2022年秋新教材高中數(shù)學課時跟蹤檢測二空間向量的數(shù)量積運算新人教A版選擇性必修第一冊

1、PAGE PAGE PAGE 6課時跟蹤檢測（二） 空間向量的數(shù)量積運算1多選下列各命題中，正確的命題是()Aeq r(aa)|a|Bm(a)b(m)ab(m，R)Ca(bc)(bc)aDa2bb2a解析：選ABCaa|a|2，eq r(aa)|a|，故A正確m(a)b(ma)bmab(m)ab，故B正確a(bc)abac，(bc)abacaabaca(bc)，故C正確a2b|a|2b，b2a|b|2a，故D不一定正確2已知e1，e2為單位向量，且e1e2，若a2e13e2，bke14e2，ab，則實數(shù)k的值為( )A6 B6C3 D3解析：選B由題意可得ab0，e1e20，|e1|e2|1，

2、(2e13e2)(ke14e2)0，2k120，k6.3已知空間四邊形ABCD的每條邊和對角線的長都等于a，點E，F(xiàn)分別是BC，AD的中點，則eq o(AE,sup7()eq o(AF,sup7()的值為( )Aa2 Beq f(1,2)a2Ceq f(1,4)a2 Deq f(r(3),4)a2解析：選Ceq o(AE,sup7()eq o(AF,sup7()eq f(1,2)(eq o(AB,sup7()eq o(AC,sup7()eq f(1,2)eq o(AD,sup7()eq f(1,4)(eq o(AB,sup7()eq o(AD,sup7()eq o(AC,sup7()eq o(

3、AD,sup7()eq f(1,4)eq blc(rc)(avs4alco1(aaf(1,2)aaf(1,2)eq f(1,4)a2.4已知正三棱柱ABCA1B1C1的各條棱的長度都為2，E，F(xiàn)分別是AB，A1C1的中點，則EF的長是( )A2 Beq r(3)Ceq r(5) Deq r(7) 解析：選C由于eq o(EF,sup7()eq o(EA,sup7()eq o(AA1,sup7()eq o(A1F,sup7()，所以|eq o(EF,sup7()|eq r(eq o(EA,sup7()eq o(AA1,sup7()eq o(A1F,sup7()2)eq r(1412blc(rc)

4、(avs4alco1(00f(1,2)eq r(5)，即EF的長是eq r(5).5.如圖，已知PA平面ABC，ABC120，PAABBC6，則PC等于( )A6eq r(2) B6C12 D144解析：選C因為eq o(PC,sup7()eq o(PA,sup7()eq o(AB,sup7()eq o(BC,sup7()，所以eq o(PC,sup7()2eq o(PA,sup7()2eq o(AB,sup7()2eq o(BC,sup7()22eq o(PA,sup7()eq o(AB,sup7()2eq o(PA,sup7()eq o(BC,sup7()2eq o(AB,sup7()eq

5、 o(BC,sup7()363636236cos 60144，所以PC12.6已知|a|13，|b|19，|ab|24，則|ab|_.解析：|ab|2a22abb21322ab192242，2ab46，|ab|2a22abb253046484，故|ab|22.答案：227.如圖，已知四棱柱ABCDA1B1C1D1的底面ABCD是矩形，AB4，AA13，BAA160，E為棱C1D1的中點，則eq o(AB,sup7()eq o(AE,sup7()_.解析：eq o(AE,sup7()eq o(AA1,sup7()eq o(AD,sup7()eq f(1,2)eq o(AB,sup7()，eq o

6、(AB,sup7()eq o(AE,sup7()eq o(AB,sup7()eq o(AA1,sup7()eq o(AB,sup7()eq o(AD,sup7()eq f(1,2)eq o(AB,sup7()243cos 600eq f(1,2)4214.答案：148已知e1，e2是夾角為60的兩個單位向量，則ae1e2與be12e2的夾角是_解析：ab(e1e2)(e12e2)eeq oal(2,1)e1e22eeq oal(2,2)111eq f(1,2)2eq f(3,2)，|a|eq r(a2)eq r(e1e22)eq r(eoal(2,1)2e1e2eoal(2,2)eq r(11

7、1)eq r(3)，|b|eq r(b2)eq r(e12e22)eq r(eoal(2,1)4e1e24eoal(2,2)eq r(124)eq r(3).cosa，beq f(ab,|a|b|)eq f(f(3,2),3)eq f(1,2).a，b120.答案：1209.如圖，在正方體ABCDA1B1C1D1中，E，F(xiàn)分別是C1D1，D1D的中點，正方體的棱長為1.(1)求eq o(CE,sup7()，eq o(AF,sup7()的余弦值；eq o(C1E,sup7()(2)求證：BD1EF.解：(1)eq o(AF,sup7()eq o(AD,sup7()eq o(DF,sup7()eq

8、 o(AD,sup7()eq f(1,2)eq o(AA1,sup7()，eq o(CE,sup7()eq o(CC1,sup7()eq o(C1E,sup7()eq o(AA1,sup7()eq f(1,2)eq o(CD,sup7()eq o(AA1,sup7()eq f(1,2)eq o(AB,sup7().因為eq o(AB,sup7()eq o(AD,sup7()0，eq o(AB,sup7()eq o(AA1,sup7()0，eq o(AD,sup7()eq o(AA1,sup7()0，所以eq o(CE,sup7()eq o(AF,sup7()eq blc(rc)(avs4alc

9、o1(eq o(AA1,sup7()f(1,2) eq o(AB,sup7() )eq blc(rc)(avs4alco1(eq o(AD,sup7()f(1,2) eq o(AA1,sup7() )eq f(1,2).又|eq o(AF,sup7()|eq o(CE,sup7()|eq f(r(5),2)，所以coseq o(CE,sup7()，eq o(AF,sup7()eq f(2,5).(2)證明：因為eq o(BD1,sup7()eq o(BD,sup7()eq o(DD1,sup7()eq o(AD,sup7()eq o(AB,sup7()eq o(AA1,sup7()，eq o(

10、EF,sup7()eq o(ED1,sup7()eq o(D1F,sup7()eq f(1,2)(eq o(AB,sup7()eq o(AA1,sup7()，所以eq o(BD1,sup7()eq o(EF,sup7()0，所以eq o(BD1,sup7()eq o(EF,sup7().即BD1EF.10.如圖，正四棱錐PABCD的各棱長都為a.(1)用向量法證明：BDPC；(2)求|eq o(AC,sup7()eq o(PC,sup7()|的值解：(1)證明：eq o(BD,sup7()eq o(BC,sup7()eq o(CD,sup7()，eq o(BD,sup7()eq o(PC,su

11、p7()(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(PC,sup7()eq o(AB,sup7()eq o(BC,sup7()eq o(PC,sup7()，|eq o(AC,sup7()eq o

12、(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.1多選在正方體ABCDA1B1C1D1中，則下列命題正確的是()A(eq o(AA1,sup7()eq o(AD,sup7()eq o(AB,sup7()23eq

13、o(AB,sup7()2Beq o(A1C,sup7()(eq o(A1B1,sup7()eq o(A1A,sup7()0Ceq o(AD1,sup7()與eq o(A1B,sup7()的夾角為60D正方體的體積為|eq o(AB,sup7()eq o(AA1,sup7()eq o(AD,sup7()|解析：選AB如圖所示，(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,s

14、up7()(eq o(A1B1,sup7()eq o(A1A,sup7()eq o(A1C,sup7()eq o(AB1,sup7()0；eq o(AD1,sup7()與eq o(A1B,sup7()的夾角是eq o(D1C,sup7()與eq o(D1A,sup7()夾角的補角，而eq o(D1C,sup7()與eq o(D1A,sup7()的夾角為60，故eq o(AD1,sup7()與eq o(A1B,sup7()的夾角為120；正方體的體積為|eq o(AB,sup7()|eq o(AA1,sup7()|eq o(AD,sup7()|.綜上可知，A、B正確2設空間上有四個互異的點A，B

15、，C，D，已知(eq o(DB,sup7()eq o(DC,sup7()2eq o(DA,sup7()(eq o(AB,sup7()eq o(AC,sup7()0，則ABC是( )A直角三角形 B等腰三角形C等腰直角三角形 D等邊三角形解析：選B因為eq o(DB,sup7()eq o(DC,sup7()2eq o(DA,sup7()(eq o(DB,sup7()eq o(DA,sup7()(eq o(DC,sup7()eq o(DA,sup7()eq o(AB,sup7()eq o(AC,sup7()，所以(eq o(AB,sup7()eq o(AC,sup7()(eq o(AB,sup7(

16、)eq o(AC,sup7()|eq o(AB,sup7()|2|eq o(AC,sup7()|20，所以|eq o(AB,sup7()|eq o(AC,sup7()|，即ABC是等腰三角形3.如圖，在長方體ABCDA1B1C1D1中，設ADAA11，AB2，P是C1D1的中點，則eq o(B1C,sup7()與eq o(A1P,sup7()所成角的大小為_，eq o(B1C,sup7()eq o(A1P,sup7()_.解析：法一：連接A1D，則PA1D就是eq o(B1C,sup7()與eq o(A1P,sup7()所成角連接PD，在PA1D中，易得PA1DA1PDeq r(2)，即PA1

17、D為等邊三角形，從而PA1D60，即eq o(B1C,sup7()與eq o(A1P,sup7()所成角的大小為60.因此eq o(B1C,sup7()eq o(A1P,sup7()eq r(2)eq r(2)cos 601.法二：根據(jù)向量的線性運算可得eq o(B1C,sup7()eq o(A1P,sup7()(eq o(A1A,sup7()eq blc(rc)(avs4alco1(eq o(AD,sup7()f(1,2) eq o(AB,sup7() )eq o(AD,sup7()21. 由題意可得PA1B1Ceq r(2)，則eq r(2)eq r(2)coseq o(B1C,sup7(

18、)，eq o(A1P,sup7()1，從而eq o(B1C,sup7()，eq o(A1P,sup7()60.答案：6014.在四面體OABC中，各棱長都相等，E，F(xiàn)分別為AB，OC的中點，求異面直線OE與BF所成角的余弦值解：取eq o(OA,sup7()a，eq o(OB,sup7()b，eq o(OC,sup7()c，且|a|b|c|1，則abbccaeq f(1,2).又eq o(OE,sup7()eq f(1,2)(ab)，eq o(BF,sup7()eq f(1,2)cb，eq o(OE,sup7()eq o(BF,sup7()eq f(1,2)(ab)eq blc(rc)(avs4alco1(f(1,2)cb)eq f(1,4)aceq f(1,4)bceq f(1,2)abeq f(1,2)|b|2eq f(1,2).又|eq o(OE,sup7()|eq f(r(3),2)，|eq o(BF,sup7()|eq f(r(3),2)，coseq o(OE,sup7()，eq o(BF,sup7()eq f(eq o(OE,sup7()eq o(BF,sup7(),|eq o(OE,sup7()|eq o(BF,sup7()|)eq f(2,3)，異面直線夾角的范圍為eq blc(rc(avs4alco1(0，f(,