選修2-1兩個(gè)向量的數(shù)量積課時(shí)作業(yè)

1、精選優(yōu)質(zhì)文檔-傾情為你奉上精選優(yōu)質(zhì)文檔-傾情為你奉上專心-專注-專業(yè)專心-專注-專業(yè)精選優(yōu)質(zhì)文檔-傾情為你奉上專心-專注-專業(yè)課時(shí)作業(yè)18兩個(gè)向量的數(shù)量積時(shí)間：45分鐘滿分：100分一、選擇題(每小題5分，共30分)1下列式子中正確的是()A|a|aa2B|ab|2a2b2C(ab)ca(bc)D|ab|a|b|【答案】D【解析】選項(xiàng)A，|a|a應(yīng)是一個(gè)向量，而a2是一個(gè)數(shù)選項(xiàng)B，|ab|2a2b2cos2a，b，而不是a2b2.選項(xiàng)C，向量運(yùn)算中沒有乘法結(jié)合律2已知空間四邊形每條邊和對(duì)角線的長等于a，點(diǎn)E、F、G分別是AB、AD、DC的中點(diǎn)，則a2等于()A2eq o(BA,sup6()eq

2、 o(AC,sup6() B2eq o(AD,sup6()eq o(BD,sup6()C2eq o(FG,sup6()eq o(CA,sup6() D2eq o(EF,sup6()eq o(CB,sup6()【答案】B【解析】2eq o(BA,sup6()eq o(AC,sup6()a2,2eq o(AD,sup6()eq o(BD,sup6()a2,2eq o(FG,sup6()eq o(CA,sup6()a2,2eq o(EF,sup6()eq o(CA,sup6()a2,2eq o(EF,sup6()eq o(CB,sup6()eq f(1,2)a2.3已知a，b是異面直線，A、Ba，C

3、、Db，ACb，BDb且AB2，CD1，則a與b所成的角是()A30 B45C60 D90【答案】C【解析】eq o(AB,sup6()eq o(AC,sup6()eq o(CD,sup6()eq o(DB,sup6()，eq o(AB,sup6()eq o(CD,sup6()(Aeq o(C,sup6()Ceq o(D,sup6()Deq o(B,sup6()Ceq o(D,sup6()Aeq o(C,sup6()Ceq o(D,sup6()eq o(CD2,sup6()Deq o(B,sup6()Ceq o(D,sup6()01201，又|Aeq o(B,sup6()|2，|Ceq o(D

4、,sup6()|1.cosAeq o(B,sup6()，Ceq o(D,sup6()eq f(Ao(B,sup6()Co(D,sup6(),|Ao(B,sup6()|Co(D,sup6()|)eq f(1,21)eq f(1,2).a與b所成的角是60.4已知向量a，b，c兩兩之間的夾角都為60，其模都為1，則|ab2c|等于()A.eq r(5) B5C6 D.eq r(6)【答案】A【解析】(ab2c)2a2b24c22ab4ac4bc1142cos605.|ab2c|eq r(5).5在正三棱柱ABCA1B1C1中，若ABeq r(2)BB1，則AB1與BC1所成角的大小為()A60 B

5、90C105 D75【答案】B【解析】設(shè)ABeq r(2)BB1eq r(2)a，則eq o(AB1,sup6()eq o(BC1,sup6()(eq o(AB,sup6()eq o(BB1,sup6()(eq o(BC,sup6()eq o(CC1,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(BB1,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(CC1,sup6()eq o(BB1,sup6()eq o(CC1,sup6()2a2cos120a20，eq o(AB1,sup6()eq o(BC1,sup6()，即AB1與BC1

6、所成角的大小為90.6正三棱柱ABCA1B1C1的各棱長都為2，E、F分別是AB、A1C1的中點(diǎn)，則EF的長是()A2 B.eq r(3)C.eq r(5) D.eq r(7)【答案】C【解析】如圖所示，設(shè)Aeq o(B,sup6()a，Aeq o(C,sup6()b，eq o(AA1,sup6()c.由題意知|a|b|c|2，且a，b60，a，cb，c90.因?yàn)镋eq o(F,sup6()Eeq o(A,sup6()eq o(AA1,sup6()eq o(A1F,sup6()eq f(1,2)eq o(AB,sup6()eq o(AA1,sup6()eq f(1,2)eq o(AC,sup6

7、()eq f(1,2)aeq f(1,2)bc，所以|EF|2|Eeq o(F,sup6()|2Eeq o(F,sup6()2eq f(1,4)a2eq f(1,4)b2c22(eq f(1,2)aeq f(1,2)beq f(1,2)bceq f(1,2)ac)eq f(1,4)22eq f(1,4)22222(eq f(1,4)22cos6011415，所以|EF|eq r(5).二、填空題(每小題10分，共30分)7已知|a|12，|b|9，ab54eq r(2)，則a，b_.【答案】135【解析】cosa，beq f(ab,|a|b|)eq f(54r(2),129)eq f(r(2)

8、,2)，a，b135.8已知i，j，k都是兩兩垂直的單位向量，a2ijk，bij3k，則ab等于_【答案】2【解析】ab(2ijk)(ij3k)2i2j23k22.9若a，b是兩個(gè)非零向量，且a2bb2a，則a與b的關(guān)系是_【答案】ab【解析】a2bb2a，|a|2b|b|2a.beq f(|b|2,|a|2)a，a與b共線且方向相同又|b|eq f(|b|2,|a|2)|a|，|a|b|.ab.三、解答題(本題共3小題，共40分解答應(yīng)寫出必要的文字說明、證明過程或演算步驟)10(13分)已知三棱錐OABC的各個(gè)側(cè)面都是正三角形，且棱長為1，求：(1)eq o(OA,sup6()eq o(OB

9、,sup6()；(2)(eq o(OA,sup6()eq o(OB,sup6()(eq o(CA,sup6()eq o(CB,sup6()；(3)|eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()|.【解析】設(shè)eq o(OA,sup6()a，eq o(OB,sup6()b，eq o(OC,sup6()c，則|a|b|c|1，a，ba，cb，c60，eq o(CA,sup6()ac，eq o(CB,sup6()bc，(1)eq o(OA,sup6()eq o(OB,sup6()|a|b|cos60eq f(1,2).(2)由(1)知，abacbceq f(1,

10、2)，則(eq o(OA,sup6()eq o(OB,sup6()(eq o(CA,sup6()eq o(CB,sup6()(ab)(ab2c)a2b22ab2ac2bc1.(3)|eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()|eq r(abc2)eq r(a2b2c22ab2ac2bc)eq r(6).11(13分)如圖所示，正方形ABCD與正方形ABEF邊長均為1，且平面ABCD平面ABEF，點(diǎn)M在AC上移動(dòng)，點(diǎn)N在BF上移動(dòng)若CMBNa(0aeq r(2)(1)求MN的長度；(2)當(dāng)a為何值時(shí)，MN的長最小【分析】根據(jù)向量的基本運(yùn)算，利用Neq o

11、(M,sup6()Neq o(F,sup6()Feq o(A,sup6()Aeq o(M,sup6()這一關(guān)系來求|Meq o(N,sup6()|2，這是求長度問題的常見方法；利用向量的坐標(biāo)形式來處理問題，使幾何問題徹底向代數(shù)問題轉(zhuǎn)化【解析】(1)ACeq r(2)，BFeq r(2)，CMBNa(0aeq r(2)Aeq o(M,sup6()(1eq f(a,r(2)Aeq o(C,sup6()，Neq o(F,sup6()(1eq f(a,r(2)Beq o(F,sup6().Neq o(M,sup6()Neq o(F,sup6()Feq o(A,sup6()Aeq o(M,sup6()(

12、1eq f(a,r(2)Beq o(F,sup6()Feq o(A,sup6()(1eq f(a,r(2)Aeq o(C,sup6()(1eq f(a,r(2)(Beq o(E,sup6()Beq o(A,sup6()Feq o(A,sup6()(1eq f(a,r(2)(Aeq o(B,sup6()Beq o(C,sup6()(1eq f(a,r(2)(Beq o(E,sup6()Beq o(A,sup6()Beq o(E,sup6()(1eq f(a,r(2)(Beq o(A,sup6()Beq o(C,sup6()(1eq f(a,r(2)Beq o(C,sup6()(eq f(a,r(

13、2)Beq o(E,sup6().|Neq o(M,sup6()|eq r(avs4al(|No(M,sup6()|2)eq r(1f(a,r(2)Bo(C,sup6()f(a,r(2)Bo(E,sup6()2)eq r(1f(a,r(2)2f(2a,r(2)1f(a,r(2)Bo(C,sup6()Bo(E,sup6()f(1,2)a2)eq r(af(r(2),2)2f(1,2)(0aeq r(2)(2)由(1)知當(dāng)aeq f(r(2),2)時(shí)，|Meq o(N,sup6()|的最小值為eq f(r(2),2)，即M，N分別為AC、BF的中點(diǎn)時(shí)，MN的長最小，最小值為eq f(r(2),2)

14、.12(14分)如圖，正方體ABCDA1B1C1D1中，M、N分別為正方形ABCD和AA1B1B的中心(1)求證：AC1平面A1BD；(2)求eq o(D1M,sup6()與Ceq o(N,sup6()夾角的余弦值【解析】(1)證明：設(shè)Aeq o(B,sup6()a，Aeq o(D,sup6()b，eq o(AA1,sup6()c，則eq o(AC1,sup6()abc，eq o(BD,sup6()ba，eq o(A1B,sup6()ac，注意到abacbc0，則有eq o(AC1,sup6()eq o(BD,sup6()(abc)(ba)b2a2bcac|b|2|a|20.同理，eq o(A

15、C1,sup6()eq o(A1B,sup6()0，eq o(AC1,sup6()eq o(BD,sup6()，eq o(AC1,sup6()eq o(A1B,sup6()，eq o(AC1,sup6()BD，AC1A1B.又A1B與BD相交，eq o(AC1,sup6()平面A1BD.(2)解：設(shè)正方體的棱長為a，eq o(D1M,sup6()eq o(D1D,sup6()Deq o(M,sup6()ceq f(1,2)(ab)，Ceq o(N,sup6()Ceq o(B,sup6()Beq o(N,sup6()beq f(1,2)(ca)，|eq o(D1M,sup6()|2ceq f(1,2)(ab)2a2eq f(1,4)a2eq f(1,4)a2eq f(3,2)a2，|eq o(D1M,s