人教A版選修2-1第三章3.1.1空間向量及其加減運算達標過關(guān)訓練

1、3.1.1空間向量及其加減運算一、選擇題1對于空間中任意四點A，B，C，D，都有eq o(DA,sup6()eq o(CD,sup6()eq o(CB,sup6()()Aeq o(DB,sup6()Beq o(AC,sup6()Ceq o(AB,sup6() Deq o(BA,sup6()解析：eq o(DA,sup6()eq o(CD,sup6()eq o(CB,sup6()eq o(DA,sup6()eq o(CD,sup6()eq o(BC,sup6()eq o(BD,sup6()eq o(DA,sup6()eq o(BA,sup6().答案：D2在正方體ABCDA1B1C1D1中，下列

2、選項中化簡后為零向量的是()Aeq o(AB,sup6()eq o(AD,sup6()eq o(AA1,sup6() Beq o(AB,sup6()eq o(AC,sup6()eq o(BB1,sup6()Ceq o(AB,sup6()eq o(A1D1,sup6()eq o(C1A1,sup6() Deq o(AC,sup6()eq o(CB1,sup6()eq o(AB,sup6()解析：eq o(AB,sup6()eq o(A1D1,sup6()eq o(C1A1,sup6()eq o(A1B1,sup6()eq o(A1D1,sup6()eq o(C1A1,sup6()eq o(A1C

3、1,sup6()eq o(A1C1,sup6()0.答案：C3如圖所示，在正方體ABCDA1B1C1D1中，下列各式中運算結(jié)果為向量eq o(AC1,sup6()的是()(eq o(AB,sup6()eq o(BC,sup6()eq o(CC1,sup6()；(eq o(AA1,sup6()eq o(A1D1,sup6()eq o(D1C1,sup6()；(eq o(AB,sup6()eq o(BB1,sup6()eq o(B1C1,sup6()；(eq o(AA1,sup6()eq o(A1B1,sup6()eq o(B1C1,sup6().A BC D解析：(eq o(AB,sup6()e

4、q o(BC,sup6()eq o(CC1,sup6()eq o(AC,sup6()eq o(CC1,sup6()eq o(AC1,sup6().(eq o(AA1,sup6()eq o(A1D1,sup6()eq o(D1C1,sup6()eq o(AD1,sup6()eq o(D1C1,sup6()eq o(AC1,sup6().(eq o(AB,sup6()eq o(BB1,sup6()eq o(B1C1,sup6()eq o(AB1,sup6()eq o(B1C1,sup6()eq o(AC1,sup6().(eq o(AA1,sup6()eq o(A1B1,sup6()eq o(B1

5、C1,sup6()eq o(AB1,sup6()eq o(B1C1,sup6()eq o(AC1,sup6().答案：D4在下列條件中，使M與A、B、C一定共面的是()Aeq o(OM,sup6()3eq o(OA,sup6()2eq o(OB,sup6()eq o(OC,sup6()Beq o(OM,sup6()eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()0Ceq o(MA,sup6()eq o(MB,sup6()eq o(MC,sup6()0Deq o(OM,sup6()eq f(1,4)eq o(OB,sup6()eq o(OA,sup6()eq

6、f(1,2)eq o(OC,sup6()解析：在選項C中，eq o(MA,sup6()eq o(MB,sup6()eq o(MC,sup6()，故M與A，B，C四點共面答案：C5下列命題中正確的個數(shù)是()若a與b共線，b與c共線，則a與c共線；向量a，b，c共面即它們所在的直線共面；若ab，則存在唯一的實數(shù)，使ab.A1 B2C3 D0解析：中，b0時，則a與c不一定共線；中，由共面向量的定義是平行于同一平面的向量，表示這些向量的有向線段所在的直線不一定共面；中，當b0，a0時，不存在，故均錯答案：D二、填空題6把所有單位向量的起點移到同一點，則這些向量的終點組成的圖形是_解析：單位向量的長度

7、都相等，起點移到同一點，則終點在以單位長度為半徑的球面上答案：球面7已知空間向量eq o(AB,sup6()，eq o(BC,sup6()，eq o(CD,sup6()，eq o(DA,sup6()，則下列結(jié)論正確的是_eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()；eq o(AB,sup6()eq o(DC,sup6()eq o(BC,sup6()eq o(AD,sup6()；eq o(AB,sup6()eq o(BC,sup6()eq o(DC,sup6()eq o(AD,sup6()；eq o(BC,sup6()eq o(BD,sup6()eq o(

8、DC,sup6().解析：eq o(BC,sup6()eq o(CD,sup6()eq o(BD,sup6()eq o(AB,sup6()，故不正確；eq o(AB,sup6()eq o(DC,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()eq o(AD,sup6()，故正確；不正確答案：8給出下列命題：方向相反的兩個向量是相反向量；若A，B，C，D是不共線的四點，則eq o(AB,sup6()eq o(DC,sup6()是四邊形ABCD為平行四邊形的充要條件；|a|b|是向量ab的必要不充分條件；eq o(AB,s

9、up6()eq o(CD,sup6()的充要條件是A與C重合，B與D重合其中正確命題的序號為_解析：模相等，方向相反的兩個向量是相反向量，故不正確，正確；模相等，方向相同的兩個向量是相等向量，故不正確答案：三、解答題9設(shè)有空間四邊形ABCD，對角線AC和BD的中點分別是E和F，求證：eq o(AB,sup6()eq o(CB,sup6()eq o(AD,sup6()eq o(CD,sup6()4eq o(EF,sup6().證明：如圖，取CD中點M.則eq o(EF,sup6()eq o(EM,sup6()eq o(MF,sup6()eq f(1,2)eq o(AD,sup6()eq f(1,

10、2)eq o(CB,sup6()，eq o(AD,sup6()eq o(CB,sup6()2eq o(EF,sup6()，同理，可證eq o(AB,sup6()eq o(CD,sup6()2eq o(EF,sup6()，由得，eq o(AB,sup6()eq o(CB,sup6()eq o(AD,sup6()eq o(CD,sup6()4eq o(EF,sup6().10. 如圖所示，在空間四邊形OABC中，G，H分別是ABC，OBC的重心，設(shè)eq o(OA,sup6()a，eq o(OB,sup6()b，eq o(OC,sup6()c.試用向量a，b，c表示向量eq o(OG,sup6()和

11、eq o(GH,sup6().解：G為ABC重心，eq o(AG,sup6()eq f(2,3)eq o(AD,sup6()，eq o(AD,sup6()eq o(OD,sup6()eq o(OA,sup6().D為BC中點，eq o(OD,sup6()eq f(1,2)(eq o(OB,sup6()eq o(OC,sup6()，eq o(OG,sup6()eq o(OA,sup6()eq o(AG,sup6()eq o(OA,sup6()eq f(2,3)eq o(AD,sup6()eq o(OA,sup6()eq f(2,3)(eq o(OD,sup6()eq o(OA,sup6()eq o(OA,sup6()eq f(2,3)eq f(1,2)(eq o(OB,sup6()eq o(OC,sup6()eq f(2,3)eq o(OA,sup6()eq f(1,3)(eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()eq f(1,3)(abc)又eq o(OH,sup6()eq f(2,3)eq o(OD,sup6()eq f(2,3)eq