(浙江專用)2020版高考數(shù)學(xué)新增分大一輪復(fù)習(xí) 第六章 平面向量、復(fù)數(shù) 6.4 平面向量的應(yīng)用(第2課時)平面向量的綜合應(yīng)用課件.ppt
第2課時平面向量的綜合應(yīng)用,第六章6.4平面向量的應(yīng)用,NEIRONGSUOYIN,內(nèi)容索引,題型分類 深度剖析,課時作業(yè),題型分類深度剖析,1,PART ONE,題型一平面向量與數(shù)列,師生共研,所以xn12xn1,又x11, 所以x23,x37,x415,故選A.,向量與其他知識的結(jié)合,多體現(xiàn)向量的工具作用,利用向量共線或向量數(shù)量積的知識進(jìn)行轉(zhuǎn)化,“脫去”向量外衣,利用其他知識解決即可.,題型二和向量有關(guān)的最值問題,多維探究,命題點(diǎn)1與平面向量基本定理有關(guān)的最值問題,解析設(shè)ABC的三個內(nèi)角A,B,C所對的邊分別為a,b,c.,解析連接MN交AC于點(diǎn)G. 由勾股定理,知MN2CM2CN2,,所以C到直線MN的距離為定值1,此時MN是以C為圓心,1為半徑的圓的一條切線(如圖所示),,命題點(diǎn)2與數(shù)量積有關(guān)的最值問題,I1I30,即I1I3.I3I1I2,故選C.,(2)(2018紹興市柯橋區(qū)質(zhì)檢)已知向量a,b,c滿足|b|c|2|a|1,則(ca)(cb)的最大值是_,最小值是_.,3,命題點(diǎn)3與模有關(guān)的最值問題,由x2y4z1,得x12y4z.,則A在以O(shè)為圓心,3為半徑的圓上運(yùn)動.,且知當(dāng)A,D在線段OE上時取等號,,和向量有關(guān)的最值問題,要回歸向量的本質(zhì)進(jìn)行轉(zhuǎn)化,利用數(shù)形結(jié)合、基本不等式或者函數(shù)的最值求解.,解析設(shè)BC中點(diǎn)為M,連接P0M,,即P0MAB,取AB的中點(diǎn)N,連接CN,,(2)(2018臺州期末)已知m,n是兩個非零向量,且|m|1,|m2n|3,則|mn|n|的最大值為,解析因?yàn)?m2n)24n24mn19,所以n2mn2, 所以(mn)2m22mnn25n2,,題型三和向量有關(guān)的創(chuàng)新題,師生共研,例5稱d(a,b)|ab|為兩個向量a,b間的“距離”.若向量a,b滿足: |b|1;ab;對任意的tR,恒有d(a,tb)d(a,b),則 A.ab B.b(ab) C.a(ab) D.(ab)(ab),解析由于d(a,b)|ab|, 因此對任意的tR,恒有d(a,tb)d(a,b),即|atb|ab|, 即(atb)2(ab)2,t22tab(2ab1)0對任意的tR都成立,因此有(2ab)24(2ab1)0,即(ab1)20,得ab10, 故abb2b(ab)1120,故b(ab).,解答創(chuàng)新型問題,首先需要分析新定義(新運(yùn)算)的特點(diǎn),把新定義(新運(yùn)算)所敘述的問題的本質(zhì)弄清楚,然后應(yīng)用到具體的解題過程之中,這是破解新定義(新運(yùn)算)信息題難點(diǎn)的關(guān)鍵所在.,解析當(dāng)a,b共線時,ab|ab|ba|ba, 當(dāng)a,b不共線時,ababbaba, 故是正確的; 當(dāng)0,b0時,(ab)0,(a)b|0b|0, 故是錯誤的; 當(dāng)ab與c共線時,存在a,b與c不共線, (ab)c|abc|,acbcacbc, 顯然|abc|acbc,故是錯誤的;,當(dāng)e與a不共線時, |ae|ae|<|a|e|<|a|1, 當(dāng)e與a共線時,設(shè)aue,uR, |ae|ae|uee|u1|u|1, 故是正確的. 綜上,結(jié)論一定正確的是.,課時作業(yè),2,PART TWO,基礎(chǔ)保分練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析在平面直角坐標(biāo)系中,由于|a|b|c|2,且ab0, 設(shè)a(2,0),b(0,2),c(2cos ,2sin ), 則(ac)(bc)(22cos ,2sin )(2cos ,22sin ) 44(sin cos)0, 即sin cos 1, 結(jié)合三角函數(shù)的性質(zhì)知1sin cos ,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,2.(2018紹興質(zhì)檢)已知不共線的兩個非零向量a,b滿足|ab|2ab|,則 A.|a|2|b| C.|b|ab|,解析設(shè)向量a,b的夾角為,則由|ab|2ab|, 得(ab)2(2ab)2, 即|a|22|a|b|cos |b|24|a|24|a|b|cos |b|2, 化簡得|a|2|b|cos . 因?yàn)橄蛄縜,b不共線,所以cos (0,1), 所以|a|<2|b|,故選A.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,3.(2018浙江名校新高考研究聯(lián)盟聯(lián)考)已知向量a,b滿足|ab|4,|ab|3,則|a|b|的取值范圍是 A.3,5 B.4,5C.3,4 D.4,7,解析由題意知|a|b|max|ab|,|ab|4(當(dāng)a,b共線時等號成立), 又(|a|b|)2|a|2|b|22|a|b|2(|a|2|b|2) |ab|2|ab|225(當(dāng)|a|b|時取等號), 所以|a|b|5, 故|a|b|的取值范圍是4,5.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由于|ab|,|ab|與|a|,|b|的大小關(guān)系與夾角大小有關(guān),故A,B錯. 當(dāng)a,b夾角為銳角時,|ab|ab|,此時,|ab|2|a|2|b|2; 當(dāng)a,b夾角為鈍角時,|ab|a|2|b|2; 當(dāng)ab時,|ab|2|ab|2|a|2|b|2,故選D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析在平面直角坐標(biāo)系xOy中,不妨令a(1,0),b(0,1),,所以問題轉(zhuǎn)化為求點(diǎn)(2,0),(0,1)與線段上點(diǎn)的距離之和的最小值,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,a22c23b2,解析設(shè)AC邊上的中點(diǎn)為D,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由|ab|2|ab|兩邊平方, 得a22abb24(a22abb2), 化簡得3a23b210ab10|a|b|,|b|210|b|90, 解得1|b|9.,9.(2019溫州模擬)設(shè)向量a,b滿足|ab|2|ab|,|a|3,則|b|的最大值是_;最小值是_.,9,1,故(12)(12),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析以C為坐標(biāo)原點(diǎn),CB所在直線為x軸建立平面直角坐標(biāo)系.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,技能提升練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,23n11,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以an113(an1). 因?yàn)閍112, 所以數(shù)列an1是以2為首項(xiàng),3為公比的等比數(shù)列, 所以an123n1, 所以an23n11.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析因?yàn)锳CAB,所以以A為坐標(biāo)原點(diǎn),以AB,AC所在的直線分別為x軸,y軸,建立平面直角坐標(biāo)系(圖略),則A(0,0),B(3,0),C(0,4). 由題意可知ABC內(nèi)切圓的圓心為D(1,1),半徑為1. 因?yàn)辄c(diǎn)P在ABC的內(nèi)切圓上運(yùn)動, 所以可設(shè)P(1cos ,1sin )(02).,(12cos ,22sin ),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1cos 2cos222sin2 1cos 112,,拓展沖刺練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,0,1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以點(diǎn)C的軌跡是以O(shè)為圓心,1為半徑的圓的劣弧 和劣弧 關(guān)于直線AB對稱的弧,即過點(diǎn)A,O,B的弧(如圖).,當(dāng)點(diǎn)C在劣弧 上時,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,當(dāng)點(diǎn)C在過點(diǎn)A,O,B的弧上時,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,