平面向量:
1.已知向量a=(1,2),b=(2,0),若向量λa+b与向量c=(1,-
2)共线,则实数λ等于( )
A.-2 B.-1 3
C.-1 D.-2 3
[答案] C
[解析] λa+b=(λ,2λ)+(2,0)=(2+λ,2λ),
∵λa+b与c共线,
∴-2(2+λ)-2λ=0,∴λ=-1.
2.(文)已知向量a=(3,1),b=(0,1),c=(k,3),若a+2b
与c垂直,则k=( )
A.-1 B.- 3
C.-3 D.1
[答案] C
[解析] a+2b=(3,1)+(0,2)=(3,3),
∵a+2b与c垂直,∴(a+2b)·c=3k+33=0,
∴k=-3.
(理)已知a=(1,2),b=(3,-1),且a+b与a-λb互相垂直,则实数λ的值为( )
A.-6
11
B.-
11
6
C.6
11D.
11
6
[答案] C
[解析] a +b =(4,1),a -λb =(1-3λ,2+λ), ∵a +b 与a -λb 垂直,
∴(a +b )·(a -λb )=4(1-3λ)+1×(2+λ)=6-11λ=0,∴λ=6
11
.
3. 设非零向量a 、b 、c 满足|a |=|b |=|c |,a +b =c ,则向量a 、b 间的夹角为( ) A .150° B .120° C .60° D .30°
[答案] B
[解析] 如图,在?ABCD 中,
∵|a |=|b |=|c |,c =a +b ,∴△ABD 为正三角形, ∴∠BAD =60°,∴〈a ,b 〉=120°,故选B.
(理)向量a ,b 满足|a |=1,|a -b |=3
2,a 与b 的夹角为60°,
则|b |=( )
A.12
B.13
C.14
D.15
[解析] ∵|a -b |=
32,∴|a |2+|b |2-2a ·b =34
, ∵|a |=1,〈a ,b 〉=60°,
设|b |=x ,则1+x 2
-x =34,∵x >0,∴x =1
2
.
4. 若AB →·BC →+AB →2=0,则△ABC 必定是( )
A .锐角三角形
B .直角三角形
C .钝角三角形
D .等腰直角三角形
[答案] B
[解析] AB →·BC →+AB →2=AB →·(BC →+AB →)=AB →·AC →=0,∴AB →⊥AC →, ∴AB ⊥AC ,∴△ABC 为直角三角形.
5. (文)若向量a =(1,1),b =(1,-1),c =(-2,4),则用a ,b 表
示c 为( ) A .-a +3b B .a -3b C .3a -b D .-3a +b
[答案] B
[解析] 设c =λa +μb ,则(-2,4)=(λ+μ,λ-μ),
∴?????
λ+μ=-2
λ-μ=4
,∴???
??
λ=1
μ=-3
,∴c =a -3b ,故选B.
(理)在平行四边形ABCD 中,AC 与BD 交于O ,E 是线段OD 的中点,AE 的延长线与CD 交于点F ,若AC →=a ,BD →=b ,则AF
→等于( ) A.14a +1
2b B.23a +13b C.12a +14
b D.13a +23
b
[解析] ∵E 为OD 的中点,∴BE →=3ED →, ∵DF ∥AB ,∴|AB ||DF |=|EB ||DE |
,
∴|DF |=13|AB |,∴|CF |=23|AB |=2
3|CD |,
∴AF →=AC →+CF →=AC →+23CD →=a +23(OD →-OC →)
=a +23(12b -12a )=23a +1
3
b .
6. 若△ABC 的三边长分别为AB =7,BC =5,CA =6,则AB →·BC →的值为
( ) A .19 B .14 C .-18 D .-19
[答案] D
[解析] 据已知得cos B =72+52-622×7×5=1935
,故AB →·BC →=|AB →|×|BC
→|×(-cos B )=7×5×? ??
??
-1935=-19.
7. 若向量a =(x -1,2),b =(4,y )相互垂直,则9x +3y 的最小值为
( ) A .12
B .2 3
C .3 2
D .6
[答案] D
[解析] a ·b =4(x -1)+2y =0,∴2x +y =2,∴9x +3y =32x +3y
≥23
2x +y
=6,等号在x =1
2
,y =1时成立.
8. 若A ,B ,C 是直线l 上不同的三个点,若O 不在l 上,存在实数
x 使得x 2OA →+xOB →+BC →=0,实数x 为( ) A .-1 B .0 C.-1+5
2
D.1+5
2
[答案] A
[解析] x 2OA →+xOB →+OC →-OB →=0,∴x 2OA →+(x -1)OB →+OC →=0,由向量共线的充要条件及A 、B 、C 共线知,1-x -x 2=1,∴x =0或-1,当x =0时,BC
→=0,与条件矛盾,∴x =-1. 9. (文)已知P 是边长为2的正△ABC 边BC 上的动点,则AP →·(AB →+
AC →)( ) A .最大值为8 B .最小值为2 C .是定值6 D .与P 的位置有关
[答案] C
[解析] 以BC 的中点O 为原点,直线BC 为x 轴建立如图坐标系,则B (-1,0),C (1,0),A (0,3),AB →+AC →=(-1,-3)+(1,-3)=(0,-23),
设P (x,0),-1≤x ≤1,则AP →=(x ,-3),
∴AP →·(AB →+AC →)=(x ,-3)·(0,-23)=6,故选C.
(理)在△ABC 中,D 为BC 边中点,若∠A =120°,AB →·AC →=-1,则|AD →|的最小值是( )
A.12
B.3
2 C. 2 D.22
[答案] D
[解析] ∵∠A =120°,AB →·AC →=-1, ∴|AB →|·|AC →|·cos120°=-1, ∴|AB →|·|AC →|=2,
∴|AB →|2+|AC →|2≥2|AB →|·|AC →|=4,
∵D 为BC 边的中点,∴AD →=12(AB →+AC →),∴|AD →|2=14(|AB →|2+|AC →|2
+2AB →·AC →)=14(|AB →|2+|AC →|2-2)≥14(4-2)=12
,
∴|AD →|≥2
2
.