WWW.EEE206.COM:高一数学课件-高一数学课件:平面向量数量积的坐标表示6 最新_图文

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高一 数学 平面向量数量积的坐标表示 王人伟 北航附中 播放时间 6月9日 1.已知两个非零向量a和b,它们的夹角为θ,则 a· b= a b cos θ . a· b称为向量a与b的数量积(或内积). 2.数量积a· b等于a的长度 a 与b在a的方向上的 投影 b cosθ 的乘积. 3. a⊥b a· b=0. 4. a· a= a 2=a2. a· b 5. cos θ = a b 6. a· b≤ a b . . 复习题1 已知: a =4,b =5,a· b=10, 求:a与b的夹角θ. 解:设a与b的夹角为θ ,则 a· b 1 cos θ = = 2 , a b θ =60°. 复习题2 已知:A(2,1),B(3,2),C(–1,4), 求证:?ABC是直角三角形. y 分析:先画图, 从图中可知, ∠A应为90°,为证明∠A =90°,只需证明 C B A O x AB · AC=0. 复习题2 已知:A(2,1),B(3,2),C(–1,4), 求证:?ABC是直角三角形. 由AB· AC= AB AC cosA 可知,为了证明AB· AC = 0, C y 需先得出 cosA = 0,需先 证明∠A为90°,而这正是 A O B x 最终要证明的结论. 5.7 平面向量 数量积的坐标表示 在坐标平面xoy内,已知a=(x1, y1),b= (x2,y2),则 a· b=x1x2+y1y2. 即 两个向量的数量积等于它们 对应坐标的乘积的和. a· b=x1x2+y1y2 证明:设x轴、y轴方向的单位向量 分别是i、j,则 a· b=(x1i+y1j)· (x2i+y2j) =x1x2i· i+ x1y2i· j+ y1x2j· i+ y1y2j· j =x1x2+y1y2. 已知:A(2,1),B(3,2),C(–1,4), 求证:?ABC是直角三角形. 证明: AB=(3 – 2,2 – 1)=(1,1), AC=(– 1 – 2,4 – 1)=(– 3,3), ∵ AB · AC=1×(– 3)+1×3=0, ∴ AB⊥AC. ∴ ?ABC是直角三角形. 由向量数量积的坐标表示,可得 (1)若A、B坐标分别为 (x1,y1)、 (x2,y2),则 ( |AB| =√(x1-x2)2+(y1-y2)2 |AB|2 = AB· AB = (x1-x2)2+(y1-y2)2 a ⊥b B x1x2+y1y2=0 (x2,y2) x1y2-x2y1=0) ) y (2)设 a=(x1,y1),b= (x2,y2),则 (a∥b (b≠0) A (x1,y1) x O 例 1 已知a=(1,√3 ),b=(– 2,2√3 ), (1)求a· b; (2)求a与b的夹角θ. 解:(1)a· b=1×(–2)+√3×2√3=4; ( 2) a = √1 +(√3 ) =2, b =√(– 2) +(2√3 ) =4, 2 2 2 2 cos θ = a· b a b = 4 =1 , 2 2× 4 ∴ θ =60? . 例 2:已知a=(5, 0),b=(–3.2, 2.4), 求证:(a+b)⊥b . 证明: ∵(a+b)· b = a· b+b2 =5×(–3.2)+0×2.4+(–3.2)2+2.42 = 0, ∴ (a+b)⊥b . 例 3:已知:A、B、C三点坐标分别为(2,0)、 (4,2)、(0,4),直线 l 过A、B两点,求 点C到 l 的距离. y 分析一:如图, 为求CH长,由 CH=AH-AC可知,关键 C 在于求出AH. B l 由AC· AB的几何意 义,AC· AB等于AB的长度 H 与AC在AB方向上的投影的 O A x 乘积. 所以 AC· AB=AH· AB. 解:AC=(0 – 2,4 – 0)=(–2,4), ∴AB AH = == (1(2 , 1). = (4(2 –m 2 , 2m –)0) , 2), y AC· AB=–2×2+4×2=4. CH=AH-AC=(3,–3), C ∵AH与AB共线, 2+(–3)2=3√2 . CH = √ 3 ∴可设AH=mAB=(2m,2m). B l x AH· AB=4m+4m=8m. 即由 CAC· 点到直线 l 的距离为 AB=AH· AB,得 3√2 . m= 1 2 . H O A 若能确定H点坐标, 分析二: y C B l CH长就易求了. 为定H点坐标(两个未 知数), 可利用H点在 l 上, 及CH⊥AB这两个条件. H O A x 练习:1.向量a、b夹角为θ, (1)a=(3,-2),b=(1,1), 1 则a· b=_________ ,cos θ =______. 26 |a|=√13,|b|=√2 (2)a=(-1,2),b=(2,-4), -10 ,θ=__________ 180? 则 a· b=_______ 2. 已知?ABC三个顶点坐标A( 1 ,4 ), 3 3 B(-2,3),C(0,1), √26 求证:?ABC是直角三角形. 小结: (1)掌握平面向量数量积的坐标表示,即 两个向量的数量积等于它们对应坐标的 乘积之和; (2)要学会运用平面向量数量积的坐标表 示解决有关长度、角度及垂直问题. (1)P121 练习; (2)P121 习题5.7 第1、2、4、5题. 今日 作业

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