高中数学选修第一册：课时跟踪检测（二） 空间向量的数量积运算

1、课时跟踪检测（二） 空间向量的数量积运算A级基础巩固1已知两异面直线的方向向量分别为a，b，且|a|b|1，abeq f(1,2)，则两直线的夹角为()A30B60C120D150解析：选B设向量a，b的夹角为，则cos eq f(ab,|a|b|)eq f(1,2)，所以120，则两个方向向量对应的直线的夹角为18012060.2已知a，b是异面直线，且ab，e1，e2分别为取自直线a，b上的单位向量，且m2e13e2，nke14e2，mn，则实数k的值为()A6B6C3D3解析：选B由mn，得mn0，(2e13e2)(ke14e2)0.2k120.k6.3(多选)如图所示，正方体ABCDA

2、1B1C1D1的棱长为a，对角线AC1和BD1相交于点O，则有()Aeq o(A1B,sup7()eq o(B1C,sup7()a2Beq o(AB,sup7()eq o(AC1,sup7()eq r(2)a2Ceq o(AB,sup7()eq o(AO,sup7()eq f(1,2)a2Deq o(BC,sup7()eq o(DA1,sup7()a2解析：选AC连接A1D，则eq o(A1B,sup7()eq o(B1C,sup7()eq o(A1B,sup7()eq o(A1D,sup7()|eq o(A1B,sup7()|eq o(A1D,sup7()|coseq o(A1B,sup7(

3、)，eq o(A1D,sup7()eq r(2)aeq r(2)acos 60a2.A正确eq o(AB,sup7()eq o(AC1,sup7()eq o(AB,sup7()(eq o(AB,sup7()eq o(BC,sup7()eq o(CC1,sup7()eq o(AB,sup7()2eq o(AB,sup7()eq o(BC,sup7()eq o(AB,sup7()eq o(CC1,sup7()a2，故B错误eq o(AB,sup7()eq o(AO,sup7()eq o(AB,sup7()eq f(1,2)eq o(AC1,sup7()eq f(1,2)eq o(AB,sup7()

4、(eq o(AB,sup7()eq o(AD,sup7()eq o(AA1,sup7()eq f(1,2)(eq o(AB,sup7()2eq o(AB,sup7()eq o(AD,sup7()eq o(AB,sup7()eq o(AA1,sup7()eq f(1,2)eq o(AB,sup7()2eq f(1,2)|eq o(AB,sup7()|2eq f(1,2)a2.C正确eq o(BC,sup7()eq o(DA1,sup7()eq o(BC,sup7()(eq o(AA1,sup7()eq o(AD,sup7()eq o(BC,sup7()eq o(AA1,sup7()eq o(BC

5、,sup7()eq o(AD,sup7()a2.D错误4在四面体OABC中，OBOC，AOBAOCeq f(,3)，则coseq o(OA,sup7()，eq o(BC,sup7()()A.eq f(1,2)B.eq f(r(2),2)Ceq f(1,2)D0解析：选Deq o(OA,sup7()eq o(BC,sup7()eq o(OA,sup7()(eq o(OC,sup7()eq o(OB,sup7()eq o(OA,sup7()eq o(OC,sup7()eq o(OA,sup7()eq o(OB,sup7()|eq o(OA,sup7()|eq o(OC,sup7()|coseq o

6、(OA,sup7()，eq o(OC,sup7()|eq o(OA,sup7()|eq o(OB,sup7()|coseq o(OA,sup7()，eq o(OB,sup7()，因为eq o(OA,sup7()，eq o(OC,sup7()eq o(OA,sup7()，eq o(OB,sup7()eq f(,3)，|eq o(OB,sup7()|eq o(OC,sup7()|，所以eq o(OA,sup7()eq o(BC,sup7()0，所以eq o(OA,sup7()eq o(BC,sup7()，所以coseq o(OA,sup7()，eq o(BC,sup7()0.5已知在平行六面体AB

7、CDA1B1C1D1中，AA1ABAD1，且这三条棱彼此之间的夹角都是60，则AC1的长为()A6B.eq r(6)C3D.eq r(3)解析：选B设eq o(AB,sup7()a，eq o(AD,sup7()b，eq o(AA1,sup7()c，则|a|b|c|1，且a，bb，cc，a60，因此abbccaeq f(1,2).由eq o(AC1,sup7()abc得|eq o(AC1,sup7()|2eq o(AC1,sup7()2a2b2c22ab2bc2ca6.|eq o(AC1,sup7()|eq r(6)，故选B.6已知空间中四点A，B，E，C，若eq o(AB,sup7()eq o

8、(BE,sup7()eq o(AB,sup7()eq o(BC,sup7()，则eq o(AB,sup7() _eq o(CE,sup7().解析：eq o(AB,sup7()eq o(BE,sup7()eq o(AB,sup7()eq o(BC,sup7()，则eq o(AB,sup7()(eq o(BE,sup7()eq o(BC,sup7()eq o(AB,sup7()eq o(CE,sup7()0.eq o(AB,sup7()eq o(CE,sup7().答案：7已知a，b是空间两个向量，若|a|2，|b|2，|ab|eq r(7)，则cosa，b_.解析：将|ab|eq r(7)两边

9、平方，得(ab)27.因为|a|2，|b|2，所以abeq f(1,2).又ab|a|b|cosa，b，故cosa，beq f(1,8).答案：eq f(1,8)8.如图，在长方体ABCDA1B1C1D1中，设ADAA11，AB2，P是C1D1的中点，则eq o(B1C,sup7()与eq o(A1P,sup7()所成角的大小为_；eq o(B1C,sup7()eq o(A1P,sup7()_.解析：法一：连接A1D，则PA1D就是eq o(B1C,sup7()与eq o(A1P,sup7()所成角连接PD，在PA1D中，易得PA1DA1PDeq r(2)，即PA1D为等边三角形，从而PA1D

10、60，即eq o(B1C,sup7()与eq o(A1P,sup7()所成角的大小为60.因此eq o(B1C,sup7()eq o(A1P,sup7()eq r(2)eq r(2)cos 601.法二：根据向量的线性运算可得eq o(B1C,sup7()eq o(A1P,sup7()(eq o(A1A,sup7()eq o(AD,sup7()eq blc(rc)(avs4alco1(eq o(AD,sup7()f(1,2)eq o(AB,sup7()eq o(AD,sup7()21.由题意可得PA1B1Ceq r(2)，则eq r(2)eq r(2)coseq o(B1C,sup7()，eq

11、 o(A1P,sup7()1，从而eq o(B1C,sup7()，eq o(A1P,sup7()60.答案：6019在长方体ABCDA1B1C1D1中，ABAA12，AD4，E为侧面AA1B1B的中心，F为A1D1的中点，求下列向量的数量积：(1)eq o(BC,sup7()eq o(ED1,sup7()；(2)eq o(BF,sup7()eq o(AB1,sup7().解：如图，设eq o(AB,sup7()a，eq o(AD,sup7()b，eq o(AA1,sup7()c，则|a|c|2，|b|4，abbcca0.(1)eq o(BC,sup7()eq o(ED1,sup7()eq o(

12、BC,sup7()(eq o(EA1,sup7()eq o(A1D1,sup7()beq blcrc(avs4alco1(f(1,2)cab)|b|24216.(2)eq o(BF,sup7()eq o(AB1,sup7()(eq o(BA1,sup7()eq o(A1F,sup7()(eq o(AB,sup7()eq o(AA1,sup7()eq blc(rc)(avs4alco1(caf(1,2)b)(ac)|c|2|a|222220.10.如图，正四棱锥PABCD的各棱长都为a.(1)用向量法证明BDPC；(2)求|eq o(AC,sup7()eq o(PC,sup7()|的值解：(1)

13、证明：eq o(BD,sup7()eq o(BC,sup7()eq o(CD,sup7()，eq o(BD,sup7()eq o(PC,sup7()(eq o(BC,sup7()eq o(CD,sup7()eq o(PC,sup7()eq o(BC,sup7()eq o(PC,sup7()eq o(CD,sup7()eq o(PC,sup7()|eq o(BC,sup7()|eq o(PC,sup7()|cos 60|eq o(CD,sup7()|eq o(PC,sup7()|cos 120eq f(1,2)a2eq f(1,2)a20.BDPC.(2)eq o(AC,sup7()eq o(P

14、C,sup7()eq o(AB,sup7()eq o(BC,sup7()eq o(PC,sup7()，|eq o(AC,sup7()eq o(PC,sup7()|2|eq o(AB,sup7()|2|eq o(BC,sup7()|2|eq o(PC,sup7()|22eq o(AB,sup7()eq o(BC,sup7()2eq o(AB,sup7()eq o(PC,sup7()2eq o(BC,sup7()eq o(PC,sup7()a2a2a202a2cos 602a2cos 605a2，|eq o(AC,sup7()eq o(PC,sup7()|eq r(5)a.B级综合运用11设A，B

15、，C，D是空间不共面的四点，且满足eq o(AB,sup7()eq o(AC,sup7()0，eq o(AC,sup7()eq o(AD,sup7()0，eq o(AB,sup7()eq o(AD,sup7()0，则BCD是()A钝角三角形B锐角三角形C直角三角形D等腰三角形解析：选Beq o(BD,sup7()eq o(AD,sup7()eq o(AB,sup7()，eq o(BC,sup7()eq o(AC,sup7()eq o(AB,sup7()，eq o(BD,sup7()eq o(BC,sup7()(eq o(AD,sup7()eq o(AB,sup7()(eq o(AC,sup7(

16、)eq o(AB,sup7()eq o(AD,sup7()eq o(AC,sup7()eq o(AD,sup7()eq o(AB,sup7()eq o(AB,sup7()eq o(AC,sup7()|eq o(AB,sup7()|2|eq o(AB,sup7()|20，cos CBDcoseq o(BC,sup7()，eq o(BD,sup7()eq f(eq o(BC,sup7()eq o(BD,sup7(),|eq o(BC,sup7()|eq o(BD,sup7()|)0，CBD为锐角，同理，BCD与BDC均为锐角，BCD为锐角三角形12已知PA平面ABC，垂足为A，ABC120，PAA

17、BBC6，则PC等于_解析：如图所示，eq o(PC,sup7()eq o(PA,sup7()eq o(AB,sup7()eq o(BC,sup7()，eq o(PC,sup7()2eq o(PA,sup7()2eq o(AB,sup7()2eq o(BC,sup7()22eq o(AB,sup7()eq o(BC,sup7()363636236cos 60144.|eq o(PC,sup7()|12.答案：1213在正方体ABCDA1B1C1D1中，有下列命题：(eq o(AA1,sup7()eq o(AD,sup7()eq o(AB,sup7()23eq o(AB,sup7()2；eq o

18、(A1C,sup7()(eq o(A1B1,sup7()eq o(A1A,sup7()0；eq o(AD1,sup7()与eq o(A1B,sup7()的夹角为60；正方体的体积为|eq o(AB,sup7()eq o(AA1,sup7()eq o(AD,sup7()|.其中正确的命题是_(填序号)解析：如图所示，(eq o(AA1,sup7()eq o(AD,sup7()eq o(AB,sup7()2(eq o(AA1,sup7()eq o(A1D1,sup7()eq o(D1C1,sup7()2eq o(AC1,sup7()23eq o(AB,sup7()2；eq o(A1C,sup7()

19、(eq o(A1B1,sup7()eq o(A1A,sup7()eq o(A1C,sup7()eq o(AB1,sup7()0；eq o(AD1,sup7()与eq o(A1B,sup7()的夹角是eq o(D1C,sup7()与eq o(D1A,sup7()夹角的补角，而eq o(D1C,sup7()与eq o(D1A,sup7()的夹角为60，故eq o(AD1,sup7()与eq o(A1B,sup7()的夹角为120；正方体的体积为|eq o(AB,sup7()|eq o(AA1,sup7()|eq o(AD,sup7()|.综上可知，正确答案：14.如图，正三棱柱ABCA1B1C1中

20、，底面边长为eq r(2).(1)设侧棱长为1，求证：AB1BC1；(2)设AB1与BC1的夹角为eq f(,3)，求侧棱的长解：(1)证明：eq o(AB1,sup7()eq o(AB,sup7()eq o(BB1,sup7()，eq o(BC1,sup7()eq o(BB1,sup7()eq o(BC,sup7().BB1平面ABC，eq o(BB1,sup7()eq o(AB,sup7()0，eq o(BB1,sup7()eq o(BC,sup7()0.又ABC为正三角形，eq o(AB,sup7()，eq o(BC,sup7()eq o(BA,sup7()，eq o(BC,sup7()

21、eq f(,3)eq f(2,3).eq o(AB1,sup7()eq o(BC1,sup7()(eq o(AB,sup7()eq o(BB1,sup7()(eq o(BB1,sup7()eq o(BC,sup7()eq o(AB,sup7()eq o(BB1,sup7()eq o(AB,sup7()eq o(BC,sup7()eq o(BB1,sup7()2eq o(BB1,sup7()eq o(BC,sup7()|eq o(AB,sup7()|eq o(BC,sup7()|coseq o(AB,sup7()，eq o(BC,sup7()eq o(BB1,sup7()2110，AB1BC1.(2)由(1)知eq o(AB1,sup7()eq o(BC1,sup7()|eq o(AB,sup7()|eq o(BC,sup7()|coseq o(AB,sup7()，eq o(BC,sup7()eq o(BB1,sup7()2eq o(BB1,sup7()21.又|eq o(AB1,sup7()| eq r(eq o(AB,sup7()2eq o(BB1,sup7()2) eq r(2eq o(BB1,sup7()2