女生影院:【金版学案】2019学年高一数学人教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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第二章 平面向量 2.4 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 题型1 向量数量积、模及夹角的坐标运算 例 1 已知向量 a=(4,3),b=(-2,1). (1)求|a|,|b|; (2)求 a· b 的值; (3)求(a+2b)·(a-b)的值. 分析:用向量的数量积、模及夹角的坐标运算. 解析:(1)|a|= 42+32=5; |b|= (-2)2+12= 5. (2)a· b=(4,3)·(-2,1)=-8+3=-5. (3)方法一 ∵a+2b=(0,5),a-b=(6,2), ∴(a+2b)·(a-b)=(0,5)·(6,2)=10. 方法二 (a+2b)·(a-b)=a2+a· b-2b2= 25-5-10=10. 点评: 求(a+2b)·(a-b)的值时,方法一用向量的 坐标法,先分别求出(a+2b)与(a-b)的坐标,再用数 量积公式求解,方法二直接用向量运算律进行运算. ?跟踪训练 1.已知向量 a=(4,3),b=(-2,1). (1)求向量 a+b 与 a-b 的夹角 θ; (2)若向量 a-λb 与 2a+b 垂直,求 λ 的值. 分析:先把向量 a+b 与 a-b 用坐标表示出来, 然后再根据夹角公式求解. 解析:(1)∵a+b=(2,4),a-b=(6,2), ∴|a+b|=2 5,|a-b|=2 10, 2 cos θ= = , 2 2 5×2 10 ∴θ=45°. (2)方法一 a-λb=(4+2λ,3-λ),2a+b=(6,7), (2,4)·(6,2) ∵向量 a-λb 与 2a+b 垂直, ∴(4+2λ,3-λ)·(6,7)=0. 解得 λ=-9. 方法二 ∵向量 a-λb 与 2a+b 垂直, ∴(a-λb)· (2a+b)=0, ∴2a2+(-2λ+1)a· b-λb2=0, 又|a|= 42+32=5,|b|= (-2)2+12= 5, a· b=(4,3)·(-2,1)=-8+3=-5, ∴50-5(-2λ+1)-5λ=0, 解得 λ=-9. 题型2 根据向量间的关系求向量的坐标 例 2 已知 a 与 b 同向,b=(1,2),a· b=10. (1)求向量 a 的坐标; c)及(a· b)c. (2)若 c=(2,-1),求 a(b· 分析:设向量 a 的坐标为(x,y),根据等量关系 列方程组求解. 解析:(1)设 a=(x,y),依题意有 ? ? ?x+2y=10, ?x=2, ? 解得? ?2x-y=0, ?y=4. ? ? ∴a=(2,4). (2)∵b· c=(1,2)·(2,-1)=0,a· b=10, c)=0,(a· b)c=10(2,-1)=(20,-10). ∴a(b· c)与(a· b)c 表示的意义不一样,故不相等. 点评:a(b· ?跟踪训练 2.已知 a=(4,2),|b|= 5,且 a⊥b,求向量 b 的坐标. 分析:设向量 b 的坐标为(x,y),根据等量关系列方程组求解. 解析:设 b=(x,y),依题意有 2 2 ? ? x + y =5, ? ?x=1, ? ?x=-1, ? 解得? 或? ? ? ?4x+2y=0, ?y=-2 ? ?y=2. ∴b=(1,-2)或 b=(-1,2). 题型3 向量的综合应用 例 3 设 a=(cos α ,sin α ),b=(cos β , sin β )(0<α<β<π ). (1)求证:(a+b)⊥(a-b); (2)若|ka+b|= 3|a-kb|(k>0),用 k 表示数量积 a· b. 分析:题目给出了向量的坐标,可以考虑坐标法解决 问题,但实际上向量法会显得更简单明了. (1) 证明:∵a=(cos α,sin α),b=(cos β, sin β)(0<α<β<π), ∴|a|=1,|b|=1. ∴(a+b)· (a-b)=a2-b2=1-1=0. ∴(a+b)⊥(a-b). (2)解析:由|ka+b|= 3|a-kb|得 2 2 (ka+b) =3(a-kb) , ∴k2a2+2ka· b+b2=3a2-6ka· b+3k2b2, ∴(k2-3)a2+8ka· b+(1-3k2)b2=0. ∵|a|=1,|b|=1, ∴(k2-3)+8ka· b+(1-3k2)=0. 2k2+2 k2+1 (k>0). ∴ a· b= = 8k 4k 点评:向量与代数中的一些问题如函数的最值问题等, 是向量作为工具的具体体现,解决此类问题应熟练掌 握向量的坐标运算法则,特别是共线、垂直、数量积 等坐标表示. ?跟踪训练 3.如下图所示,以原点 O 和 A(5,2)为两个顶点作 等腰直角△AOB,使∠B=90°,求点 B 的坐标. 分析:向量数量积的坐标运算;化归与转化的思想. → =(x,y),AB → =(x-5,y-2), 解析:设点 B(x,y),则OB ∵∠B=90°,∴x(x-5)+y(y-2)=0.① → |=|OB → |, 又|AB ∴x2+y2=(x-5)2+(y-2)2.② 2 2 ? ?x +y -5x-2y=0, 由①,②整理得? ?10x+4y=29, ? 7 3 ? ? ?x=2, ?x=2, 解得? 或? 3 7 ? ?y=-2 ? ?y=2. ?7 3? ?3 7? ? ? ∴B 2,-2 或 B?2,2?. ? ? ? ?

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