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(完整word版)高考数学空间向量与立体几何总复习.pdf

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1、空间向量与立体几何总复习一、知识网络构建二、课标及考纲要求空间向量与立体空间向量及其运算经历向量及其运算由平面向空间推广的过程了解空间向量的概念、基本定理及其意义，掌握空间向量的正交分解及其坐标表示掌握空间向量的线性运算及其坐标表示掌握空间向量的数量积及其坐标表示，能运用向量的数量积判断向量的共线与垂直空间理解直线的方向向量与平面的法向量空间向量的定义及其运算空间向量运算的几何表示（如平行四边形法则）用空间向量表示点、线、面等元素建立空间图形与空间向量的联系利用空间向量运算解决立体几何问题空间向量运算的坐标表示（加减法、数乘、数量积）空间向量定义运算坐标表示加法减法数量积立体几何中的向量方法垂

2、直关系平行关系空间距离空间角2几何向量的运用能用向量语言表述线线、线面、面面的垂直和平行关系能用向量方法证明有关线、面位置关系的一些定理能用向量方法解决线线、线面、面面的夹角的计算问题，体会向量方法在研究几何问题中的作用三、知识要点及考点精析（一）空间向量及其运算1空间向量的概念在空间，我们把具有大小和方向的量叫做空间向量，向量的大小叫做向量的长度或模还需要掌握的几个相关的概念包括相等向量、零向量、共线向量等2空间向量的线性运算（1）空间向量的加法、减法和数乘运算平面向量中的三角形法则和平行四边形法则同样适用于空间向量的加（减）法运算加法运算对于有限个向量求和，交换相加向量的顺序其和不变三个

3、不共面的向量的和等于以这三个向量为邻边的平行六面体的对角线所表示的向量加法和数乘运算满足运算律：交换律，即a+b=b+a；结合律，即()()a+bcab+c；分配律，即()a=a+a及()a+bab（其中，均为实数）（2）空间向量的基本定理共线向量定理：对空间向量，ab(0)，bab的充要条件是存在实数，使a=b 共面向量定理：如果空间向量，ab不共线，则向量c与向量a,b共面的充要条件是，存在惟一的一对实数xy，使c=xya+b 空间向量基本定理：如果三个向量a,b,c不共面，那么对空间任一向量p，存在有序实数组x，y，z，使xyzp=a+b+c其中，abc是空间的一个基底，a,b,c都叫做

4、基向量，该定理可简述为：空间任一向量p都可以用一个基底，abc惟一线性表示（线性组合）（3）两个向量的数量积文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9

5、文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H

6、7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5

7、K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6

8、Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q

9、8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9

10、A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9文档编码:CI8P1Q8H7I9 HR6M3X9A5K7 ZC10K9T7R6Y9两个向量的数量积是ab=|a|b|cos，数量积有如下性质：a,b,ca e=|a|cos（e为单位向量）；aaab=0；a a=|a|2；|a b|a|b

11、|数量积运算满足运算律：交换律，即a b=b a；与数乘的结合律，即（a）b=（ab）；分配律，即（a+b）c=a c+b c3空间向量的坐标运算（1）给定空间直角坐标系xyzO和向量a，存在惟一的有序实数组使123aaaa=i+j+k，则123()aaa，叫作向量a在空间的坐标，记作123()aaa，a=（2）空间向量的直角坐标运算律若123123()()aaabbb，a=b=，则a+b112233()ababab，ab112233()ababab，123()aaa，a，a b),(332211bababa112233()abababR，ab，1 1223 30a ba ba bab若111

12、222()()A xyzB xyz，则212121()ABxxyyzz，即一个向量在直角坐标系中的坐标等于表示这个向量的有向线段的终点的坐标减去起点的坐标4直线的方向向量与向量方程（）位置向量：已知向量a，在空间固定一个基点O，作向量OAa，则点A在空间的位置被a所惟一确定，a称为位置向量（）方向向量与向量方程：给定一个定点A和一个向量a，再任给一个实数t，以A为起点作向量tAPa，则此向量方程称为动点P对应直线l的参数方程，向量a称为直线l的方向向量典型例题分析：例 1若AB=(x2,1,3),CD=(1,-y2,9),如果AB与CD为共线向量,则()文档编码:CD2F8V2Z3J2 HP1

13、S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7

14、D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:

15、CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 H

16、P1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 Z

17、E7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编

18、码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2

19、 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K44A 1x，1yB21x,21yC61x,23yD61x,23y答案:C 例 2已知向量a(1，1，0)，b(-1，0，2)，且kab与 2 a-b互相垂直，则k的值是()A 1 B51 C53 D57答案:D 例 3已知AB=(2,2,1),AC=(4,5,3),求平面 ABC的单位法向量解:设平面

20、ABC的法向量n=(x,y,1),则nAB且nAC,即nAB=0,且nAC=0,即,0354,0122yxyx即,1,21yxn=(21,-1,1),单位法向量n=(31,-32,32)（二）立体几何中的向量方法1利用向量法确定直线、平面间的平行、垂直等位置关系设直线1l 的方向向量是1u111()，abc，直线2l 的方向向量是2u222()abc，平面的法向量是1v111()xyz，平面的法向量是2v222()xyz，则有如下结论成立：（1）12llu1u2u1k2u212121,kcckbbkaa；（2）12ll12120uuu u1 2121 20a ab bc c；（3）1l1111

21、0uvu v11111 10a xb yc z；（4）1l111uvuk1v111111,kzckybkxa；（5）121vvvk2v121212，xkxykyzkz；（6）12120vvv v1212120 x xy yz z第一部分：平行问题利用空间向量解决线线平行问题（06 山东模拟）已知直线OA平面，直线BD平面，OB，为垂足求证：OABD文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4

22、文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z

23、3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A

24、6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9

25、K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V

26、2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z

27、2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7

28、X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4证明：以点O为原点，以射线OA为非负 z 轴，如图 1，建立空间直角坐标系Oxyz，ijk为沿，xyz轴的单位向量，且设BD()xyz，BD，BDi，BDj，()(100)0BDxyzx，i，BD j()(01 0)0 xyzy，(0 0)BDz，BDz k BDk，即OABD点评：由向量的共线的充要条件知，只要证明OABD 即可利用空间向量解决线面平行问题（06 山西模拟）已知111ABCA BC 是正三棱柱，D是AC的中点，求证：1AB 平面1DBC 证法 1：建立如图2的空间直角坐标系Axyz设正

29、三棱柱的底面边长为a，侧棱长为b，则1133(0 0 0)0(0)0022222aaaABaCabBabD，设平面1DBC 的法向量为()xyz，n，则11330 002222aaABabBDaDCb，由BDn，1DCn，得130202BDaxaDCybz，nn02xazyb，取得1y，得012ab，n由130 10222aaABabb，n，得1ABn，即1AB 平面1DBC 证法 2：如图 3，记1ABACAA，abc，则1111122ABDBABADDCDCCC，b+cacab11DBDCABac，11DB DC AB，共面又1B平面1C BD，1AB 平面1DBC 点评：用向量证明线面平

30、行问题通常有两种方法：向量p与两个不共线的向量，ab 共面文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2

31、A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X

32、9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8

33、V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10

34、Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G

35、7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2

36、F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K46的充要条件是存在惟一的有序实数对()，xy，使xypab利用共面向量定理可证明线面平行问题，如证法2设 n 为平面的法向量，要证明a，只需证明0a n，如证法1 利用空间向量解决面面平行问题例题：已知正方

37、体1AC 的棱长为1，EFG，分别为1ABADAA，的中点，求证：平面EFG平面11B CD 证明：建立空间直角坐标系Dxyz，则111(10 0)(11 0)(01 0)(0 0 0)(101)(111)(0 01)ABCDABD，得111100 010222EFG，设1111()xyz，n为平面EFG的法向量，设2222()xyz，n为平面11B CD 的法向量空间计算：12(111)(111)，nn由12nn，得平面EFG平面11B CD 点评：设12，nn 分别为平面，的法向量，要证，只需证明：存在一个非零常数，满足12nn，则其实本题也可转化为线线平行，则面面平行即用向量先证明1D

38、CGE，11D BEF，则有线面平行，从而平面EFG平面11B CD 第二部分：垂直问题利用空间向量解决线线垂直问题（2003 年高考题）已知正四棱1111ABCDABC D，112ABAA，点E为1CC 中点，点F为1BD 中点证明：EF为1BD 与1CC 的公垂线证明：如图1，在以C为的原点的空间直角坐标系中，111 1(010)(10 2)(0 0 2)(0 01)12 2BDCEF，由1 102 2EF，11(0 0 2)(112)CCBD，得111100EF BDEF CCEFBDEFCC，EF为1BD 与1CC 的公垂线点评：把推理论证（1EFCC）用向量运算（10EF CC）来

39、代替，减少了构造辅助图形，降低了思维量利用空间向量解决线面垂直问题（2005 年高考题）如图 2，在四棱锥PABCD中，底面ABCD为矩形，侧棱PA底面ABCD，文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z

40、3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A

41、6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9

42、K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V

43、2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z

44、2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7

45、X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4312ABBCPAE，为PD的中点，在侧面PAB内找一点N，使NE面PAC解：如图2，在以A为原点的

46、空间直角坐标系中，1(31 0)(0 10)(0 0 2)012CDPE，设11(31 0)(0 0 2)2NExzACAP，由NE面PAC，得00NE ACNE AP，即133026101xxzz，3016N，点评：按照传统方法，要构造三条辅助线，多解两个三角形，画图、看图以及计算都增加了难度用空间向量的观点处理立体几何中的线面关系，把几何问题代数化，降低了难度利用空间向量解决面面垂直问题（07 北京海淀）如图 3，在正方体1111ABCDA BC D 中，O为AC与BD的交点，G为1CC 的中点，求证：平面1A BD平面GBD分析：要证明平面1A BD平面GBD，只要证明平面内的一条直线

47、1AO 垂直于平面GBD中的两条相交直线即可，而从图中观察，证11AOBDAOOG，较容易成功证明：设11111A BADA A，abc 则000a bb ca c，而11111()()22cabAOA AAOA AABAD，baBDADAB，11111()()2222abcOGOCCGABADCC，1AO BD221()02c bc aba，1AO OG22211()042abc1AOBD，1AOOG 又BDOGO，1AO平面BDG又1AO平面1ABD，文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1

48、 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文

49、档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3

50、J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6Z1 ZE7D2G7X9K4文档编码:CD2F8V2Z3J2 HP1S10Z2A6

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