平面向量典型例题
- 格式:doc
- 大小:263.50 KB
- 文档页数:14
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 、