第一章向量代数.16.2解：(1)由|API:|PB|=3：2可得APPB.利用例3.1的定比分点2141 12)3.可得P点坐标公式.取k=25-552PA，用定比分点公式算得P点坐标(2）由已知条件可得BP-3(10, 7,0).5已知A(1,-1)，B(-4,5),将线段AB延长至C使|AC=5|ABl.求点C的坐标解:(I)当AC=5.AB,则 AB+BC=5.AB.因此AB=IBC,即 B是线段AC的比值为的定比分点，所以11+1rcCR1+#解得：C(-24,29)-1 + ycYB=1+1(2)当 AC =-5AB,则 CA=5AB.所以C+5aBZA=31 + 5解得：C(26,-31)yc+5yBYA1+56.已知线段AB被点C(2,0,2)和D(5,-2,0)三等分，试求出这线段的两个端点A,B的坐标解：不妨设A,B,C,D四点如图所示，A,B两点的坐标分别为（&A,9A，ZA)与(B,YB,zB),则AC=CD,CD=DB.所以( A,C-A, ZC- ZA) = (D - AC,D -C, ZDzc),即:CA=2C-AD=-1YA=2yC-YD=2ZA=2ZC - ZD = 4.同理，CB=2D-AC=8YB=2yD-Yc=-4ZB=22-ZC=-2
· 16 · 1Ù þê ): (1) d | −→AP| : | −−→PB| = 3 : 2 −→AP = 3 2 −−→PB. |^~ 3.1 ½'©: úª, k = 3 2 , P :I 14 5 , − 1 5 , 12 5 . (2) d®^ −−→BP = − 2 3 −→PA, ^½'©:úª P :I (10, 7, 0). 5 ® A(1, −1), B(−4, 5), òã AB ò C ¦ |AC| = 5|AB|. ¦ : C I. ): (1) −→AC = 5 −−→AB, K −−→AB + −−→BC = 5 −−→AB. Ïd −−→AB = 1 4 −−→BC, = B ´ ã AC ' 1 4 ½'©:. ¤± xB = 1 + 1 4 xC 1 + 1 4 yB = −1 + 1 4 yC 1 + 1 4 ): C(−24, 29). (2) −→AC = −5 −−→AB, K −→CA = 5 −−→AB. ¤± xA = xC + 5xB 1 + 5 yA = yC + 5yB 1 + 5 ): C(26, −31). 6. ®ã AB : C(2, 0, 2) Ú D(5, −2, 0) n ©, Á¦Ñùã üà: A,B I. ): Ø A,B,C,D o:X㤫, A,B ü:I©O (xA,yA,zA) (xB,yB,zB), K −→AC = −−→CD, −−→CD = −−→DB. ¤± (xC − xA,yC − yA,zC − zA) = (xD − xC ,yD − yC,zD − zC ), =: xA = 2xC − xD = −1 yA = 2yC − yD = 2 zA = 2zC − zD = 4. Ón, xB = 2xD − xC = 8 yB = 2yD − yC = −4 zB = 2zD − zC = −2.
.17 :S3用坐标表示向量因此A,B两点的坐标分别为（-1,2,4)与(8,-4,-2)（两种可能)ABABCDCDEF:第6题图第7题图7.设A,B两点的坐标分别为（-6,5,-8)，（4,0,7)，试确定点C,D,E,F使C,D,E,F将线段AB五等分解：不妨设 A,B,C,D,E,F如图.所以ICB,2DBAC=AD=34PEB.AE=AF=4FB2利用定比分点公式算得C点坐标为（-4,4.-5).D点坐标为（-2.3.-2).E点坐标为(0,2,1)，F点坐标为(2,1,4)8.ABCD为平行四边形已知A，B及对角线交点的坐标分别为(-3,1,5)，(2,3,4)，(1,-1,2).试确定点C，D的坐标解：设对角线交点为M,C,D的坐标分别为（c,yc,zc),（aD,D,zD)由于M是A,C的中点，因此(-3+rc)=1(1 + yc) = -((5 + zc) = 2,解得C点坐标为(5,-3,-1).由于M也是B,D的中点，同理可得D点坐标为(0,1,0)9.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GAD与CF交于G.则kAG=kAD=k(AB+IACAB+-AC2222若建立仿射标架[A;AB,AC],则点 G坐标为2.2BG=mBB=m(BA+BC)=m[-IAB+(AC-AD)
§ 3 ^IL«þ · 17 · Ïd A,B ü:I©O (−1, 2, 4) (8, −4, −2) (ü«U). b b b b A C B D 1 6 Kã b b b b b b A C B D E F 1 7 Kã 7. A,B ü:I©O (−6, 5, −8), (4, 0, 7), Á(½: C,D,E,F, ¦ C,D,E,F òã AB Ê ©. ): Ø A,B,C,D,E,F Xã. ¤± −→AC = 1 4 −−→CB, −−→AD = 2 3 −−→DB, −→AE = 3 2 −−→EB, −→AF = 4 −−→FB. |^½'©:úª C :I (−4, 4, −5), D :I (−2, 3, −2), E : I (0, 2, 1), F :I (2, 1, 4). 8. ABCD ²1o>/. ® A, B 9é:I©O (−3, 1, 5), (2, −3, 4), (1, −1, 2). Á(½: C, D I. ): é: M, C,D I©O (xC ,yC ,zC ),(xD,yD,zD). du M ´ A,C ¥:, Ïd 1 2 (−3 + xC ) = 1 1 2 (1 + yC) = −1 1 2 (5 + zC ) = 2, ) C :I (5, −3, −1). du M ´ B,D ¥:, Ón D :I (0, 1, 0). 9. y²n/n^¥u: (%). y²: D,E,F ©O´> BC,CA,AB þ¥:. AD BE u G, AD CF u G′ . K −→AG = k −−→AD = k 1 2 −−→AB + 1 2 −→AC = k 2 −−→AB + k 2 −→AC. eïáIe [A; −−→AB, −→AC], K: G I k 2 , k 2 . q −−→BG = m −−→BE = m 1 2 −−→BA + 1 2 −−→BC = m − 1 2 −−→AB + 1 2 ( −→AC − −−→AB)
.18.第一章向量代数"AC-mAB,所以 BG 坐标为(-m,号).但 AG=AB+BG,所以,kk(鲁) =(1,0) + (-m,号),解方程组k=1-m122m(222(11所以G的坐标为得k=m:同理，可以推得G的坐标为3331证得G=G'3.3ABFRA第9题图第10题图*10，证明三角形的三条角平分线交于一点证明：设△ABC的三条角平分线分别为AD,BE和CF.且设AD与BE交于T点令kAT - kAD(IAC|AB +AB|AC)ABI+AC建立仿射标架[A;AB,AC],且令AB=，AC=6.则T点坐标是k/b6a我们还知道a|+61la|+6]1BE(IBCIBA+BAIBC),BA| +|BC所以mBT= mBE(-b -a+a(6 -))[al+/b-al
· 18 · 1Ù þê = m 2 −→AC − m −−→AB, ¤± −−→BG I −m, m 2 . −→AG = −−→AB + −−→BG, ¤±, k 2 , k 2 = (1, 0) + −m, m 2 , )§| k 2 = 1 − m k 2 = m 2 k = m = 2 3 . ¤± G I 1 3 , 1 3 . Ón, ±í G′ I 1 3 , 1 3 . y G = G′ . A F B C E D M 1 9 Kã A F B C D E M 1 10 Kã ∗10. y²n/n^²©u:. y²: △ABC n^²©©O AD,BE Ú CF. AD BE u T :. - −→AT = k −−→AD = k | −−→AB| + | −→AC| (| −→AC| −−→AB + | −−→AB| −→AC). ïáIe [A; −−→AB, −→AC], - −−→AB = −→a , −→AC = −→ b . K T :I´ 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 ))
83用坐标表示向量.19 .m|al6-ma-：[a]+/6 -a]由于AT=AB+BT,所以k/6k/alm/al(1.0)(/a|+161a/+/6]al+/6 -al即:K161-m[a]+/6]m/aa(/a|+/6)[al+/6 -a]解得：1a1+/6]k:[al+/6l+b -al又设AD与CF交于T点，s161saAT = SADh2[a|+ /6][a]+/6]s[6]sal得T点的坐标为[al+1611al+16]tttCFa-tb.CT!ICBCA2ICAICBCAI+ICB6l+/a- 6]由AT=AC+CT,得：8/6tbs/a(0.1)(/a]+/61[a]+/6]161+la-b即：s/6t/61[a]±[b][61+/a - 6]sa:1-t([a]+/6]解得：[a| +[6][al+i61+a- 60
§ 3 ^IL«þ · 19 · = −m−→a + m| −→a | | −→a | + | −→b − −→a | −→b . du −→AT = −−→AB + −→BT, ¤± k| −→b | | −→a | + | −→b | , k| −→a | | −→a | + | −→b | ! = (1, 0) + −m, m| −→a | | −→a | + | −→b − −→a | ! , =: k| −→b | | −→a | + | −→b | = 1 − m k| −→a | | −→a | + | −→b | = m| −→a | | −→a | + | −→b − −→a | ): k = | −→a | + | −→b | | −→a | + | −→b | + | −→b − −→a | . q AD CF u T ′ :, −−→ AT′ = s −−→AD = s| −→b | | −→a | + | −→b | −→a + s| −→a | | −→a | + | −→b | −→b . T ′ :I s| −→b | | −→a | + | −→b | , s| −→a | | −→a | + | −→b | ! . −−→ CT′ = t −−→CF = t | −→CA| + | −−→CB| (| −−→CB| −→CA+| −→CA| −−→CB) = t| −→b | | −→b | + | −→a − −→b | −→a −t −→b , d −−→AT′ = −→AC + −−→CT′ , : s| −→b | | −→a | + | −→b | , s| −→a | | −→a | + | −→b | ! = (0, 1) + t| −→b | | −→b | + | −→a − −→b | , −t ! , =: s| −→b | | −→a | + | −→b | = t| −→b | | −→b | + | −→a − −→b | s| −→a | | −→a | + | −→b | = 1 − t ): s = | −→a | + | −→b | | −→a | + | −→b | + | −→a − −→b | .
.20第一章向量代数由此可见s=k.即T=T11在△ABC中，点P由AP=kAB+mtAC所确定，其中实数k,m,t+-1<t<1.若使P为△ABC的重心,则满足 k+m=1, k≥3.m>3°k,m,t各应取什么值？解：建立仿射坐标系[A;AB,AC].则A(0,0)，B(1,0),C(0,1),P(k,mt)，从而若P为△ABC的重心则k=3mt：而k+m=1,推知m=3'3'1t=2.12.如图，已知平行六面体OABC-O,A,B,C中，点P在棱AA,上，且AP=2PA.点S在棱CC1上.且CS=SC，点Q,R分别是棱O,C1，AB的中点.求证：直线PQ与直线RS平行QO1第13题图第12题图证明：建立坐标系[O;OA,OC,OO].因为AP=2PAi],所以2AAI,AP=2PAi=3OP- OA + AP- OA + 2AA, = OA + 2001,--FY0Q =00 +01Q=00 + 0C,从而PQ=OQ-OP=-OA+200 +:10c23类似地，OR-OA+AR=AB+1002oS=OC+Cs=oC+00,3所以2001 ++100RS-OS-OR--OA+232
· 20 · 1Ù þê dd s = k, = T = T ′ . 11 3 △ABC ¥, : P d −→AP = k −−→AB + mt −→AC ¤(½, Ù¥¢ê k,m,t ÷v k + m = 1, k ≥ 1 3 , m ≥ 1 3 , −1 ≤ t ≤ 1. e¦ P △ABC %, K k,m,t Ao? ): ïáIX [A; −−→AB, −→AC]. K A(0, 0), B(1, 0), C(0, 1), P(k,mt). e P △ABC %, K k = 1 3 , mt = 1 3 . k + m = 1, í m = 2 3 , l t = 1 2 . 12. Xã, ®²18¡N OABC −O1A1B1C1 ¥, : P 3c AA1 þ, −→AP = 2 −−→PA1, : S 3c CC1 þ, −→CS = 1 2 −−→SC1, : Q,R ©O´c O1C1,AB ¥:. ¦y: PQ RS ²1. A B C O A1 B1 O1 C1 P Q R S 1 12 Kã b b b b A B D C E F H G 1 13 Kã y²: ïáIX [O; −→OA, −−→OC, −−→OO1]. Ï | −→AP| = 2| −−→PA1|, ¤± −→AP = 2 −−→PA1 = 2 3 −−→AA1, −−→OP = −→OA + −→AP = −→OA + 2 3 −−→AA1 = −→OA + 2 3 −−→OO1, −−→OQ = −−→OO1 + −−→O1Q = −−→OO1 + 1 2 −−→OC, l −−→PQ = −−→OQ − −−→OP = − −→OA + 2 3 −−→OO1 + 1 2 −−→OC. aq/, −−→OR = −→OA + −→AR = −−→AB + 1 2 −−→OC, −→OS = −−→OC + −→CS = −−→OC + 2 3 −−→OO1, ¤± −→RS = −→OS − −−→OR = − −→OA + 2 3 −−→OO1 + 1 2 −−→OC.