gq2784:18学年高中数学17平面向量数量积的坐标表示练习(含解析)北师大版必修4

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17 平面向量数量积的坐标表示

时间:45 分钟 满分:80 分

班级________ 姓名________ 分数________ 一、选择题:(每小题 5 分,共 5×6=30 分) 1.已知 a=(-3,4),b=(5,2),则 a·b=( )

A.23 B.7 C.-23 D.-7

答案:D 2.若向量 a=(2,0),b=(1,1),则下列结论正确的是( ) A.a·b=1 B.|a|=|b| C.(a-b)⊥b D.a∥b

答案:C

解析:a·b=2,选项 A 错误;|a|=2,|b|= 2,选项 B 错误;(a-b)·b=(1,-1)·(1,1)

=0,选项 C 正确,故选 C.

3.已知向量 a=(0,-2 3),b=(1, 3),则向量 a 在 b 方向上的投影为( )

A. 3 B.3

C.- 3 D.-3

答案:D

解析:向量

a



b

a·b -6 方向上的投影为 |b| = 2 =-3.选

D.

4.若 A(1,2),B(2,3),C(-3,5),则△ABC 为( )

A.直角三角形 B.锐角三角形

C.钝角三角形

D.不等边三角形

答案:C 解析:∵A(1,2),B(2,3),C(-3,5),

∴A→B=(1,1),A→C=(-4,3),

cosA=|A→ A→BB· ||→A→ ACC|=

- +1×3=- 1 <0,∴∠A 为钝角,△ABC 为钝角三角

2× 25

52

形. 5.若向量A→B=(3,-1),n=(2,1),且 n·→AC=7,那么 n·B→C等于( )

A.-2

B.2

C.-2 或 2 D.0 答案:B

解析:n·(A→B+B→C)=n·→AB+n·B→C=7,所以 n·→BC=7-n·→AB=7-(6-1)=2.

6.与向量 a=(72,12),b=(12,-72)的夹角相等,且模为 1 的向量是(

)

A.(45,-35)

1

B.(45,-35)或(-45,35)

C.(2 3 2,-13)

D.(2 3 2,-13)或(-2 3 2,13) 答案:B 解析:设与向量 a=(72,12),b=(12,-72)的夹角相等,且模为 1 的向量为 e=(x,y), 则

??x2+y2=1 ???72x+12y=21x-72y

??x=45, ,解得???y=-35,

??x=-45, 或???y=35,

故选 B.

二、填空题:(每小题 5 分,共 5×3=15 分) 7.已知点 A(4,0),B(0,3),OC⊥AB 于点 C,O 为坐标原点,则→OA·→OC=________. 答案:14454

?? O→C·A→B=0 解 析: 设点 C 的坐标 为 (x , y) ,因为 OC ⊥ AB 于 点 C ,∴???A→C∥A→B

,即

?? x,y

-4,

???3x+4y-12=0

=-4x+3y=0

??x=3265 ,解得???y=4285

,∴→OA·→OC=4x=12454.

8.已知向量 a=(1, 3),2a+b=(-1, 3),a 与 2a+b 的夹角为 θ ,则 θ =________.

答案:π3

解析:∵a=(1, 3),2a+b=(-1, 3),∴|a|=2,|2a+b|=2,a·(2a+b)=2, ∴cosθ =a|a||2aa++bb| =12,∴θ =π3 .

9.若平面向量 a=(log2x,-1),b=(log2x,2+log2x),则满足 a·b<0 的实数 x 的取 值集合为________.

答案:?????x???12<x<4

?? ? ??

解析:由题意可得(log2x)2-log2x-2<0? (log2x+1)(log2x-2)<0,所以-1<log2x<2,

所以12<x<4.

三、解答题:(共 35 分,11+12+12) 10.在平面直角坐标系中,A,B 两点的坐标分别为(1,2),(3,8),向量→CD=(x,3). (1)若A→B∥C→D,求实数 x 的值; (2)若A→B⊥C→D,求实数 x 的值. 解析:(1)依题意,A→B=(3,8)-(1,2)=(2,6). ∵A→B∥C→D,C→D=(x,3),

2

∴2×3-6x=0,∴x=1. (2)∵A→B⊥C→D,C→D=(x,3), ∴2x+6×3=0,∴x=-9. 11.已知:a=(4,3),b=(-1,2),m=a-λ b,n=2a+b.按照下列条件求 λ 的值: (1)m 与 n 的夹角为钝角; (2)|m|=|n|. 解析:(1)因为 m 与 n 的夹角为钝角,所以 m·n<0,且 m 与 n 不共线. 因为 m=a-λ b=(4+λ ,3-2λ ),n=2a+b=(7,8).

所以???
??

+λ + -λ -

解得 λ >592.

-2λ <0 .
-2λ

(2)因为|m|=|n|,所以 +λ 2+ -2λ 2= 72+82.整理可得 5λ 2-4λ -88

=0.解之得 λ =2±25 111.

12.已知平面向量 a=(sinα ,1),b=(1,cosα ),-π2 <α <π2 . (1)若 a⊥b,求 α ; (2)求|a+b|的最大值. 解析:(1)由已知,得 a·b=0, 即 sinα +cosα =0,∴tanα =-1. ∵-π2 <α <π2 ,∴α =-π4 . (2)由已知得|a+b|2=a2+b2+2a·b=sin2α +1+cos2α +1+2(sinα +cosα )=3+
2 2sin???α +π4 ???.
∵-π2 <α <π2 ,

∴-π4 <α +π4 <3π4 ,∴- 22<sin???α +π4 ???≤1, 即 1<|a+b|2≤3+2 2,∴1<|a+b|≤1+ 2,

即|a+b|的最大值为 1+ 2.

3


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