2022版人教A版高中数学必修第二册练习题--平面向量基本定理及坐标表示综合拔高练

2022版人教A 版高中数学必修第二册--6.3综合拔高练
五年高考练
考点1 平面向量的坐标运算 1.(2019课标全国Ⅱ,3,5分,
)已知AB ⃗⃗⃗⃗⃗ =(2,3),AC ⃗⃗⃗⃗⃗ =(3,t ),|BC ⃗⃗⃗⃗⃗ |=1,则AB ⃗⃗⃗⃗⃗ ·BC
⃗⃗⃗⃗⃗ = ( ) A.-3 B.-2 C.2 D.3 2.(2019课标全国Ⅱ文,3,5分,
)已知向量a =(2,3),b =(3,2),则|a-b |= ( )
A.√2
B.2
C.5√2
D.50 3.(2021全国乙文,13,5分,)已知向量a =(2,5),b =(λ,4),若a ∥b ,则λ= . 4.(2021全国甲理,14,5分,)已知向量a =(3,1),b =(1,0),c =a +k b .若a ⊥c ,则
k = .
5.(2019课标全国Ⅲ,13,5分,
)已知向量a =(2,2),b =(-8,6),则
cos<a ,b >= . (注:<a ,b >表示向量a 与b 的夹角) 考点2 平面向量坐标运算的应用 6.(多选)(2021新高考Ⅰ,10,5分,
)已知O 为坐标原点,点P 1(cos α,sin α),P 2(cos
β,-sin β),P 3(cos (α+β),sin (α+β)),A (1,0),则 ( ) A.|OP 1⃗⃗⃗⃗⃗⃗⃗ |=|OP 2⃗⃗⃗⃗⃗⃗⃗ | B.|AP 1⃗⃗⃗⃗⃗⃗⃗ |=|AP 2⃗⃗⃗⃗⃗⃗⃗ |
C.OA ⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ =OP 1⃗⃗⃗⃗⃗⃗⃗ ·OP 2⃗⃗⃗⃗⃗⃗⃗
D.OA ⃗⃗⃗⃗⃗ ·OP 1⃗⃗⃗⃗⃗⃗⃗ =OP 2⃗⃗⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ 7.(2017课标全国Ⅱ,12,5分,
)已知△ABC 是边长为2的等边三角形,P 为平面ABC
内一点,则PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC
⃗⃗⃗⃗⃗ )的最小值是 ( )
A.-2
B.-32
C.-4
3
D.-1 8.(2020北京,13,5分,
)已知正方形ABCD 的边长为2,点P 满足AP ⃗⃗⃗⃗⃗ =12
(AB ⃗⃗⃗⃗⃗ +AC
⃗⃗⃗⃗⃗ ),则|PD
⃗⃗⃗⃗⃗ |= ;PB ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ = . 9.(2020天津,15,5分,
)如图,在四边形ABCD 中,∠B =60°,AB =3,BC =6,且AD ⃗⃗⃗⃗⃗ =
λBC ⃗⃗⃗⃗⃗ ,AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =−32
,则实数λ的值为 ,若M ,N 是线段BC 上的动点,且
|MN ⃗⃗⃗⃗⃗⃗⃗ |=1,则DM ⃗⃗⃗⃗⃗⃗ ·DN
⃗⃗⃗⃗⃗⃗ 的最小值为 .
10.(2019天津,14,5分,
)在四边形ABCD 中,AD ∥BC ,AB =2√3,AD =5,∠A =30°,点E
在线段CB 的延长线上,且AE =BE ,则BD ⃗⃗⃗⃗⃗⃗ ·AE
⃗⃗⃗⃗⃗ = . 三年模拟练
1.(多选)(2020山东德州高一下期中,)设向量a =(k ,2),b =(1,-1),则下列叙述错误的
是 ( )
A.若k <2,则a 与b 的夹角为钝角
B.|a |的最小值为2
C.与b 共线的单位向量只有一个,为(√2
2
,-
√2
2
)
D.若|a |=2|b |,则k =2√2或k =−2√2 2.(2021四川成都外国语学校高一下月考,
)勾股定理最早的证明是东汉数学家
赵爽在为《周髀算经》作注时给出的,被后人称为“赵爽弦图”.“赵爽弦图”是数形结合思想的体现,是中国古代数学的瑰宝,还被用作第24届国际数学家大会的
会标.如图,大正方形ABCD 是由4个全等的直角三角形和中间的小正方形组成的,若AB ⃗⃗⃗⃗⃗ =a ,AD ⃗⃗⃗⃗⃗ =b ,E 为BF 的中点,则AE
⃗⃗⃗⃗⃗ = ( )
A.45a +2
5b B.25a +4
5b C.43a +23
b
D.23a +43
b 3.(2020湖南长沙长郡中学高一上期末,)如图,圆O 是边长为2的正方形ABCD
的内切圆,若P ,Q 是圆O 上的两个动点,则AP ⃗⃗⃗⃗⃗ ·CQ
⃗⃗⃗⃗⃗ 的最小值为 ( )
A.-6
B.-3-2√2
C.−3+√2
D.-4 4.(2020浙江嘉兴高一上期末,
)已知向量OA ⃗⃗⃗⃗⃗ =(k ,12),OB ⃗⃗⃗⃗⃗ =(4,5),OC
⃗⃗⃗⃗⃗ =(-k ,10),若|AB ⃗⃗⃗⃗⃗ |=|BC
⃗⃗⃗⃗⃗ |,则k = ;若A ,B ,C 三点共线,则k = . 5.(2020江西南昌第十中学高一上期末,)已知向量a =(cos θ,sin θ),向量b =(√3,-1),
则|2a -b |的最大值为 . 6.(2021浙江杭州学军中学高一上期末,
)已知向量a =(1,2),b =(1,1),若a 与a +λb 的夹角为直角,则实数λ= ;若a 与a +λb 的夹角为锐角,则实数λ的取值范围是 .
7.(2021北京丰台高一下期中,
)已知向量a =(1,0),b =(-2,1).
(1)求向量2a-b;
(2)设a与b的夹角为θ,求cos θ的值;
(3)若向量k a+b与a+k b互相平行,求k的值.
8.(2021浙北G2高一下期中联考,)在平面直角坐标系中,BC⃗⃗⃗⃗⃗ =(2-k,3),AC⃗⃗⃗⃗⃗ =(2,4),k ∈R.
(1)若A,B,C三点不能构成三角形,求实数k的值;
(2)若△ABC为直角三角形,求实数k的值.
9.(2020江西抚州高一上期末,)已知向量a=(cos3x
2,sin3x
2
),b=(-cos x
2
,sin x
2
),且
x∈[π,3π
2
].
(1)求a·b及|a+b|;
(2)求函数f(x)=a·b+|a+b|的最小值,并求使函数f(x)取得最小值时x的值.
10.(2021天津一中高一下期中,
)在三角形ABC 中,AB =2,AC =1,∠ACB =π2
,D 是线段
BC 上一点,且BD ⃗⃗⃗⃗⃗⃗ =12
DC ⃗⃗⃗⃗⃗ ,F 为线段AB 上一点. (1)设AB ⃗⃗⃗⃗⃗ =a ,AC ⃗⃗⃗⃗⃗ =b ,AD ⃗⃗⃗⃗⃗ =x a +y b ,求x -y ; (2)求CF ⃗⃗⃗⃗⃗ ·FA
⃗⃗⃗⃗⃗ 的取值范围; (3)若F 为线段AB 的中点,直线CF 与AD 相交于点M ,求CM ⃗⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ .
答案全解全析 五年高考练
1.C ∵BC ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ =(1,t -3), ∴|BC ⃗⃗⃗⃗⃗ |=√12+(t -3)2=1,∴t =3, ∴AB ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ =(2,3)·(1,0)=
2. 2.A ∵a =(2,3),b =(3,2), ∴a -b =(-1,1),
∴|a -b |=√(-1)2+12=√2.故选A . 3.答案 8
5
解析 由a ∥b 得2×4=5λ,∴λ=8
5
.
4.答案 -10
3
解析 由题意知c =a +k b =(3,1)+k (1,0)=(3+k ,1), 结合a ⊥c 得3(3+k )+1×1=0,解得k =-10
3.
5.答案 -√2
10
解析 由题意知cos<a ,b >=
a ·b
|a |·|b |
=
√22√22
=−√2
10
.
6.AC ∵|OP 1⃗⃗⃗⃗⃗⃗⃗ |=√cos 2α+sin 2α=1,|OP 2⃗⃗⃗⃗⃗⃗⃗ |=√cos 2β+(-sinβ)2=1,∴|OP 1⃗⃗⃗⃗⃗⃗⃗ |=|OP 2⃗⃗⃗⃗⃗⃗⃗ |,A 选项正确.
易知|AP 1
⃗⃗⃗⃗⃗⃗⃗ |=√(cosα-1)2+sin 2α =√2-2cosα,
|AP 2
⃗⃗⃗⃗⃗⃗⃗ |=√(cosβ-1)2+(-sinβ)2
=√2-2cosβ,
由于α,β的大小关系不确定,所以不能确定|AP 1⃗⃗⃗⃗⃗⃗⃗ |=|AP 2⃗⃗⃗⃗⃗⃗⃗ |是否成立,B 选项不正确. ∵OA ⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ =(1,0)·(cos (α+β),sin (α+β))=cos (α+β),
OP 1⃗⃗⃗⃗⃗⃗⃗ ·OP 2⃗⃗⃗⃗⃗⃗⃗ =(cos α,sin α)·(cos β,-sin β)=cos αcos β-sin αsin β=cos (α+β), ∴OA ⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ =OP 1⃗⃗⃗⃗⃗⃗⃗ ·OP 2⃗⃗⃗⃗⃗⃗⃗ ,C 选项正确. ∵OA ⃗⃗⃗⃗⃗ ·OP 1⃗⃗⃗⃗⃗⃗⃗ =(1,0)·(cos α,sin α)=cos α, OP 2⃗⃗⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ =(cos β,-sin β)·(cos (α+β),sin (α+β)) =cos βcos (α+β)-sin βsin (α+β) =cos (β+α+β)=cos (α+2β),
∴OA ⃗⃗⃗⃗⃗ ·OP 1⃗⃗⃗⃗⃗⃗⃗ =OP 2⃗⃗⃗⃗⃗⃗⃗ ·OP 3⃗⃗⃗⃗⃗⃗⃗ 不一定成立, D 选项不正确.故选AC .
7.B 以AB 所在直线为x 轴,AB 的中点为原点建立平面直角坐标系,如图,
则A (-1,0),B (1,0),C (0,√3),设P (x ,y ),取BC 的中点D ,则D (1
2,√3
2).
PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )=2PA ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =2(-1-x ,-y )·(1
2-x ,
√3
2
-y)
=2(x +1)·(x -12)+y ·(y -√3
2)
=2(x +1
4)2
+(y -√3
4)2
−3
4
. 因此,当x =-14
,y =√3
4
时,PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )取得最小值,为2×(-34
)=−32
,故选B .
8.答案 √5;-1
解析 解法一:∵AP ⃗⃗⃗⃗⃗ =12
(AB ⃗⃗⃗⃗⃗ +AC
⃗⃗⃗⃗⃗ ),∴P 为BC 的中点.以A 为原点,AB ,AD 所在直线为x 轴,y 轴建立平面直角坐标系,则A (0,0),B (2,0),C (2,2),D (0,2),P (2,1),
∴|PD
⃗⃗⃗⃗⃗ |=√(2-0)2+(1-2)2=√5,PB ⃗⃗⃗⃗⃗ =(0,-1),PD ⃗⃗⃗⃗⃗ =(-2,1),∴PB ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =(0,-1)·(-2,1)=-1.
解法二:在正方形ABCD 中,由AP ⃗⃗⃗⃗⃗ =1
2
(AB ⃗⃗⃗⃗⃗ +AC
⃗⃗⃗⃗⃗ )得点P 为BC 的中点,∴|PD ⃗⃗⃗⃗⃗ |=√5,PB ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =PB ⃗⃗⃗⃗⃗ ·(PC ⃗⃗⃗⃗⃗ +CD ⃗⃗⃗⃗⃗ )=PB ⃗⃗⃗⃗⃗ ·PC ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ ·CD ⃗⃗⃗⃗⃗ =PB ⃗⃗⃗⃗⃗ ·PC ⃗⃗⃗⃗⃗ =1×1×cos 180°=-1. 9.答案 16;13
2
解析 以B 为原点,BC 所在直线为x 轴,过B 且垂直于BC 的直线为y 轴建立平面直角坐标系,如图所示,
则B (0,0),A (3
2,
3√32
),C (6,0),则AD ⃗⃗⃗⃗⃗ =λBC ⃗⃗⃗⃗⃗ =λ(6,0)=(6λ,0),AB ⃗⃗⃗⃗⃗ =(-32
,-3√32
),
∵AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =6λ×(-32
)+0×(-3√32
)=−9λ=−32
,∴λ=1
6
.
∴AD ⃗⃗⃗⃗⃗ =(1,0),∴D (52
,
3√32
),
不妨设M (x ,0),N (x +1,0),且x ∈[0,5], ∴DM ⃗⃗⃗⃗⃗⃗ =(x -52,-3√32
),
DN
⃗⃗⃗⃗⃗⃗ =(x -32
,-3√32). ∴DM ⃗⃗⃗⃗⃗⃗ ·DN
⃗⃗⃗⃗⃗⃗ =(x -5
2)(x -3
2
)+(-3√32
)2
=x2−4x +
154
+274
=(x -2)2+13
2
,∴当且仅当x =2
时,DM ⃗⃗⃗⃗⃗⃗ ·DN
⃗⃗⃗⃗⃗⃗ 取最小值132
.
10.答案 -1
解析 解法一:∵∠BAD =30°,AD ∥BC , ∴∠ABE =30°,
又EA =EB ,∴∠EAB =30°, 在△EAB 中,AB =2√3,∴EA =EB =2.
以A 为坐标原点,AD 所在直线为x 轴建立如图所示的平面直角坐标系.
则A (0,0),D (5,0),E (1,√3),B (3,√3), ∴BD ⃗⃗⃗⃗⃗⃗ =(2,-√3),AE ⃗⃗⃗⃗⃗ =(1,√3), ∴BD ⃗⃗⃗⃗⃗⃗ ·AE ⃗⃗⃗⃗⃗ =(2,-√3)·(1,√3)=-1. 解法二:同解法一,得EB =EA =2,
则BD ⃗⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ ,AE ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +BE ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ −2
5
AD ⃗⃗⃗⃗⃗ ,
∴BD ⃗⃗⃗⃗⃗⃗ ·AE ⃗⃗⃗⃗⃗ =(AD ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ -25
AD ⃗⃗⃗⃗⃗ )
=AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ 2+25
AB ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ −2
5
AD ⃗⃗⃗⃗⃗ 2
=75AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ 2−25
AD ⃗⃗⃗⃗⃗ 2 =75
×5×2√3×√3
2
−12−2
5
×25=-1.
三年模拟练
1.ACD 对于A 选项,若a 与b 的夹角为钝角,则a ·b <0,且a 与b 不共线, 所以{
a ·
b =k -2<0,
-k ≠2,
解得k <2且k ≠-2,
所以A 中叙述错误;
对于B 选项,|a |=√k 2+4≥√4=2,当且仅当k =0时,等号成立,所以B 中叙述正确; 对于C 选项,|b |=√2,与b 共线的单位向量为±b
|b |,即(√22,-√22)或(-√22
,√2
2
),所以C 中叙
述错误;
对于D 选项,因为|a |=2|b |=2√2,所以√k 2+4=2√2,解得k =±2,所以D 中叙述错误.故选ACD .
2.A 如图所示,建立平面直角坐标系.
不妨设AB =1,BE =x ,则AE =BF =2x ,∴x 2+4x 2
=1,∴x =
√5
5
.
设∠BAE =θ,则sin θ=√5
5
,cos θ=
2
√5
5
,
∴x E =
2
√5
5
cos θ=45
,y E =
2
√5
5sin θ=25
.
设AE ⃗⃗⃗⃗⃗ =mAB ⃗⃗⃗⃗⃗ +nAD ⃗⃗⃗⃗⃗ ,则(45
,25
)=m (1,0)+n (0,1),∴m =45
,n =25
, ∴AE
⃗⃗⃗⃗⃗ =45
a +25
b .故选A . 3.B 以O 为坐标原点建立如图所示的平面直角坐标系,则A (-1,-1),C (1,1),P ,Q 在以O 为圆心的单位圆上.
设P (cos α,sin α),Q (cos β,sin β),
∴AP ⃗⃗⃗⃗⃗ =(cos α+1,sin α+1),CQ
⃗⃗⃗⃗⃗ =(cos β-1,sin β-1), ∴AP ⃗⃗⃗⃗⃗ ·CQ
⃗⃗⃗⃗⃗ =(cos α+1)·(cos β-1)+(sin α+1)·(sin β-1) =cos αcos β+cos β-cos α-1+sin αsin β+sin β-sin α-1
=(cos αcos β+sin αsin β)+(sin β+cos β)-(sin α+cos α)-2
=cos (α-β)+√2sin (β+π4)−√2sin α+π
4
-2. 当cos (α-β)=-1,sin (β+π4)=-1,且sin (α+π4)=1时,AP ⃗⃗⃗⃗⃗ ·CQ ⃗⃗⃗⃗⃗ 有最小值,最小值为-3-2√2,故选B .
4.答案 32;-2
3 解析 ∵向量OA ⃗⃗⃗⃗⃗ =(k ,12),OB ⃗⃗⃗⃗⃗ =(4,5),OC
⃗⃗⃗⃗⃗ =(-k ,10), ∴AB ⃗⃗⃗⃗⃗ =OB ⃗⃗⃗⃗⃗ −OA ⃗⃗⃗⃗⃗ =(4-k ,-7),BC ⃗⃗⃗⃗⃗ =OC
⃗⃗⃗⃗⃗ −OB ⃗⃗⃗⃗⃗ =(-k -4,5). 若|AB ⃗⃗⃗⃗⃗ |=|BC
⃗⃗⃗⃗⃗ |, 则√(4-k )2+49=√(-k -4)2+25,
解得k =3
2. 若A ,B ,C 三点共线,则向量AB ⃗⃗⃗⃗⃗ ,BC
⃗⃗⃗⃗⃗ 共线, ∴5(4-k )=-7(-k -4),解得k =-2
3. 5.答案 4
解析 设a ,b 的夹角为α,α∈[0,π],
因为a 2=|a |2=1,b 2=|b |2=4,
所以|2a -b |=√(2a -b )2
=√4a 2-4a ·b +b 2
=√4-4×1×2×cosα+4=√8-8cosα,
因为α∈[0,π],所以-1≤cos α≤1,
所以0≤8-8cos α≤16,
所以0≤√8-8cosα≤4,
所以√8-8cosα的最大值为4,
即|2a -b |的最大值为4.
6.答案 -53;(-53
,0)∪(0,+∞) 解析 a +λb =(1+λ,2+λ),
若a 与a +λb 的夹角为直角,
则a ·(a +λb )=λ+1+2(λ+2)=0,解得λ=-53. 若a 与a +λb 的夹角为锐角,
则a ·(a +λb )>0,且a 与a +λb 不共线,
∴{λ+1+2(λ+2)>0,2+λ-2(1+λ)≠0,
解得λ>−53,且λ≠0, ∴λ的取值范围是(-53,0)∪(0,+∞). 7.解析 (1)因为a =(1,0),b =(-2,1),
所以2a -b =(4,-1).
(2)cos θ=a ·b
|a ||b |=1×√5=−2√55.
(3)k a +b =(k -2,1),a +k b =(1-2k ,k ),
由题意可得,k (k -2)+2k -1=0,
整理可得,k 2-1=0,解得k =±1.
8.解析 (1)当A ,B ,C 三点共线时,A ,B ,C 三点不能构成三角形,
即BC ⃗⃗⃗⃗⃗ ∥AC ⃗⃗⃗⃗⃗ ,即4(2-k )=6,解得k =12
. (2)AB ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ −BC
⃗⃗⃗⃗⃗ =(k ,1). ①若C 为直角,则BC ⃗⃗⃗⃗⃗ ⊥AC ⃗⃗⃗⃗⃗ ,∴BC ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =2(2-k )+3×4=0,解得k =8.
②若B 为直角,则BC ⃗⃗⃗⃗⃗ ⊥AB ⃗⃗⃗⃗⃗ ,∴BC ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =k (2-k )+3×1=0,解得k =-1或k =3.
③若A 为直角,则AC ⃗⃗⃗⃗⃗ ⊥AB ⃗⃗⃗⃗⃗ ,∴AC ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =2k +4=0,解得k =-2.
经检验,均满足题意.
综上,k 的值为-1,-2,3,8.
易错警示
解决本题的关键是要判断△ABC 中哪个内角为直角,故应进行分类讨论.
9. 解析 (1)由题意得,a ·b =-cos
3x 2·cos x 2+sin 3x 2sin x 2=-cos 2x , |a +b |=√(cos 3x 2-cos x 2
)2+(sin 3x 2+sin x 2)2 =√2-2(cos 3x 2cos x 2-sin 3x 2sin x 2) =√2-2cos2x =2|sin x |,
∵x ∈[π,3π2],∴-1≤sin x ≤0, ∴|a +b |=-2sin x.
(2)由(1)得, f (x )=a ·b +|a +b |=-cos 2x -2sin x =2sin 2x -2sin x -1=2(sinx -12)2−32.∵-1≤sin x ≤0,
∴当sin x =0,即x =π时, f (x )取得最小值,且f (x )min =-1.
10.解析 解法一:(1)因为AD ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ +23CB ⃗⃗⃗⃗⃗ =AC ⃗⃗⃗⃗⃗ +23(AB ⃗⃗⃗⃗⃗ −AC ⃗⃗⃗⃗⃗ )=23AB ⃗⃗⃗⃗⃗ +13
AC ⃗⃗⃗⃗⃗ =23
a +13
b =x a +y b , 所以x =23,y =13,所以x -y =13.
(2)由题意得∠CAB =π3
,BC =√3 , 设|AF
⃗⃗⃗⃗⃗ |=t ,t ∈[0,2], 则CF ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ =(CA ⃗⃗⃗⃗⃗ +AF ⃗⃗⃗⃗⃗ )·FA ⃗⃗⃗⃗⃗ =CA ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ +AF ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ =1·t ·cos π3+t2·cos π=−t2+12t =−(t -14)2+116
,t ∈[0,2], 所以CF ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ 的取值范围是[-3,116]. (3)因为F 为线段AB 的中点,
所以CF ⃗⃗⃗⃗⃗ =12CA ⃗⃗⃗⃗⃗ +12
CB ⃗⃗⃗⃗⃗ , 设CM ⃗⃗⃗⃗⃗⃗ =λCF ⃗⃗⃗⃗⃗ ,则CM ⃗⃗⃗⃗⃗⃗ =λ2CA ⃗⃗⃗⃗⃗ +λ2CB ⃗⃗⃗⃗⃗ , 所以AM ⃗⃗⃗⃗⃗⃗ =CM ⃗⃗⃗⃗⃗⃗ −CA ⃗⃗⃗⃗⃗ =(λ2-1)CA ⃗⃗⃗⃗⃗ +λ2
CB ⃗⃗⃗⃗⃗ , 又AD ⃗⃗⃗⃗⃗ =CD ⃗⃗⃗⃗⃗ −CA ⃗⃗⃗⃗⃗ =23
CB ⃗⃗⃗⃗⃗ −CA ⃗⃗⃗⃗⃗ ,且A ,M ,D 三点共线, 所以存在μ∈R,使得AM
⃗⃗⃗⃗⃗⃗ =μAD ⃗⃗⃗⃗⃗ , 即(λ2-1)CA ⃗⃗⃗⃗⃗ +λ2CB ⃗⃗⃗⃗⃗ =μ(23
CB ⃗⃗⃗⃗⃗ -CA ⃗⃗⃗⃗⃗ ), ∴{λ2
-1=-μ,λ2=23
μ,∴λ=45, ∴CM ⃗⃗⃗⃗⃗⃗ =25CA ⃗⃗⃗⃗⃗ +25
CB ⃗⃗⃗⃗⃗ , ∴CM ⃗⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =(25CA ⃗⃗⃗⃗⃗ +25
CB ⃗⃗⃗⃗⃗ )·(CB ⃗⃗⃗⃗⃗ −CA ⃗⃗⃗⃗⃗ ) =25CB ⃗⃗⃗⃗⃗ 2−25CA ⃗⃗⃗⃗⃗ 2=45
. 解法二:(1)以点C 为坐标原点,CB
⃗⃗⃗⃗⃗ ,CA ⃗⃗⃗⃗⃗ 的方向分别为x 轴,y 轴的正方向建立平面直角坐标系,则C (0,0),A (0,1),B (√3,0),D (
2√33,0), 所以AD ⃗⃗⃗⃗⃗ =(2√33,-1),AB ⃗⃗⃗⃗⃗ =a =(√3,-1),AC
⃗⃗⃗⃗⃗ =b =(0,-1), AD ⃗⃗⃗⃗⃗ =x a +y b =x (√3,-1)+y (0,-1)=(√3x ,-x -y ),
所以{2√3 3=√3 x ,-1=-x -y ,解得{x =23,y =13
, 所以x -y =13. (2)由题意可设F (m ,-
√3 3m +1),m ∈[0,√3], 所以FA
⃗⃗⃗⃗⃗ =(-m ,√3 3m), CF ⃗⃗⃗⃗⃗ =(m ,-√3
3m +1),
所以CF ⃗⃗⃗⃗⃗ ·FA
⃗⃗⃗⃗⃗ =−43m2+√3 3m =-43(m -√38)2+116
,m ∈[0,√3
], 所以CF ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ 的取值范围是[-3,116]. (3)因为F 为线段AB 的中点,
所以F (√3 2,12
), 过点F 作FG ⊥CD ,交CD 于点G , 过点M 作MH ⊥CD ,交CD 于点H , 设M (p ,q ),
易得△CMH ∽△CFG ,△MHD ∽△ACD , CH =p ,MH =q ,CG =√32,FG =12,CD =2√33,AC =1, 则CH CG
=MH FG ,MH AC =HD CD , 即√32=q 12,q 1=2√33-p 2√33
, 即p =√3q ,2q =2-√3p ,
解得p =
2√35,q =25, 即M (2√35,25),则CM ⃗⃗⃗⃗⃗⃗ =(2√35,25
), 又AB ⃗⃗⃗⃗⃗ =(√3
,-1),所以CM ⃗⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ =45.。