【高考必备】2017年高考数学（文）热点题型和提分秘籍专题33空间向量及其运算word版含解析

1、PAGE 1.了解空间直角坐标系，会用空间直角坐标表示点的位置2.会简单应用空间两点间的距离公式3.了解空间向量的概念，了解空间向量的基本定理及其意义，掌握空间向量的正交分解及其坐标表示4.掌握空间向量的线性运算及其坐标表示5.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热点题型一 空间向量的运算例1、如图，在长方体ABCDA1B1C1D1中，O为AC的中点。(1)化简：eq o(A1O,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AD,sup10()；(2)设E是棱DD1上的点，且eq o(DE,sup10()eq f

2、(2,3)eq o(DD1,sup10()，若eq o(EO,sup10()xeq o(AB,sup10()yeq o(AD,sup10()zeq o(AA1,sup10()，试求x、y、z的值。eq f(2,3)eq o(D1D,sup10()eq f(1,2)(eq o(DA,sup10()eq o(AB,sup10()eq f(2,3)eq o(A1A,sup10()eq f(1,2)eq o(DA,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AD,sup10()eq f(2,3)eq o(AA

3、1,sup10()，xeq f(1,2)，yeq f(1,2)，zeq f(2,3)。【提分秘籍】空间向量的表示方法用已知不共面的向量表示某一向量时，应结合图形，将已知向量和未知向量转化至三角形或平行四边形中，然后利用三角形法则或平行四边形法则，把所求向量用已知向量表示出来。【举一反三】 如图，在平行六面体ABCDA1B1C1D1中，M为AC与BD的交点，若eq o(A1B1,sup10()a，eq o(A1D1,sup10()b，eq o(A1A,sup10()c，则下列向量中与eq o(B1M,sup10()相等的向量是()Aeq f(1,2)aeq f(1,2)bcB.eq f(1,2)

4、aeq f(1,2)bcC.eq f(1,2)aeq f(1,2)bcDeq f(1,2)aeq f(1,2)bc【答案】A热点题型二 共线、共面向量定理的应用 例2、已知E、F、G、H分别是空间四边形ABCD的边AB、BC、CD、DA的中点，(1)求证：E、F、G、H四点共面；(2)求证：BD平面EFGH。由共面向量定理的推论知：E、F、G、H四点共面。(2)因为eq o(EH,sup10()eq o(AH,sup10()eq o(AE,sup10()eq f(1,2)eq o(AD,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)(eq o(AD,sup10

5、()eq o(AB,sup10()eq f(1,2)eq o(BD,sup10()，所以EHBD。又EH平面EFGH，BD平面EFGH，所以BD平面EFGH。【提分秘籍】 应用共线向量定理、共面向量定理证明点共线、点共面的方法比较：三点(P，A，B)共线空间四点(M，P，A，B)共面eq o(PA,sup10()eq o(PB,sup10()eq o(MP,sup10()xeq o(MA,sup10()yeq o(MB,sup10()对空间任一点O，eq o(OP,sup10()eq o(OA,sup10()teq o(AB,sup10()(t为参数)对空间任一点O，eq o(OP,sup10

6、()eq o(OM,sup10()xeq o(MA,sup10()yeq o(MB,sup10()对空间任一点O，eq o(OP,sup10()(1t)eq o(OA,sup10()teq o(OB,sup10()(t为参数)对空间任一点O，eq o(OP,sup10()(1xy)eq o(OM,sup10()xeq o(OA,sup10()yeq o(OB,sup10() 【举一反三】 已知A、B、C三点不共线，对平面ABC外的任一点O，若点M满足eq f(1,3)(eq o(OA,sup10()eq o(OB,sup10()eq o(OC,sup10()。(1)判断eq o(MA,sup1

7、0()，eq o(MB,sup10()，eq o(MC,sup10()三个向量是否共面；(2)判断点M是否在平面ABC内。 (2)由(1)知eq o(MA,sup10()，eq o(MB,sup10()，eq o(MC,sup10()共面，且共过同一点M，四点M，A，B，C共面。从而点M在平面ABC内。热点题型三 空间向量数量积的运算例3如图所示，已知空间四边形ABCD的每条边和对角线长都等于1，点E，F，G分别是AB，AD，CD的中点，计算： (1)eq o(EF,sup10()eq o(BA,sup10()。(2)eq o(EG,sup10()eq o(BD,sup10()。 (2)eq

8、o(EG,sup10()eq o(EB,sup10()eq o(BC,sup10()eq o(CG,sup10()eq f(1,2)eq o(AB,sup10()(eq o(AC,sup10()eq o(AB,sup10()eq f(1,2)(eq o(AD,sup10()eq o(AC,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AC,sup10()eq f(1,2)eq o(AD,sup10()eq f(1,2)aeq f(1,2)beq f(1,2)c，eq o(BD,sup10()eq o(AD,sup10()eq o(AB,sup10()

9、ca。所以eq o(EG,sup10()eq o(BD,sup10()eq blc(rc)(avs4alco1(f(1,2)af(1,2)bf(1,2)c)(ca)eq f(1,2)a2eq f(1,2)abeq f(1,2)bceq f(1,2)c2aceq f(1,2)eq f(1,4)eq f(1,4)eq f(1,2)eq f(1,2)eq f(1,2)。【提分秘籍】1空间向量数量积计算的两种方法(1)基向量法：ab|a|b|cosa，b。(2)坐标法：设a(x1，y1，z1)，b(x2，y2，z2)，则abx1x2y1y2z1z2。2利用数量积可解决有关垂直、夹角、长度问题(1)a0

10、，b0，abab0。(2)|a|eq r(a2)。(3)cosa，beq f(ab,|a|b|)。【举一反三】 已知A(1,0,0)，B(0，1,1)，eq o(OA,sup10()eq o(OB,sup10()与eq o(OB,sup10()的夹角为120，则的值为()Aeq f(r(6),6) B.eq f(r(6),6) Ceq f(r(6),6) Deq r(6)【解析】eq o(OA,sup10()eq o(OB,sup10()(1，)，cos120eq f(,r(122)r(2)eq f(1,2)，得eq f(r(6),6)。经检验eq f(r(6),6)不合题意，舍去，eq f(

11、r(6),6)。【答案】C 1.【2016高考浙江文数】如图，已知平面四边形ABCD，AB=BC=3，CD=1，AD=，ADC=90沿直线AC将ACD翻折成，直线AC与所成角的余弦的最大值是_【答案】则，与平行的单位向量为，所以，所以时，取得最大值，为1若a，b，c为空间的一组基底，则下列各项中，能构成基底的一组向量是()Aa，ab，ab Bb，ab，abCc，ab，ab Dab，ab，a2b【答案】C2在正方体ABCDA1B1C1D1中，给出以下向量表达式：(eq o(A1D1,sup10()eq o(A1A,sup10()eq o(AB,sup10()； (eq o(BC,sup10()e

12、q o(BB1,sup10()eq o(D1C1,sup10()；(eq o(AD,sup10()eq o(AB,sup10()2eq o(DD1,sup10()； (eq o(B1D1,sup10()eq o(A1A,sup10()eq o(DD1,sup10()。其中能够化简为向量eq o(BD1,sup10()的是()A B C D【解析】(eq o(A1D1,sup10()eq o(A1A,sup10()eq o(AB,sup10()eq o(AD1,sup10()eq o(AB,sup10()eq o(BD1,sup10()；(eq o(BC,sup10()eq o(BB1,sup1

13、0()eq o(D1C1,sup10()eq o(BC1,sup10()eq o(D1C1,sup10()eq o(BD1,sup10()；(eq o(AD,sup10()eq o(AB,sup10()2eq o(DD1,sup10()eq o(BD,sup10()2eq o(DD1,sup10()eq o(BD1,sup10()；(eq o(B1D1,sup10()eq o(A1A,sup10()eq o(DD1,sup10()eq o(B1D,sup10()eq o(DD1,sup10()eq o(B1D1,sup10()eq o(BD1,sup10()，所以选A。 【答案】A3若向量a(

14、1，2)，b(2，1,2)且a与b的夹角的余弦值为eq f(8,9)，则等于()A2 B2C2或eq f(2,55) D2或eq f(2,55)【答案】C4平行六面体ABCDABCD中，若eq o(AC,sup10()xeq o(AB,sup10()2yeq o(BC,sup10()3zeq o(CC,sup10()，则xyz()A1 B.eq f(7,6)C.eq f(5,6) D.eq f(2,3)【解析】eq o(AC,sup10()eq o(AC,sup10()eq o(CC,sup10()eq o(AD,sup10()eq o(AB,sup10()eq o(CC,sup10()eq

15、o(AB,sup10()eq o(BC,sup10()eq o(CC,sup10()xeq o(AB,sup10()2yeq o(BC,sup10()3zeq o(CC,sup10()，故x1，yeq f(1,2)，zeq f(1,3)，xyz1eq f(1,2)eq f(1,3)eq f(7,6)。 【答案】B5已知直线AB、CD是异面直线，ACCD，BDCD，且AB2，CD1，则异面直线AB与CD夹角的大小为()A30 B45C60 D75【解析】coseq o(AB,sup10()，eq o(CD,sup10()eq f(o(AB,sup10()o(CD,sup10(),|o(AB,su

16、p10()|o(CD,sup10()|)eq f(o(AC,sup10()o(CD,sup10()o(DB,sup10()o(CD,sup10(),21)eq f(o(CD,sup10()2,2)eq f(1,2)，eq o(AB,sup10()与eq o(CD,sup10()所成角为60。 【答案】C6正方体ABCDA1B1C1D1的棱长为a，点M在eq o(AC1,sup10()上且eq o(AM,sup10()eq f(1,2)eq o(MC1,sup10()，N为B1B的中点，则|eq o(MN,sup10()|为()A.eq f(r(21),6)a B.eq f(r(6),6)aC.

17、eq f(r(15),6)a D.eq f(r(15),3)a点M在eq o(AC1,sup10()上且eq o(AM,sup10()eq f(1,2)eq o(MC1,sup10()，(xa，y，z)eq f(1,2)(x，ay，az)，xeq f(2,3)a，yeq f(a,3)，zeq f(a,3)，得Meq blc(rc)(avs4alco1(f(2a,3)，f(a,3)，f(a,3)，|eq o(MN,sup10()|eq r(blc(rc)(avs4alco1(af(2,3)a)2blc(rc)(avs4alco1(af(a,3)2blc(rc)(avs4alco1(f(a,2)f

18、(a,3)2)eq f(r(21),6)a。 【答案】A7如图所示，已知空间四边形ABCD，F为BC的中点，E为AD的中点，若eq o(EF,sup10()(eq o(AB,sup10()eq o(DC,sup10()，则_。【解析】如图所示，取AC的中点G，连接EG、GF，则eq o(EF,sup10()eq o(EG,sup10()eq o(GF,sup10()eq f(1,2)(eq o(AB,sup10()eq o(DC,sup10()eq f(1,2)。 【答案】eq f(1,2)8已知O(0,0,0)，A(1,2,3)，B(2,1,2)，P(1,1,2)，点Q在直线OP上运动，当e

19、q o(QA,sup10()eq o(QB,sup10()最小时，点Q的坐标是_。【答案】eq blc(rc)(avs4alco1(f(4,3)，f(4,3)，f(8,3)9在正方体ABCDA1B1C1D1中，下面给出四个命题：(eq o(A1A,sup10()eq o(A1D1,sup10()eq o(A1B1,sup10()23(eq o(A1B1,sup10()2；eq o(A1C,sup10()(eq o(A1B1,sup10()eq o(A1A,sup10()0；eq o(AD1,sup10()与eq o(A1B,sup10()的夹角为60；此正方体的体积为|eq o(AB,sup1

20、0()eq o(AA1,sup10()eq o(AD,sup10()|。则正确命题的序号是_(填写所有正确命题的序号)。【解析】|eq o(A1A,sup10()eq o(A1D1,sup10()eq o(A1B1,sup10()|eq o(A1C,sup10()|eq r(3)|eq o(A1B1,sup10()|，正确；eq o(A1C,sup10()(eq o(A1B1,sup10()eq o(A1A,sup10()eq o(A1C,sup10()eq o(A1B1,sup10()eq o(A1C,sup10()eq o(A1A,sup10()；(eq o(A1C,sup10()，eq

21、o(A1B1,sup10()(eq o(A1C,sup10()，eq o(A1A,sup10()，|eq o(A1B1,sup10()|eq o(A1A,sup10()|eq o(A1C,sup10()eq o(A1B1,sup10()eq o(A1C,sup10()eq o(A1A,sup10()0.正确；AD1与A1B两异面直线的夹角为60，但eq o(AD1,sup10()与eq o(A1B,sup10()的夹角为120，eq o(A1B,sup10()eq o(D1C,sup10()，注意方向，eq o(AB,sup10()eq o(AA1,sup10()0，正确的应是|eq o(AB

22、,sup10()|eq o(AA1,sup10()|eq o(AD,sup10()|。 【答案】10如图所示，在平行六面体ABCDA1B1C1D1中，设eq o(AA1,sup10()a，eq o(AB,sup10()b，eq o(AD,sup10()c，M、N、P分别是AA1、BC、C1D1的中点，试用a，b，c表示以下各向量：(1)eq o(AP,sup10()；(2)eq o(A1N,sup10()；(3)eq o(MP,sup10()eq o(NC1,sup10()。 (3)M是AA1的中点，eq o(MP,sup10()eq o(MA,sup10()eq o(AP,sup10()eq f(1,2)eq o(A1A,sup10()eq o(AP,sup10()eq f(1,2)aeq blc(rc)(avs4alco1(acf(1,2)b)eq f(1,2)aeq f(1,2)bc。又eq o(NC1,sup10()eq o(NC,sup10()eq o(CC1,sup10()eq f(1,2)eq o(BC,sup10()eq o(AA1,sup10()eq f(1,2)eq o(AD,sup10()eq o(AA1,sup10()eq f(1,2)ca，eq o(MP,sup10()eq o(NC1,sup10()eq blc(rc)(avs