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www.xpj9899.com: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

.

1

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°,

2

∴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;

3

(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.

4

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:

5

6.已知 a=( 最小值.

,-1),b=

,且存在实数 k 和 t,使得 x=a+(t -3)b,y=-ka+tb,且 x⊥y,试求

2



解:∵a=(

,-1),b=

,

∴a·b=

-1×

=0.

∵|a|=

=2,

|b|=

=1,a·b=0,∴a⊥b.

∵x⊥y,∴[a+(t2-3)b]·(-ka+tb)=0,
即-ka +(t -3t)b +(t-t k+3k)a·b=0.
2 3 2 2

∴k=

.



(t +4t-3)=

2

(t+2) -

2

.

故当 t=-2 时,

有最小值-

. =(5,1), =(2,1),点 Q 为直线 OP 上的一个动点.

7. 导学号 08720072 平面内有向量 (1)当 取最小值时,求

=(1,7),

的坐标;

(2)当点 Q 满足(1)的条件和结论时,求 cos∠AQB 的值. 解:(1)设

=(x,y).
上,∴向量 共线.

∵点 Q 在直线


=(2,1),∴x=2y,∴

=(2y,y).

6



=(1-2y,7-y), =(5-2y,1-y),



=(1-2y)(5-2y)+(7-y)(1-y)

=5y2-20y+12=5(y-2)2-8,
故当 y=2 时, (2)由(1)知, 有最小值-8,此时

=(4,2).

=(-3,5), =-8,| |=

=(1,-1),
,|

|=

,

cos∠AQB=

=-

.

7


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