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ak361: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课后习题新人教A版必修4

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2.4.2 平面向量数量积的坐标表示、模、夹角 一、A 组 2 1.已知 a=(-4,3),b=(1,2),则 a -(a-b)·b=( ) A.8 2 B.3+ 2 2 C.28 D.32 解析:a -(a-b)·b=a -a·b+b =25-(-4+6)+5=28. 答案:C 2.若 a=(3,4),则与 a 共线的单位向量是( A.(3,4) ) B. C. D.(1,1) 解析:与 a 共线的单位向量是± =± (3,4),即与 a 共线的单位向量是 答案:C 3.已知 a=(1,2),b=(-2,m),若 a∥b,则|2a+3b|等于 A. C.3 B.4 D.2 ( ) . 解析:∵a∥b,∴m=-4,b=(-2,-4). ∴2a+3b=2(1,2)+3(-2,-4)=(-4,-8). ∴|2a+3b|= 答案:B 4.(2016·广东深圳南山期末)已知向量 a=(1,1),b=(2,-3),若 ka-2b 与 a 垂直,则实数 k 的值为 ( A.-1 C.2 ) B.1 D.-2 =4 . 解析:∵向量 a=(1,1),b=(2,-3), ∴ka-2b=k(1,1)-2(2,-3)=(k-4,k+6). ∵ka-2b 与 a 垂直, ∴(ka-2b)·a=k-4+k+6=0,解得 k=-1.故选 A. 答案:A 5.已知 a=(1, ),b=(x,2 ),且 b 在 a 方向上的投影为 2,则 a 与 b 的夹角为( ) A. B. C. D. 解析:∵b 在 a 方向上的投影为 2,则 =2, ∴ ∴b=(-2,2 =2,解得 x=-2, ). 设 a,b 的夹角为 θ ,则 cos θ = . ∵0≤θ ≤π ,∴θ = 答案:D . 6.已知向量 a=(1,n),b=(-1,n),a·b=2,则|a|= 解析:∵a=(1,n),b=(-1,n),a·b=2, . ∴-1+n2=2,∴n2=3. ∴|a|2=1+n2=4,∴|a|=2. 答案:2 7.已知 a=(-1,3),b=(1,y).若 a 与 b 的夹角为 45°,则 y= 解析:a·b=-1+3y,|a|= ,|b|= , . ∵a 与 b 的夹角为 45°, ∴cos 45°= . 解得 y=2 或 y=答案:2 8.在△ABC 中,∠C=90°, 解析:∵∠C=90°,∴ 又 (舍去). =(k,1), ,∴ =(2,3),则 k 的值是 =0. . =(2-k,2), ∴2(2-k)+6=0,k=5. 答案:5 9.在平面直角坐标系内,已知三点 A(1,0),B(0,1),C(2,5),求: (1) (2)| 的坐标; |的值; (3)cos∠BAC 的值. 解:(1) =(0,1)-(1,0)=(-1,1), =(2,5)-(1,0)=(1,5). (2)因为 所以| (3)因为 =(-1,1)-(1,5)=(-2,-4), |= =(-1,1)·(1,5)=4, |= ,| =2 . | |= , 所以 cos∠BAC= . 10. 导学号 08720071 已知向量 a=(1,x),b=(2x+3,-x)(x∈R). (1)若 a⊥b,求 x 的值; (2)若 a∥b,求|a-b|. 解:(1)因为 a⊥b,所以 a·b=0, 即(1,x)·(2x+3,-x)=0. 所以 x -2x-3=0,解得 x=3 或 x=-1. (2)因为 a∥b,所以 1×(-x)-(2x+3)×x=0, 即 x +2x=0,所以 x=0 或 x=-2. 当 x=0 时,a=(1,0),b=(3,0), a-b=(-2,0),所以|a-b|=2; 当 x=-2 时,a=(1,-2),b=(-1,2), a-b=(2,-4),所以|a-b|=2 2 2 . 二、B 组 1.在平行四边形 ABCD 中, A.4 B.-4 =(1,0), =(2,2),则 C.2 等于( D.-2 ) 解析:如图所示,由向量的加减,可得 =(1,2), 故 答案:A 2.已知向量 a=(2,0),|b|=1,|a+2b|=2 A.30° 2 2 -2 =(0,2). =(1,2)·(0,2)=0+4=4. ,则 a 与 b 的夹角为( D.150° ) B.60° 2 C.120° 解析:∵|a+2b| =a +4a·b+4b =4+4a·b+4=12, ∴a·b=1. 设 a 与 b 的夹角为 θ , 则 cos θ = , 又 0°≤θ ≤180°,∴θ =60°. 答案:B 3.设 a=(2,3),a 在 b 方向上的投影为 3,b 在 x 轴上的投影为 1,则 b=( ) A. B. C. D. 解析:由 b 在 x 轴上的投影为 1,设 b=(1,y). ∵a 在 b 方向上的投影为 3, ∴ 答案:A =3,解得 y= ,则 b= .故选 A. 4.已知向量 a=(2,4),b=(-1,2),若 c=a-(a·b)b,则|c|= 解析:∵a=(2,4),b=(-1,2), . ∴a·b=-2+8=6. ∴c=(2,4)-6(-1,2)=(8,-8). ∴|c|=8 答案:8 5.已知向量 a=( 解析:设 b=(x,y). ,1),b 是不平行于 x 轴的单位向量,且 a·b= ,则 b = . . ∵|b|= ∵a·b= x+y= =1,∴x2+y2=1. ,∴x +[ 2 (1-x)] =1. 2 ∴4x2-6x+2=0.∴2x2-3x+1=0. ∴x1=1,x2= ,∴y1=0,y2= . ∵(1,0)是与 x 轴平行的向量, ∴b= . 答案: 6.已知 a=( 最小值. ,-1),b= ,且存在实数 k 和 t,

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