高中數(shù)學(xué)必修二 第六章 6.2 6.2.1 課后課時(shí)精練

1、A級(jí)：“四基”鞏固訓(xùn)練一、選擇題1在平行四邊形ABCD中，下列式子：eq o(AD,sup16()eq o(AB,sup16()eq o(BD,sup16()；eq o(AD,sup16()eq o(AC,sup16()eq o(CD,sup16()；eq o(AD,sup16()eq o(AB,sup16()eq o(AC,sup16()；eq o(AB,sup16()eq o(BC,sup16()eq o(AC,sup16()；eq o(AD,sup16()eq o(AB,sup16()eq o(BC,sup16()eq o(CD,sup16()；eq o(AD,sup16()eq o(D

2、C,sup16()eq o(CA,sup16().其中不正確的個(gè)數(shù)是()A1 B2 C4 D6答案A解析eq o(DC,sup16()eq o(CA,sup16()eq o(DA,sup16()，故不正確；其他都正確2設(shè)a(eq o(AB,sup16()eq o(CD,sup16()(eq o(BC,sup16()eq o(DA,sup16()，b是任一非零向量，則在下列結(jié)論中，正確的是()ab；aba；abb；|ab|a|b|；|ab|a|b|.A B C D答案C解析a(eq o(AB,sup16()eq o(CD,sup16()(eq o(BC,sup16()eq o(DA,sup16(

3、)eq o(AB,sup16()eq o(BC,sup16()eq o(CD,sup16()eq o(DA,sup16()0，易知正確故選C.3已知D，E，F(xiàn)分別是ABC的邊AB，BC，CA的中點(diǎn)，則下列等式中不正確的是()A.eq o(FD,sup16()eq o(DA,sup16()eq o(FA,sup16()B.eq o(FD,sup16()eq o(DE,sup16()eq o(EF,sup16()0C.eq o(DE,sup16()eq o(DA,sup16()eq o(EC,sup16()D.eq o(DA,sup16()eq o(DE,sup16()eq o(FD,sup16(

4、)答案D解析由向量加法的平行四邊形法則可知，eq o(DA,sup16()eq o(DE,sup16()eq o(DF,sup16()eq o(FD,sup16().4已知四邊形ABCD是菱形，則下列等式中成立的是()A.eq o(AB,sup16()eq o(BC,sup16()eq o(CA,sup16() B.eq o(AB,sup16()eq o(AC,sup16()eq o(BC,sup16()C.eq o(AC,sup16()eq o(BA,sup16()eq o(AD,sup16() D.eq o(AC,sup16()eq o(AD,sup16()eq o(DC,sup16()答

5、案C解析由四邊形ABCD是菱形知Ceq o(D,sup16()Beq o(A,sup16()，則Aeq o(C,sup16()Beq o(A,sup16()Aeq o(C,sup16()Ceq o(D,sup16()Aeq o(D,sup16().故選C.5已知ABC的三個(gè)頂點(diǎn)A，B，C及平面內(nèi)一點(diǎn)P滿足eq o(PA,sup16()eq o(PB,sup16()eq o(PC,sup16()，則下列結(jié)論中正確的是()AP在ABC的內(nèi)部BP在ABC的邊AB上CP在AB邊所在的直線上DP在ABC的外部答案D解析由eq o(PA,sup16()eq o(PB,sup16()eq o(PC,sup1

6、6()可得四邊形PBCA為平行四邊形，所以點(diǎn)P在ABC的外部二、填空題6根據(jù)圖示填空(1)eq o(AB,sup16()eq o(OA,sup16()_；(2)eq o(BO,sup16()eq o(OD,sup16()eq o(DO,sup16()_；(3)eq o(AO,sup16()eq o(OB,sup16()eq o(BC,sup16()_.答案(1)eq o(OB,sup16()(2)eq o(BO,sup16()(3)eq o(AC,sup16()解析由三角形法則知(1)eq o(AB,sup16()eq o(OA,sup16()eq o(OA,sup16()eq o(AB,su

7、p16()eq o(OB,sup16().(2)eq o(BO,sup16()eq o(OD,sup16()eq o(DO,sup16()eq o(BD,sup16()eq o(DO,sup16()eq o(BO,sup16().(3)eq o(AO,sup16()eq o(OB,sup16()eq o(BC,sup16()eq o(AB,sup16()eq o(BC,sup16()eq o(AC,sup16().7已知eq o(AB,sup16()a，eq o(BC,sup16()b，eq o(CD,sup16()c，eq o(DE,sup16()d，eq o(AE,sup16()e，則ab

8、cd_.答案e解析abcdeq o(AB,sup16()eq o(BC,sup16()eq o(CD,sup16()eq o(DE,sup16()eq o(AE,sup16()e.8若P為ABC的外心，且eq o(PA,sup16()eq o(PB,sup16()eq o(PC,sup16()，則ACB_.答案120解析如圖，因?yàn)閑q o(PA,sup16()eq o(PB,sup16()eq o(PC,sup16()，則四邊形APBC是平行四邊形又P為ABC的外心，所以|eq o(PA,sup16()|eq o(PB,sup16()|eq o(PC,sup16()|.因此，ACB120.三、

9、解答題9已知矩形ABCD中，寬為2，長(zhǎng)為2eq r(3)，eq o(AB,sup16()a，eq o(BC,sup16()b，eq o(AC,sup16()c，試作出向量abc，并求出其模的大小解如圖，作eq o(CE,sup16()eq o(AC,sup16()，則abceq o(AE,sup16()，abceq o(AB,sup16()eq o(BC,sup16()eq o(AC,sup16()2eq o(AC,sup16()2c，|abc|2eq o(AC,sup16()|2eq r(222r(3)2)8.B級(jí)：“四能”提升訓(xùn)練1在某地大地震后，一架救援直升飛機(jī)從A地沿北偏東60方向飛行

10、了40 km到B地，再由B地沿正北方向飛行40 km到達(dá)C地，此時(shí)直升飛機(jī)位于A地_方向，與A地的距離為_(kāi)km.答案北偏東3040eq r(3)解析如圖所示，設(shè)eq o(AB,sup16()，eq o(BC,sup16()分別是直升飛機(jī)兩次位移，則eq o(AC,sup16()表示兩次位移的合位移，即eq o(AC,sup16()eq o(AB,sup16()eq o(BC,sup16().在RtABD中，|eq o(DB,sup16()|20 km，|eq o(AD,sup16()|20eq r(3) km，在RtACD中，|eq o(AC,sup16()|eq r(avs4al(|o(AD,sup16()|2|o(DC,sup16()|2)40eq r(3) km，CAD60，即此時(shí)直升飛機(jī)位于A地北偏東30，與A地的距離為40eq r(3) km.2已知船在靜水中的速度為20 m/min，水流的速度為10 m/min，如果船從岸邊出發(fā)沿垂直于水流的航線到達(dá)對(duì)岸，求船行進(jìn)的方向解作eq o(AB,sup16()v水，eq o(AD,sup16()v船，以eq o(AB,sup16()，eq o(AD,sup16()為鄰邊作ABCD，則eq o(AC,sup16()v實(shí)際，如圖由題意可知CAB90，在RtABC中，|