.31 :86几何空间向量的内积求a,b 的角解：由sa+tbl=ta-sbl得(sa+tb)2=(ta-sb)2.推出s2+t2+2sta.6=t2+82-2sta..故2sta.6=0.又因st+0,得.=0,夹角为12.如图，已知长方体OABC-O,A,B,C中，OA=8，OC=6,JOO=1.P是棱OC上的点，且PC=2OPl，M是棱AB上的点，且[AM|=2|MBI,N是棱B,Ci的中点.求直线A,P与直线MN所成的角解：把向量0A,OC,O0的单位向量记为7，了，k，建立直角坐标系[O; 7,了,]. 则0Ai=87 +K =(8,0,1);OP= IC= 27 = (0,2,0);+20C = 87 +47 = (8,4,0);OM=8i+3ON-10A+67 += (4,6,1);2因此A,P= OP_ OA =(-8,2, -1);MN =ON-OM = (-4,2,1)所以355V161cos(A,P,MN)69V69V21DB第12题图第13题图13.计算正方体的对角线与它的任一个面的对角线之间的夹角解：建立直角坐标系[A;AB,AD,AAi],以对角线AC,来计算此题.AC = AB +BC +CC = AB +AD +AA = (1,1,1)
§ 6 AÛmþSÈ · 31 · ¦ −→a , −→b Y. ): d |s −→a + t −→b | = |t −→a − s −→b | (s −→a + t −→b ) 2 = (t −→a − s −→b ) 2 . íÑ s 2 + t 2 + 2st−→a · −→b = t 2 + s 2 − 2st−→a · −→b . 2st−→a · −→b = 0. qÏ st 6= 0, −→a · −→b = 0, Y π 2 . 12. Xã, ®N OABC–O1A1B1C1 ¥, |OA| = 8, |OC| = 6, |OO1| = 1. P ´c OC þ:, |PC| = 2|OP|, M ´c AB þ:, |AM| = 2|MB|, N ´c B1C1 ¥:. ¦ A1P MN ¤¤. ): rþ −→OA, −−→OC, −−→OO1 ü þP −→i , −→j , −→k , ïáIX [O; −→i , −→j , −→k ]. K −−→OA1 = 8−→i + −→k = (8, 0, 1); −−→OP = 1 3 −−→OC = 2−→j = (0, 2, 0); −−→OM = 8−→i + 2 3 −−→OC = 8−→i + 4−→j = (8, 4, 0); −−→ON = 1 2 −→OA + 6−→j + −→k = (4, 6, 1); Ïd −−→A1P = −−→OP − −−→OA1 = (−8, 2, −1); −−→MN = −−→ON − −−→OM = (−4, 2, 1). ¤± cosh −−→A1P, −−→MNi = 35 √ 69√ 21 = 5 √ 161 69 . O A B C C1 A1 O1 B1 P M N 1 12 Kã A B D C A1 B1 D1 C1 1 13 Kã 13. ONé§?¡émY. ): ïáIX [A; −−→AB, −−→AD, −−→AA1], ±é AC1 5OdK. −−→AC1 = −−→AB + −−→BC + −−→CC1 = −−→AB + −−→AD + −−→AA1 = (1, 1, 1).
.32第一章向量代数(a) ABI = AB + BB = AB + AA = (1, 0, 1), 所以 cos(AC, AB) =V62由对称性，AC,与A,CI,AD1BCi,DCi,AC的夹角余弦也V3.V23V6A30(b) BD = AD - AB = (-1, 0, 1), 所以 cos(ACI,BD) = -V3. V2 = 0,即(AC,BD)=。同理。AC 与B,D,DAI,CB,BAi,CD,的夹角也为元214.试问（ab）=a（6）一定成立吗？请给出该向量等式成立的条件.解：等式左端是与共线的向量，右端是与共线的向量如果两端都不等于0.则a与共线，即存在k≠0.使a=k.反之若a=kc,左边=k(.b)=(k)(b .)=右边若等式两边都等于0则或者己，b，中至少有一个零向量；或者三个向量都不等于0,但.=.=0,即与，均正交15.求解向量方程a=b元.解：因为（-)=0,分两种情况：(a)若-0,则解为任意与-b垂直的向量；(b)若α-b=0,则任意向量都是解向量16.若向量a.6.不共面.而且a元=0.6±=0.元=0.则至=0.试证之证明：因为，6，不共面，所以它们线性无关，且立可由它们线性表示，即：=kia+k2b+k3.于是2=ki(a.)+k2(b.)+k3(.)=0即：至=017.证明三个向量己，6，共面的充分必要条件是aa abac6a 6 = 0.cacbc证明：a,6,共面当且仅当有不全为零的实数ki,k2,ks使：kia+kzb+k3=0.从而[kia?+k2(. 6)+ks(.) =0ki(b .a)+k 62+k3(b .)= 0ki(.a)+k2(.b)+k32=0
· 32 · 1Ù þê (a) −−→AB1 = −−→AB + −−→BB1 = −−→AB + −−→AA1 = (1, 0, 1), ¤± cosh −−→AC1, −−→AB1i = 2 √ 3 · √ 2 = √ 6 3 . dé¡5, −−→AC1 −−−→ A1C1, −−→AD1, −−→BC1, −−→DC1, −→AC Y{u √ 6 3 . (b) −−→BD = −−→AD − −−→AB = (−1, 0, 1), ¤± cosh −−→AC1, −−→BDi = 0 √ 3 · √ 2 = 0, = h −−→AC1, −−→BDi = π 2 . Ón. −−→AC1 −−→B1D, −−→DA1, −−→CB1, −−→BA1, −−→CD1 Y π 2 . 14. Á¯ ( −→a −→b ) −→c = −→a ( −→b −→c ) ½¤áí? ÑTþ ª¤á^ . ): ªà´ −→c þ, mà´ −→a þ. XJüàÑ Ø u 0, K −→a −→c , =3 k 6= 0, ¦ −→a = k −→c . e −→a = k −→c , > = k( −→c · −→b ) −→c = (k −→c )(−→b · −→c ) = m>. e ªü>Ñ u 0, K½ö −→a , −→b , −→c ¥k"þ; ½ön þÑØ u 0, −→a · −→b = −→b · −→c = 0, = −→b −→a , −→c þ. 15. ¦)þ§ −→a −→x = −→b −→x . ): Ï ( −→a − −→b ) −→x = 0, ©ü«¹: (a) e −→a − −→b 6= 0, K) −→x ? ¿ −→a − −→b Rþ; (b) e −→a − −→b = 0, K?¿þ −→x Ñ´)þ. 16. eþ −→a , −→b , −→c Ø¡, −→a −→x = 0, −→b −→x = 0, −→c −→x = 0. K −→x = 0. Áy. y²: Ï −→a , −→b , −→c Ø¡, ¤±§5Ã', −→x d§5L«, =: −→x = k1 −→a +k2 −→b +k3 −→c . u´ −→x 2 = k1( −→a · −→x )+k2( −→b · −→x )+k3( −→c · −→x ) = 0, =: −→x = 0. 17. y²nþ −→a , −→b , −→c ¡¿©7^´ −→a −→a −→a −→b −→a −→c −→b −→a −→b −→b −→b −→c −→c −→a −→c −→b −→c −→c = 0. y²: −→a , −→b , −→c ¡ =kØ "¢ê k1,k2,k3 ¦: k1 −→a + k2 −→b + k3 −→c = 0. l k1 −→a 2 + k2( −→a · −→b ) + k3( −→a · −→c ) = 0 k1( −→b · −→a ) + k2 −→b 2 + k3( −→b · −→c ) = 0 k1( −→c · −→a ) + k2( −→c · −→b ) + k3 −→c 2 = 0,
.33 :86几何空间向量的内积也即齐次线性方程组ra?+y(a.6)+z(α.)=0r(6 .±)+y62+z(6 .)=0(*)r(.a)+y(.b)+z?=0有非零解=ki，y=k2，z=k3.根据引理4.1,系数行列式a.bd2a.b.a72b.= 0..b2r.a反之，若a?a.ba.b.a62b.= 0,.b72.a则(*)必有非零解，设为=k1,y=k2，z=k3,令=ki+k2+k,那么．从(*)知：p?=kip.a+kp.b+k3p.=0即节=0.，6,线性无关，必定共面18.三角形ABC中，已知BC边上的高为AH.试用ABAC表示AHHB第18题图解：将AH=AB+kBC代人AH.BC=0,得AB.BC+kBC=0因此-AB.BCk:BC2得AH = AB_ (AB-BC)BCBC2
§ 6 AÛmþSÈ · 33 · =àg5§| x −→a 2 + y( −→a · −→b ) + z( −→a · −→c ) = 0 x( −→b · −→a ) + y −→b 2 + z( −→b · −→c ) = 0 x( −→c · −→a ) + y( −→c · −→b ) + z −→c 2 = 0 (∗) k") x = k1, y = k2, z = k3. âÚn 4.1, Xê1ª −→a 2 −→a · −→b −→a · −→c −→b · −→a −→b 2 −→b · −→c −→c · −→a −→c · −→b −→c 2 = 0. , e −→a 2 −→a · −→b −→a · −→c −→b · −→a −→b 2 −→b · −→c −→c · −→a −→c · −→b −→c 2 = 0, K (∗) 7k"), x = k1, y = k2, z = k3, - −→p = k1 −→a + k2 −→b + k3 −→c , @o. l (∗) : −→p 2 = k1 −→p · −→a + k2 −→p · −→b + k3 −→p · −→c = 0, = −→p = 0, −→a , −→b , −→c 5Ã', 7½¡. 18. n/ ABC ¥, ® BC >þp AH. Á^ −−→AB, −→AC L« −−→AH. B H C A 1 18 Kã ): ò −−→AH = −−→AB + k −−→BC \ −−→AH · −−→BC = 0, −−→AB · −−→BC + k −−→BC2 = 0. Ïd k = − −−→AB · −−→BC −−→BC2 , −−→AH = −−→AB − ( −−→AB · −−→BC) −−→BC −−→BC2 .
:34第一章向量代数再用BC=AC-AB代入，整理后得1AH =-[(AC. (AC-AB)AB - (AB. (AC -AB))AC)(AC-AB)219在平面四边形ABCD中，设AB=a，BC=,CD=.DA=.那么，当.=.=.=.时，ABCD是什么四边形？为什么？解：由于=0,所以(a+b=-(+a)(+=-(+)即:a2+62+2a.6=?+7?+2.d12+7?+2a.7-62+2+26.所以2=222=22.推知2=2从而AB=CDI，BCI=ADI，即ABCD是平行四边形，且+=0,b+a=0.由a.b-6.=-a.b知0=a.b=b.z=.d7a.因此ABCD是矩形.*20.设一个四边形各边之长分别是a,b,c,d，且其对角线互相垂直.求证各边之长也是a,b,c,d的任一四边形的两条对角线也相互垂直.证明：如图，有AC =AB+BC-AD+DCBD=BC+CD=BA+AD考虑以下内积AC.BD=(AB+BC)-(BC+CD)=BC2+AB.BC+AB.CD+BC.CDAC.BD = (AB+BC)-(BA+AD) = -AB?+AB.AD-AB.BC+AD.BC:AC.BD=(AD+DC).(BC+CD)=-CD?+AD.BC+AD.CD-BC.CDAC.BD = (AD+DC)·(BA+AD) = AD?-AB.AD+AB.CD-AD.CD将上述4式相加，可得：4AC.BD=AD2+BC2-AB2-CD2+2AB.CD+2AD.BC
· 34 · 1Ù þê 2^ −−→BC = −→AC − −−→AB \, n −−→AH = 1 ( −→AC − −−→AB) 2 [(−→AC · ( −→AC − −−→AB))−−→AB − ( −−→AB · ( −→AC − −−→AB))−→AC]. 19 3²¡o>/ ABCD ¥, −−→AB = −→a , −−→BC = −→b , −−→CD = −→c , −−→DA = −→d . @o, −→a · −→b = −→b · −→c = −→c · −→d = −→d · −→a , ABCD ´oo>/? o? ): du −→a + −→b + −→c + −→d = 0, ¤± −→a + −→b = −( −→c + −→d ) −→a + −→d = −( −→b + −→c ) =: −→a 2 + −→b 2 + 2−→a · −→b = −→c 2 + −→d 2 + 2−→c · −→d −→a 2 + −→d 2 + 2−→a · −→d = −→b 2 + −→c 2 + 2 −→b · −→c ¤± −→a 2 + −→b 2 = −→c 2 + −→d 2 , −→a 2 + −→d 2 = −→b 2 + −→c 2 . í −→a 2 = −→c 2 , −→b 2 = −→d 2 . l |AB| = |CD|, |BC| = |AD|, = ABCD ´²1o>/, −→a + −→c = 0, −→b + −→d = 0. d −→a · −→b = −→b · −→c = − −→a · −→b 0 = −→a · −→b = −→b · −→c = −→c · −→d = −→d · −→a . Ïd ABCD ´Ý/. ∗20. o>/>©O´ a,b,c,d, ÙépR. ¦y >´ a,b,c,d ?o>/ü^épR. y²: Xã, k −→AC = −−→AB + −−→BC = −−→AD + −−→DC, −−→BD = −−→BC + −−→CD = −−→BA + −−→AD. ıeSÈ: −→AC · −−→BD = (−−→AB+ −−→BC)·( −−→BC + −−→CD) = −−→BC2+ −−→AB· −−→BC + −−→AB· −−→CD+ −−→BC · −−→CD; −→AC· −−→BD = (−−→AB+ −−→BC)·( −−→BA+ −−→AD) = − −−→AB2+ −−→AB· −−→AD− −−→AB· −−→BC+ −−→AD· −−→BC; −→AC· −−→BD = (−−→AD+ −−→DC)·( −−→BC+ −−→CD) = − −−→CD2+ −−→AD· −−→BC+ −−→AD· −−→CD− −−→BC· −−→CD; −→AC · −−→BD = (−−→AD+ −−→DC)·( −−→BA+ −−→AD) = −−→AD2− −−→AB· −−→AD+ −−→AB· −−→CD− −−→AD· −−→CD. òþã 4 ª\, : 4 −→AC · −−→BD = −−→AD2 + −−→BC2 − −−→AB2 − −−→CD2 + 2 −−→AB · −−→CD + 2 −−→AD · −−→BC.
.35.86几何空间向量的内积由AB+ BC+CD+ DA= 0可得AB+CD=AD-BC.由(AB +CD)? =(AD - BC)2整理得AD2 + BC2 - AB2 - CD2 = 2(AB .CD + AD. BC)代入上式得4AC.BD = 2(AD2+ BC2- AB2-CD2) = 2(62 + d - α2 - c2)这说明对角线垂直的充分必要条件是α2+c2=62+d，只与四边形的边长有关.第20题图第21题图21有一个三角形△ABC和一个圆.三角形的三边之长分别是BC|=a，CAI=b,AB|=C，圆的圆心在点A，半径为.作圆的一条直径PQ使BP.CQ达到：(1）最大，(2)最小.试分别求出PQ，并用a,b,c及r分别表示这最大值和最小值解：设PQ为圆的一条直径，则AQ=-AP,AQI=API=r.由于BP-BA+AP.CQ=CA+AQ,得BP.CQ=(BA+AP)·(CA+AQ) = BA.CA+ (CA-BA)·AP- r2=BA.CA+CB.AP-r?
§ 6 AÛmþSÈ · 35 · d −−→AB + −−→BC + −−→CD + −−→DA = 0 −−→AB + −−→CD = −−→AD − −−→BC. d ( −−→AB + −−→CD) 2 = (−−→AD − −−→BC) 2 n −−→AD2 + −−→BC2 − −−→AB2 − −−→CD2 = 2(−−→AB · −−→CD + −−→AD · −−→BC). \þª 4 −→AC · −−→BD = 2(−−→AD2 + −−→BC2 − −−→AB2 − −−→CD2 ) = 2(b 2 + d 2 − a 2 − c 2 ). ù`²éR¿©7^´ a 2 + c 2 = b 2 + d 2 , o>/>k '. A B D C O 1 20 Kã A B C Q P 1 21 Kã 21 kn/ △ABC Ú . n/n>©O´ |BC| = a, |CA| = b, |AB| = c, %3: A, » r. ^» PQ ¦ −−→BP · −−→CQ : (1) , (2) . Á©O¦Ñ −−→PQ, ¿^ a,b,c 9 r ©OL« ùÚ. ): −−→PQ ^», K −→AQ = − −→AP, | −→AQ| = | −→AP| = r. du −−→BP = −−→BA + −→AP, −−→CQ = −→CA + −→AQ, −−→BP · −−→CQ = (−−→BA + −→AP) · ( −→CA + −→AQ) = −−→BA · −→CA + (−→CA − −−→BA) · −→AP − r 2 = −−→BA · −→CA + −−→CB · −→AP − r 2 .