(2019新教材)人教A版高中數(shù)學(xué)必修第二冊全冊學(xué)案

（2019A6．1平面向量的概念念P2－P4→→A→．→→-1-→AB，點AB→→→01))))→→)))a)→AC．起點是MBM指向NDM-2-→OA)ABDC→和→→→→A，B，C，D→→；→→→→→→→→a與b的b與ca與c)AB-3-CDACD)→→→→ABCD→→→→AB→→A在點O→→B在點A→→C在點BA在點OA距點O→A距點O→A→B在點AB距點A→為B→C在點B距點B3C-4-→A2到達BB2到達CC12到達D→→→→；D地在A地距A→→→→＝12DAA地12→→-5-aa→→→→a.→→→→→→→→→a.→→→→→→→ab，F(xiàn)A；與c→→→.→→→→→→→→→→→.→→→)→→A→→→B→→CD→→→，BC和-6-→→和，→→→→→→.→→→→→→→→→)B．2D．4→)→→→，→→.O，→→→→→→→→.→→→→.a)a.A．3C．1B．2D．0a共aa)A．若a與b平行，b與c平行，則a與c一定平行BD1a與cO)→→→→→→→→C|→→→→→→-8-→→→O)ACBDOA，B，C的→→→aa)ABCDaaab→→22→5D→53.2532→→A，B，CmmA，B，C→→→→又m-9-C→→→→→→→→→→，→→→→→.綊BC.→→.→→→→→→C→→與P，點E，F(xiàn)P，且)→→→→→→→→→→→→→→→.→→交ABD→→A作BCE.，→3A5BCCD→，＝90°，BC＝102米，CD＝10＝5＝2255→AO九128128→→i8Oi8A128AAAAAAAA135724686．2.1向量的加法運算-12-P7－P10→A→a與b→→→即OOb→就是a與b的和則-13-))())A．2B．3D．5C．4→→B．b→→→O→→→→O＝→→→，→→→→→→→→→→→→-15-→→；→→→；→→→→→＋FA.→→→→→.→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→＋FA＝AF＋FA＝0.)→→→→→→.AC．①BD．③→→→→→→→→→-16-→→→→→→→→→→→→→→→→→→→.→→→→→→→→→→→→→→→→→B為→→→→Akm到達BB東km送往C→→→→→又β→→→|22＝222→→→→→→→→→→→→→→.→→→)ABCD→→→→→的→1→→→→O)→→→→→→→→O＋→→→與O→)→→→→→→→→→→→→→→→→→→a131α＝3B．2C．3D．23→→，→→→→→→→，→→→→)Aa與b與aB．若a與b與bC．若a與b同向，則a＋b與a同向D．若a與b同向，則a＋b與b同向與b與aB與b與a同向，也與b→→→→→O→→→→→→→→→→→M.→→→→，→→→→→→→→→b→→→→a→→；-20-→→，→→→→→→→→→→→→→又P→→→→→→→→→→→→222，→→→→→→→→222→→→→→O，BA11→BN，→→→→aaa與baba與b→→→Oba)))))→→→→→→→→→→-24-設(shè)b是a)A與bC．a(chǎn)與bBD是b→→→→→→→→→→.→→→→→→→→→→→M.→→→→→→→→→→→→→.→→→.→→→→→→.→)→→→→→→→→→.ABCD→→→→→→→→C→→→-25-→→→→→→→→→→→→→→→→→→→→→→→→→→→→.→→→O，→過點A作綊，連接OD，→→→O→→→→→→→→→→O→→B→→→→→.→→→→→→→→→→→→→→→→＝________．-27-→→→→→→→→→→→→→→→→→→→→O與BD→→→→→→，→→→→→→即.→→→→→→→→→→→→→→→→→→→又|→→→→→→→→→O→→→→→→→→→→→.→→→→→→→→→→→→→)ACBD→→→→→O，E，F(xiàn))→→→→→→→→→→→→→→→→→→→→→→→→→)ACBD→→→→→→→；→→→→；→→→；→→→→.0)A．4B．3-29-D．1→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→)B．2D．4→→→→→→→→→→→→→→→→→→→→→→2→→和O→→→→→→→→→→→→→→，→→→→→→→→→→→→→→→→→→→→→→f表示以下向量：→→；→→→→→→→→→→→→→→→→→→→與→→.與與；→→→；→→；→→→→→→→得點B→→→→→→→→→→→→→→→→→→→A.→→→d和→→→→→→→→→→＝→→M→→6．2.3向量的數(shù)乘運算P13－P16aaaaa如設(shè)μμ121212-34-ba與ba)與a與a)m)))ACB．a(chǎn)Da與b)ABDC12231134③.1323－b-35-23137324b25＝b325＝313231323＝b555(2(33353＝－－j5333532513222244224b＋－＋－＋5533535324112a－322113322272112322-36-7211232241177e12→→→121212和ek1212→→→→→.12121212→→B，A、B、D與e12121212，12由于e與e12b與a→→b與a→→→→A，從而A，B，Ca與b1，313.13-37-→→1212→→→→→21→→→→12→→→14e－e＋e＝e－e.12112212→→→1212，12→21→)11→→11→→A．AB＋22BAB－2211→→11→→C．－＋22D．AB－2211→→→→→＝AB－.2211)32ACBD16131442333332→→Oe－e＝()1221→→→→13→→22→→→→→＝＝e－e.2121→→→e，e12121212證A，B，D→→，1212→→→.12→→→→→1212與BDB，所以A，B，Da)λ-39-ABDC．a(chǎn)與λa2ABDa與λa22m與m)A．－1或3C．－1或43D．3或4m與以m＝m＝－1或m＝3.→→→O為BC)→→→→→→→→D→→→D為BC，→→→→→→.→→)ABDC→→A，B，C→→m→→→→)13→→13→→A．－AB＋22AB－2211→→11→→AB－22D．－AB＋22121→→→2→→→→→→→→→＝＝PA＋AB＝-40-11→→AB－.22→→→→P＝λ131→→→→→→＝PB.313與與131412()＋122371376276－a.13121172＋－－＋342322317237367－b71716262＝→→→a與b→→→kA，B，Ck→→→→→→→→→(2)AB＝OBA1→→在.2→→→)-41-2121→→→→→→1112與OBO)AC→交AB＝→，λ>1，注意到λ3131115311＋2＝3>1，＋>＋＋＝<1，＋＝<1，故選434423645→→→m→→→→→→13→→→→→→→→→→→→→→→→→→→→→→→→2→→→→→1233A，B，C123312→→→＝＋，332313→→→→→→即2，→→→→C，A，B，C→→A，B，C，→→→→→→λB→→→→→→→→則6．2.4向量的數(shù)量積算a在b-43-P17－P22→→a與bπa與b2→→a與a與b.→→→→AB，分別作CDA，→11→111→→OM→Mab11→b與bθ1π2→→→111π2→→11設(shè)b.當(dāng)a與ba2.c共a.222或))))與n的夾角為45°，則m)AB2D2C221ba與b)5ABDC解析：選B.設(shè)a與b1b515151.2[]，與與即a222222即-46-與b→→→→→→.222222→→→→→→→→→→→→→→→→1－2→→.→→→→→→，→→→→→→)→→22-47-)A．4C．2B．3D．02→→32→→→→→→→→→→·BA＝(BA＋BC·BA＝a＝a.22223a22a與b)3C．4B．23D3與b)213121145＝a222＝2212＝343412.2222a22a222a與bθ2222b2222-48-222234得a2222a＝2212222222123a與ba與b的a與b2222221＝.2π3[].-49-b2a與b＝.212＝＝.22π.3ππ33＝a＋tb222222222－22|2與互相垂直，則k)3A32B．2C32D．1πb3與223.2即2222而則a222π則，23-50-即或2或5a與b＝π3e，則λ121212132414e·e＝＋222212121.212a與b)ππ64ππ321＝2π3.與b與dk的)AC．3B．6D22eba在b解析：設(shè)a與b的夾角θ，則-51-4＝＝，54a在b－e55e5與a垂直，求a與b解：設(shè)向量a與b222a22a＝＝2→→)ABCD→→)3C．35D．522a與b)ππ6363-52-122.3a與b)A．2C．6B．4D222222→→)332A．－2B．C32D．32→→＝3.2aba與b則213.2232πa與b3π23＝3-53-→→1→→→→＝21212.12b，1212所以a＝，2222221212＝，2＝212，22222.2bab(1)求a與b的夾角θ；又a在bθ123.2與224.7-54-→→→→→→→)2ACBD→→→→→→→→→→→→→→，22→→→→→→→→→→→→2→→與b)ππ636322223向量a－b與bθ＝＝222.6→→→→1→→→→→→→＝－AB，3→→→→→13→→→→→＝＋－AB）)21→→→→＝＋)331312→→→→＝AB·AC＋－223313132→→1311223383→→＝＋|－|＝－＋.2223-55-83ee121212121＋2與e＋te1212）|得1212<0，1212121222若21則t∈.21212設(shè)<0，可得1212.21∪－.222→→.→→→→→→→→122→3→→→→→→＝＝.33()()→→＝→→→→·1323→→→→＝＋·－-56-132929→→→→＝－－221313→→→→→→→＋＋AB，2323→→→→→→→BP＝BC＋－AB，1323→→→→→→＝＋·－1329→→→→＝－－22131→→3→→.→→1→→3→→→→→→→→，2＝.323→→.6．3.1平面向量基本定理P25－P27-57-e，e1212若e，e12121212λ1212)μ()1122121212A12121212112→1111→12→→→→→＝2設(shè)e，e12與e與e與與e－e.112122112211212λ①設(shè)e＋e＝λe，則121所以e與ee與e＋e121112e122112則e與ee與e122112211e21221e與1221即e與1221ee與e－121212121ee與e－e21212③x＝x，12a與bx1122y＝y(tǒng).12O是)→→→→→→→→.ACBD-59-→→→→→→→→O)→→→→→→→→→→→→→→→→→→→→G.→→→→1→→→＋21212→→→＋→→→→1212→→→＋→.2＝BF，32→→→→→＋321a32-60-21223333→→→→→→→→.→→→→→→→→→→→→→→12→→→→→D＝)1221b33b333443b55b55121→31323→→→→→→＝13，E，F(xiàn)＝→→→→.1FA，DF＝BC，311→→11→→＝＝＝＝3326121212→→→→→→＝＝b，所以FA＝BA161123→→→→→－FA＝－b＝131126→→→→→b＝-61-M是BCN在ACM與BNP，求與BP∶PN.→→＝e，12→→→→→→.2112A，P，M和B，P，N→→，12→→.12→→→→→＋PA＝BP.12→→→12得4＝，53＝.54535→→→→＝＝BN，→→→.2→→＝，5252→→5→→→→→＋)42345555N為AC與BP∶PN.-62-→→＝e，12→→→→→→.2112A，P，M和B，P，N→→，12→→.12→→→→→＋PA＝BP.12→→→12得2＝，32＝.32323→→→→＝＝BN，e121212121233e＝，112113312e＝2121123333313故e＋e＝b＋b＝－1221－331→→→→→為AB＝3-63-→→，→→→→→→→→＋CA＋CB，2.32→→→)1212)121212)212112→→→→→→→→→→→→→11→→→＝22→→→→→→→，又→→→，11221122→11221122→＝＝e)12e1212λ，μλμ與11221112212211122122使12ACBD．②1λ212→→→)121212(e＋e)(e－e)12121212)(e－e)21211→→→→→O＝)2121＝212→→→，若A，B，D三12121212k)A．2BD．3C→→→→→是1212-65-→→→D)→→→34→→→→→→→→→→→→＋＝AB＋AB＋AC.→→→→D＝rAB＋sAC)5844→→→→→→45.－＝.555a與bOC＝→→→→→→→→O→→→μ→→→→→→→→＝＝＝＋＝，14＝.3e.121212.12121212則.12121212131313→→→→→AC－→→→→121212k12121→→→→為CD，→→→→→→→→→→→→→→xAB－yAD12→→→→→－＋，λ1x1113→→→→→→→→→→→→，11＝211→b55→→→1111→→→→22＝＋AC＝＋1115→→→b＝666615→→→＝66116623b＝1121623→＝＋112＋＝，231111＋＝＋26＝.22314→→＝＝.→→；交于O1411→→441423→→→＝＝AB＝＝＝BC，222→→→33323→→→＝＝＝→→→→→→→→→2323→→→bD＝1＝43＝，2314→μ267＝.33→→→→＝＝-69-6.3.2平面向量的正交分解及坐標(biāo)表示6.3.3平面向量加、減運算的坐標(biāo)表示6.3.4平面向量數(shù)乘運算的坐標(biāo)表示第1課時P27－P331和e2a＝bx＝x且y12121122-70-112212121212y11→→11222121))))→A)ACBD→)ACBD設(shè)與→已知OA→-71-→→→→即A(2→00→→x＝＝00333－，22xy→→→→→→→，→→c)ACBD→→→→AM的c滿足A→→→→→→CB，→→設(shè)M1122→11→22xxxx1212yyyy1212所以M→→→→OCB，→→→→→→→→→→→→→→.→→所以M-73-→→1→→－2A→→→1→→→→－2mmma＋nb＝(2m，m＋n，m－mmm→→→OP在x軸上？點P在y軸上？點P→→→2P在x.31若點P在y.3若點P2313.→→→→-74-→→→→→→→M，→)32323－23－211→→221232＝＝.2．已知在非平行四邊形ABCDA，B，DC→→→C-75-)ABCD→→A→→→→又32，2.2aOaa設(shè)a→則A即ax→→→→)ACBD→→→→)A．－22-76-21，2所→→→→AD)22→→→)2→A→→λ→AB→→→→→OBa和b故1323→→→→→A＝11→→＝AB＝332323→→→→→→→→→Ammm→→AM→→y11→-78-111y11y122222221→→→→3.7mmxy112244ACπ4→→→π→→→→C作⊥x軸于點＝E42→→→2→→→→P在BC，點Q是AC＝→→→→→→Q是，→→→→→→→→→＝O＝→Ox32設(shè)x－，111y＝，1設(shè)λ121212λ，λ122λ－，2λ1211222A→→→→→→→(m，n∈RPm→→→P→→→P→P00A→→→→→，00x0y0m，00Pym00第2課時P31－P33-81-設(shè)xy－xy11221221xy1＝1≠0，yxy2222axyyxy－122112xy＝021)xyy)11221221)A．a(chǎn)11B．a(chǎn)22C．a(chǎn)33D．a(chǎn)44→Aa)ABCD→→以D23→→A-82-1313.→→→→→→2→→→→＝31，313131＝＝33與a則)52AC．7B．12D．－與a1.2-83-→→A→→，→→→→→→→→→A→→→k為何值時，A，B，C→→→3→→→→→＝AB，2→→→→A，所以點A，B，C→→A，B，C→→→→→→→→即或或時，A，B，C→→→→→→→→k或或時，A，B，C-84-A，B，CACC)AB．9D．3CC→→A，B，C，→→→→AxA，B，C→→→→→x→→→→→→→→而A，B，CA，B，C-85-1→→252→設(shè)M→→→，72→→→→即77與F，求DFA→→121212→→→D是＝＋AC7－.21→→→12因為MF為AD27274－＝.)ABCDmA)ABCD→→A1212mmmk5m＝，98.9-87-又.)ACBDm)12B．2C12Dα1＝＝.α212u∥v，則實數(shù)k()A7B122C43D．－38u∥v，1.2→→→→)ACBD→→→→即B1AA，B，CC)2-88-ABCDCA，B，C→→.1272→＝，→72xA→→→；→→→；→→→.→→→→→→→→→→＝→→→→mmnmmmm-89-k與→→b且A，B，Cm與1.21與2A，B，C→→λ∈R，即3解得m＝.2→→→→AM及→2→→1PPP|點P312122→A→→→→→設(shè)M1122→→1122xM1122→O→→→→→→→→