2.1.2向量的幾何表示 (3)

1、 第 PAGE 7 頁 共 NUMPAGES 7 頁 雙基限時練1若非零向量a，b互為相反向量，則下列說法錯誤的是()Aab BabC|a|b| Dba解析根據相反向量的定義：大小相等，方向相反，可知|a|b|.答案C2給出下列四個結論：eq o(AB,sup16()eq o(AO,sup16()eq o(OB,sup16()； eq o(AB,sup16()eq o(AC,sup16()eq o(BC,sup16()；eq o(AB,sup16()eq o(BC,sup16()eq o(CA,sup16()0; |ab|ab|.其中錯誤的有()A1個 B2個C3個 D4個解析正確，錯誤，eq

2、 o(AB,sup16()eq o(AC,sup16()eq o(AB,sup16()eq o(CA,sup16()eq o(CB,sup16()eq o(BC,sup16().錯誤，eq o(AB,sup16()eq o(BC,sup16()eq o(CA,sup16()00.錯誤，當a與b方向相反時，有|ab|eq o(AC,sup16()|eq o(AB,sup16()|可得|ab|a|b|；當a與b反向時，如圖2，知abeq o(CB,sup16()，|eq o(CB,sup16()|eq o(AB,sup16()|eq o(AC,sup16()|，|ab|a|b|.當a與b同向時，如

3、圖3，abeq o(CB,sup16()，|eq o(CB,sup16()|eq o(AB,sup16()|eq o(AC,sup16()|，|ab|a|b|.答案相同10給出下列命題：若eq o(OD,sup16()eq o(OE,sup16()eq o(OM,sup16()，則eq o(OM,sup16()eq o(OE,sup16()eq o(OD,sup16()；若eq o(OD,sup16()eq o(OE,sup16()eq o(OM,sup16()，則eq o(OM,sup16()eq o(DO,sup16()eq o(OE,sup16()；若eq o(OD,sup16()eq

4、o(OE,sup16()eq o(OM,sup16()，則eq o(OD,sup16()eq o(EO,sup16()eq o(OM,sup16()；若eq o(OD,sup16()eq o(OE,sup16()eq o(OM,sup16()，則eq o(DO,sup16()eq o(EO,sup16()eq o(MO,sup16().其中所有正確命題的序號為_答案11如圖，解答下列各題：(1)用a，d，e表示eq o(DB,sup16()；(2)用b，c表示eq o(DB,sup16()；(3)用a，b，e表示eq o(EC,sup16()；(4)用d，c表示eq o(EC,sup16().

5、解eq o(AB,sup16()a，eq o(BC,sup16()b，eq o(CD,sup16()c，eq o(DE,sup16()d，eq o(EA,sup16()e，(1)eq o(DB,sup16()eq o(DE,sup16()eq o(EA,sup16()eq o(AB,sup16()dea.(2)eq o(DB,sup16()eq o(CB,sup16()eq o(CD,sup16()eq o(BC,sup16()eq o(CD,sup16()bc.(3)eq o(EC,sup16()eq o(EA,sup16()eq o(AB,sup16()eq o(BC,sup16()abe

6、.(4)eq o(EC,sup16()eq o(CE,sup16()(eq o(CD,sup16()eq o(DE,sup16()cd.12.如圖所示，O為ABC內一點，eq o(OA,sup16()a，eq o(OB,sup16()b，eq o(OC,sup16()c，求作向量bca.解以eq o(OB,sup16()，eq o(OC,sup16()為鄰邊作OBDC，連接OD，AD，則eq o(OD,sup16()eq o(OB,sup16()eq o(OC,sup16()bc，eq o(AD,sup16()eq o(OD,sup16()eq o(OA,sup16()bca.13已知|a|6，|b|8，且|ab|ab|，求|ab|.解如下圖，設eq o(AB,sup16()a，eq o(AD,sup16()b，以AB，AD為鄰邊作ABCD，則eq o(AC,sup16()eq o(AB,sup16()eq o(AD,sup16()ab，eq o(DB,sup16()