高中数学必修4同步演习(2.3-2.4平面向量的基础定理及坐标
现及数量积ab卷谜底[宝典]
,,2,,,,220参考
A 1-7 aabbaa,,,,,,,cos60672,2240,
,,,一.选择
(每题5分) (4)(2)0,4aaa,,,,1.A 25.,,0,2 2.C 3.B
2.3-2.4平面向量基本定理、坐标表4.D
5.B 示、数量积检测题参考答案B 1-7 6.C
.D 7一.选择题(每题5分)
8.B 1.B 9.B 2.A 10.D 3.B 二.填空题(每题5分) 4.D 11.(8,-6);10;(1,1) 5.C
7,,6.C 12. 2,,,7.B 2,,
8.A 13.矩
14.4 9.D解:
315. 410.B 16.; 62二.填空题(每题5分) 3m,6且m,,17. 31311. 2
12.,,AB,3,53,18.; 13. y,,sinx,22
19.2; ,,,,525525,,,,14.或 ,,,,20.2 ,,2,4,,,,5555,,,,三.解答题(每题10分) ,,,,15.; (,4,2)21.解:设BxyABxy(,),(,)(3,2)(8,0).,,,, 1016. x,3,8x,5,,?, ,,10y,2,0y,2,,17.???? ?B(5,2),x,1,y,2,C(1,2) CC1018.2; ,,,,
22.解:,kabkk,,,,(3,22)ab,,,3(10,4)19.-25 ,,,,o20.45 kab,,ab,3(1),则
三.解答题(每题10分) k,19kk,,,,,,,31022(4)0,得,,,, ,,,,OACBOB,b21.证明:设在中,,对角线OC,m,kab,//ab,3kk,,,,,,,3(4)22100(2),则,,,,,m,a,bn,a,bBA,n,则,, 1222222?,即,,,,m,n,a,b,a,b,2,,a,bk,,得 32222,,m,n,2a,b,,. ,,1,,abab,3此时与反向 2x22.解:
数与轴的交点为A(,1,0)和y,x,5x,43,,B(,4,0)故依题意有两种情形: ab23.解:(I)?//, ,,,,,,2x(1)将函数的图象沿轴平移,使y,x,5x,4a,b2abab?若,共向,则,||•||, ,,,,,,A(,1,0)与原点重合,这时 a,b2abab?若,异向,则,,||•||,, ,,,,,,;AO,(m,n),(0,(,1),0,0),(1,0)a,babab(II)?,的夹角为135?,?,||•||•cos1 235?,,1 P(x,y)设是图象上任一点,平移后对y,x,5x,4,,,,,,,,2222abababab ?|,|,(,) ,,,2,1,2,2,,,P(x,y)应点的坐标为,则 ,,
,1,? ||1ab,,,,,,,,,,2224.解:(2)(3)672ababaabb,,,,,,,
35,,,x,x,1,2,,,,因此,-1,于是cos,可知0,,,,代入中得y,x,5x,4,4,y,y ,,,3,53,512,,,,,即 cos,,得 ,解得,1,,1,,,y,(x,1),5(x,1),44,,3422,,,.即 y,x,3xy,x,3x
2(2):依(1)可求使函数的图象平移,使y,x,5x,4
2点与原点重合时的解析式为B(,4,0)y,x,3x
3223.解:(?)=3 sinωcosω,cosω=sixxxfx()2
π11n2ωx, (1+cos2ωx) =sin(2ωx+ )+ 262
2π?ω>0,?T=π= ,?ω=1( 2ω
π1(?)由(1),得=sin(2x+ )+ , fx()62
πππ5π?0,x? ,? ,2x+ ? ( 3666
3??[1, ]( fx()2
,,
3cos,,1,(sin,,m)24.解:(1)??,?=ab
0(
,m,sin,3cos,2sin(,),,,?,3
,=–2.又又??R,?时,m,,,,[0,2,)sin(,),,1min,3
11 ,,,?6
m,03sin,,cos,,0(2)?,且,?a,b
3,tan,, ,3
,,,,cos(,),sin(,2),,sin,(,sin2)2, ,cos,cos(,,,)
1,2tan,,,tan, 221,tan,
CA,,BC25.解:由已知,可得
,3sin,(2cos,+1,)=(-1-2cos,-),,3sin,
,,,,1,,2cos,,,,,2cos,1,,,2cos,,cos,,,,2,,3sin,3,sin,,,,,sin,,,,sin,, 22sin,,cos,由=1,得
2,1,,2cos,,,,,2,,,sin,,1,,,4 21,,,,21cos1,,,,即,,,,,,,,,4
,CA,,BCAB,0若=-1,则,得,这与A,B两点不
重合矛盾,