.36第一章向量代数(1)要使BP.CQ达到极大当且仅当CB.AP达到极大，当且仅当AP=kCB, k>0;(2)要使BP.CQ达到极小当且仅当CB.AP达到极小，当且仅当AP=-mCB, m > 0.二.PQ=-2AP.最后得到：而|AP=r，所以k=m=(1) PQ=-CB,最大值为acbcos LBAC + ar -r2_ b2 + c2 - α?+ar-r?:2(2) PQ=CB,最小值为ar2-62 + c2- α2cbcosZBAC-ar-rarr2222设有一向量集合S=2+a.11是一个非零向量证明：对于任意,E及实数0≤t≤1,t+(1-t)eS.解：d2a≤1+42d2a，则ES当且仅当立落在以为圆心，半径为r的圆若记r-1-内（包括圆周).因此对0t<1及任意，yES,只需验证92at+(1-t)+≤r=124+(1-)+=(+号)+(1-)(寸+号+≤tr +(1-t)r =r.所以t+(1-t)s.*23设四边形A,A2AA为圆C的内接四边形、H,H.H,H依次是AAAAAAAAl.AA4AA2.AAA2A的垂心求证：H.H.H3.H四点共圆.（提示：以圆C的圆心为原点建立直角坐标系）解：以圆C的圆心为原点O建立直角坐标系.设圆C的半径为r，则OA/=r.在△A2A3A,中令0H=0A2+0A+0A4，则A,H·A3A = (OH1-OA2)(OA, -OA3) = (OA+OA3)(OA -OA3) = 0
· 36 · 1Ù þê (1) ¦ −−→BP · −−→CQ 4 = −−→CB · −→AP 4, = −→AP = k −−→CB, k > 0; (2) ¦ −−→BP · −−→CQ 4 = −−→CB · −→AP 4, = −→AP = −m −−→CB, m > 0. | −→AP| = r, ¤± k = m = r a , −−→PQ = −2 −→AP. : (1) −−→PQ = − 2r a −−→CB, cb cos ∠BAC + ar − r 2 = b 2 + c 2 − a 2 2 + ar − r 2 ; (2) −−→PQ = 2r a −−→CB, cb cos ∠BAC − ar − r 2 = b 2 + c 2 − a 2 2 − ar − r 2 . 22 kþ8Ü S = { −→x | |−→x | 2 + −→a · −→x ≤ 1}, −→a ´"þ. y²: éu?¿ −→x , −→y ∈ S 9¢ê 0 ≤ t ≤ 1, t −→x + (1 − t) −→y ∈ S. ): | −→x | 2 + −→a · −→x ≤ 1 ⇐⇒ −→x + −→a 2 2 ≤ 1 + −→a 2 4 . eP r = 1 + −→a 2 4 , K −→x ∈ S = −→x á3± − −→a 2 %, » r S () ±). Ïdé 0 ≤ t ≤ 1 9?¿ −→x , −→y ∈ S, Iy t −→x + (1 − t) −→y + −→a 2 ≤ r = 1 + −→a 2 4 . t −→x + (1 − t) −→y + −→a 2 = t −→x + −→a 2 + (1 − t) −→y + −→a 2 ≤ t −→x + −→a 2 + (1 − t) −→y + −→a 2 ≤ tr + (1 − t)r = r. ¤± t −→x + (1 − t) −→y ∈ S. ∗23 o>/ A1A2A3A4 C So>/, H1,H2,H3,H4 g´ △A2A3A4, △A3A4A1, △A4A1A2, △A1A2A3 R%. ¦y: H1,H2,H3,H4 o: . (J«: ± C %:ïáIX) ): ± C %: O ïáIX. C » r, K |OAi | = r. 3 △A2A3A4 ¥, - −−→OH1 = −−→OA2 + −−→OA3 + −−→OA4, K −−−→ A2H1 · −−−→ A3A4 = (−−→OH1− −−→OA2)(−−→OA4 − −−→OA3) = (−−→OA4+ −−→OA3)(−−→OA4 − −−→OA3) = 0.
.37 :S6几何空间向量的内积同理，AH·AA=AHAA=0,即H是△AAA的垂心.其余同理.因此有OH,=(OA+OA+0A+0A)-OA,(i=1,2,3,4)令OH。=-,OA，这是一个常向量，则JOH,-OH=[OA/=r.所以H1,H2,H3,H在以Ho为圆心、半径为r的圆周上*24设0<a<1,0<b<1.用几何方法证明：Va2 + 62+ V(1-a)2 + 62 + Va2 + (1 -b)2 + V(1-a)2 + (1 -b)2 ≥2V2.解：建立平面直角坐标系[O；i，了].取A(0,0)，B(1,0),C(1,1)，D(0,1)四个点构成一个正方形，边长为1.设P(a,6)，则P在正方形ABCD之内，且 [PA|= Va2 + 62, IPB|= V(1 -a)2 + 62, IPC = V(1 - a)2 + (1- b)2,[PD|= Vα2 +(1-b)2 由于 [PA| +[PCI ≥ IAC| = V2, IPB| +IPDI ≥[BD|=V2,得证.*25设P1,P2，·，P6是中心在原点O、半径为1的圆上相异的6点.证明：总可以在OP,OP2，.，OP中找出两个向量OP和OP(1≤i≠j≤6)使得JOP,+OP≥V3解：令a表示OP与OP的夹角（≤)，即ag=POP,.设?元：不妨设从OPao=min[ai1≤i≠j≤6],我们先证0<ao≤3T开始依逝时针转动时依次得到OFO，若a412>，23>多，将导致2元=Q12+…·+461>6×元=2元，矛盾。因此不妨设a16 >33.利用余弦定理便有a12=ZPOP≤"OP+OP/2=2+2cOsa12≥2+1=3.即OP+OP≥V3.*26设实数a,b,c满足：a+b+c=0,a2+b2+c2=1.如果记=(yiz）（i=1,，6)，其中【a,yi=[a,b,c].则必存在，使T·T≥解：设=(1,1,1).则对任意(1≤≤6)，总有，节=0.因此，，r均与P垂直，从而它们共面.再由=1（1≤i≤6)，可知ri，..，r是单位圆上的6个相异点.由习题25可知，必存在相异的两个向量，设为，使得+≥3.两边平方之后可知++2(·)≥3,即·≥
§ 6 AÛmþSÈ · 37 · Ón, −−−→ A3H1 · −−−→ A2A4 = −−−→ A4H1 · −−−→ A2A3 = 0, = H1 ´ △A2A3A4 R%. Ù{Ó n. Ïdk −−→OHi = (−−→OA1 + −−→OA2 + −−→OA3 + −−→OA4) − −−→OAi , (i = 1, 2, 3, 4) - −−→OH0 = P4 i=1 −−→OAi , ù´~þ, K | −−→OHi − −−→OH0| = | −−→OAi | = r. ¤± H1,H2,H3,H4 3± H0 %!» r ±þ. ∗24 0 < a < 1, 0 < b < 1. ^AÛ{y²: √ a 2 + b 2 + p (1 − a) 2 + b 2 + p a 2 + (1 − b) 2 + p (1 − a) 2 + (1 − b) 2 ≥ 2 √ 2. ): ïᲡIX [O; −→i , −→j ]. A(0, 0), B(1, 0), C(1, 1), D(0, 1) o:¤/, > 1. P(a,b), K P 3/ ABCD S, |PA| = √ a 2 + b 2 , |PB| = p (1 − a) 2 + b 2 , |PC| = p (1 − a) 2 + (1 − b) 2 , |PD| = p a 2 + (1 − b) 2 . du |PA| + |PC| ≥ |AC| = √ 2, |PB| + |PD| ≥ |BD| = √ 2, y. ∗25 P1,P2, · · · ,P6 ´¥%3: O!» 1 þÉ 6 :. y ²: o±3 −−→OP1, −−→OP2, · · · , −−→OP6 ¥éÑüþ −−→OPi Ú −−→OPj (1 ≤ i 6= j ≤ 6) ¦ | −−→OPi + −−→OPj | ≥ √ 3. ): - aij L« −−→OPi −−→OPj Y (≤ π), = aij = ∠PiOPj 6= π. a0 = min{aij | 1 ≤ i 6= j ≤ 6}, ·ky 0 < a0 ≤ π 3 . Øl −−→OP1 m©_=Äg −−→OP2, · · · , −−→OP6, e a12 > π 3 , a23 > π 3 , . . . , a16 > π 3 , ò 2π = a12 + · · · + a61 > 6 × π 3 = 2π, gñ. ÏdØ a12 = ∠P1OP2 ≤ π 3 . |^{u½nBk | −−→OP1 + −−→OP2| 2 = 2 + 2 cos a12 ≥ 2 + 1 = 3. = | −−→OP1 + −−→OP2| ≥ √ 3. ∗26 ¢ê a,b,c ÷v: a + b + c = 0, a 2 + b 2 + c 2 = 1. XJP −→ri = (xi ,yi ,zi) (i = 1, · · · , 6), Ù¥ {xi ,yi ,zi} = {a,b,c}. K73 −→ri 6= −→rj , ¦ −→ri · −→rj ≥ 1 2 . ): −→p = (1, 1, 1). Ké?¿ −→ri (1 ≤ i ≤ 6), ok −→ri · −→p = 0. Ï d −→r1 , · · · , −→r6 þ −→p R, l §¡. 2d | −→ri | = 1 (1 ≤ i ≤ 6), −→r1 , · · · , −→r6 ´ü þ 6 É:. dSK 25 , 73Éüþ, −→ri 6= −→rj , ¦ | −→ri + −→rj | ≥ √ 3. ü>² | −→ri | 2 + | −→rj | 2 + 2(−→ri · −→rj ) ≥ 3, = −→ri · −→rj ≥ 1 2 .
:38.第一章向量代数87几何空间向量的外积1.在直角坐标系中，已知a,6,的坐标分别是(1,0,1)，(1,-2,0)，(-1,2,1),求(3+）×()的坐标.解: 3a + 6 =(4,-2,3),6 - =(2,-4,-1),所以( 3-13 3(3a +)×(6 - ) = (= (14,10,-12)2.证明(×)≤262.并说明等式何时成立.证明:(a× )=[a×=a6sin(a,)≤a12, 等号成立当且仅当sin(a，b）=0.即：a//b.3.已知a，是两个互不平行的向量，求证（a_）×（α+6）2(a×b)，并说明它的几何意义证明(a-b)x(a+b)=axa+ax-xa-x2(a×b).几何意义：若以a，b构成一个平行四边形的相邻两边，则_ba+b为此平行四边形的两对角线．上式说明：以对角线构成的平行四边形面积为原平行四边形面积2倍.4.求向量,使,其中，(1)=-27+3,=4-5(2)=37-+,=-+2-解：令=x.则a,.计算得：(1)=×=-2+5+4(2)=×=+2+5（本题答案不唯)5.计算由向量a，6所张成的平行四边形的面积(1)=37+47+2,6=27++;(2)=-+2+, 6 =27-2.解：(1)×=2+了-5,×=30.所以,张成的平行四边形面积为V30.(2)×=27+27-2ka×=2V3所以,6张成的平行四边形面积为2V3
· 38 · 1Ù þê § 7 AÛmþ È 1. 3IX¥, ® −→a , −→b , −→c I©O´ (1, 0, 1), (1, −2, 0), (−1, 2, 1), ¦ (3−→a + −→b ) × ( −→b − −→c ) I. ): 3−→a + −→b = (4, −2, 3), −→b − −→c = (2, −4, −1), ¤± (3−→a + −→b ) × ( −→b − −→c ) = −2 3 −4 −1 , − 4 3 2 −1 , 4 −2 2 −4 ! = (14, 10, −12). 2. y² ( −→a × −→b ) 2 ≤ −→a 2−→b 2 . ¿`² ªÛ¤á. y²: (−→a × −→b ) 2 = | −→a × −→b | 2 = | −→a | 2 | −→b | 2 sin2 h −→a , −→b i ≤ |−→a | 2 | −→b | 2 , Ò ¤á = sinh −→a , −→b i = 0. =: −→a //−→b . 3. ® −→a , −→b ´üpز1þ, ¦y ( −→a − −→b ) × ( −→a + −→b ) = 2(−→a × −→b ), ¿`²§AÛ¿Â. y²: (−→a − −→b ) × ( −→a + −→b ) = −→a × −→a + −→a × −→b − −→b × −→a − −→b × −→b = 2(−→a × −→b ). AÛ¿Â: e± −→a , −→b ¤²1o>/ü>, K −→a − −→b , −→a + −→b d²1o>/üé. þª`²: ±é¤²1o>/¡ Ȳ1o>/¡È 2 . 4. ¦þ −→c , ¦ −→c ⊥ −→a , −→c ⊥ −→b , Ù¥, (1) −→a = −→i − 2 −→j + 3 −→k , −→b = 4−→j − 5 −→k ; (2) −→a = 3−→i − −→j + −→k , −→b = − −→i + 2−→j − −→k . ): - −→c = −→a × −→b . K −→c ⊥ −→a , −→c ⊥ −→b . O: (1) −→c = −→a × −→b = −2 −→i + 5−→j + 4 −→k . (2) −→c = −→a × −→b = − −→i + 2−→j + 5 −→k . ( KYØ) 5. Odþ −→a , −→b ¤Ü¤²1o>/¡È: (1) −→a = 3−→i + 4−→j + 2 −→k , −→b = 2−→i + −→j + −→k ; (2) −→a = − −→i + 2−→j + −→k , −→b = 2−→i − 2 −→j . ): (1) −→a × −→b = 2−→i + −→j − 5 −→k , | −→a × −→b | = √ 30. ¤± −→a , −→b ܤ ²1o>/¡È √ 30. (2) −→a × −→b = 2−→i + 2−→j − 2 −→k , | −→a × −→b | = 2√ 3, ¤± −→a , −→b ܤ²1 o>/¡È 2 √ 3.
.39.S7几何空间向量的外积Bi第6题图6.如图，已知ABCD-A,B,CiD1是单位正方体，P是棱DD1上任意一个点.线段CiC2的中点是B1.请指出下列各个向量积所确定的向量：(1) AP× AA; (2) PC ×A1A; (3) AC ×A1A解：建立直角标架[A1;A,B1,A,D1,A,A]设P为DD1上任一点.则A,P- AD +D,P= A,DI + kD,D - A,D+ kAA= (0, 1,k), A,C =(1, 1,1).(1) A,P × A,A= (0, 1,k) × (0, 0, 1) = (1, 0,0) = ABI(2) PC = A,C-AP= (1,0,1-k), PC×AA= (1,0,1-k)×(0,0, 1) =(0, -1,0) = -A,D, = D,AI(3) AC×A1A= (1,1,1)×(0, 0, 1) = (1, -1, 0) = AB- AD = AiC2.7.设=27+37-，=-87-5+3.求，,使= +2, 1 , //.解：令”(2,3,-1),P=()=(23,-1),7令=(27+3)，==(-117+87+2k)则：/，且=0.8.设为给定的非零向量，～为任一向量(1)证明：可唯一分解为=其中，，(2)具体写出，的表达式证明：（1)若有两种分解法：==，其中,,,则-=. 但-)=0()=0所以0推0-=0即=,=(2) 含-p,4-(可.3)2-(44,可=可-7, 则或/,12=0,即且知=+，由(1)知[u2
§ 7 AÛmþ È · 39 · D A B C C1 A1 D1 B1 P C2 1 6 Kã 6. Xã, ® ABCD−A1B1C1D1 ´ü N, P ´c DD1 þ?¿ :. ã C1C2 ¥:´ B1. ÑeþȤ(½þ: (1) −−→A1P × −−→A1A; (2) −−→PC × −−→A1A; (3) −−→A1C × −−→A1A. ): ïáIe [A1; −−−→ A1B1, −−−→ A1D1, −−→A1A]. P DD1 þ?:. K: −−→A1P = −−−→ A1D1 + −−→D1P = −−−→ A1D1 + k −−→D1D = −−−→ A1D1 + k −−→A1A = (0, 1,k), −−→A1C = (1, 1, 1). (1) −−→A1P × −−→A1A = (0, 1,k) × (0, 0, 1) = (1, 0, 0) = −−−→ A1B1. (2) −−→PC = −−→A1C− −−→A1P = (1, 0, 1−k), −−→PC× −−→A1A = (1, 0, 1−k)×(0, 0, 1) = (0, −1, 0) = − −−−→ A1D1 = −−−→ D1A1. (3) −−→A1C× −−→A1A = (1, 1, 1)×(0, 0, 1) = (1, −1, 0) = −−−→ A1B1− −−−→ A1D1 = −−−→ A1C2. 7. −→u = 2−→i + 3−→j − −→k , −→v = −8 −→i − 5 −→j + 3 −→k . ¦ −→v1 , −→v2 , ¦−→v = −→v1 + −→v2 , −→v1 ⊥ −→u , −→v2 //−→u . ): - −→ u 0 = −→u | −→u | = 1 √ 14 (2, 3, −1), pr−→ u 0 −→v = (−→v · −→ u 0 ) −→ u 0 = −17 7 (2, 3, −1). - −→v2 = −17 7 (2−→i + 3−→j − −→k ), −→v1 = −→v − −→v2 = 2 7 (−11−→i + 8−→j + 2 −→k ), K: −→v2 //−→u , −→v1 · −→u = 0. 8. −→u ½"þ, −→v ?þ. (1) y²: −→v ©) −→v = −→v1 + −→v2 , Ù¥, −→v1 ⊥ −→u , −→v2 //−→u ; (2) äNÑ −→v1 , −→v2 Lª. y²: (1) e −→v kü«©){: −→v = −→v1 + −→v2 = −→v ′ 1 + −→v ′ 2 , Ù¥. −→v1 ⊥ −→u , −→v ′ 1 ⊥ −→u , −→v2 //−→u , −→v ′ 2 //−→u , K −→v1 − −→v ′ 1 = −→v ′ 2 − −→v2 . ( −→v1 − −→v ′ 1 ) · −→u = 0, ( −→v ′ 2 − −→v2 )× −→u = 0, ¤± −→v1 − −→v ′ 1 ⊥ −→u , −→v1 − −→v ′ 1 //−→u , −→u 6= 0, íÑ: −→v1 − −→v ′ 1 = 0, −→v ′ 2 − −→v2 = 0, = −→v1 = −→v ′ 1 , −→v2 = −→v ′ 2 . (2) - −→v2 = pr−→ u 0 −→v = (−→v · −→ u 0 ) −→ u 0 = ( −→v · −→u ) | −→u | 2 −→u , −→v1 = −→v − −→v2 . K−→v2 //−→u , −→v1 · −→u = −→v · −→u − ( −→v · −→u ) | −→u | 2 −→u 2 = 0, =−→v1 ⊥ −→u . : −→v = −→v1 + −→v2 , d (1)
第一章向量代数:40这种分解是唯一的，故表达式为w.uU12u=-U2.9.设，,为两两不共线的向量证明：++=0当且仅当axb=bx=xa.证明：若++=0,则此式与和作外积后可得×+×=0以及a×+×=0,即a×=×=×a反之，设=++.由上述等式可得×=××=0以及×=×+×=0.如果0,则由,b共线以及P，b共线可得b与共线，与假设矛盾.10.设,为两不共线的向量，AB=+,BC=2+8,CD=3(a-b).证明：A,B,D三点共线证明：要证A,B,D三点共线，只须证明：AB×BD=0即可.由BD=BC+CD=5α+56=5(α+6)=5AB可得AB×BD=0,即：A,B,D三点共线11.三个向量0AOB,OC满足OB×OC+OC×OA+OA×OB= 0求证：三点A,B,C共线证明：要证A,B,C共线只须证明：AB×AC=0.因AB=OB-OA. AC-OC-OA所以AB×AC=(OB-OA)×(OC-OA)=OB×OC+OAxOB+OC×OA=0故A,B,C三点共线12.如果=×，b=可×，=×，则..共面试证明之证明：由于n.a=n.b=n.c=0,若n≠0,则a，b,c共面；否则，由n=0可得a=b==0,也共面
· 40 · 1Ù þê ù«©)´, Lª −→v2 = −→v · −→u | −→u | 2 −→u −→v1 = −→v − −→v2 . 9. −→a , −→b , −→c üüØþ. y²: −→a + −→b + −→c = 0 = −→a × −→b = −→b × −→c = −→c × −→a . y²: e −→a + −→b + −→c = 0, Kdª −→b Ú −→c È −→a × −→b + −→c × −→b = 0 ±9 −→a × −→c + −→b × −→c = 0, = −→a × −→b = −→b × −→c = −→c × −→a . , −→p = −→a + −→b + −→c . dþã ª−→p × −→b = −→a × −→b + −→c × −→b = 0 ±9 −→p × −→c = −→a × −→c + −→b × −→c = 0. XJ −→p 6= 0, Kd −→p , −→b ±9 −→p , −→b −→b −→c , bgñ. 10. −→a , −→b üØþ, −−→AB = −→a + −→b , −−→BC = 2−→a + 8 −→b , −−→CD = 3(−→a − −→b ). y²: A,B,D n:. y²: y A,B,D n:, Ly²: −−→AB × −−→BD = 0 =. d −−→BD = −−→BC + −−→CD = 5−→a + 5 −→b = 5(−→a + −→b ) = 5−−→AB, −−→AB × −−→BD = 0, =: A,B,D n:. 11. nþ −→OA, −−→OB, −−→OC ÷v −−→OB × −−→OC + −−→OC × −→OA + −→OA × −−→OB = 0. ¦y: n: A,B,C . y²: y A,B,C Ly²: −−→AB × −→AC = 0. Ï −−→AB = −−→OB − −→OA, −→AC = −−→OC − −→OA, ¤± −−→AB× −→AC = (−−→OB− −→OA)×( −−→OC− −→OA) = −−→OB× −−→OC+ −→OA× −−→OB+ −−→OC× −→OA = 0, A,B,C n:. 12. XJ −→a = −→p × −→n , −→b = −→q × −→n , −→c = −→r × −→n , K −→a , −→b , −→c ¡. Áy². y²: du −→n · −→a = −→n · −→b = −→n · −→c = 0, e −→n 6= 0, K −→a , −→b , −→c ¡; ÄK, d −→n = 0 −→a = −→b = −→c = 0, ¡.