华师大附中2011届数学复习教学案平面向量数量积的坐标表示课题：平面向量数量积的坐标表示教学目的：⑴要求学生掌握平面向量数量积的坐标表示⑵掌握向量垂直的坐标表示的充要条件，及平面内两点间的距离公式⑶能用所学知识解决有关综合问题教学重点：平面向量数量积的坐标表示教学难点：平面向量数量积的坐标表示的综合运用授课类型：新授课课时安排：1课时教具：多媒体、实物投影仪教学过程：一、复习引入：1．两个非零向量夹角的概念已知非零向量«Skip Record If...»与«Skip Record If...»，作«Skip Record If...»＝«Skip Record If...»，«Skip Record If...»＝«Skip Record If...»，则∠A ＯB＝θ（０≤θ≤π）叫«Skip Record If...»与«Skip Record If...»的夹角.2．平面向量数量积（内积）的定义：已知两个非零向量«Skip Record If...»与«Skip Record If...»，它们的夹角是θ，则数量|«Skip Record If...»||«Skip Record If...»|c osθ叫«Skip Record If...»与«Skip Record If...»的数量积，记作«Skip Record If...»⋅«Skip Record If...»，即有«Skip Record If...»⋅«Skip Record If...» = |«Skip Record If...»||«Skip Record If...»|c osθ，（０≤θ≤π）.并规定«Skip Record If...»与任何向量的数量积为03．向量的数量积的几何意义：数量积«Skip Record If...»⋅«Skip Record If...»等于«Skip Record If...»的长度与«Skip Record If...»在«Skip Record If...»方向上投影|«Skip Record If...»|c osθ的乘积4．两个向量的数量积的性质：设«Skip Record If...»、«Skip Record If...»为两个非零向量，«Skip Record If...»是与«Skip Record If...»同向的单位向量1︒«Skip Record If...»⋅«Skip Record If...» = «Skip Record If...»⋅«Skip Record If...»=|«Skip Record If...»|c osθ；2︒«Skip Record If...»⊥«Skip Record If...»⇔«Skip Record If...»⋅«Skip Record If...» = 03︒当«Skip Record If...»与«Skip Record If...»同向时，«Skip Record If...»⋅«Skip Record If...» = |«Skip Record If...»||«Skip Record If...»|；当«Skip Record If...»与«Skip Record If...»反向时，«Skip Record If...»⋅«Skip Record If...» = -|«Skip Record If...»||«Skip Record If...»|特别的«Skip Record If...»⋅«Skip Record If...» = |«Skip Record If...»|2或«Skip Record If...»4︒c osθ =«Skip Record If...»；5︒|«Skip Record If...»⋅«Skip Record If...»| ≤ |«Skip Record If...»||«Skip Record If...»|5．平面向量数量积的运算律交换律：«Skip Record If...»⋅«SkipRecord If...» = «Skip Record If...»⋅«SkipRecord If...»数乘结合律：(«Skip Record If...»«SkipRecord If...»)⋅«Skip Record If...» =«Skip Record If...»(«Skip Record If...»⋅«Skip Record If...») = «Skip Record If...»⋅(«Skip Record If...»«Skip Record If...»)分配律：(«Skip Record If...» + «Skip Record If...»)⋅«Skip Record If...» = «Skip Record If...»⋅«Skip Record If...» + «Skip Record If...»⋅«Skip Record If...»二、讲解新课：⒈平面两向量数量积的坐标表示已知两个非零向量«Skip Record If...»，«Skip Record If...»，试用«Skip Record If...»和«Skip Record If...»的坐标表示«Skip Record If...»设«Skip Record If...»是«Skip Record If...»轴上的单位向量，«Skip Record If...»是«Skip Record If...»轴上的单位向量，那么«Skip Record If...»，«Skip Record If...»所以«Skip Record If...»«Skip Record If...»又«Skip Record If...»，«Skip Record If...»，«Skip Record If...»所以«Skip Record If...»«Skip Record If...»这就是说：两个向量的数量积等于它们对应坐标的乘积的和即«Skip Record If...»«Skip Record If...»2.平面内两点间的距离公式（1）设«Skip Record If...»，则«Skip Record If...»或«Skip Record If...»（2）如果表示向量«Skip Record If...»的有向线段的起点和终点的坐标分别为«Skip Record If...»、«Skip Record If...»，那么«Skip Record If...»(平面内两点间的距离公式)3.向量垂直的判定设«Skip Record If...»，«Skip Record If...»，则«Skip Record If...»«Skip Record If...»«Skip Record If...»4.两向量夹角的余弦（«Skip Record If...»）c o s θ =«Skip RecordIf...»222221212121y x y x y y x x +++=三、讲解范例：例1 设«Skip Record If...» = (5, -7)，«Skip Record If...» = (-6, -4)，求«SkipRecord If...»⋅«Skip Record If...»解：«Skip Record If...» = 5×(-6) + (-7)×(-4) = -30 + 28 = -2例2 已知«Skip Record If...»(1, 2)，«Skip Record If...»(2, 3)，«Skip Record If...»(-2, 5)，求证：△ABC是直角三角形证明：∵«Skip Record If...»=(2-1, 3-2) = (1, 1), «Skip Record If...»= (-2-1, 5-2) = (-3, 3)∴«Skip Record If...»⋅«Skip Record If...»=1×(-3) + 1×3 = 0 ∴«Skip Record If...»⊥«Skip Record If...»∴△ABC是直角三角形例3已知«Skip Record If...» = (3, -1)，«Skip Record If...»= (1, 2)，求满足«Skip Record If...»⋅«Skip Record If...»= 9与«Skip Record If...»⋅«Skip Record If...» = -4的向量«Skip Record If...»解：设«Skip Record If...»= (t, s)，由«Skip Record If...»«Skip Record If...»∴«Skip Record If...»= (2, -3)例4已知«Skip Record If...»＝（１，«Skip Record If...»），«Skip Record If...»＝（«Skip Record If...»＋１，«Skip Record If...»－１），则«Skip Record If...»与«Skip Record If...»的夹角是多少?分析：为求«Skip Record If...»与«Skip Record If...»夹角，需先求«Skip Record If...»及｜«Skip Record If...»｜·｜«Skip Record If...»｜，再结合夹角θ的范围确定其值.解：由«Skip Record If...»＝（１，«Skip Record If...»），«Skip Record If...»＝（«Skip Record If...»＋１，«Skip Record If...»－１）有«Skip Record If...»·«Skip Record If...»＝«Skip Record If...»＋１＋«Skip Record If...»（«Skip Record If...»－１）＝４，｜«Skip Record If...»｜＝２，｜«Skip Record If...»｜＝２«Skip Record If...»．记«Skip Record If...»与«Skip Record If...»的夹角为θ，则cosθ＝«Skip Record If...»又∵０≤θ≤π，∴θ＝«Skip Record If...»评述：已知三角形函数值求角时，应注重角的范围的确定.例5如图，以原点和A (5, 2)为顶点作等腰直角△ABC，使∠«Skip Record If...» = 90︒，求点«Skip Record If...»和向量«Skip Record If...»的坐标解：设«Skip Record If...»点坐标(x, y)，则«Skip Record If...»= (x, y)，«Skip Record If...»= (x-5, y-2)∵«Skip Record If...»⊥«Skip Record If...» ∴x (x -5) + y (y -2) = 0即：x 2 + y 2 -5x- 2y = 0又∵|«Skip Record If...»| = |«Skip Record If...»| ∴x 2 + y 2 = (x -5)2 + (y -2)2即：10x+ 4y = 29由«Skip Record If...»∴«Skip Record If...»点坐标«Skip Record If...»或«SkipRecord If...»；«Skip Record If...»=«Skip Record If...»或«SkipRecord If...»例6 在△ABC 中，«Skip Record If...»=(2, 3)，«Skip Record If...»=(1, k )，且△ABC 的一个内角为直角， 求k 值解：当«Skip Record If...» = 90︒时，«Skip Record If...»⋅«Skip Record If...»= 0，∴2×1 +3×k = 0 ∴k =23- 当«Skip Record If...» = 90︒时，«Skip Record If...»⋅«Skip Record If...»= 0，«Skip Record If...»=«Skip Record If...»-«Skip Record If...»= (1-2, k -3) = (-1, k -3)∴2×(-1) +3×(k -3) = 0 ∴k =«Skip Record If...»当C = 90︒时，«Skip Record If...»⋅«Skip Record If...»= 0，∴-1 + k (k -3) = 0∴k =«Skip Record If...»四、课堂练习：1.若«Skip Record If...»=(-4,3),«Skip Record If...»=(5,6),则3|«Skip RecordIf...»|２－４«Skip Record If...»＝（ ）A .23B .57C .63 D.832.已知«Skip Record If...»(1,2),«Skip Record If...»(2,3),«Skip Record If...»(-2,5)，则△«Skip Record If...»«Skip Record If...»«Skip Record If...»为（ ）A .直角三角形B .锐角三角形C .钝角三角形 D.不等边三角形3.已知«Skip Record If...»=(4,3),向量«Skip Record If...»是垂直«Skip Record If...»的单位向量，则«Skip Record If...»等于（ ）A .«Skip Record If...»或«Skip Record If...»B .«Skip Record If...»或«Skip Record If...»C.«Skip Record If...»或«Skip Record If...»D.«Skip Record If...»或«Skip Record If...»4.«Skip Record If...»=(2,3),«Skip Record If...»=(-2,4),则(«Skip Record If...»+«Skip Record If...»)·(«Skip Record If...»-«Skip Record If...»)= .5.已知«Skip Record If...»(3，2)，«Skip Record If...»(-1，-1)，若点P(x,-«Skip Record If...»)在线段«Skip Record If...»«Skip Record If...»的中垂线上，则x= .6.已知«Skip Record If...»(1，0)，«Skip Record If...»(3，1)，«Skip Record If...»(2，0)，且«Skip Record If...»=«Skip Record If...»,«Skip Record If...»=«Skip Record If...»,则«Skip Record If...»与«Skip Record If...»的夹角为 .参考答案：1.D 2.A 3.D 4. –7 5.«Skip Record If...» 6.45°五、小结两向量数量积的坐标表示长度、夹角、垂直的坐标表示六、课后作业：1.已知«Skip Record If...»=(2,3),«Skip Record If...»=(-4,7),则«Skip Record If...»在«Skip Record If...»方向上的投影为（）A.«Skip Record If...»B.«Skip Record If...»C.«Skip Record If...»D.«Skip Record If...»2.已知«Skip Record If...»=(λ，２)，«Skip Record If...»=(-3,5)且«Skip Record If...»与«Skip Record If...»的夹角为钝角，则λ的取值范围是（）A.λ＞«Skip Record If...»B.λ≥«Skip Record If...»C.λ＜«Skip Record If...»D.λ≤«Skip Record If...»3.给定两个向量«Skip Record If...»=(3,4),«Skip Record If...»=(2,-1)且(«Skip Record If...»+x«Skip Record If...»)⊥(«Skip Record If...»-«Skip Record If...»),则x等于（）A.23B.«Skip Record If...»C. «Skip Record If...»D. «Skip Record If...»4.已知|«Skip Record If...»|=«Skip Record If...»,«Skip Record If...»=(1,2)且«Skip Record If...»∥«Skip Record If...»,则«Skip Record If...»的坐标为 .5.已知«Skip Record If...»=(1,2),«Skip Record If...»(1,1),«Skip Record If...»=«Skip Record If...»-k«Skip Record If...»,若«Skip Record If...»⊥«Skip Record If...»，则«Skip Record If...»＝ .6.已知«Skip Record If...»=(3,0),«Skip Record If...»=(k,5)且«Skip Record If...»与«Skip Record If...»的夹角为«Skip Record If...»，则k的值为 .7.已知«Skip Record If...»=(3,-1),«Skip Record If...»=(1,2),求满足条件x·«Skip Record If...»=9与x·«Skip Record If...»=-4的向量x.8.已知点A (1，2)和B (4，-1)，问能否在y轴上找到一点C，使∠ABC＝90°，若不能，说明理由；若能，求C点坐标.9.四边形ABC D中=«Skip Record If...»(6,1), «Skip Record If...»=(x,y)，«Skip Record If...»=(-2,-3),(1)若«Skip Record If...»∥«Skip Record If...»，求x与y间的关系式；(2)满足(1)问的同时又有«Skip Record If...»⊥«Skip Record If...»,求x,y的值及四边形ABC D的面积.参考答案：1.C 2.A 3.C 4.（«Skip Record If...»，２«Skip Record If...»）或（-«Skip Record If...»，－２«Skip Record If...»）5.(«Skip Record If...»)6.-57.(2,-3)8.不能(理由略)9.(1)x+2y=0 (2)«Skip Record If...»S四边形ABC D=16七、板书设计（略）八、课后记及备用资料：已知«Skip Record If...»＝（3，4），«Skip Record If...»＝（4，3），求x,y的值使(x«Skip Record If...»+y«Skip Record If...»)⊥«Skip Record If...»，且｜x«Skip Record If...»+y«Skip Record If...»｜=1.分析：这里两个条件互相制约，注意体现方程组思想.解：由«Skip Record If...»＝（3，4），«Skip Record If...»＝（4，3），有x«Skip Record If...»+y«Skip Record If...»=(3x+4y,4x+3y)又（x«Skip Record If...»+y«Skip Record If...»）⊥«Skip Record If...»«Skip Record If...»(x«Skip Record If...»+y«Skip Record If...»)·«Skip Record If...»＝０«Skip Record If...»3(3x+4y)+4(4x+3y)=0即25x+24y＝０①又｜x«Skip Record If...»+y«Skip Record If...»｜=1«Skip Record If...»｜x«Skip Record If...»+y«Skip Record If...»｜２＝１«Skip Record If...»（３x+4y）２＋（４x+3y）２＝１整理得：25x２＋48xy+25y２＝１即x(25x+24y)+24xy+25y２＝１②由①②有24xy+25y２＝１③将①变形代入③可得：y=±«Skip Record If...»再代回①得：«Skip Record If...»。