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[2014创新设计二轮专题复习常考问题8.ppt

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[2014创新设计二轮专题复习常考问题8

常考问题8 平面向量的线性运算及 综合应用 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.       [真题感悟]  [考题分析] Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2.两非零向量平行、垂直的充要条件 设a=(x1,y1),b=(x2,y2), (1)若a∥b?a=λb(λ≠0);a∥b?x1y2-x2y1=0. (2)若a⊥b?a·b=0;a⊥b?x1x2+y1y2=0. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 5.根据平行四边形法则,对于非零向量a,b,当|a+b|=|a-b|时,平行四边形的两条对角线长度相等,此时平行四边形是矩形,条件|a+b|=|a-b|等价于向量a,b互相垂直,反之也成立. 6.两个向量夹角的范围是[0,π],在使用平面向量解决问题时要特别注意两个向量夹角可能是0或π的情况,如已知两个向量的夹角为钝角时,不单纯就是其数量积小于零,还要求不能反向共线. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 热点与突破 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 答案 -12 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. [规律方法] 求数量积的最值,一般要先利用向量的线性运算,尽可能将所求向量转化为长度和夹角已知的向量,利用向量的数量积运算建立目标函数,利用函数知识求解最值. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides

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