高中新课标数学必修课件本章回顾 38页

本章回顾Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n一､知识结构Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n二､方法总结1.公理的应用(1)证明共面问题证明共面问题,一般有两种证法.一是由某些元素确定一个平面,再证明其余元素在这个平面内;二是分别由不同元素确定若干个平面,再证明这些平面重合.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(2)证明三点共线问题证明空间三点共线问题,通常证明这些点都在两个面的交线上,即先确定出某两点在某两个平面的交线上,再证明第三点是两个平面的公共点.当然必在两个平面的交线上.(3)证明三线共点问题证明空间三线共点问题,先证两条直线交于一点,再证明第三条直线经过这点,把问题转化为证明点在直线上的问题.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2.判定空间两条直线是异面直线的方法(1)根据异面直线的定义.(2)反证法.3.求异面直线所成角的方法求异面直线所成的角是通过平移直线,把异面问题转化为共面问题来解决.根据等角定理及推论,异面直线所成的角的大小与顶点位置无关,将角的顶点取在一些特殊点上(如线段端点,中点等),以便于计算,具体步骤如下:Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(1)利用定义构造角;(2)证明所作出的角为异面直线所成的角;(3)解三角形求角.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n4.线线平行的判定方法(1)定义:同一平面内没有公共点的两条直线是平行直线;(2)公理4:a∥b,b∥ca∥c;(3)平面几何中判定两直线平行的方法;(4)线面平行的性质:a∥α,aβ,α∩β=ba∥b;(5)线面垂直的性质:a⊥α,b⊥αa∥b;(6)面面平行的性质:α∥β,α∩γ=a,β∩γ=ba∥b.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n5.直线和平面平行的判定方法(1)定义:;(2)判定定理:a∥b,aα,bαa∥α;(3)线面垂直的性质:b⊥a,b⊥α,aαa∥α;(4)面面平行的性质:α∥β,aαa∥β.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n6.两个平面平行的判定方法(1)依定义采用反证法;(2)利用判定定理:a∥β,b∥β,aα,bα,a∩b=Aα∥β;(3)垂直于同一条直线的两个平面平行:a⊥α,a⊥βα∥β;(4)平行于同一平面的两个平面平行:α∥γ,β∥γα∥β.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n7.平行关系的转化由上面的框图易知三者之间可以进行任意转化,因此要判定某一平行的过程就是从一平行出发不断转化的过程,在解题时把握这一点,灵活确定转化的思路和方向.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n8.线线垂直的判定方法(1)定义:两条直线所成的角为90°;(2)平面几何中证明线线垂直的方法;(3)线面垂直的性质:a⊥α,bαa⊥b;(4)线面垂直的性质:a⊥α,b∥αa⊥b.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n9.线面垂直的判定方法(1)线面垂直的定义:a与α内任意直线垂直a⊥α;(3)判定定理2:a∥b,a⊥αb⊥α;(4)面面平行的性质:α∥β,a⊥αa⊥β;(5)面面垂直的性质:α⊥β,α∩β=l,aα,a⊥la⊥β.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n10.两个平面垂直的判定方法(1)利用定义:两个平面相交,所成的二面角是直二面角;(2)判定定理:aα,a⊥βα⊥β.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n11.垂直关系的转化在证明两平面垂直时一般先从现有直线中寻找平面的垂线,若这样的直线图中不存在,则可通过作辅助线来解决.如有平面垂直时,一般要用性质定理,在一个平面内作交线的垂线,使之转化为线面垂直,然后进一步转化为线线垂直.故熟练掌握“线线垂直”､“面面垂直”间的转化条件是解决这类问题的关键.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n三､数学思想1.转化的思想例1:如下图,点P是△ABC所在平面外一点,A′､B′､C′分别是△PBC､△PCA､△PAB的重心.(1)求证:平面A′B′C′∥平面ABC;(2)求A′B′:AB.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解:(1)如右图,连结并延长PA′､PB′､PC′,其延长线分别交BC､AC､AB于M､N､Q.∵A′､B′是△PBC､△PAC的重心,Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n同理由B′､C′是△PAC､△PAB的重心,可知B′C′∥QN.∵A′B′∥MN,MN平面ABC,A′B′平面ABC,∴A′B′∥平面ABC,同理B′C′∥平面ABC.又A′B′面A′B′C′,B′C′面A′B′C′,A′B′∩B′C′=B′.∴平面A′B′C′∥平面ABC.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(2)平面A′B′C′∥平面ABC,平面PMN∩平面ABC=MN,平面PMN∩平面A′B′C′=A′B′,∴A′B′∥MN,2.数形结合思想Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n例2:某几何体的三视图如下图所示,P是正方形ABCD对角线的交点,G是PB的中点.(1)根据三视图,画出该几何体的直观图;(2)在直观图中,①证明:PD∥面AGC;②证明:面PBD⊥面AGC.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解:(1)该几何体是底面为2的正方形,侧面为全等的三角形的四棱锥,直观图如下图所示Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(2)①连结AC,BD交于点O,连结OG,因为G为PB的中点,O为BD的中点,所以OG∥PD.又OG面AGC,PD面AGC,所以PD∥面AGC.②连结PO,由三视图,PO⊥面ABCD,所以AO⊥PO.又AO⊥BO,所以AO⊥平面PBD.∵AO平面AGC,∴面PBD⊥面AGC.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n3.分类讨论思想Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n例3:如果二面角α-l-β的平面角是锐角,点P到α､β和棱l的距离分别为求二面角的大小.分析:点P可能在二面角α-l-β的内部,也可能在外部,应分类解答.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解:如图所示,P在二面角α-l-β的内部时,图(1),点P在α-l-β的外部时,图(2).在图(1)中,∵PA⊥α,∴PA⊥l,Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n∵AC⊥l,∴l⊥面PAC,同理,l⊥面PBC而面PAC∩面PBC=PC,∴面PAC与面PBC应重合,即A､C､B､P在同一平面内,∠ACB是二面角α-l-β的平面角,在Rt△APC中,Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n在Rt△BCP中,∴∠BCP=45°∴∠ACB=30°+45°=75°在图(2)中,应有∠ACB=45°-30°=15°.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n4.函数的思想Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n例4:在长方体A1B1C1D1—ABCD中,点M在AB1上移动,点N在BC1上移动,求点M和点N的最短距离.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解:如右图所示,在BB1上取动点P,作PM⊥AB1于M,PN∥BC交BC1于N,连结MN,因为BC垂直于平面A1ABB1内的所有直线,所以BC⊥BB1,BC⊥PM.又PN∥BC,∴PN⊥BB1,PN⊥PM.设BP=x,Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n四､数学方法1.利用平移求异面直线所成的角例5:如右图,在空间四边形ABCD中,E､F分别是AB､CD的中点,且AD=BC,AD⊥BC,求EF与BC所成的角.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解:连结EF,取BD的中点G,连结GE､GF.∵E､F分别是AB､CD的中点,∴EG和FG分别是△ABD和△DBC的中位线.∴GE.故∠EFG(或其补角)就是EF和BC所成的角.∵AD=BC,∴GE=GF.又∵AD⊥BC,∴∠EGF=90°.∴∠EFG=45°.故EF与BC所成的角为45°.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2.反证法例6:求证:过直线外一点有且只有一条直线和这条直线平行.已知:点P直线a.求证:过点P和直线a平行的直线b有且只有一条.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n证明:①存在性:如下图,∵Pa,∴点P和直线a确定一个平面α,在平面α内过点P作直线b与直线a平行,故这样的直线b存在.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n②唯一性:假设过点P还有一条直线c与a平行.∵a∥b,a∥c,∴b∥c,这与b､c共点于P矛盾.故假设不成立,因此直线b唯一.∴过直线外一点有且只有一条直线和这条直线平行.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n规律技巧:对于“有且唯一”性命题的证明,既要证明“有”即存在性,又要证明唯一性,其中唯一性的证明大多采用反证法.反证法的证明过程是:从否定结论入手,进行推理,直至推出矛盾(可以与假设相矛盾,也可以与已知､定理､公理相矛盾)矛盾产生的原因是假设不成立,故原命题成立.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n3.同一法例7:已知平面α⊥平面β,直线AB过α内一点A,且AB⊥平面β.求证:ABα.分析:此题直接证明,不易表达,可用同一法证明.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n证明:在平面α内过点A作直线AC垂直于平面α､β的交线OE.如下图,则AC⊥平面β.因为AB和AC都过点A,且垂直于平面β,所以AB､AC重合.也就是AB平面α.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.