2019-2020学年高中数学第二章平面向量2.7向量应用举例课后导练北师大版必修4
基础达标
1.已知A(1,2),B(3,4),则AB中点的坐标是( )
A.(2,3) B.(-2,-3)
C.(word/media/image2_1.png,word/media/image3_1.png) D.(3,2)
解析:设AB中点为C(x,y),
则x=word/media/image4_1.png=2,y=word/media/image5_1.png=3,
∴C(2,3).
答案:A
2.某人用50 N的力(与水平方向成30°角,斜向下)推动一质量为8 kg的木箱沿水平平面运动了20 m,若动摩擦因数μ=0.02,g取10 m/s2,则摩擦力f所做的功为( )
A.42 J B.-42 J C.22 J D.-22 J
解析:由数量积的物理意义,只需求出摩擦力f的大小,及它与位移的夹角即可.
|f|=(80+50×sin30°)×0.02 N=2.1 N,又f与位移所成的角为180°,
∴f·s=|f||s|cos180°=2.1×20×(-1) J=-42 J.
答案:B
3.三点A(x1,y1),B(x2,y2),C(x3,y3)共线,则有…( )
A.x1y2-x2y1=0 B.x1y3-x3y1=0
C.(x2-x1)(y3-y1)=(x3-x1)(y2-y1) D.(x2-x1)(x3-x1)=(y2-y1)(y3-y1)
解析:word/media/image6_1.png=(x2-x1,y2-y1),word/media/image7_1.png=(x3-x1,y3-y1),
∵AB∥AC,
∴(x2-x1)(y3-y1)-(x3-x1)(y2-y1)=0.
答案:C
4.已知a=(1,2),a⊥b,则b可以是( )
A.(-4,2) B.(2,-4)
C.(2,1) D.(-2,-1)
解析:把选项通过x1x2+y1y2=0检验知b可以是(-4,2).
答案:A
5.某人向正东走x km后,又向右转150°,然后朝新方向走3 km.结果他离出发点恰好word/media/image8_1.png km,那么x的值等于( )
A.3 B.word/media/image9_1.png C.word/media/image10_1.png或word/media/image9_1.png D.3
解析:设向量a为“向东走x m”,则|a|=x,设向量b为“朝新方向走word/media/image8_1.png km”,则|b|=3,且a与b的夹角为150°,离出发点为word/media/image8_1.png km,即|a+b|=word/media/image8_1.png.
由分析知|a+b|=word/media/image8_1.png,∴a2+2a·b+b2=word/media/image8_1.png.
∴x2+6x·cos150°+9-3=0,
即x2-word/media/image11_1.pngx+6=0.
解得x=word/media/image8_1.png或word/media/image9_1.png.
答案:C
6.已知向量word/media/image12_1.png=(k,12),word/media/image13_1.png=(4,5),word/media/image14_1.png=(-k,10),且A、B、C三点共线,则k=________.
解析:
word/media/image15_1.png=(k,12),word/media/image13_1.png=(4,5),word/media/image16_1.png
=(-k,10).
∵A、B、C三点共线,
∴word/media/image17_1.png∥word/media/image18_1.png.
∵word/media/image17_1.png=(k-4,12-5),word/media/image18_1.png=(4+k,5-10),
∴(k-4)·(5-10)-(12-5)(4+k)=0,
解之得k=word/media/image19_1.png.
答案:word/media/image19_1.png
7.以原点和点A(4,2)为顶点作等腰直角三角形OAB,∠B=90°,则向量word/media/image20_1.png的坐标为________.
解析:利用长度公式和垂直条件列出关于向量坐标的方程,然后求解.
设word/media/image21_1.png=(x,y),则word/media/image6_1.png=(x-4,y-2).
由已知word/media/image22_1.png
word/media/image23_1.png
故B(1,3)或B(3,-1).
∴word/media/image20_1.png=(-3,1)或(-1,-3).
答案:(-3,1)或(-1,-3)
8.如右图所示,在△ABC中,D、E、F分别是边AB、BC、CA的中点,G是它的重心,已知D点的坐标是(1,2),E点的坐标是(3,5),F点的坐标是(2,7),求A、B、C、G的坐标.
解析:设A(x1,y1),由已知得EF平行且等于AD.
∴word/media/image25_1.png=word/media/image26_1.png.
∴(x1-1,y1-2)=(2-3,7-5)=(-1,2).
∴word/media/image27_1.png
∴A(0,4).同理可得B(2,0),C(4,10).连结AE,则AE过点G.
设G(x2,y2),由word/media/image28_1.png=2word/media/image29_1.png得(x2,y2-4)=2(3-x2,5-y2),
∴word/media/image30_1.png
∴G(2,word/media/image31_1.png).
9.如右图所示,在细绳O处用水平力F2缓慢拉起所受重力为G的物体,绳子与铅垂方向的夹角为θ,绳子所受到的拉力为F1,求:
(1)|F1|、|F2|随角θ的变化而变化的情况;
(2)当|F1|≤2|G|时,θ角的取值范围.
解析:(1)如右图所示,由力的平衡及向量加法的平行四边形法则知:G=F1+F2.
解直角三角形得
|F1|=word/media/image34_1.png,
|F2|=|G|·tanθ,
当θ从0°趋向于90°时,
|F1|、|F2|皆逐渐增大.
(2)令|F1|=word/media/image35_1.png=2|G|,
得cosθ≥word/media/image36_1.png,又0°≤θ<90°,∴0°≤θ≤60°.
10.在四边形ABCD中(A、B、C、D顺时针排列),word/media/image6_1.png=(6,1),word/media/image37_1.png=(-2,-3).若有word/media/image38_1.png∥word/media/image25_1.png,又有word/media/image7_1.png⊥word/media/image39_1.png,求word/media/image40_1.png的坐标.
解析:设word/media/image40_1.png=(x,y),则word/media/image7_1.png=(6+x,1+y),word/media/image41_1.png=(4+x,y-2),word/media/image42_1.png=(-x-4,2-y),word/media/image39_1.png=(x-2,y-3).
又word/media/image40_1.png∥word/media/image42_1.png及word/media/image7_1.png⊥word/media/image39_1.png,
∴x(2-y)-(-x-4)y=0,①
(6+x)(x-2)+(1+y)(y-3)=0,②
解得word/media/image43_1.png
∴word/media/image40_1.png=(-6,3)或(2,-1).
综合运用
11.已知word/media/image44_1.png=λword/media/image45_1.png+μword/media/image46_1.png,若M、P、N三点共线,则λ与μ的关系为( )
A.λ-μ=0 B.λ+μ=0
C.λ-μ=1 D.λ+μ=1
解析:可根据教材中的例题解此题,也可据M、P、N三点共线推导λ与μ的关系.
∵M、P、N三点共线,故存在实数k,使word/media/image47_1.png,
∴word/media/image44_1.png-word/media/image45_1.png=kword/media/image46_1.png-kword/media/image45_1.png,即word/media/image44_1.png=kword/media/image46_1.png+(1-k)word/media/image45_1.png.又word/media/image44_1.png=λword/media/image45_1.png+μword/media/image46_1.png,
∴word/media/image48_1.png∴λ+μ=1.
答案:D
12.若ABCD为正方形,E是CD的中点,且word/media/image49_1.png=a,word/media/image50_1.png=b,则word/media/image51_1.png等于( )
A.b+word/media/image52_1.pnga B.b-word/media/image52_1.pnga
C.a+word/media/image52_1.pngb D.a-word/media/image36_1.pngb
解析:word/media/image51_1.png=word/media/image53_1.png-word/media/image49_1.png=word/media/image50_1.png+word/media/image54_1.png-word/media/image49_1.png=word/media/image50_1.png+word/media/image36_1.pngword/media/image49_1.png-word/media/image49_1.png=b-word/media/image36_1.pnga.
答案:B
13.在水流速度为word/media/image55_1.png km/h的河水中,一艘船以12 km/h的速度垂直对岸行驶,求这艘船实际航行速度的大小_______,方向_______.
解析:如右图,设word/media/image49_1.png表示水流速度,word/media/image57_1.png表示船垂直对岸行驶的速度,以word/media/image49_1.png为一边、word/media/image57_1.png为一对角线作ABCD,则word/media/image50_1.png就是船实际航行的速度.
∵|word/media/image49_1.png|=word/media/image55_1.png,|word/media/image57_1.png|=12,
∴|word/media/image50_1.png|=|word/media/image59_1.png|=word/media/image60_1.png;
tan∠ACB=word/media/image61_1.png,∠CAD=∠ACB=30°,∠BAD=120°.
答案:word/media/image62_1.png km/h 与水流速度方向的夹角为120°
14.已知线段AB的长度为4,点M在线段AB上,若点P(P与AB不共线)满足word/media/image63_1.png=word/media/image52_1.png(word/media/image64_1.png+word/media/image65_1.png)且|word/media/image66_1.png|=2,则word/media/image64_1.png与word/media/image67_1.png的夹角为___________.
解析:∵word/media/image66_1.png=word/media/image52_1.png(word/media/image64_1.png+word/media/image67_1.png),|word/media/image66_1.png|=2,
∴4word/media/image66_1.png2=word/media/image64_1.png2+2word/media/image64_1.png·word/media/image67_1.png+word/media/image67_1.png2.①
又|word/media/image64_1.png-word/media/image67_1.png|=|word/media/image68_1.png|=4,
∴word/media/image64_1.png2-2word/media/image64_1.png·word/media/image67_1.png+word/media/image67_1.png2=16.②
由①②可知,word/media/image64_1.png·word/media/image67_1.png=0,故word/media/image64_1.png与word/media/image67_1.png的夹角为word/media/image69_1.png.
答案:word/media/image70_1.png
15.如右图,已知A、B、C是不共线的三点,O是△ABC内的一点,若word/media/image12_1.png+word/media/image13_1.png+word/media/image14_1.png=0,求证:O是△ABC的重心.
证明:如右图,由于word/media/image15_1.png+word/media/image13_1.png+word/media/image16_1.png=0,
∴word/media/image15_1.png=-(word/media/image13_1.png+word/media/image16_1.png),即word/media/image13_1.png+word/media/image16_1.png是word/media/image15_1.png的相反向量.以word/media/image13_1.png,word/media/image16_1.png为邻边构造平行四边形OBDC,则有word/media/image72_1.png=-word/media/image15_1.png.在平行四边形BOCD中,设BC与OD交于E点,则word/media/image73_1.png=word/media/image74_1.png,word/media/image75_1.png=word/media/image76_1.png,∴AE是△ABC的中线,且|word/media/image15_1.png|=2|word/media/image77_1.png|,故O是△ABC的重心.
拓展探究
16.美国不顾国际社会的强烈反对,于2001年7月14日进行导弹防御系统拦截技术的第四次实验,军方先从加利福尼亚州的危登堡空军基地发射一枚作为标靶的洲际弹道导弹和诱弹,再从马绍尔群岛的夸贾林环礁发射另一枚导弹对前一枚导弹进行拦截,实施拦截时必须准确计算标靶的飞行速度、瞬时位置.现假设标靶与拦截导弹的飞行轨迹均在同一平面内,标靶飞行速度为|v|=10n km/h.令ν=λ1e1+λ2e2,基底e1、e2是平面内的单位向量.若标靶的飞行方向为北偏东30°, e1方向为正东,e2方向为北偏东60°,试求λ1、λ2的值.
解析:建立如右图所示的直角坐标系,则e1=(1,0),e2=(word/media/image80_1.png,word/media/image52_1.png),v=(5n,word/media/image81_1.pngn).
∵e1,e2不共线,
∴v=λ1 e1+λ2 e2=λ1(1,0)+λ2(word/media/image82_1.png,word/media/image52_1.png),
(5n,word/media/image81_1.pngn)=(λ1+word/media/image82_1.pngλ2,word/media/image52_1.pngλ2).
∴word/media/image83_1.png∴λ1=-10n,λ2=word/media/image84_1.pngn.
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