1、A級：“四基”鞏固訓練一、選擇題1下列各式計算正確的個數(shù)是()(7)5a35a；a2b2(ab)3a；ab(ab)0.A0 B1 C2 D3答案C解析根據向量數(shù)乘的運算律可驗證正確；錯誤，因為向量的和、差及數(shù)乘運算的結果仍為一個向量，而不是實數(shù)2如圖所示，D是ABC的邊AB上的中點，則向量eq o(CD,sup16()()A.eq o(BC,sup16()eq f(1,2)eq o(BA,sup16()Beq o(BC,sup16()eq f(1,2)eq o(BA,sup16()Ceq o(BC,sup16()eq f(1,2)eq o(BA,sup16()D.eq o(BC,sup16()

2、eq f(1,2)eq o(BA,sup16()答案B解析解法一：D是AB的中點，eq o(BD,sup16()eq f(1,2)eq o(BA,sup16()，eq o(CD,sup16()eq o(CB,sup16()eq o(BD,sup16()eq o(BC,sup16()eq f(1,2)eq o(BA,sup16().解法二：由eq o(CD,sup16()eq f(1,2)(eq o(CB,sup16()eq o(CA,sup16()eq f(1,2)eq o(CB,sup16()(eq o(CB,sup16()eq o(BA,sup16()eq o(CB,sup16()eq f

3、(1,2)eq o(BA,sup16()eq o(BC,sup16()eq f(1,2)eq o(BA,sup16().3設Aeq o(B,sup16()eq f(r(2),2)(a5b)，Beq o(C,sup16()2a8b，Ceq o(D,sup16()3(ab)，則共線的三點是()AA，B，D BA，B，C CB，C，D DA，C，D答案A解析Beq o(D,sup16()Beq o(C,sup16()Ceq o(D,sup16()a5b，Aeq o(B,sup16()eq f(r(2),2)Beq o(D,sup16()，A，B，D三點共線故選A.4若eq o(AB,sup16()3

4、e1，eq o(CD,sup16()5e1，且|eq o(AD,sup16()|eq o(BC,sup16()|，則四邊形ABCD是()A平行四邊形 B菱形C等腰梯形 D不等腰的梯形答案C解析因為eq o(AB,sup16()eq f(3,5)eq o(CD,sup16()，所以ABCD，且|eq o(AB,sup16()|eq o(CD,sup16()|.而|eq o(AD,sup16()|eq o(BC,sup16()|，所以四邊形ABCD為等腰梯形5在平行四邊形ABCD中，AC與BD交于點O，E是線段OD的中點，AE的延長線與CD交于點F.若eq o(AC,sup16()a，eq o(B

5、D,sup16()b，則eq o(AF,sup16()等于()A.eq f(1,4)aeq f(1,2)b B.eq f(2,3)aeq f(1,3)bC.eq f(1,2)aeq f(1,4)b D.eq f(1,3)aeq f(2,3)b答案B解析如圖所示，E是OD的中點，eq o(OE,sup16()eq f(1,4)eq o(BD,sup16()eq f(1,4)b.又ABEFDE，eq f(AE,FE)eq f(BE,DE)eq f(3,1).eq o(AE,sup16()3eq o(EF,sup16()，eq o(AE,sup16()eq f(3,4)eq o(AF,sup16()

6、，在AOE中，eq o(AE,sup16()eq o(AO,sup16()eq o(OE,sup16()eq f(1,2)aeq f(1,4)b，eq o(AF,sup16()eq f(4,3)eq o(AE,sup16()eq f(2,3)aeq f(1,3)b.故選B.二、填空題6設e1，e2是兩個不共線的向量，若向量ke12e2與8e1ke2方向相反，則k_.答案4解析ke12e2與8e1ke2共線，ke12e2(8e1ke2)8e1ke2.eq blcrc (avs4alco1(k8，,2k，)解得eq blcrc (avs4alco1(f(1,2)，,k4)或eq blcrc (av

7、s4alco1(f(1,2)，,k4.)ke12e2與8e1ke2反向，eq f(1,2)，k4.7若ae13e2，b4e12e2，c3e112e2，則向量a寫為1b2c的形式是_答案eq f(1,18)beq f(7,27)c解析若a1b2c，則e13e21(4e12e2)2(3e112e2)，e13e2(4132)e1(21122)e2.eq blcrc (avs4alco1(41321，,211223.)解得eq blcrc (avs4alco1(1f(1,18)，,2f(7,27).)8如圖，在ABC中，點O是BC的中點，過點O的直線分別交直線AB，AC于不同的兩點M，N，若eq o(

8、AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，則mn的值為_答案2解析解法一：因為eq o(AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，所以eq o(AM,sup16()eq f(1,m)eq o(AB,sup16()，eq o(AN,sup16()eq f(1,n)eq o(AC,sup16()，則eq o(MN,sup16()eq o(AN,sup16()eq o(AM,sup16()eq f(1,n)eq o(AC,sup16()eq f(1,

9、m)eq o(AB,sup16().因為點O為BC的中點，連接AO，所以eq o(AO,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()，則eq o(MO,sup16()eq o(AO,sup16()eq o(AM,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()eq f(1,m)eq o(AB,sup16()eq blc(rc)(avs4alco1(f(1,2)f(1,m)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()，因為M，O，N三點

10、共線，所以可設eq o(MO,sup16()eq o(MN,sup16()，即eq blc(rc)(avs4alco1(f(1,2)f(1,m)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()eq f(,n)eq o(AC,sup16()eq f(,m)eq o(AB,sup16()，則eq blc(rc)(avs4alco1(f(1,2)f(1,m)f(,m)eq o(AB,sup16()eq blc(rc)(avs4alco1(f(1,2)f(,n)eq o(AC,sup16()0，由于eq o(AB,sup16()，eq o(AC,sup16()不共線，所以

11、eq blcrc (avs4alco1(f(1,2)f(1,m)f(,m)0，,f(1,2)f(,n)0，)消去得eq f(1,2)eq f(1,m)eq f(n,2m)0，變形整理可得mn2.解法二：在ABC中，連接AO.由于O是BC的中點，因此eq o(AO,sup16()eq f(1,2)(eq o(AB,sup16()eq o(AC,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16().由于eq o(AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，則eq o(

12、AO,sup16()eq f(1,2)meq o(AM,sup16()eq f(1,2)neq o(AN,sup16().由于M，O，N三點共線，則eq f(1,2)meq f(1,2)n1，從而mn2.三、解答題9設e1，e2是兩個不共線的向量，如果eq o(AB,sup16()2e1e2，eq o(BC,sup16()3e1e2，eq o(CD,sup16()7e16e2.(1)求證：A，B，D三點共線；(2)試確定的值，使2e1e2和e1e2共線；(3)若e1e2與e1e2不共線，試求的取值范圍解(1)證明：因為eq o(BD,sup16()eq o(BC,sup16()eq o(CD,

13、sup16()3e1e27e16e210e15e25(2e1e2)5eq o(AB,sup16()，所以eq o(AB,sup16()與eq o(BD,sup16()共線因為eq o(AB,sup16()與eq o(BD,sup16()有公共點B，所以A，B，D三點共線(2)因為2e1e2與e1e2共線，所以存在實數(shù)，使2e1e2(e1e2)因為e1，e2不共線，所以eq blcrc (avs4alco1(2，,1.)所以eq f(r(2),2).(3)假設e1e2與e1e2共線，則存在實數(shù)，使e1e2(e1e2)因為e1，e2不共線，所以eq blcrc (avs4alco1(1，,，)所以

14、1.所以當1時，e1e2與e1e2不共線B級：“四能”提升訓練1如圖所示，向量eq o(OA,sup16()，eq o(OB,sup16()，eq o(OC,sup16()的終點A，B，C在一條直線上，且eq o(AC,sup16()3eq o(CB,sup16().設eq o(OA,sup16()p，eq o(OB,sup16()q，eq o(OC,sup16()r，則以下等式中成立的是()Areq f(1,2)peq f(3,2)q Brp2qCreq f(3,2)peq f(1,2)q Drq2p答案A解析eq o(OC,sup16()eq o(OB,sup16()eq o(BC,sup

15、16()，eq o(AC,sup16()3eq o(CB,sup16()3eq o(BC,sup16()，eq o(BC,sup16()eq f(1,3)eq o(AC,sup16().eq o(OC,sup16()eq o(OB,sup16()eq f(1,3)eq o(AC,sup16()eq o(OB,sup16()eq f(1,3)(eq o(OC,sup16()eq o(OA,sup16()rqeq f(1,3)(rp)req f(1,2)peq f(3,2)q.2設D，E分別是ABC的邊AB，BC上的點，ADeq f(1,2)AB，BEeq f(2,3)BC.若eq o(DE,sup16()1eq o(AB,sup16()2eq o(AC,sup16()(1，2為實數(shù))，則12的值為_答案eq f(1,2)解析由已知eq