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单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,1.4,平面向量,-,2,-,平面向量的线性运算、平面向量基本定理,高考真题体验,对方向,1,.,(2018,全国,7),在,ABC,中,AD,为,BC,边上的中线,E,为,AD,的中,答案,:,A,-,3,-,2,.,(2014,全国,6),设,D,E,F,分别为,ABC,的三边,BC,CA,AB,的中点,答案,:,A,3,.,(2014,福建,10),设,M,为平行四边形,ABCD,对角线的交点,O,为平行,答案,:,D,-,4,-,新题演练提能,刷高分,1,.,(2018,重庆二诊,),已知两个非零向量,a,b,互相垂直,若向量,m,=,4,a,+,5,b,与,n,=,2,a,+,b,共线,则实数,的值为,(,),A.5B.3C.2,.,5D.2,答案,:,C,解析,:,向量,m,=,4,a,+,5,b,与,n,=,2,a,+,b,共线,存在实数,t,使得,m,=,t,n,即,4,a,+,5,b,=t,(2,a,+,b,),又向量,a,b,互相垂直,故,a,b,不共线,.,-,5,-,2,.,(2018,山西一模,),在平行四边形,ABCD,中,点,E,为,CD,的中点,BE,与,答案,:,C,-,6,-,3,.,(2018,安徽安庆二模,),在,ABC,中,点,D,是边,BC,上任意一点,M,是,答案,:,B,-,7,-,4,.,(2018,福建福州,3,月质量检测,),如图,在平面四边形,ABCD,中,-,8,-,答案,:,-,1,解析,:,过点,D,作,BC,的垂线,交,BC,的延长线于点,M,设,BAC=,则,ACD=,2,ACB=,90,-,DCM=,180,-,2,-,(90,-,),=,90,-,.,Rt,ABC,Rt,DMC,-,9,-,答案,:,1,-,10,-,6,.,(2018,四川,“,联测促改,”),在平面向量中有如下定理,:,设点,O,P,Q,R,为同一平面内的点,则,P,、,Q,、,R,三点共线的充要条件是,:,存在实数,t,使,-,11,-,-,12,-,平面向量的坐标运算,高考真题体验,对方向,ABC=,(,),A.30,B.45,C.60,D.120,答案,:,A,-,13,-,2,.,(2018,全国,13),已知向量,a,=,(1,2),b,=,(2,-,2),c,=,(1,),.,若,c,(2,a,+,b,),则,=,.,解析,:,2,a,+,b,=,2(1,2),+,(2,-,2),=,(4,2),c,=,(1,),由,c,(2,a,+,b,),得,4,-,2,=,0,得,=,.,3,.,(2017,全国,13),已知向量,a,=,(,-,2,3),b,=,(3,m,),且,a,b,则,m=,.,答案,:,2,解析,:,a,b,a,b,=,(,-,2,3)(3,m,),=-,2,3,+,3,m=,0,解得,m=,2,.,-,14,-,4,.,(2017,全国,13),已知向量,a,=,(,-,1,2),b,=,(,m,1),若向量,a,+,b,与,a,垂直,则,m=,.,答案,:,7,解析,:,因为,a,=,(,-,1,2),b,=,(,m,1),所以,a,+,b,=,(,m-,1,3),.,因为,a,+,b,与,a,垂直,所以,(,a,+,b,),a,=,0,即,-,(,m-,1),+,2,3,=,0,解得,m=,7,.,5,.,(2016,全国,13),已知向量,a,=,(,m,4),b,=,(3,-,2),且,a,b,则,m=,.,答案,:,-,6,解析,:,因为,a,b,所以,-,2,m-,4,3,=,0,解得,m=-,6,.,-,15,-,新题演练提能,刷高分,1,.,(2018,东北三省三校二模,),已知向量,a,=,(1,1),b,=,(,-,1,2),若,(,a,-,b,),(2,a,+t,b,),则,t=,(,),答案,:,C,解析,:,因为,a,-,b,=,(2,-,1),2,a,+t,b,=,(2,-t,2,+,2,t,),又因为,(,a,-,b,),(2,a,+t,b,),所以,2(2,+,2,t,),=-,(2,-t,),t=-,2,故选,C,.,2,.,(2018,山东济南一模,),已知向量,a,=,(1,1),b,=,(2,x,),若,a,+,b,与,3,a,-,b,平行,则实数,x,的值是,.,答案,:,2,解析,:,a,=,(1,1),b,=,(2,x,),a,+,b,与,3,a,-,b,平行,a,+,b,=,(3,x+,1),3,a,-,b,=,(1,3,-x,),所以,3(3,-x,),-,(,x+,1),=,0,解得,x=,2,.,-,16,-,3,.,(2018,江西上饶二模,),已知向量,a,=,(2,-,1),b,=,(6,x,),且,a,b,则,|,a,-,b,|=,.,4,.,(2018,贵州凯里二模,),已知,a,=,(,-,1,1),b,=,(2,-,1),c,=,(1,2),若,a,=,b,+,c,答案,:,-,3,解析,:,由,a,=,b,+,c,可知,(,-,1,1),=,(2,-,1),+,(1,2),=,(2,+,-,+,2,),-,17,-,答案,:,(3,6),或,(,-,1,-,2),-,18,-,计算平面向量的数量积,高考真题体验,对方向,1,.,(2018,全国,4),已知向量,a,b,满足,|,a,|=,1,a,b,=-,1,则,a,(2,a,-,b,),=,A.4B.3C.2D.0,答案,:,B,解析,:,a,(2,a,-,b,),=,2,a,2,-,a,b,=,2,-,(,-,1),=,3,.,-,19,-,2,.,(2016,天津,7),已知,ABC,是边长为,1,的等边三角形,点,D,E,分别是边,AB,BC,的中点,连接,DE,并延长到点,F,使得,DE=,2,EF,则,的值为,(,),答案,:,B,-,20,-,方法,2(,坐标法,):,建立如图所示的平面直角坐标系,-,21,-,3,.,(2017,北京,12),已知点,P,在圆,x,2,+y,2,=,1,上,点,A,的坐标为,(,-,2,0),O,为,答案,:,6,-,22,-,4,.,(2017,天津,14),在,ABC,中,A=,60,AB=,3,AC=,2,若,-,23,-,新题演练提能,刷高分,1,.,(2018,河北石家庄一模,),点,B,是以线段,AC,为直径的圆上的一点,其,A.1B.2C.3D.4,答案,:,D,解析,:,由圆的性质知,ABC=,90,-,24,-,答案,:,B,-,25,-,答案,:,B,-,26,-,-,27,-,4,.,(2018,河北唐山一模,),在,ABC,中,C=,90,|AB|=,6,点,P,满足,A.9B.16C.18D.25,答案,:,B,-,28,-,5,.,(2018,湖南衡阳二模,),如图,在正方形,ABCD,中,AB=,2,点,E,为,BC,的,答案,:,0,-,29,-,6,.,(2018,吉林长春质量监测,),已知菱形,ABCD,的一条对角线,BD,长为,答案,:,-,7,解析,:,如图,建立平面直角坐标系,-,30,-,平面向量数量积的应用,高考真题体验,对方向,1,.,(2017,全国,4),设非零向量,a,b,满足,|,a,+,b,|=|,a,-,b,|,则,(,),A,.,a,b,B,.|,a,|=|,b,|,C,.,a,b,D,.|,a,|,b,|,答案,:,A,解析,:,由,|,a,+,b,|=|,a,-,b,|,平方得,a,2,+,2,a,b,+,b,2,=,a,2,-,2,a,b,+,b,2,即,a,b,=,0,.,又,a,b,为非零向量,故,a,b,故选,A,.,-,31,-,a,b,=,(,),A.1B.2C.3D.5,答案,:,A,|,a,|,2,+|,b,|,2,-,2,a,b,=,6,由,-,得,a,b,=,1,故选,A,.,-,32,-,新题演练提能,刷高分,1,.,(2018,河南八市学评,),已知,a,b,c,均是单位向量,若,a,-,b,+,c,=,0,则向量,a,b,的夹角为,(,),答案,:,D,解析,:,a,-,b,+,c,=,0,c,=,b,-,a,.,1,=,1,+,1,-,2,a,b,.,-,33,-,2,.,(2018,安徽宣城二调,),已知在,ABC,中,A=,120,且,AB=,3,AC=,4,答案,:,A,-,34,-,A.,等边三角形,B.,锐角三角形,C.,钝角三角形,D.,直角三角形,答案,:,D,-,35,-,-,36,-,6,.,(2018,湖南、江西十四校第二次联考,),已知向量,a,b,满足,|,a,|=,5,|,a,-,b,|=,6,|,a,+,b,|=,4,则向量,b,在向量,a,上的投影为,.,答案,:,-,1,解析,:,|,a,-,b,|=,6,a,2,-,2,a,b,+,b,2,=,36,同理,a,2,+,2,a,b,+,b,2,=,16,
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