www.22mmqq.com:高一数学人教a版必修4课件2.4.2平面向量数量积的坐标表示模夹角_图文

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2.4 读教材·填要点 课前预习·巧设计 第 二 章 平 面 向 量 平 面 向 量 的 数 量 积 2.4.2 平面向 量数量 积的坐 标表示 、模、 夹角 小问题·大思维 考点一 名师课堂·一点通 考点二 考点三 解题高手 NO.1课堂强化 创新演练·大冲关 NO.2课下检测 返回 返回 返回 返回 [读教材·填要点] 1.两向量的数量积与两向量垂直的坐标表示 设向量a=(x1,y1),b=(x2,y2),a与b的夹角为θ. 两个向量的数量积等于它们 对应坐标的乘积 的 数量积 和,即a· b= x1x2+y1y2 两个向 量垂直 a⊥b? x1x2+y1y2=0 返回 2.三个重要公式 返回 [小问题·大思维] 1.已知向量a=(x,y),与向量a共线的单位向量a0的坐 标是什么? a 1 提示:∵a0=± =± 2 2(x,y), |a| x +y x y ∴a0=(- 2 2,- 2 2) x +y x +y x y 或 a0=( 2 2, 2 2). x +y x +y 返回 2.向量a=(x1,y1),b=(x2,y2),则向量a在向量b方向 上的投影怎样用a,b的坐标表示? 提示: 向量 a 在向量 b 方向上的投影为|a|cos θ(θ 为向 a· b 量 a 与 b 的夹角),而 cos θ= , |a||b| a· b x1x2+y1y2 ∴|a|cos θ= = 2 2 . |b| x2+y2 返回 返回 [研一题] [例1] 已知向量a=(1,3),b=(2,5),c=(2,1),求: (1)2a· (b-a); (2)(a+2b)· c. [自主解答] 法一:(1)∵2a=2(1,3)=(2,6), b-a=(2,5)-(1,3)=(1,2), ∴2a· (b-a)=(2,6)· (1,2) =2×1+6×2=14. 返回 (2)∵a+2b=(1,3)+2(2,5) =(1,3)+(4,10)=(5,13), ∴(a+2b)· c=(5,13)· (2,1) =5×2+13×1=23. 返回 法二:(1)2a· (b-a) =2a· b-2a2 =2(1×2+3×5)-2(1+9) =14. (2)(a+2b)· c =a· c+2b· c =1×2+3×1+2(2×2+5×1) =23. 返回 本例条件中“c=(2,1)”若变为“c=(2,k)”,且“(a- c)⊥b”,求k. 解:∵a=(1,3),c=(2,k), ∴a-c=(-1,3-k), 又(a-c)⊥b,∴-1×2+(3-k)×5=0, 13 ∴k= . 5 返回 [悟一法] 1.通过向量的坐标表示可实现向量问题的代数化,应 注意与函数、方程等知识的联系. 2.向量数量积的运算有两种思路:一种是向量式,另 一种是坐标式,两者相互补充. 返回 [通一类] 1.若向量a=(4,-3),|b|=1,且a· b=5,求向量b. 解:法一:设 b=(x,y),则由|b|=1 可得 x2+y2=1. 由 a· b=5,a=(4,-3),可得 4x-3y=5. 4 3 4 3 联立①②,解得 x= ,y=- ,故 b=( ,- ). 5 5 5 5 法二:由于 a=(4,-3),∴|a|=5,设 a 与 b 的夹角为 θ, a· b 则 cos θ= =1,∴θ=0° ,从而 a,b 同向且共线, |a||b| a 1 4 3 又|b|=1,∴b= = (4,-3)=( ,- ). |a| 5 5 5 返回 ① ② [研一题] [例 2] 平面直角坐标系 xOy 中,O 是原点(如图).已知点 A(16,12)、B(-5,15). (1)求| OA |,| AB |; (2)求∠OAB. [自主解答] (1)由 OA =(16,12), AB =(-5-16,15-12)=(-21,3),得 | OA |= 162+122=20, | AB |= ?-21?2+32=15 2. 返回 AO · AB (2)cos∠OAB=cos〈 AO , AB 〉= . | AO || AB | 其中 AO · AB =- OA · AB =-(16,12)· (-21,3) =-[16× (-21)+12× 3]=300. 300 2 故 cos∠OAB= = . 20×15 2 2 ∴∠OAB=45° . 返回 [悟一法] 利用向量的数量积求两向量夹角的一般步骤为: (1)利用向量的坐标求出这两个向量的数量积. (2)利用|a|= x2+y2求两向量的模. (3)代入夹角公式求 cos θ,并根据 θ 的范围确定 θ 的值. 返回 [通一类] 2.已知 a,b 为平面向量,a=(4,3),2a+b=(3,18),则 a,b 的夹角 θ 的余弦值等于 8 A. 65 8 B.- 65 ( ) 16 16 C. D.- 65 65 解析:b=(2a+b)-2a=(3,18)-(8,6)=(-5,12), cos θ= a· b 16 = . |a||b| 65 答案:C 返回 [研一题] [例 3] 已知△ABC 中,A(2,-1), B(3,2),C(-3,-1),BC 边上的高为 AD, 如图,求 D 点及 AD 的坐标. [自主解答] 设 D(x,y), ∴ AD =(x-2,y+1), BC =0, 又 BC =(-6,-3), AD · ∴-6(x-2)-3(y+1)=0,即 2x+y=3. ① 返回 又 BD 与 BC 为共线向量, BD =(x-3,y-2), BC =(-6,-3), ∴-3(x-3)+6(y-2)=0,即 x-2y=-1. 联立①②,解得 x=1,y=1. ∴D(1,1), AD =(-1,2). ② 返回 [悟一法] 利用向量数量积的坐标表示解决垂直问题

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