1、1.1.1空間向量及其線性運(yùn)算基 礎(chǔ) 練 鞏固新知 夯實(shí)基礎(chǔ)1判斷下列各命題的真假：向量a與b平行，則a與b的方向相同或相反；兩個(gè)有共同起點(diǎn)而且相等的向量，其終點(diǎn)必相同；兩個(gè)有公共終點(diǎn)的向量，一定是共線向量；有向線段就是向量，向量就是有向線段其中假命題的個(gè)數(shù)為()A2 B3C4 D52已知空間向量eq o(AB,sup11()、eq o(BC,sup11()、eq o(CD,sup11()、eq o(AD,sup11()，則下列結(jié)論正確的是()A.eq o(AB,sup11()eq o(BC,sup11()eq o(CD,sup11() B.eq o(AB,sup11()eq o(DC,sup

2、11()eq o(BC,sup11()eq o(AD,sup11()C.eq o(AD,sup11()eq o(AB,sup11()eq o(BC,sup11()eq o(DC,sup11() D.eq o(BC,sup11()eq o(BD,sup11()eq o(DC,sup11()3在正方體ABCDA1B1C1D1中，向量表達(dá)式eq o(DD1,sup6()eq o(AB,sup6()eq o(BC,sup6()化簡(jiǎn)后的結(jié)果是()A.eq o(BD1,sup6() B.eq o(D1B,sup6()C.eq o(B1D,sup6() D.eq o(DB1,sup6()4已知正方形ABCD

3、的邊長(zhǎng)為1，設(shè)eq o(AB,sup6()a，eq o(BC,sup6()b，eq o(AC,sup6()c，則|abc|等于()A0 B3 C2eq r(2) D2eq r(2)5在平行六面體ABCDA1B1C1D1中，向量eq o(D1A,sup6()，eq o(D1C,sup6()，eq o(A1C1,sup6()是()A有相同起點(diǎn)的向量 B等長(zhǎng)向量C共面向量 D不共面向量6已知向量a，b，且eq o(AB,sup6()a2b，eq o(BC,sup6()5a6b，eq o(CD,sup6()7a2b，則一定共線的三點(diǎn)是()AA，B，D BA，B，CCB，C，D DA，C，D7已知P為空

4、間中任意一點(diǎn)，A，B，C，D四點(diǎn)滿足任意三點(diǎn)均不共線，但四點(diǎn)共面，且eq o(PA,sup6()eq f(4,3)eq o(PB,sup6()xeq o(PC,sup6() eq f(1,6)eq o(DB,sup6()，則實(shí)數(shù)x的值為()A.eq f(1,3) Beq f(1,3) C.eq f(1,2) Deq f(1,2)8如圖，在正方體ABCDA1B1C1D1中，M、N分別為AB、B1C的中點(diǎn)如何用eq o(AB,sup6()、eq o(AD,sup6()、eq o(AA1,sup6()表示向量eq o(MN,sup6()？ 能 力 練 綜合應(yīng)用 核心素養(yǎng)9.已知非零向量e1，e2不共

5、線，如果eq o(AB,sup6()e1e2，eq o(AC,sup6()2e18e2，eq o(AD,sup6()3e13e2，則A，B，C，D四點(diǎn)()A一定共線B恰是空間四邊形的四個(gè)頂點(diǎn)C一定共面D一定不共面10.在平行六面體ABCDEFGH中，若eq o(AG,sup6()xeq o(AB,sup6()2yeq o(BC,sup6()3zeq o(DH,sup6()，則xyz等于()A.eq f(7,6) B.eq f(2,3) C.eq f(3,4) D.eq f(5,6)11在直三棱柱ABCA1B1C1中，若eq o(CA,sup6()a，eq o(CB,sup6()b，eq o(C

6、C1,sup6()c，則eq o(A1B,sup6()_12設(shè)e1，e2是空間兩個(gè)不共線的向量，已知eq o(AB,sup6()e1ke2，eq o(BC,sup6()5e14e2，eq o(DC,sup6()e12e2，且A，B，D三點(diǎn)共線，實(shí)數(shù)k_.13在長(zhǎng)方體ABCDA1B1C1D1中，化簡(jiǎn)eq o(DA,sup6()eq o(DB,sup6()eq o(B1C,sup6()eq o(B1B,sup6()eq o(A1B1,sup6()eq o(A1B,sup6().14 . 如圖，設(shè)O為ABCD所在平面外任意一點(diǎn)，E為OC的中點(diǎn)，若eq o(AE,sup6()eq f(1,2)eq o

7、(OD,sup6()xeq o(OB,sup6()yeq o(OA,sup6()，求x，y的值15如圖所示，已知空間四邊形ABCD，連接AC，BD，E，F(xiàn)，G分別是BC，CD，DB的中點(diǎn)，請(qǐng)化簡(jiǎn)(1)eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()；(2)eq o(AB,sup6()eq o(GD,sup6()eq o(EC,sup6()，并標(biāo)出化簡(jiǎn)結(jié)果的向量16已知點(diǎn)G是ABC的重心，O是空間任意一點(diǎn)，若eq o(OA,sup11()eq o(OB,sup11()eq o(OC,sup11()eq o(OG,sup11()，求的值【參考答案】1. B 解析

8、假命題，若a與b中有一個(gè)為零向量時(shí)，其方向是不確定的；真命題；假命題，終點(diǎn)相同并不能說明這兩個(gè)向量的方向相同或相反；假命題，向量可用有向線段來表示，但并不是有向線段2B【解析】eq o(AB,sup11()eq o(DC,sup11()eq o(BC,sup11()eq o(AB,sup11()eq o(BC,sup11()eq o(CD,sup11()eq o(AC,sup11()eq o(CD,sup11()eq o(AD,sup11().3. A解析如圖所示，因?yàn)閑q o(DD1,sup6()eq o(AA1,sup6()，eq o(DD1,sup6()eq o(AB,sup6()eq

9、o(AA1,sup6()eq o(AB,sup6()eq o(BA1,sup6()，eq o(BA1,sup6()eq o(BC,sup6()eq o(BD1,sup6()，eq o(DD1,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(BD1,sup6().4D【解析】利用向量加法的平行四邊形法則結(jié)合正方形性質(zhì)求解，|abc|2|eq o(AC,sup6()|2eq r(2).5.C 解析因?yàn)閑q o(D1C,sup6()eq o(D1A,sup6()eq o(AC,sup6()，且eq o(AC,sup6()eq o(A1C1,sup6()，所以eq o(D

10、1C,sup6()eq o(D1A,sup6()eq o(A1C1,sup6()，即eq o(D1C,sup6()eq o(D1A,sup6()eq o(A1C1,sup6().又eq o(D1A,sup6()與eq o(A1C1,sup6()不共線，所以eq o(D1C,sup6()，eq o(D1A,sup6()，eq o(A1C1,sup6()三向量共面6.A 解析因?yàn)閑q o(AD,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()3a6b3(a2b)3eq o(AB,sup6()，故eq o(AD,sup6()eq o(AB,sup6()

11、，又eq o(AD,sup6()與eq o(AB,sup6()有公共點(diǎn)A，所以A，B，D三點(diǎn)共線7. A 解析eq o(PA,sup6()eq f(4,3)eq o(PB,sup6()xeq o(PC,sup6()eq f(1,6)eq o(DB,sup6()eq f(4,3)eq o(PB,sup6()xeq o(PC,sup6()eq f(1,6)(eq o(PB,sup6()eq o(PD,sup6()eq f(3,2)eq o(PB,sup6()xeq o(PC,sup6()eq f(1,6)eq o(PD,sup6().又P是空間任意一點(diǎn)，A，B，C，D四點(diǎn)滿足任意三點(diǎn)均不共線，但四

12、點(diǎn)共面，eq f(3,2)xeq f(1,6)1，解得xeq f(1,3).8. 解 eq o(MN,sup6()eq o(MB,sup6()eq o(BC,sup6()eq o(CN,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)(eq o(CB,sup6()eq o(BB1,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)(eq o(AD,sup6()eq o(AA1,sup6()eq f(1,2)eq o(AB,sup6()eq f(1,2)eq o(AD,sup6()eq

13、f(1,2)eq o(AA1,sup6().9.C 解析因?yàn)榉橇阆蛄縠1，e2不共線，eq o(AB,sup6()e1e2，eq o(AC,sup6()2e18e2，eq o(AD,sup6()3e13e2，所以5eq o(AB,sup6()eq o(AD,sup6()5e15e23e13e22e18e2eq o(AC,sup6()，所以eq o(AC,sup6()5eq o(AB,sup6()eq o(AD,sup6().由向量共面的充要條件可知，A，B，C，D四點(diǎn)共面10.D 解析由于eq o(AG,sup6()eq o(AB,sup6()eq o(AD,sup6()eq o(CG,sup

14、6()eq o(AB,sup6()eq o(BC,sup6()eq o(DH,sup6()，對(duì)照已知式子可得x1，2y1,3z1，故x1，yeq f(1,2)，zeq f(1,3)，從而xyzeq f(5,6).11.abc解析eq o(A1B,sup6()eq o(A1A,sup6()eq o(AB,sup6()eq o(C1C,sup6()(eq o(CB,sup6()eq o(CA,sup6()eq o(CC1,sup6()eq o(CB,sup6()eq o(CA,sup6()cba.12.1 解析eq o(AD,sup6()eq o(AB,sup6()eq o(BC,sup6()eq

15、 o(CD,sup6()7e1(k6)e2，且eq o(AB,sup6()與eq o(AD,sup6()共線，故eq o(AD,sup6()xeq o(AB,sup6()，即7e1(k6)e2xe1xke2，故(7x)e1(k6xk)e20，又e1，e2不共線，eq blcrc (avs4alco1(7x0，,k6kx0，)解得eq blcrc (avs4alco1(x7，,k1，)故k的值為1.13. 解eq o(DA,sup6()eq o(DB,sup6()eq o(B1C,sup6()eq o(B1B,sup6()eq o(A1B1,sup6()eq o(A1B,sup6()(eq o(

16、DA,sup6()eq o(DB,sup6()(eq o(B1C,sup6()eq o(B1B,sup6() (eq o(A1B1,sup6()eq o(A1B,sup6()eq o(BA,sup6()eq o(BC,sup6()eq o(BB1,sup6()eq o(BD,sup6()eq o(BB1,sup6()eq o(BD1,sup6().14. 解因?yàn)閑q o(AE,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(CE,sup6()eq o(OB,sup6()eq o(OA,sup6()eq o(OC,sup6()eq o(OB,sup6()eq f(1

17、,2)eq o(OC,sup6()eq o(OA,sup6()eq f(1,2)eq o(OC,sup6()eq o(OA,sup6()eq f(1,2)(eq o(OD,sup6()eq o(DC,sup6()eq o(OA,sup6()eq f(1,2)(eq o(OD,sup6()eq o(AB,sup6()eq o(OA,sup6()eq f(1,2)eq o(OD,sup6()eq f(1,2)(eq o(OB,sup6()eq o(OA,sup6()eq f(3,2)eq o(OA,sup6()eq f(1,2)eq o(OD,sup6()eq f(1,2)eq o(OB,sup6

18、()，所以xeq f(1,2)，yeq f(3,2).解(1)eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()eq o(AC,sup6()eq o(CD,sup6()eq o(AD,sup6().(2)E，F(xiàn)，G分別為BC，CD，DB的中點(diǎn)eq o(BE,sup6()eq o(EC,sup6()，eq o(EF,sup6()eq o(GD,sup6().eq o(AB,sup6()eq o(GD,sup6()eq o(EC,sup6()eq o(AB,sup6()eq o(BE,sup6()eq o(EF,sup6()eq o(AF,sup6().故所求向量eq o(AD,sup6()，eq o(AF,sup6()如圖所示16解：連接CG并延長(zhǎng)交AB于D，則D為AB中點(diǎn)，且CG2GD，所以eq o(OA,sup11()eq o(OB