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WWW.P2P555.CC:新人教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 1.已知 a=(-4,3),b=(1,2),则 a2-(a-b)·b=()

A.8

B.3+

C.28

解析:a2-(a-b)·b=a2-a·b+b2=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)

D.32

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.

B.4

C.3

D.2

解析:∵a∥b,∴m=-4,b=(-2,-4).

∴2a+3b=2(1,2)+3(-2,-4)=(-4,-8).

∴|2a+3b|=

=4 .

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

B.1

C.2

D.-2

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.

1

C.

D.

解析:∵b 在 a 方向上的投影为 2,则 =2,

∴ =2,解得 x=-2, ∴b=(-2,2 ).

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=- (舍去). 答案:2
8.在△ABC 中,∠C=90°, =(k,1), =(2,3),则 k 的值是.

解析:∵∠C=90°,∴

,∴

=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)

的坐标;

2

(2)|

|的值;

(3)cos∠BAC 的值.

解:(1) =(0,1)-(1,0)=(-1,1),

=(2,5)-(1,0)=(1,5).

(2)因为

=(-1,1)-(1,5)=(-2,-4),

所以|

|=

=2 .

(3)因为

=(-1,1)·(1,5)=4,

| |= ,| |= ,

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 x2-2x-3=0,解得 x=3 或 x=-1.

(2)因为 a∥b,所以 1×(-x)-(2x+3)×x=0, 即 x2+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 .

二、B 组

1.在平行四边形 ABCD 中, =(1,0), =(2,2),则

A.4

B.-4

C.2

等于() D.-2

解析:如图所示,由向量的加减,可得 =(1,2),

-2 =(0,2). 3



=(1,2)·(0,2)=0+4=4.

答案:A

2.已知向量 a=(2,0),|b|=1,|a+2b|=2 ,则 a 与 b 的夹角为()

A.30°

B.60°

C.120°

D.150°

解析:∵|a+2b|2=a2+4a·b+4b2=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,



=3,解得 y= ,则 b=

.故选 A.

答案: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=( ,1),b 是不平行于 x 轴的单位向量,且 a·b= ,则 b=. 解析:设 b=(x,y).

∵|b|=

=1,∴x2+y2=1.

∵a·b= x+y= ,∴x2+[ (1-x)]2=1. ∴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,

4

∴b=

.

答案:

6.已知 a=( ,-1),b= 的最小值.

,且存在实数 k 和 t,使得 x=a+(t2-3)b,y=-ka+tb,且 x⊥y,试求

解:∵a=( ,-1),b=

,

∴a·b= ∵|a|=

-1× =0. =2,

|b|=

=1,a·b=0,∴a⊥b.

∵x⊥y,∴[a+(t2-3)b]·(-ka+tb)=0, 即-ka2+(t3-3t)b2+(t-t2k+3k)a·b=0.

∴k=

.



(t2+4t-3)= (t+2)2- .

故当 t=-2 时,

有最小值- .

7.导学号 08720072 平面内有向量 =(1,7), =(5,1), =(2,1),点 Q 为直线 OP 上的一个动点.

(1)当

取最小值时,求 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).

5

∵点 Q 在直线 上,∴向量

共线.

又 =(2,1),∴x=2y,∴ =(2y,y).



=(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,

有最小值-8,此时 =(4,2).

(2)由(1)知, =(-3,5), =(1,-1),

=-8,| |= ,| |= ,

cos∠AQB=

=- .

6


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