平面向量经典例题:
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 .-13
C .-1
D .-23
[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 +3
3=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 .-
611
B .-116
C.611
D.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=611
.
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 |=32
,a 与b 的夹角为60°,则|b |=( )
A.1
2 B.1
3 C.1
4 D.15
[答案] A [解析] ∵|a -b |=
3
2,∴|a |2+|b |2-2a ·b =3
4,∵|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.
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=-2λ-μ=4
,∴??
?
λ=1μ=-3
,∴c =a -3b ,故选B.
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 +12
b B.23a +1
3
b
C.12a +14b
D.13a +2
3
b [答案] 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 +13b .
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 ≥232x +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+52
D.1+52
[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,-2
3), 设P (x,0),-1≤x ≤1,则AP →
=(x ,-3),
∴AP →·(AB →+AC →)=(x ,-
3)·(0,-2
3)=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.32
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 →
|≥22
.
10. 如图,一直线EF 与平行四边形ABCD 的两边AB ,AD
分别交于E 、F 两点,且交其对角线于K ,其中AE →=13
AB →,
AF →
=12
AD →
,AK →=λAC →
,则λ的值为( )
A.15
B.14
C.
13
D.12
[答案] A
[解析] 如图,取CD 的三等分点M 、N ,BC 的中点Q ,则EF ∥DG ∥BM ∥NQ ,易知AK →=15AC →
,∴λ=15
.
11. 已知向量a =(2,3),b =(-1,2),若m a +4b 与a -2b 共线,则m 的值为( )
A.1
2 B .2 C .-2 D .-1
2
[答案] C
[解析] m a +4b =(2m -4,3m +8),a -2b =(4,-1), 由条件知(2m -4)·(-1)-(3m +8)×4=0,∴m =-2,故选C.
12. 在△ABC 中,C =90°,且CA =CB =3,点M 满足BM →=2MA →,则CM →·CB →等于( )
A .2
B .3
C .4
D .6 [答案] B
[解析] CM →·CB →=(CA →+AM →)·CB →
=(CA →+13AB →)·CB →=CA →·CB →+13AB →·CB →
=13|AB →|·|CB →|·cos45°=1
3
×32×3×
22
=3.
13. 在正三角形ABC 中,D 是BC 上的点,AB =3,BD =1,则AB →·AD →
=________.
[答案]
152
[解析] 由条件知,|AB →|=|AC →|=|BC →|=3,〈AB →,AC →
〉=60°, 〈AB →,CB →〉=60°,CD →=23
CB →,
∴AB →·AD →=AB →·(AC →+CD →)=AB →·AC →+AB →·23CB →
=3×3×cos60°+23×3×3×cos60°=152.
14. 已知向量a =(3,4),b =(-2,1),则a 在b 方向上的投影等于________.
[答案] -
25
5。[解析] a 在b 方向上的投影为
a ·
b |b |=
-2
5
=-
25
5
.
15. 已知向量a 与b 的夹角为2π
3
,且|a |=1,|b |=4,若(2a +λb )⊥a ,则实数λ=________.
[答案] 1
[解析] ∵〈a ,b 〉=
2π3
,|a |=1,|b |=4,∴a ·b =|a |·|b |·cos 〈a ,b 〉=1×4×cos
2π3
=-2,∵(2a
+λb )⊥a ,∴a ·(2a +λb )=2|a |2+λa ·b =2-2λ=0,∴λ=1. 16. 已知:|OA →|=1,|OB →
|=
3,OA →·OB →=0,点C 在∠AOB 内,且∠AOC =30°,设OC →=mOA →+nOB
→(m ,n ∈R +),则m n
=________.
[答案] 3
[解析] 设mOA →=OF →,nOB →=OE →,则OC →=OF →+OE →
,
∵∠AOC =30°,∴|OC →|·cos30°=|OF →|=m |OA →
|=m , |OC →|·sin30°=|OE →|=n |OB →|=
3n ,
两式相除得:
m
3n =|OC →
|cos30°|OC →|sin30°=1tan30°
=3,∴m n
=3.
17. (文)设i 、j 是平面直角坐标系(坐标原点为O )内分别与x 轴、y 轴正方向相同的两个单位向量,且OA
→
=-2i +j ,OB →
=4i +3j ,则△OAB 的面积等于________. [答案] 5
[解析] 由条件知,i 2=1,j 2=1,i ·j =0,∴OA →·OB →=(-2i +j )·(4i +3j )=-8+3=-5,又OA →·OB →
=|OA →|·|OB →|·cos 〈OA →,OB →〉=5
5cos 〈OA →,OB →
〉, ∴cos 〈OA →,OB →
〉=-
55
,∴sin 〈OA →,OB →
〉=
255, ∴S △OAB =12|OA →|·|OB →|·sin 〈OA →,OB →
〉=12
×
5×5×
25
5=5.
(理)三角形ABC 中,a ,b ,c 分别是角A ,B ,C 所对的边,能得出三角形ABC 一定是锐角三角形的条件是________(只写序号)
①sin A +cos A =15 ②AB →·BC →
<0 ③b =3,c =3
3,B =30° ④tan A +tan B +tan C >0.
[答案] ④
[解析] 若A 为锐角,则sin A +cos A >1,∵sin A +cos A =15
,∴A 为钝角,∵AB →·BC →<0,∴BA →·BC →
>0,
∴∠B 为锐角,由∠B 为锐角得不出△ABC 为锐角三角形;由正弦定理
b sin B =
c
sin C 得,3sin30°=33
sin C
,∴sin C
=32
,∴C =60°或120°,∵c ·sin B =33
2,3<33
2<33,∴△ABC 有两解,故①②③都不能得出△ABC
为锐角三角形.
④由tan A +tan B +tan C =tan(A +B )(1-tan A tan B )+tan C =-tan C (1-tan A tan B )+tan C =tan A tan B tan C >0,及A 、B 、C ∈(0,π),A +B +C =π知A 、B 、C 均为锐角,
∴△ABC 为锐角三角形.
18. 已知平面向量a =(1,x ),b =(2x +3,-x ).
(1)若a ⊥b ,求x 的值. (2)若a ∥b ,求|a -b |. [解析] (1)若a ⊥b ,
则a ·b =(1,x )·(2x +3,-x )=1×(2x +3)+x (-x )=0, 整理得x 2-2x -3=0,解得x =-1或x =3.
(2)若a ∥b ,则有1×(-x )-x (2x +3)=0,则x (2x +4)=0,解得x =0或x =-2, 当x =0时,a =(1,0),b =(3,0), ∴|a -b |=|(1,0)-(3,0)|=|(-2,0)|=
(-2)2+02=2,
当x =-2时,a =(1,-2),b =(-1,2),