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www.6681yy.com:新课标高一数学平面向量数量积的坐标表示教学课件_图文

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复习与回顾 一、向量的数量积的定义: ? ?, 则a ? b ? ?a ? 0, b ? 0其 夹 角 为 ? 0 ? 0 ? a ? ? 二、平面向量数量积的运算律: a b cos? 向量a, b, c 和实数 ? ,则向量的数量积满足: ? ? ? ? ? ? ? ? ? ? ? (1) 交换律: a?b ? b? a (2) 数乘结合律: (? a) ? b ? a ? (? b) ? ? (a ? b) ? ? a ? b (3) 分配律: (a ? b) ? c ? a ? c ? b ? c 数量积重要性质: a·b=|a||b| cosθ 设 a, b 都是非零向量, e 是与 b 方向相同的单 位向量,θ是 a 与 b 的夹角,则: (1) e ? a ? a ? e ? |a| cosθ a ? b ? 0 ( 2) a b ? ⊥ (3)当 a 与 b同向时,a · b = 当 a 与 b 反向时,a · b = 特别地, a?a ? a 2 a b ? a b 或a ? a?a ? a 2 (4)cosθ= 2 2 a ? b a ? b (5)| a · b |≤ 2 a b 2 2 (6)(a ? b) ? a ? 2a ? b ? b (7)(a ? b)(a ? b) ? a - b 二、新课讲授 问题展示: 已知 a ? ( x1, y1 ),b ? ( x2 , y2 ), 怎样用 a, b 的坐标表示 a ? b 呢?请同学们看下 列问题. ? 设x轴上单位向量为 i ,Y轴上单位向量为 请计算下列式子: ? ? ① i ?i = 1 ? j ? ? ③ i?j = 0 ? ? ② j? j = 1 ? ? ④ j ?i = 0 ? ? 那么如何推导出 a ? b的坐标公式? ? ? ? ? ? ? ? 已知: a ? x1i ? y1 j , b ? x2i ? y2 j , ? ? ? ? ? 解: a ? b ? ( x1i ? y1 j ) ? ( x2i ? y2 j ) ?2 ? ? ? ? ?2 ? x1x2i ? x1 y2i ? j ? x2 y1i ? j ? y1 y2 j ? x1 x2 ? y1 y2 这就是向量数量积的坐标表示。由此我 们得到:两个向量的数量积等于它们对坐 标的乘积之和。 探讨合作1: 已知a ? (x, y),如何将 a 用其坐标表示? 结论1: ? a ? x2 ? y2 . 探讨合作2: 若设A( x1 , y1 ), B( x2 , y2 ), 如何将 AB 用A、 B的坐标表示? 结论2: AB ? ( x2 ? x1 )2 ? ( y2 ? y1 )2 , 这就是A、B两点间的距离公式. 探讨合作3:非零向量 a ? ( x1, y1 ),b ? ( x2 , y2 ), 它们的 夹角 ? ,如何用坐标表示cos ? .若 a ? b 你又能 得到什么结论? 结论3: (1) cos ? ? x1 x 2 ? y1 y2 x1 ? y1 ? 2 2 x2 ? y2 2 2 (2)a ? b ? x1 x2 ? y1 y2 ? 0 : (2)a ? b ? x1 x2 ? y1 y2 ? 0 与 a // b ? x1y2 ? x2 y1 ? 0 的区别。 三、例题分析 ? ? ? ? 设a ? (5,?7), b ? (?6,?4), 求a ? b . 例1: 解:a ? b ? 5 ? (?6) ? (?7) ? (?4) ? ?2 想一想 a, b 的夹角有多大? 例2:已知A(1, 2),B(2,3),C(-2,5),求证 △ABC是直角三角形. 证明:AB ? (2 ? 1,3 ? 2) ? (1,1) BC ? (?2 ? 2,5 ? 3) ? (?4,2) AC ? (?2 ? 1,5 ? 2) ? (?3,3) ? AB ? AC ? 1? (?3) ? 1? 3 ? 0 想一想:还 有其他证明 方法吗? 所以△ABC是直角三角形 ? 例3:求与向量 a ? ( 3 ? 1, 3 ? 1) 的夹角为45o的 单位向量. 分析: 可设x(m, n),只需求m, n. 易知 m ? n ? 1 ? ? 再利用 a ? x(定义) ? a ? x(数量积 的坐标 法)即可! 2 2 ? 解:设所求向量为 x ? (m, n) ,由定义知: 2 ? ? ? ? a ? x ? a ? x ? cos 45 ? 8 ? ?2 2 另一方面 ? ? a ? x ? ( 3 ? 1) ? m ? ( 3 ? 1) ? n ……① ……② ∴由①,②知 ( 3 ? 1) ? m ? ( 3 ? 1) ? n ? 2 由 ( 3 ? 1) ? m ? ( 3 ? 1) ? n ? 2 m2 ? n 2 ? 1 解得: 3 m1 ? ? 2 ∴ 1 n1 ? 2 ? 3 1 x ? (? , ) 2 2 或 1 m2 ? ? 2 3 n2 ? 2 ? 1 3 ) 或 x ? (? , 2 2 说明:可设 x ? (cos? , sin ? ) 进行求解. 四、演练反馈 ? A. 63 65 B. 33 65 ? ? ? 1、若a ? (?3,4), b ? (5,12), 则 a 与 b 夹角的余弦值 为 ( B ) C. ? ? ? 2、已知: a ? (cos? , sin ? ),b ? (cos? , sin ? ) ? ? ? ? 求证:(a ? b ) ⊥ (a ? b ) 答案: ? ? 33 65 D. ? 63 65 ? ? (a ? b ) ? (a ? b ) ? (cos? ? cos ? , sin ? ? sin ?

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