所以G的坐标为(,3)同解可以推得C的坐标为(3,3)证得G=C 第4题图 第5题图 5.证明三角形的三条角平分线证于一点 证明设△ABC的三条角平分线分别为AD,BE和CF.且设AD与BE证于T点.令 AT=kAD k (AC AB+JABJAC 1AB|+|4 建你法,且令=、正则7坐(同下)数的是 知道 BA (BCIBA+BABC 4+|BC 所以 =mBE la1+ b-al (-|b-da+|a(b-a) -ma'+-ma16 la+b-a 由于AT=AB+BT,所以 k =(1,0)+-m d+|b1|a+b1 la+ b-a 明 klbl la+b a+b-a 解得: Ial+b|+b-a 又设AD与CF证于T点, r=s万=-4b17+_7Lb la+b 得点的坐标为( +|b1’|+|b1 可面师麻+ 10

-@AB ( k 2 = 1 − m k 2 = m 2 P k = m = 2 3 . #$ G WU" ³ 1 3 , 1 3 ´ . C, >$^P G0 WU" ³ 1 3 , 1 3 ´ . SP G = G0 .       o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  666666666666 V 6 M 66666676666666 NMMMNMMMNMMMNMMMNMM A F B C DE M ￾ 4
                
o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  565565565565565565565565565 %YSSSSSSSSSSSSSSSSS A F B C D E M ￾ 5
5. ST456415t<H. :  4ABC 415t" AD, BE : CF. ? AD B BE < T . I −→AT = k −→AD = k | −→AB| + | −→AC| (| −→AC| −→AB + | −→AB| −→AC). [+\]UV [A; −→AB, −→AC], ?I −→AB = −→a , −→AC = −→b . J T WU Ã k| −→b | | −→a | + | −→b | , k| −→a | | −→a | + | −→b | ! . _ ` −→BE = 1 | −→BA| + | −→BC| (| −→BC| −→BA + | −→BA| −→BC), #$ −→BT = m −→BE = m | −→a | + | −→b − −→a | (−|−→b − −→a | −→a + | −→a |( −→b − −→a )) = −m−→a + m| −→a | | −→a | + | −→b − −→a | −→b . N< −→AT = −→AB + −→BT, #$ Ã k| −→b | | −→a | + | −→b | , k| −→a | | −→a | + | −→b | ! = (1, 0) + Ã −m, m| −→a | | −→a | + | −→b − −→a | ! , :    k| −→b | | −→a | + | −→b | = 1 − m k| −→a | | −→a | + | −→b | = m| −→a | | −→a | + | −→b − −→a | -P: k = | −→a | + | −→b | | −→a | + | −→b | + | −→b − −→a | . Q AD B CF < T 0 , −−→ AT0 = s −→AD = s| −→b | | −→a | + | −→b | −→a + s| −→a | | −→a | + | −→b | −→b . P T 0 WU" Ã s| −→b | | −→a | + | −→b | , s| −→a | | −→a | + | −→b | ! . −−→ CT0 = t −→CF = t | −→CA| + | −→CB| (| −→CB| −→CA + | −→CA| −→CB) = t| −→b | | −→b | + | −→a − −→b | −→a − t −→b , · 10 ·