- 高等数学习题全解与学习指导(下册)
- 张弢 殷俊锋
- 2681字
- 2024-11-02 02:35:11
四、习题详解
习题5-1 向量及其运算
1. 填空题.
(1)已知点A(2,-1,1),则点A与z轴的距离是_____,与x轴的距离是_____,与y轴的距离是_____.
(2)向量a=(-2,6,-3)的模为|a|=_____,方向余弦为cosα=_____,cosβ=_____,cosγ=_____,其同方向的单位向量ea=_____.
(3)设α、β、γ是向量a的三个方向角,则sin2α+sin2β+sin2γ=_____.
(4)设向量a=(2,-1,4)与向量b=(1,k,2)平行,则k=_____.
(5)已知三点M1(1,-2,3),M2(1,1,4),M3(2,0,2),则=_____,
(6)设点A(2,-1,-2),B(0,2,1),C(2,3,0),以AB,AC为邻边,作平行四边形,此平行四边形的面积等于_____.
(7)向量a=(4,-3,1)在b=(2,1,2)上的投影Prjba=_____,b在a上的投影Prjab=_____.
(8)设a=(1,2,3),b=(-2,k,4),而a⊥b,则k=_____.
解 (1)A与z轴的距离
与y轴的距离
与x轴的距离
![012-07](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/012-07.jpg?sign=1734435894-rCCD5dGCXxVD1ijlq5J3c7P0hTeNRPRP-0-4814aef1aa59e7baea574fdd0924d7f9)
(4)因为两向量平行,故对应坐标成比例,即
![012-08](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/012-08.jpg?sign=1734435894-7HS0vcgbD78onTh9AWAWIHzVQ9KuwN9m-0-23a1660880dcddb68dd146a76157ff8f)
从而
![013-01](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/013-01.jpg?sign=1734435894-YkPQzRuMeDDa91mHuV9L7pdkGgsai8i4-0-b3569d7c858077d49c98398d26548bf5)
(8)因为a⊥b,所以a与b的数量积为零,即
a·b=(1,2,3)·(-2,k,4)=-2+2k+12=0,
从而k=-5.
2. 一向量与x轴和y轴的夹角相等,而与z轴的夹角是与x轴、y轴夹角的两倍,求向量的方向角.
解 已知α=β,γ=2α,故由cos2α+cos2α+cos2(2α)=1得
2cos2α(2cos2α-1)=0,
解得cosα=0及于是
![013-03](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/013-03.jpg?sign=1734435894-Hp6jtOfKmMI4hdllNawmclnO51CdtysK-0-b60b563d93d5fd5dcd67dd0e8d4c9e49)
3. 给定M(-2,0,1),N(2,3,0)两点,在x轴上有一点A,满足|AM|=|AN|,求点A的坐标.
解 因为点A在x轴上,可设所求点为A(x,0,0),依题设|AM|=|AN|,即
![013-04](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/013-04.jpg?sign=1734435894-cWrxjLQ11ZFM7RlZr3uiqoqfZne0rpb7-0-674e15c9bee7400f31fab4a71755c46e)
所以x=1,从而所求点为A(1,0,0).
4. 从点A(2,-1,7)沿向量a=(8,9,-12)方向取长为34的线段AB,求点B的坐标.
解 设点B的坐标为(x,y,z),故设
即x-2=8λ,y+1=9λ,z-7=-12λ,
![013-07](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/013-07.jpg?sign=1734435894-qZqgE3HMnS7zpsDCZ2OJXj7TQ5fkakhX-0-1df9f1e2d1df7676be567692939fbad9)
解得λ=2,从而点B的坐标为(18,17,-17).
5. 设点P在y轴上,它到点的距离为到点P2(1,0,-1)的距离的两倍,求点P的坐标.
解 因为点P在y轴上,设点P的坐标为(0,y,0),
![013-09](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/013-09.jpg?sign=1734435894-hzS01LAxi4aXiRWYF3Oj7hrVljmCYf6G-0-3dfe637364a5dc5737c2919263070c67)
因为|PP1|=2|PP2|,即
![014-01](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/014-01.jpg?sign=1734435894-bZ3lNs50QFMNFcbCDaRfuShX2YdLj91T-0-60f735fc4b903e58f2ecbf08912f3a87)
解得y=±1,所求点为(0,1,0),(0,-1,0).
6. 设点A位于第Ⅰ卦限,向径与x轴、y轴的夹角依次为
和
且
求点A的坐标.
解 由关系式cos2α+cos2β+cos2γ=1,
得
因为点A在第Ⅰ卦限,知cosγ>0,故
于是点A的坐标为
7. 证明:Prju(λa)=λPrjua.
证明 记
当λ>0时,φ1=φ,Prju(λa)=|λa|cosφ1=λ|a|cosφ=λPrjua;
当λ<0时,φ1=π-φ,
Prju(λa)=|λa|cosφ1=-λ|a|cos(π-φ)=λ|a|cosφ=λPrjua.
当λ=0时,显然成立.
因此Prju(λa)=λPrjua.
8. 记ea为非零向量a的同向单位向量,证明:
证明 由于ea、a同向,故a=λea(λ>0),且|ea|=1,因此|a|=λ|ea=λ,即a=|a|ea,注意到|a|≠0,故结论成立.
9. 求平行于向量a=6i+7j-6k的单位向量.
解 所求向量有两个,一个与a同向,一个与a反向.
![014-14](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/014-14.jpg?sign=1734435894-R1Y08OsLr7I0ZKRKHl4qJvonNPIM1PGN-0-d2fbd7943d26776e215de359ae8aa1c3)
10. 设向量a与各坐标轴成相等的锐角,求向量a的坐标表达式.
解 因为向量a与各坐标轴成相等的锐角,所以a的三个方向角α=β=γ,又因为
cos2α+cos2β+cos2γ=1,
因此3cos2α=1,
因为和a同方向的单位向量从而
![014-18](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/014-18.jpg?sign=1734435894-hyrLHyfgScLqwyWOOk58ZmYCfeqAF1KB-0-56b0d3effa7640498d5468681a9c364c)
11. 已知a=(1,1,-4),b=(1,-2,2),求:
(1)a·b;
(2)a与b的夹角θ;
(3)a在b上的投影.
解 (1)a·b=1·1+1·(-2)+(-4)·2=-9.
(2)从而
(3)因为a·b=|b|Prjba,所以
12. 已知两点和M2(1,3,0),计算向量
的模、方向余弦和方向角.
![015-06](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/015-06.jpg?sign=1734435894-PILnKr2AvTs3eaSLmalwGYaZ9hw2pR7U-0-e1799373941d27024200b72c0f1da259)
13. 设|a|=3,|b|=2,求:
(1)(3a+2b)·(2a-5b);
(2)|a-b|2.
![015-08](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/015-08.jpg?sign=1734435894-zbeXHMmcHhkcOwVhNJveN69M5Cpup48S-0-9c557a00380d14e5fa3f04f3c7fedca5)
14. 已知点A(1,-3,4),B(-2,1,-1),C(-3,-1,1),求:
(1)∠BAC;
(2)上的投影.
![015-10](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/015-10.jpg?sign=1734435894-Nv7ZOGQ77iji0nR8AjtoP7EexfAKVyuy-0-85d24f744139cbe635fe76ea05a95e22)
15. 已知a=(2,3,1),b=(1,-2,1),求a×b及b×a.
![016-02](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/016-02.jpg?sign=1734435894-igwJGO1VUnDi1nbrMTUXjrNjB57VC5AD-0-32226f13329f19a522c4a1acd199441c)
16. 已知向量a=(2,-3,1),b=(1,-1,3),c=(1,-2,0),求:
(1)(a+b)×(b+c);(2)(a×b)·c;(3)(a×b)×c;(4)(a·b)c-(a·c)b.
解 (1)a+b=(2+1,-3+(-1),1+3)=(3,-4,4),
b+c=(1+1,-1+(-2),3+0)=(2,-3,3),
![016-03](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/016-03.jpg?sign=1734435894-aSe8exlLiRcOHzrNs7OdceZbintATOMV-0-01b23ab418b0bc9b16c13dc0b61a4ba0)
(4)a·b=(2,-3,1)·(1,-1,3)=2+(-3)(-1)+3=8,
a·c=(2,-3,1)·(1,-2,0)=2+(-3)(-2)+0=8,
(a·b)c-(a·c)b=8(1,-2,0)-8(1,-1,3)=(0,-8,-24).
17. 求与a=3i-2j+4k,b=i+j-2k都垂直的单位向量.
![016-04](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/016-04.jpg?sign=1734435894-3FOcmiA7omEP8w4bknYi3kO3LAU3h7kC-0-72d632ab76e022ef1801353d0e79167f)
18. 已知空间四点A(-1,0,3),B(0,2,2),C(2,-2,-1),D(1,-1,1),求与都垂直的单位向量.
![016-06](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/016-06.jpg?sign=1734435894-VD0QpkSGvN25tsJMikZKszt1jGIN5kLx-0-08e4802f7ac9b6588c83adefa9f423a0)
则与
垂直的单位向量为
![017-03](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/017-03.jpg?sign=1734435894-ZgUaiuVlHX3Tvxd3ZOXv5ooaxo9vuASt-0-f858dca44080bcfa3a4e175528de67f0)
19. 设向量a=2i+j,b=-i+2k,求以a、b为邻边的平行四边形的面积.
![017-04](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/017-04.jpg?sign=1734435894-IUM6e8WDDPUTxRgaDP9A7vT4NUkWrXkZ-0-01d87b2d3d5291876def2c16e9cb6f03)
20. 求以点A(1,2,3),B(0,0,1),C(3,1,0)为顶点的三角形的面积.
![017-05](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/017-05.jpg?sign=1734435894-vFagQeJTSdaBEZA9nUPkK7ChgVvQrlB8-0-62e6de2c3464bff8d4f79ba2ee1fb1bd)
21. 设A=2a+b,B=ka+b,其中|a|=1,|b|=2,a⊥b,问:
(1)k为何值时,A⊥B;
(2)k为何值时,以A与B为邻边的平行四边形的面积为6.
解 (1)因为a⊥b,故a·b=0,所以
0=A·B=(2a+b)·(ka+b)=2k|a|2+|b|2=2k+4,k=-2;
(2)A×B=(2a+b)×(ka+b)=(2-k)(a×b),
平行四边形面积为A×B的模,所以
![017-06](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/017-06.jpg?sign=1734435894-eKlYOEL3bQ887OqrW08lpurl1AosT381-0-353272b36bf6792cde3f1efe65d5521b)
所以 k1=5,k2=-1.
22. 已知a=2m+3n,b=3m-n,m、n是两个互相垂直的单位向量,求:
(1)a·b;(2)|a×b|.
![017-07](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/017-07.jpg?sign=1734435894-clreZnwtqdsdO0Caq3oC6JGcJtjiwB1n-0-267be84fe3248848a574592e1e7e3a35)
23. 设a、b、c满足a+b+c=0.
(1)证明:
(2)若还满足|a|=3,|b|=4,|c|=5,求|a×b+b×c+c×a|.
证明 (1)因a+b+c=0,所以
0=(a+b+c)·(a+b+c)=a·a+b·b+c·c+2(a·b+b·c+c·a),
即有
![018-01](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-01.jpg?sign=1734435894-KKOXcxdVcsSetjGdF41K8zUG0ve0uwQ2-0-b3edf8fc13a077c9f86c11c811f8417a)
(2)因c=-(a+b),故
b×c=-b×(a+b)=a×b,c×a=-(a+b)×a=a×b,
由于
![018-02](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-02.jpg?sign=1734435894-vrLRKbIdaNKlQOWzjK2rdI8pMsiafwUO-0-9d5ada303e1d36295e6891bc746b3309)
所以
![018-03](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-03.jpg?sign=1734435894-gqRqF3Raudv9jGlEwxcpUBGwGbkQHDCu-0-6f7e441c3d40b13f71920524afe9df1a)
24. 设a+3b与7a-5b垂直,a-4b与7a-2b垂直,求a与b之间的夹角θ.
解 由于(a+3b)⊥(7a-5b),所以(a+3b)·(7a-5b)=0,即
7|a|2-15|b|2+16a·b=0. (1)
又(a-4b)⊥(7a-2b),所以(a-4b)·(7a-2b)=0,即
7|a|2+8|b|2-30a·b=0. (2)
联立方程(1)、(2)得
|a|2=|b|2=2a·b,
所以
![018-04](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-04.jpg?sign=1734435894-cHFSrwUuouV36AKgjOyMSHMIyWvGuxjq-0-963edd0bf5d684e45a0ac20ca09beef2)
25. 试用向量方法证明三角形的余弦定理.
证明 设在△ABC中,∠BCA=θ,|CB|=a,|CA|=b,|AB|=c,
现要证c2=a2+b2-2abcosθ. 记则有c=a-b,从而|c|2=c·c=(a-b)·(a-b)=a·a+b·b-2a·b=|a|2+|b|2-2|a||b|cosθ.
由|a|=a,|b|=b,|c|=c,即得
c2=a2+b2-2abcosθ.
26. 试用向量积证明三角形正弦定理.
证明 设△ABC的三个内角为α、β、γ,三边长为a、b、c,
![018-06](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-06.jpg?sign=1734435894-h6Mx9yc6ZOkJx4BgKJBJrcifIymCWxhw-0-21fab7fb1f903f1cdd12a91ac4ef496b)
同理可证
![018-08](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/018-08.jpg?sign=1734435894-d3Hb6E4MEaFNjjyOq4GCxlpEnfJ4GinZ-0-ac2db7bfe5eb2ad7df5cdb9aaf00bd94)
因此三角形正弦定理得证.
27. 已知向量a≠0,b≠0,证明
|a×b|2=|a|2|b|2-(a·b)2.
![019-01](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-01.jpg?sign=1734435894-SnAJ9EtkZfjbRdplcr8wMo6SUFzOVXIg-0-cacb847da9a22942904692cf961a7680)
28. 已知a、b、c两两垂直,且|a|=1,|b|=2,|c|=3,求s=a+b+c的长度与它和a、b、c的夹角.
解 |s|2=|a+b+c|2=(a+b+c)·(a+b+c)=a·a+b·b+c·c+2(a·b+b·c+c·a)=|a|2+|b|2+|c|2=12+22+32=14.
![019-02](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-02.jpg?sign=1734435894-CukiIGlfqVp3bqZgaH3Bq99tBMJB9fpn-0-80281c901200bc3a14b0211d1b21be01)
29. 已知a=(7,-4,-4),b=(-2,-1,2),向量c在向量a与b的角平分线上,且求c的坐标.
![019-05](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-05.jpg?sign=1734435894-rKFkv8Cx7qEdUYkdZPweX0IXCmWqsn2R-0-19c72cc7517ec97a17d91f495e99ab29)
注 这里向量ea+eb在a与b的角平分线上(见图5-1).
![019-06](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-06.jpg?sign=1734435894-gwINW9kUXDcLAUI7bURmemrmbgdpHPQz-0-24357d7f5c88540dc305b678357a24e1)
图5-1
或设c=λ(|b|a+|a|b)=λ(3(7,-4,-4)+9(-2,-1,2))=3λ(1,-7,2),
![019-07](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-07.jpg?sign=1734435894-KjsMeonvpSd9WeB8wFQDjuBl7ksfw6y6-0-e4e6b57a0a2d8083a0a71875561a0b3b)
30. 设向量x与j成60°角,与k成120°角,且求x.
解 由题意知,则
![019-10](https://epubservercos.yuewen.com/7CD0D8/18130706308305906/epubprivate/OEBPS/Images/019-10.jpg?sign=1734435894-EDEznMvNZK2WV2D7nInNxrYEbVPpAsEB-0-de71ccf9e7a47a99f638c6e66c0085ea)