高一数学同步学
名校期末考题每日一练(124)
1.(2008·安徽)若AB
→
=(2,4),AC
→
=(1,3),则BC
→
=()
A.(1,1)B.(-1,-1)
C.(3,7)D.(-3,-7)
2.已知MA
→
=(-2,4),MB
→
=(2,6),则12AB
→
=()
A.(0,5)B.(0,1)
C.(2,5)D.(2,1)
3.已知平面向量a=(x,1),b=(-x,x2),则向量a+b()
A.平行于y轴B.平行于第一、第三象限的角平分线
C.平行于x轴D.平行于第三、第四象限的角平分线
4.若M(4,-1),AB
→
=(4,-1),则有()
A.点M与点A重合B.点M与点B重合
C.点M在AB
→
上D.OM
→
=AB
→
(O为坐标原点)
5.若向量a=(3,2),b=(0,-1),则向量2b-a的坐标是()
A.(3,-4)B.(-3,4)
C.(3,4)D.(-3,-4)
6.已知M(3,-2),N(-5,-1),MP
→
=12MN
→
,则P点坐标为________.
7.平面上三点分别为A(2,-5),B(3,4),C(-1,-3),D为线段
BC中点,则向量DA
→
的坐标为________.
8.已知O为坐标原点,点A在第一象限,|OA
→
|=43,∠xOA=60°,
则向量OA
→
=________.
9.已知2a+b=(-4,3),a-2b=(2,4),求a,b.
10.已知A,B,C三点坐标分别为(-1,0),(3,-1),(1,2),AE
→
=13AC
→
,
BF
→
=13BC
→
.
(1)求点E,F及向量EF
→
的坐标;(2)求证:EF
→
∥AB
→
.
1.解析BC
→
=AC
→
-AB
→
=(1,3)-(2,4)=(-1,-1).
答案B
2.解析12AB
→
=12(MB
→
-MA
→
)=12(4,2)=(2,1).
答案D
3.解析a+b=(x,1)+(-x,x2)=(0,1+x2),
由1+x2≠0及向量的性质可知,a+b平行于y轴.
答案A
4.解析M(4,-1),即OM
→
=(4,-1),又AB
→
=(4,-1),∴OM
→
=AB
→
.
答案D
5.解析a=(3,2),b=(0,-1),∴2b-a=2(0,-1)-(3,2)=(0,-
2)-(3,2)=(-3,-4).
答案D
6.解析OM
→
=(3,-2),ON
→
=(-5,-1),
∴12MN
→
=12(ON
→
-OM
→
)=12(-8,1)=??????-4,12.
设P(x,y),则
MP
→
=OP
→
-OM
→
=(x-3,y+2),由MP
→
=12MN
→
,得
?
??
x-3=-4,
y+2=12,
∴x=-1,y=-32,∴P(-1,-32).
答案(-1,-32)
7.解析依题意知OD
→
=12(OB
→
+OC
→
)=12(2,1)=??????1,12,则DA
→
=OA
→
-
OD
→
=(2,-5)-(1,12)=??????1,-112.
答案??????1,-112
8.解析设OA
→
=(x,y),则x=43cos60°=23,
y=43sin60°=6,∴OA
→
=(23,6).
答案(23,6)
9.解∵2a+b=(-4,3),
∴4a+2b=(-8,6).
又a-2b=(2,4),
∴(4a+2b)+(a-2b)=(-8,6)+(2,4).
∴5a=(-6,10).
∴a=??????-65,2.
又b=(2a+b)-2a=(-4,3)-2??????-65,2=??????-85,-1,
∴a=??????-65,2,b=??????-85,-1.
10.解(1)设O(0,0),则OE
→
=OA
→
+AE
→
=OA
→
+13AC
→
=(-1,0)+13(2,2)=??????-13,23,
OF
→
=OB
→
+BF
→
=OB
→
+13BC
→
=(3,-1)+13(-2,3)=??????73,0,
∴E??????-13,23,F??????73,0.
∴EF
→
=OF
→
-OE
→
=??????83,-23.
(2)证明:∵AB
→
=OB
→
-OA
→
=(4,-1),
EF
→
=??????83,-23
∴AB
→
=32??????83,-23=32EF
→
.
∴EF
→
∥AB
→
.
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