.26.第一章向量代数因此当k=1或2时这3个向量共面.要使a，共线必须使它们的相应坐标1-l-k_k成比例，即二，解得k=2.因此当k=2时可，z共线11-16.设i，，e为基.间向量能否表为向量，b,的线性组合？如能，则写出表达式(1)=+2+4,=-+,=++3,=6+3+15;(2)=23=+2+2,=+6+ = 2ei +e2解：设=+26+3,问题归结为解线性方程组(1)方程组为1+22+3=621—2+3=34r1+a2+33=15,11系数行列式0，不能断定方程组是否有解。用加减消去法解13123=3-3r得令3=3k可以得到线性表示式=(3-2k)a+（3-T2=3-33,k)b+3k，其中k为任意数+-=2(2）方程组为<系数行列式-21+222+ 6r3=1(31+22+53=0,11-16-2 2= 36,253方程组有解：221门266121302501116C1329363636
· 26 · 1Ù þê Ïd k = 1 ½ 2 ù 3 þ¡. ¦ −→a , −→c 7L¦§AI ¤'~, = 1 1 = 1 − k −1 = k 2 , ) k = 2. Ïd k = 2 −→a , −→c . 6. −→e1 , −→e2 , −→e3 Ä. ¯þ −→v UÄLþ −→a , −→b , −→c 5|Ü? X U, KÑLª. (1) −→a = −→e1 + 2−→e2 + 4−→e3 , −→b = −→e1 − −→e2 + −→e3 , −→c = −→e1 + −→e2 + 3−→e3 , −→v = 6−→e1 + 3−→e2 + 15−→e3 ; (2) −→a = −→e1 − 2 −→e2 + 3−→e3 , −→b = −→e1 + 2−→e2 + 2−→e3 , −→c = − −→e1 + 6−→e2 + 5−→e3 , −→v = 2−→e1 + −→e2 . ): −→v = x1 −→a + x2 −→b + x3 −→c , ¯K8()5§|. (1) §| x1 + x2 + x3 = 6 2x1 − x2 + x3 = 3 4x1 + x2 + 3x3 = 15, Xê1ª 1 1 1 2 −1 1 4 1 3 = 0, ØU佧|´Äk). ^\~{) x1 = 3 − 2 3 x3 x2 = 3 − 1 3 x3, - x3 = 3k ±5L«ª −→v = (3 − 2k) −→a + (3 − k) −→b + 3k −→c , Ù¥ k ?¿ê. (2) §| x1 + x2 − x3 = 2 −2x1 + 2x2 + 6x3 = 1 3x1 + 2x2 + 5x3 = 0, Xê1ª 1 1 −1 −2 2 6 3 2 5 = 36, §|k): x1 = 2 1 −1 1 2 6 0 2 5 36 = − 11 36 , x2 = 1 2 −1 −2 1 6 3 0 5 36 = 16 9 ,
.27 :s5n维向量空间112-2 2132019T3:363619+b线性表示式为=C367.当a为何值时，下列四点共面：M(1,a,a2), M2(1,-1,1),M3(2,1, -2),M(-1,2,2)解：根据推论4.5，此4点共面的充分必要条件是1-12-1-1-1-1-a=-7a2-5a+2=01-a2-a2-α21-α2-2a2解得 -1 学85n维向量空间1.根据n维向量的定义证明：对任意n维向量α.有(1) 0α = 0;(2) (-1)α = -α;(3)k0=0(任意数k);(4)从kQ=0推出k=0或α=0.证明：对任意的α=（a1，.，an），则(1) 0α=(0a1,-.,0an)= (0, -..,0) = 0.(2) (-1)α = ((-1)a1,.. , (-1)an) =(-a1, ..., -an) =-α(3)k0=(k0,.,k0)=(0,*,0)=0(4)ka=（kai,**,kan）=(0,.*,0).若α≠0,则存在ai0,由ka,=0可得k=02.证明：任一数域都包含有理数域.证明：设K为一个数域，则1EK所以对任意的正整数n有n=1+...+1EK，并且n的负元-nEK.因此K含有全部整数.又因对任意n11为n的逆元，则eK,所以对任意的有理数"其的整数n≠O,neK,nmn中m,n是整数),有=mxnEK，故有理数域QCKn
§ 5 n þm · 27 · x3 = 1 1 2 −2 2 1 3 2 0 36 = − 19 36 . 5L«ª −→v = − 11 36 −→a + 16 9 −→b − 19 36 −→c . 7. a Û, eo:¡: M1(1,a,a2 ), M2(1, −1, 1), M3(2, 1, −2), M4(−1, 2, 2). ): âíØ 4.5, d 4 :¡¿©7^´ 1 − 1 2 − 1 −1 − 1 −1 − a 1 − a 2 − a 1 − a 2 −2 − a 2 2 − a 2 = −7a 2 − 5a + 2 = 0, ) a = −1 ½ 2 7 . § 5 n þm 1. â n þ½Ây²: é?¿ n þ α, k (1) 0α = 0; (2) (−1)α = −α; (3) k0 = 0 (?¿ê k); (4) l kα = 0 íÑ k = 0 ½ α = 0. y²: é?¿ α = (a1, · · · ,an), K: (1) 0α = (0a1, · · · , 0an) = (0, · · · , 0) = 0. (2) (−1)α = ((−1)a1, · · · ,(−1)an) = (−a1, · · · , −an) = −α. (3) k0 = (k0, · · · ,k0) = (0, · · · , 0) = 0. (4) kα = (ka1, · · · ,kan) = (0, · · · , 0). e α 6= 0, K3 ai 6= 0, d kai = 0 k = 0. 2. y²: ?êѹknê. y²: K ê, K 1 ∈ K. ¤±é?¿ê n k n = 1 + · · · + 1 | {z } n ∈ K, ¿ n K −n ∈ K. Ïd K ¹k Üê. qÏé?¿ ê n 6= 0, n ∈ K, 1 n n _, K 1 n ∈ K, ¤±é?¿knê m n (Ù ¥ m,n ´ê), k m n = m × n ∈ K, knê Q ⊆ K.
.28.第一章向量代数3.证明：全体形如a+bv2, a,beQ的数组成的集合构成一个数域证明：把这个集合记为K，设a1+biV2,a2+baV2eK，则(a1 +b1V2)±(a2 + b2V2) = (a1±a2) + (b1± b2)V2 E K(因为有理数的和与差仍是有理数);(a1 +biV2)(a2 + b22) = (aia2 + 2bib2) +(aib2 + a2bi)2 e K(因为有理数的和、差与乘积仍是有理数)；当a2+b2V2≠0时ai +biV2_ (a1 +biV2)(a2-b2V2) - a1a2-2brb2 + a2bi-aib22EKa2+b2V2-262a-263-262(有理数关于除法也是封闭的)．因此集合K关于加减乘除法都封闭，成为一个数域4.设K为数域，V为K上的n维向量空间.证明：对所有的kEK,a,βeV,有(1) k(α- β) = ka-kβ;(2)α+α+..+α=na;个(3)若α+β=α+%则β=%证明：设α=a1,*，an)，β=(br,,bn）（ai,b,EK).则对任意的keK,(1) k(q - β) = k(ai - bi, .., an - bn) = (k(ai - b1), ..., k(an - bn)) =(ka1 -kbi,..,kan -kbn) =k(a1,...,an) -k(bi,"..,bn)= ko -kβ.(2)α+..+α= (a1+..-a1,a2+...+a2),an +...+an) = (na1,",nan)=na.(3)若设=（ci,Cn）且α+β=α+%即：（a1+b1，***an+bn）=(ai+c.an+c),则有a+b=a+c即b,=c,所以β=86几何空间向量的内积1.将下列向量单位化：(2) --(1) =57-67+3k;32
· 28 · 1Ù þê 3. y²: N/X a + b √ 2, a,b ∈ Q ê|¤8ܤê. y²: rù8ÜP K, a1 + b1 √ 2,a2 + b2 √ 2 ∈ K, K (a1 + b1 √ 2) ± (a2 + b2 √ 2) = (a1 ± a2) + (b1 ± b2) √ 2 ∈ K (ÏknêÚ E´knê); (a1 + b1 √ 2)(a2 + b2 √ 2) = (a1a2 + 2b1b2) + (a1b2 + a2b1) √ 2 ∈ K (ÏknêÚ! ¦ÈE´knê); a2 + b2 √ 2 6= 0 , a1 + b1 √ 2 a2 + b2 √ 2 = (a1 + b1 √ 2)(a2 − b2 √ 2) a 2 2 − 2b 2 2 = a1a2 − 2b1b2 a 2 2 − 2b 2 2 + a2b1 − a1b2 a 2 2 − 2b 2 2 ∈ K (knê'uØ{´µ4). Ïd8Ü K 'u\~¦Ø{ѵ4, ¤ ê. 4. K ê, V K þ n þm. y²: é¤k k ∈ K, α,β ∈ V , k (1) k(α − β) = kα − kβ; (2) α + α + · · · + α | {z } n = nα; (3) e α + β = α + γ, K β = γ. y²: α = (a1, · · · ,an), β = (b1, · · · ,bn) (ai ,bi ∈ K). Ké?¿ k ∈ K, (1) k(α − β) = k(a1 − b1, · · · ,an − bn) = (k(a1 − b1), · · · ,k(an − bn)) = (ka1 − kb1, · · · ,kan − kbn) = k(a1, · · · ,an) − k(b1, · · · ,bn) = kα − kβ. (2) α + · · · + α | {z } n = (a1 + · · · a1 | {z } n ,a2 + · · · + a2 | {z } n , · · · ,an + · · · + an | {z } n ) = (na1, · · · ,nan) = nα. (3) e γ = (c1, · · · ,cn), α + β = α + γ, =: (a1 + b1, · · · ,an + bn) = (a1 + c1, · · · ,an + cn), Kk ai + bi = ai + ci , =bi = ci , ¤±β = γ. § 6 AÛmþSÈ 1. òeþü z: (1) −→a = 5−→i − 6 −→j + 3 −→k ; (2) −→b = 1 2 −→i − 1 3 −→k .
.29.86几何空间向量的内积dV70解: (1) a0(57-67+3k)70al6V(37 -2)(2) 60_1362.计算下列向量的夹角：(1) = (1, -2,3), 6 = (2,1, -2); (2) a=(-2,1, -1), 6-(1, -1, 4),解：(1)[a=V14,/6=3,.6=-6,所以-V14-6V14cos(a, b)(a, b)=π-arccos7V14.37(2) [= V6, |= 3V2, . = -7, 所以-77V37V3(a, 6)=元- arccoscos(a.b)=6V3=-18183.求向量在e0上的投影：(1) = (1, -1,2), e = (1,1,1); (2) a = (-2,1,3), = (1,2,0)解: (1) V同=(1,1,1),则prg = ( )e =(.)e _ 2V3t2V33.1(1, 1, 1) =(1,1,1)33(2)因·=0, 所以pr==0.4.证明：以A(3,-1,2)，B(0,-4,2)，C(-3,2,1)为顶点的三角形是等腰三角形证明: AB = (-3, -3,0), AC = (-6,3, -1), BC = (-3,6, -1), AB3V2.AC|=BCI=V46≠ABI,所以△ABC是等腰三角形5.证明：以A(3,-2,1)，B(7,6,9)，C(9,1，-5)为顶点的三角形是直角三角形证明: AB = (4,8,8), AC = (6,3, -6), AB AC =12(2 +2 - 4) = 0, 所以△ABC是直角三角形6.设有三个向量，6,两两构成60°角，且知a=4,6=2[]=6. 求 a + b + 的长度.解: ++=(++).(++)=?+?+2+2.6+2a.+26.=16+4+36+28+24+12)cos60°=100.所以a+b +=10
§ 6 AÛmþSÈ · 29 · ): (1) −→ a 0 = −→a | −→a | = √ 70 70 (5−→i − 6 −→j + 3 −→k ). (2) −→ b 0 = −→b | −→b | = √ 13 13 (3−→i − 2 −→k ). 2. OeþY: (1) −→a = (1, −2, 3), −→b = (2, 1, −2); (2) −→a =(−2, 1, −1), −→b =(1, −1, 4). ): (1) | −→a | = √ 14, | −→b | = 3, −→a · −→b = −6, ¤± cosh −→a , −→b i = −6 √ 14 · 3 = − √ 14 7 , h −→a , −→b i = π − arccos √ 14 7 . (2) | −→a | = √ 6, | −→b | = 3√ 2, −→a · −→b = −7, ¤± cosh −→a , −→b i = −7 6 √ 3 = − 7 √ 3 18 , h −→a , −→b i = π − arccos 7 √ 3 18 . 3. ¦þ −→a 3 −→ e 0 þÝK: (1) −→a = (1, −1, 2), −→e = (1, 1, 1); (2) −→a = (−2, 1, 3), −→e = (1, 2, 0). ): (1) −→ e 0 = −→e | −→e | = √ 3 3 (1, 1, 1), K pr−→ e 0 −→a = (Π−→ e 0 −→a ) −→ e 0 = (−→a · −→ e 0 ) −→ e 0 = 2 √ 3 3 −→ e 0 = 2 √ 3 3 · √ 3 3 (1, 1, 1) = 2 3 (1, 1, 1). (2) Ï −→a · −→e = 0, ¤± pr−→ e 0 −→a = 0. 4. y²: ± A(3, −1, 2), B(0, −4, 2), C(−3, 2, 1) º:n/´ n/. y²: −−→AB = (−3, −3, 0), −→AC = (−6, 3, −1), −−→BC = (−3, 6, −1), | −−→AB| = 3 √ 2, | −→AC| = | −−→BC| = √ 46 6= | −−→AB|, ¤± △ABC ´ n/. 5. y²: ± A(3, −2, 1), B(7, 6, 9), C(9, 1, −5) º:n/´n /. y²: −−→AB = (4, 8, 8), −→AC = (6, 3, −6), −−→AB · −→AC = 12(2 + 2 − 4) = 0, ¤ ± △ABC ´n/. 6. knþ −→a , −→b , −→c üü¤ 60◦ , | −→a | = 4, | −→b | = 2, | −→c | = 6. ¦ −→a + −→b + −→c Ý. ): | −→a + −→b + −→c | 2 = (−→a + −→b + −→c )·( −→a + −→b + −→c ) = −→a 2 + −→b 2 + −→c 2 + 2 −→a · −→b + 2−→a · −→c + 2 −→b · −→c = 16 + 4 + 36 + 2(8 + 24 + 12) cos 60◦ = 100. ¤ ± | −→a + −→b + −→c | = 10.
:30 第一章向量代数7.已知=3,=2,,)=试求3+2与2-5的内积.解：(3a+26(2-56)=62-102-11.=54-40-11V3=14-33V33×2×28.在直角坐标系中，，6，的坐标分别是(3,5,7)，(0,4,3)，（-1,2,-4)求3+4-5与26+的夹角解：记=3a+46-5=(14,21,53)，=26+= (-1,10,2),则P=3446,72=105，P.q=302所以cos(P, 7)= 151V3618301809159.求下列向量的方向余弦：(2) = (2,3, -10).(1) a = (2, -3, -6);2-3-6解：(1)cosα=,cosβ=，COS=73V1132V11310/113，cosB=(2) cosQ :COSY11311311310设向量=(1,2,4),=(1,1,1),=-k (k是实数)(2)求与,都垂直的(1)求k使a;解：(1)由已知，laa.=0a.b-ka=01而.=7,?=21,所以k=3(2)垂直于,等价于.==0,由假设=,也等价于.=.=0.设=(,,2),得a.a=a+2y+4z=0,.=++z=0.解得r=2z,y=-3z,即d=(2k,-3k,k)(k是实数)11设a，6是两个单位向量，8,t是两个非零实数，使得Isa+tbl=[ta-sb]
· 30 · 1Ù þê 7. ® | −→a | = 3, | −→b | = 2, h −→a , −→b i = π 6 . Á¦ 3 −→a + 2 −→b 2 −→a − 5 −→b SÈ. ): (3−→a + 2 −→b )·(2−→a − 5 −→b ) = 6−→a 2 − 10 −→b 2 − 11−→a · −→b = 54− 40− 11× 3 × 2 × √ 3 2 = 14 − 33√ 3. 8. 3IX¥, −→a , −→b , −→c I©O´(3, 5, 7), (0, 4, 3), (−1, 2, −4). ¦ 3 −→a + 4 −→b − 5 −→c 2 −→b + −→c Y. ): P −→p = 3−→a + 4 −→b − 5 −→c = (14, 21, 53), −→q = 2 −→b + −→c = (−1, 10, 2), K −→p 2 = 3446, −→q 2 = 105, −→p · −→q = 302. ¤± cosh −→p , −→q i = 151√ 361830 180915 . 9. ¦eþ{u: (1) −→a = (2, −3, −6); (2) −→b = (2, 3, −10). ): (1) cos α = 2 7 , cos β = −3 7 , cos γ = −6 7 . (2) cos α = 2 √ 113 113 , cos β = 3 √ 113 113 , cos γ = − 10√ 113 113 . 10 þ −→a = (1, 2, 4), −→b = (1, 1, 1), −→c = −→b − k −→a (k ´¢ê). (1) ¦ k ¦ −→c ⊥ −→a ; (2) ¦ −→a , −→c ÑR −→d . ): (1) d®, −→c ⊥ −→a ⇐⇒ −→a · −→c = 0 ⇐⇒ −→a · −→b − k −→a 2 = 0. −→a · −→b = 7, −→a 2 = 21, ¤± k = 1 3 . (2) −→d Ru −→a , −→c du −→d · −→a = −→d · −→c = 0, db −→c = −→b − k −→a , du −→d · −→a = −→d · −→b = 0. −→d = (x,y,z), −→d · −→a = x + 2y + 4z = 0, −→d · −→b = x + y + z = 0. ) x = 2z,y = −3z, = −→d = (2k, −3k,k) (k ´¢ê). 11 −→a , −→b ´üü þ, s,t ´ü"¢ê, ¦ |s −→a + t −→b | = |t −→a − s −→b |,