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ag1519com:18版高中数学第二章平面向量2.4.2平面向量数量积的坐标表示模夹角学案41707241136

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2.4.2 平面向量数量积的坐标表示、模、夹角 1.掌握平面向量数量积的坐标表示及其运算.(重点) 2.会运用向量坐标运算求解与向量垂直、夹角等相关问题.(难点) 3.分清向量平行与垂直的坐标表示.(易混点) [基础·初探] 教材整理 平面向量数量积的坐标表示、模、夹角 阅读教材 P106“探究”以下至 P107 例 6 以上内容,完成下列问题. 1.平面向量数量积的坐标表示: 设向量 a=(x1,y1),b=(x2,y2),a 与 b 的夹角为 θ . 数量积 向量垂直 a·b=x1x2+y1y2 a⊥b?x1x2+y1y2=0 2 2 2.向量模的公式:设 a=(x1,y1),则|a|= x1+y1. → 3.两点间的距离公式:若 A(x1,y1),B(x2,y2),则AB= x2-x1 2 + y2-y1 2 . 4.向量的夹角公式:设两非零向量 a=(x1,y1),b=(x2,y2),a 与 b 夹角为 θ ,则 a·b x1x2+y1y2 cos θ = = 2 . 2 2 |a|·|b| x1+y1 x2 2+y2 判断(正确的打“√”,错误的打“×”) (1)两个非零向量 a=(x1,y1),b=(x2,y2),满足 x1y2-x2y1=0,则向量 a,b 的夹角为 0°.( ) ) ) (2)已知 a=(x1,y1),b=(x2,y2),a⊥b?x1x2-y1y2=0.( (3)若两个向量的数量积的坐标和小于零,则两个向量的夹角一定为钝角.( 【解析】 (1)×.因为当 x1y2-x2y1=0 时,向量 a,b 的夹角也可能为 180°. (2)×.a⊥b?x1x2+y1y2=0. (3)×.因为两向量的夹角有可能为 180°. 【答案】 (1)× (2)× (3)× 1 [小组合作型] 平面向量数量积的坐标运算 (1)已知向量 a=(1,2),b=(2,x),且 a·b=-1,则 x 的值等于( A. C. 1 2 3 2 1 B.- 2 3 D.- 2 ) (2)已知向量 a=(-1,2),b=(3,2),则 a·b=________,a·(a-b)=________. (3)已知 a=(2,-1),b=(3,2),若存在向量 c,满足 a·c=2,b·c=5,则向量 c= ________. 【精彩点拨】 根据题目中已知的条件找出向量坐标满足的等量关系,利用数量积的坐 标运算列出方程(组)来进行求解. 【自主解答】 (1)因为 a=(1,2),b=(2,x), 所以 a·b=(1,2)·(2,x)=1×2+2x=-1, 3 解得 x=- . 2 (2)a·b=(-1,2)·(3,2)=(-1)×3+2×2=1, a·(a-b)=(-1,2)·[(-1,2)-(3,2)]=(-1,2)·(-4,0)=4. (3)设 c=(x,y),因为 a·c=2,b·c=5, 9 ? ?x=7, 解得? 4 ? ?y=7, ? ?2x-y=2, 所以? ?3x+2y=5, ? ?9 4? 所以 c=? , ?. ?7 7? ?9 4? 【答案】 (1)D (2)1 4 (3)? , ? ?7 7? 1.进行数量积运算时,要正确使用公式 a·b=x1x2+y1y2,并能灵活运用以下几个关系: |a|2=a·a;(a+b)(a-b)=|a|2-|b|2; 2 (a+b) =|a| +2a·b+|b| . 2.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. 3.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. 2 2 2 [再练一题] 1.设向量 a=(1,-2),向量 b=(-3,4),向量 c=(3,2),则向量(a+2b)·c=( A.(-15,12) C.-3 【解析】 B.0 D.-11 依题意可知,a+2b=(1,-2)+2(-3,4)=(-5,6),∴(a+2b)·c=(- ) 5,6)·(3,2)=-5×3+6×2=-3. 【答案】 C 向量模的坐标表示 (1)设平面向量 a=(1,2),b=(-2,y),若 a∥b,则|2a-b|等于( A.4 C.3 5 B.5 D.4 5 ) (2)已知向量 a=(1,2),b=(-3,2),则|a+b|=________,|a-b|=________. 【精彩点拨】 (1)两向量 a=(x1,y1),b=(x2,y2)共线的坐标表示:x1y2-x2y1=0. (2)已知 a=(x,y),则|a|= x +y . 【自主解答】 (1)由 y+4=0 知 2 2 y=-4,b=(-2,-4), ∴2a-b=(4,8),∴|2a-b|=4 5.故选 D. (2)由题意知,a+b=(-2,4),a-b=(4,0), 因此|a+b|= - 2 +4 =2 5,|a-b|=4. 4 2 【答案】 (1)D (2)2 5 向量模的问题的解题策略: (1)字母表示下的运算,利用|a| =a 将向量模的运算转化为向量的数量积的运算. (2)坐标表示下的运算,若 a=(x,y),则|a|= x +y . 2 2 2 2 [再练一题] 2. 已知向量 a = (2x + 3,2 - x) , b = ( - 3 - x,2x)(x ∈ R) ,则 |a + b| 的取值范围为 ________. 3 【导学号:00680057】 【解析】 ∵a+b=(x,x+2), ∴|a+b|= x + x+ = 2 2 = 2x +4x+4 2 x+ 2 +2≥ 2, ∴|a+b|∈[ 2,+∞). 【答案】 [ 2,+∞) [探究共研型] 向量的夹角与垂直问题 探究 1 设 a,b 都是非零向量,a=(x1,y1),b=(x2,y2),θ 是 a 与 b 的夹角,那么 cos θ 如何用坐标表示? 【提示】 cos θ =

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