人教A版选修2-1第三章3.1.3空间向量的数量积运算达标过关训练

1、3.1.3空间向量的数量积运算一、选择题1若a，b均为非零向量，则ab|a|b|是a与b共线的()A充分不必要条件B必要不充分条件C充要条件D既不充分也不必要条件解析：若ab|a|b|cosa，b|a|b|，则cosa，b1，a与b共线，反之，若a，b共线也可能ab|a|b|.答案：A2下列命题中，正确的命题个数为()eq r(aa)|a|；m(a)b(m)ab(m，R)；a(bc)(bc)a；(ab)2a22abb2.A1 B2C3 D4解析：由向量的数量积及其运算律可知，四个命题都正确答案：D3(2019蚌埠第二中学高二月考)已知空间四边形ABCD中，ACDBDC90，且AB2，CD1，则

2、AB与CD所成的角是()A30 B45C60 D90解析：根据已知ACDBDC90，得eq o(AC,sup6()eq o(CD,sup6()eq o(DB,sup6()eq o(CD,sup6()0，eq o(AB,sup6()eq o(CD,sup6()(eq o(AC,sup6()eq o(CD,sup6()eq o(DB,sup6()eq o(CD,sup6()eq o(AC,sup6()eq o(CD,sup6()|eq o(CD,sup6()|2eq o(DB,sup6()eq o(CD,sup6()|eq o(CD,sup6()|21，coseq o(AB,sup6()，eq o

3、(CD,sup6()eq f(o(AB,sup6()o(CD,sup6(),|o(AB,sup6()|o(CD,sup6()|)eq f(1,2)，AB与CD所成的角为60.答案：C4正三棱柱ABCA1B1C1的各棱长均为2，E，F分别是AB，A1C1的中点，则EF的长为()A2 Beq r(3)Ceq r(5) Deq r(7)解析：如图所示，设eq o(AB,sup6()a，eq o(AC,sup6()b，eq o(AA1,sup6()c，则|a|b|c|2，且a，b60，a，c90，b，c90.eq o(EF,sup6()eq o(EA,sup6()eq o(AA1,sup6()eq o

4、(A1F,sup6()eq f(1,2)eq o(AB,sup6()AA1eq f(1,2)eq o(A1C1,sup6()eq f(1,2)aceq f(1,2)b，|eq o(EF,sup6()|2eq f(1,4)a2c2eq f(1,4)b22eq f(1,2)aceq f(1,2)bceq f(1,4)abeq f(1,4)44eq f(1,4)42eq blc(rc)(avs4alco1(00f(1,4)22f(1,2)14115，|eq o(EF,sup6()|eq r(5)，故EF的长为eq r(5).答案：C5如图，在直三棱柱ABCA1B1C1中，ABBCAA1，ABC90，

5、点E，F分别是棱AB，BB1的中点，则直线EF和BC1所成的角是()A45 B60C90 D120解析：eq o(EF,sup6()eq o(BC1,sup6()eq f(1,2)eq o(AB1,sup6()(eq o(BC,sup6()eq o(BB1,sup6()eq f(1,2)(eq o(AB,sup6()eq o(BB1,sup6()(eq o(BC,sup6()eq o(BB1,sup6()eq f(1,2)(eq o(AB,sup6()eq o(BC,sup6()eq o(BB1,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(BB1,sup6()

6、eq o(BB1,sup6()eq o(BB1,sup6()eq f(1,2)(000eq o(BB1,sup6()2)eq f(1,2)|eq o(BB1,sup6()|2.设|eq o(BB1,sup6()|1，则eq o(EF,sup6()eq o(BC1,sup6()eq f(1,2)，|eq o(EF,sup6()|eq f(r(2),2)，|eq o(BC1,sup6()|eq r(2)，coseq o(EF,sup6()，eq o(BC1,sup6()eq f(o(EF,sup6()o(BC1,sup6(),avs4al(|o(EF,sup6()|o(BC1,sup6()|)eq

7、 f(f(1,2),f(r(2),2)r(2)eq f(1,2).异面直线EF与BC1所成的角是60.答案：B二、填空题6如图，已知ABC在平面内，BAC90，DA平面，则直线CA与DB的位置关系是_解析：eq o(CA,sup6()eq o(BD,sup6()eq o(CA,sup6()(eq o(BA,sup6()eq o(AD,sup6()eq o(CA,sup6()eq o(BA,sup6()eq o(CA,sup6()eq o(AD,sup6()000，eq o(CA,sup6()eq o(DB,sup6().答案：垂直7(2019晋中市高二期末)如图所示，在一个直二面角AB的棱上有

8、A，B两点，AC，BD分别是这个二面角的两个面内垂直于AB的线段，且AB4，AC6，BD8，则CD的长为_解析：eq o(CD,sup6()eq o(CA,sup6()eq o(AB,sup6()eq o(BD,sup6()eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()，eq o(CD,sup6()2(eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()2eq o(AB,sup6()2eq o(AC,sup6()22eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()22eq o(AB,sup

9、6()eq o(BD,sup6()2eq o(AC,sup6()eq o(BD,sup6()163664116，|eq o(CD,sup6()|2eq r(29).答案：2eq r(29)8(2019鄂东南九校联考)在四面体OABC中，棱OA，OB，OC两两垂直，且OA1，OB2，OC3，G为ABC的重心，则eq o(OG,sup6()(eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()_.解析：eq o(OG,sup6()eq o(OA,sup6()eq o(AG,sup6()eq o(OA,sup6()eq f(1,3)(eq o(AB,sup6()eq

10、o(AC,sup6()eq o(OA,sup6()eq f(1,3)(eq o(OB,sup6()eq o(OA,sup6()(eq o(OC,sup6()eq o(OA,sup6()eq f(1,3)eq o(OB,sup6()eq f(1,3)eq o(OC,sup6()eq f(1,3)eq o(OA,sup6().eq o(OG,sup6()(eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()eq f(1,3)eq o(OB,sup6()eq f(1,3)eq o(OC,sup6()eq f(1,3)eq o(OA,sup6()(eq o(OA,sup

11、6()eq o(OB,sup6()eq o(OC,sup6()eq f(1,3)eq o(OB,sup6()2eq f(1,3)eq o(OC,sup6()2eq f(1,3)eq o(OA,sup6()2eq f(14,3).答案：eq f(14,3)三、解答题9(2019鄂东南九校联考)已知在正四面体DABC中，所有棱长都为1，ABC的重心为G，求DG的长解：如图，连接AG并延长交BC于点M，连接DM，G是ABC的重心，eq o(AG,sup6()eq f(2,3)eq o(AM,sup6()，又eq o(DG,sup6()eq o(DA,sup6()eq o(AG,sup6()eq o(

12、DA,sup6()eq f(2,3)eq o(AM,sup6()eq o(DA,sup6()eq f(2,3)(eq o(DM,sup6()eq o(DA,sup6()eq o(DA,sup6()eq f(2,3)eq f(1,2)(eq o(DB,sup6()eq o(DC,sup6()eq o(DA,sup6()eq f(1,3)(eq o(DA,sup6()eq o(DB,sup6()eq o(DC,sup6()，而(eq o(DA,sup6()eq o(DB,sup6()eq o(DC,sup6()2eq o(DA,sup6()2eq o(DB,sup6()2eq o(DC,sup6(

13、)22eq o(DA,sup6()eq o(DB,sup6()2eq o(DB,sup6()eq o(DC,sup6()2eq o(DC,sup6()eq o(DA,sup6()1112(cos 60cos 60cos 60)6，|eq o(DG,sup6()|eq f(r(6),3).10如图，PA垂直于矩形ABCD所在的平面，M，N分别为AB，PC的中点(1)求证：MNCD；(2)若PDA45，求证：MN平面PCD.证明：(1)设eq o(AB,sup6()a，eq o(AD,sup6()b，eq o(AP,sup6()c，则eq o(MN,sup6()eq o(MB,sup6()eq o

14、(BC,sup6()eq o(CN,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)eq o(PC,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)(eq o(PA,sup6()eq o(AB,sup6()eq o(BC,sup6()eq f(1,2)(eq o(AD,sup6()eq o(AP,sup6()eq f(1,2)(bc)，eq o(CD,sup6()eq o(AB,sup6()a，eq o(MN,sup6()eq o(CD,sup6()eq f(1,2)(bc)(a)eq f(1,2)(abac)四边形ABCD是矩形，PA平面ABCD，ab，ac，ab0，ac0.eq o(MN,sup6()eq o(CD,sup6()0，eq o(MN,sup6()eq o(CD,sup6()，即MNCD.(2)由(1)知，MNCD，eq o(MN,sup6()eq f(1,2)(bc)，eq o(