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bojiuyulechengpingtai:18学年高中数学第二章平面向量6平面向量数量积的坐标表示教学案北师大版必修4

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6 平面向量数量积的坐标表示 [核心必知] 1.向量数量积的坐标表示 设 a=(x1,y1),b=(x2,y2),则 a·b=x1x2+y1y2. 即两个向量的数量积等于相应坐标乘积的和. 2.度量公式 (1)长度公式:设 a=(x,y),则|a|= x +y . (2)夹角公式: 设 a=(x1, y1), b=(x2, y2), a 与 b 的夹角为 θ , 则 cos θ = 3.两向量垂直的坐标表示 设 a=(x1,y1),b=(x2,y2),则 2 2 x1x2+y1y2 . 2 2 x +y1 x2 2+y2 2 1 a⊥b?a·b=0?x1x2+y1y2=0. 4.直线的方向向量 给定斜率为 k 的直线 l,则向量 m=(1,k)与直线 l 共线,把与直线 l 共线的非零向量 m 称 为直线 l 的方向向量. [问题思考] 1.由向量长度的坐标表示,你能否得出平面内两点间的距离公式? 提示:设 A(x1,y1),B(x2,y2), 则 AB =(x2-x1,y2-y1),由向量长度的坐标表示可得 |AB|=| AB |= x2-x1 2 + y2-y1 2 . 2.坐标形式下两向量垂直与平行的条件有何区别? 提示:设 a=(x1,y1),b=(x2,y2),则: a⊥b?x1x2+y1y2=0,即“相应坐标相乘和为 0”; -1- a∥b?x1y2-x2y1=0,即“坐标交叉相乘差为 0”. 3.直线 l 的方向向量唯一吗? 提示:直线 l 的方向向量即是与 l 平行的向量,意指表示该向量的有向线段所在的直线与 l 平行或重合,所以直线 l 的方向向量不唯一(有无数个),但它们都是共线向量. 讲一讲 1.已知向量 a=(4,-2),b=(6,-3),求: (1)(2a-3b)·(a+2b); (2)(a+b) . [尝试解答] 法一:(1)∵2a-3b=(8,-4)-(18,-9)=(-10,5), 2 a+2b=(4,-2)+(12,-6)=(16,-8), ∴(2a-3b)·(a+2b)=-160-40=-200. (2)∵a+b=(10,-5) ∴(a+b) =(10,-5)×(10,-5)=100+25=125. 法二:由已知可得:a =20,b =45,a·b=30 (1)(2a-3b)·(a+2b)=2a +a·b-6b =2×20+30-6×45=-200. (2)(a+b) =a +2a·b+b =20+60+45=125. 2 2 2 2 2 2 2 2 进行向量的数量积的坐标运算关键是把握向量数量积的坐标表示,运算时常有两条途径: (1)根据向量数量积的坐标表示直接运算; (2)先利用数量积的运算律将原式展开,再依据已知计算. -2- 练一练 1.已知 a=(2,1),b=(-1,3),向量 c 满足 a·c=4,b·c=-9. (1)求向量 c 的坐标; (2)求(a+b)·c 的值. 解:(1)设 c=(x,y), 由? ? ?a·c=4, ? ?2x+y=4, 得,? ?b·c=-9, ?-x+3y=-9. ? ? 解得 x=3,y=-2. ∴c=(3,-2). (2)法一:∵a+b=(2,1)+(-1,3)=(1,4), ∴(a+b)·c=(1,4)·(3,-2) =1×3+4×(-2) =-5. 法二:(a+b)·c =a·c+b·c =(2,1)·(3,-2)+(-1,3)·(3,-2) =2×3+1×(-2)+(-1)×3+3×(-2) =-5. 讲一讲 2.已知 a=(1,2),b=(-2,-4),|c|= 5. (1)求|a+2b|; 5 (2)若(a+b)·c= ,求向量 a 与 c 的夹角. 2 [尝试解答] (1)a+2b=(1,2)+2(-2,-4)=(-3,-6) ∴|a+2b|= (-3) +(-6) =3 5. (2)∵b=(-2,-4)=-2(1,2)=-2a ∴a+b=-a, 5 ∴(a+b)·c=-a·c= 2 设 a 与 c 的夹角为 θ , -32 2 a·c 1 则 cos θ = = =- |a||c| 2 5× 5 2 ∵0≤θ ≤π ,∴θ = π 3 2 即 a 与 c 的夹角为 π . 3 - 5 2 1.已知向量的坐标和向量的模(长度)时,可直接运用公式|a|= x +y 进行计算. 2.求向量的夹角时通常利用数量积求解,一般步骤为: (1)先利用平面向量数量积的坐标表示求出两向量的数量积; (2)再求出两向量的模; (3)由公式 cos θ = 2 2 a·b 计算 cos θ 的值; |a||b| (4)在[0,π ]内,由 cos θ 的值确定角 θ . 练一练 2.已知向量 a=(x1,y1),b=(x2,y2),e=(0,1),若 a≠b,|a-b|=2,且 a-b 与 e 的 π 夹角为 ,则 x1-x2=( 3 A.2 C.± 2 ) B.± 3 D.±1 解析:选 B a-b=(x1-x2,y1-y2). ∴(a-b)·e=(x1-x2)×0+(y1-y2)×1=y1-y2. π ∵|a-b|=2,|e|=1,a-b 与 e 的夹角为 , 3 ∴cos π (a-b)·e y1-y2 1 = = = ,∴y1-y2=1, 3 |a-b||e| 2 2 2 2 又由|a-b|=2 知,(x1-x2) +(y1-y2) =4, ∴(x1-x2) =3.∴x1-x2=± 3. 2 讲一讲 3? ?1 3.已知 a=( 3,-1),b=? , ?. ?2 2 ? (1)求证:a⊥b; -4- (2)是否存在实数 k,使 x=a-2b,y=-ka+b,且 x⊥y,若存在,求 k 的值;不存在,请 说明理由. 1 3 [尝试解答] (1)证明:∵a·b= 3× +(-1)× =0. 2 2 ∴a⊥b. 3? ?1 (2)∵x=( 3,-1)-

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