2.3.1向量数量积的物理背景与定义

2.3.3向量数量积的坐标运算与度量公式课时过关能力提升1.已知a=(2,-3),b=(1,-2),且ca,bc=1,则c的坐标为()A.(3,-2)B.(3,2)C.(-3,-2)D.(-3,2)解析:设c=(x,y),则有2x-3y=0,x-2y=1,解得x=-3,y=-2.故c=(-3,-2).答案:C2.已知m=(a,b),向量n与m垂直,且|m|=|n|,则n的坐标为()A.(b,-a)B.(-a,b)C.(-a,b)或(a,-b)D.(b,-a)或(-b,a)答案:D3.已知点A(1,2),B(4,0),C(8,6),D(5,8),则四边形ABCD是()A.梯形B.矩形C.菱形D.正方形解析:由已知得AB=(3,-2),BC=(4,6),CD=(-3,2),所以AB=DC,且ABBC=0,即ABBC,所以四边形ABCD是矩形.答案:B4.已知向量a=(cos ,sin ),b=(3,0),则|2a-b|的最大值为()A.42B.22C.25D.5解析:|2a-b|=(2a-b)2=4-43cos+9=13-12cos,因此当cos=-1时,|2a-b|取得最大值5.答案:D5.在RtABC中,C=2,AC=3,取点D使BD=2DA,则CDCA等于()A.3B.4C.5D.6解析:以C为原点,分别以CA,CB所在直线为x轴、y轴,建立如图所示的平面直角坐标系,设CB=a,C(0,0),A(3,0),B(0,a).设D点坐标为(m,n),BD=2DA,即(m,n-a)=2(3-m,-n),得m=2,n=a3.CDCA=2,a3(3,0)=6,故选D.答案: D6.已知O为坐标原点,OA=(3,1),OB=(-1,2),OCOB,BCOA,则满足OD+OA=OC的向量OD的坐标为.答案:(11,6)7.设O为原点,已知点A(a,0),B(0,a)(a0),点P在线段AB上,且AP=tAB(0t1),则OAOP的最大值为.解析:OAOP=OA(OA+AP)=OA(OA+tAB)=OA2+tOAAB=a2+t(a,0)(-a,a)=a2+t(-a2+0)=(1-t)a2.0t1,01-t1,OAOP的最大值为a2.答案:a28.以原点及点A(5,2)为顶点作等腰直角三角形OAB,使B=90,求点B和向量AB的坐标.解:如图,设点B的坐标为(x,y),则OB=(x,y),AB=(x-5,y-2).OBAB,x(x-5)+y(y-2)=0,即x2+y2-5x-2y=0.|OB|=|AB|,x2+y2=(x-5)2+(y-2)2,即10x+4y=29.解方程组x2+y2-5x-2y=0,10x+4y=29,得x1=72,y1=-32或x2=32,y2=72.点B的坐标为72,-32或32,72;当点B的坐标为72,-32时,AB=-32,-72;当点B的坐标为32,72时,AB=-72,32.综上,点B的坐标为72,-32,AB=-32,-72,或点B的坐标为32,72,AB=-72,32.9.已知a=(cos ,sin ),b=(cos ,sin ),且|ka+b|=3|a-kb|(k0).(1)用k表示数量积ab;(2)求ab的最小值,并求此时a,b的夹角.解:(1)由|ka+b|=3|a-kb|,得(ka+b)2=3(a-kb)2,k2a2+2kab+b2=3a2-6kab+3k2b2.(k2-3)a2+8kab+(1-3k2)b2=0.|a|=1,|b|=1,k2-3+8kab+1-3k2=0,ab=2k2+28k=k2+14k.(2)由(1),得ab=k2+14k=14k+1k,由函数的单调性的定义,易知f(k)=14k+1k在(0,1上单调递减,在1,+)上单调递增,故当k=1时,ab的最小值为f(1)=14(1+1)=12.此时a,b的夹角为,则cos =ab|a|b|=121=12,=60.10.如图,AB=(6,1),BC=(x,y),CD=(-2,- 3),BCDA.(1)求x与y的关系式;(2)若ACBD,求x,y的值及四边形ABCD的面积.解:(1)AD=AB+BC+CD=(6,1)+(x,y)+(-2,-3)=(x+4,y-2),DA=-AD=(-x-4,2-y).BCDA,BC=(x,y),x(2-y)-(-x-4)y=0,x与y的关系式为x+2y=0.(2)因为AC=AB+BC=(6,1) +(x,y)=(x+6,y+1),BD=BC+CD=(x,y)+(-2,-3)=(x-2,y-3).ACBD,ACBD=0,即(x+6)(x-2)+(y+1)(y-3)=0.又由(1)的结论x+2y=0,得(6-2y)(-2y-2)+(y+1)(y-3)=0.化简,得y2-2y-3=0.y=3或y=-1.当y=3时,x=-6,于是BC=(-6,3),AC=(0,4),BD=(-8,0).|AC|=4,|BD|=8.S四边形ABCD=12|AC|BD|=16.当y=-1时