高中数学课件空间向量的坐标运算高二数学课件

——空间直角坐标系.向量的直角坐标运算.xyzOA(x,y,z)ijka复习提问：平面直角坐标系中１、2111,,,yxByxAOAOB1212,yyxxABayxC,点21,yxB11,yxAyxaOCAB,2、即OAxyBC一、空间直角坐标系１、单位正交基底正交：三个基向量互相垂直{,,}ijkikj其中单位基底：长都为1２、空间直角坐标系yzxOxyza123(,,)AaaajOikkajaiaOAa321321,,aaazyxOA,,即zyxA,,点xyzB1A1D1C1BDCA练习1、1111ABCDABCD正方体的棱长为2,则各顶点的坐标为:11________________AB11________,________.CDA________,B_________C________,D_________(0,0,0)(2,0,0)(2,2,0)(0,2,0)(0,0,2)(2,0,2)(2,2,2)(0,2,2)练习２、建立直角坐标系，求作点G(1,3,0)，点Q(0,2,3)QＧDzxyB1A1D1C1BCA11111如图建立直角坐标系,长方体中,已知AB=3,BC=5,AA=2,ABCDABCD则各顶点的坐标为:11________________AB11________,________.CDA________,B_________C________,D_________(0,0,0)(3,0,0)(3,5,0)(0,5,0)(0,0,2)(3,0,2)(3,5,2)(0,5,2)0,0;在x轴上的点的坐标不为其余均为x0,0;在平面xoy的点的z坐标为其余均不为总结:练习３、练习４、点B是点A(3,4,5)在坐标平面内的射影，求5A(3,4,5)与y轴垂直的坐标平面是____xozA’(3)点A(3,4,5)关于原点成中心对称的点A’的坐标是xOy||OBxyOz34B(3,4,0)练习５、(1)与x轴垂直的坐标yoz与z轴垂直的坐标平面是______xoy(2)点P(2,3,4)在平面内的射影是______xoy(2,3,0)在平面内的射影是______xoz(2,0,4)在平面内的射影是______yoz(0,3,4)'(3,4,5)A平面是______１、三个坐标二、向量的直角坐标运算a121212,,zzyyxx221322,,,,zyxzyxABOAOB111,,zyxA222,,zyxB练习６ABBAOBAkjiABCODCD则，，，，则坐标则点210,1213,3,1,2,5432,2,4,0,5,1,213,1,13,5,28,5,1//;.ab123123(,,),(,,)aaaabbbb设则_____________________;ab______________________;ab_________________;a;ab;ab112233(,,)ababab112233(,,)ababab123(,,),()aaaR112233ababab112233,,()abababR1122330abababab２、例1、(2,3,5),(3,1,4),,,8,.abababaab已知求(1)ppbpayxpccanmc求且若求且若,,,,13,//,2,2练习7写出下列各题中向量的坐标(1)23______,(2)54_______,ijkijk(3)52______,(4)3______.ijj(1,2,3)(-1,5,-4)(-5,-2,0)(0,3,0)6.(3,2,5),(1,5,1).ab已知求(1)_________,(2)______abab(3)6_________,(4)______aab(-2,7,4)(-4,-3,6)(-18,12,30)-3+10-5=27.(2,3,1),(2,0,3),(0,0,2).abc已知求(1)()_________abc(2)68_________abc(2,-3,1)·(2,0,5)=4+0+5=9.(2,-3,1)+(12,0,18)+(0,0,-16)=(14,-3,3)(3)(1)(2)练习8判定下列各题中的向量是否平行(1)(1,2,-2)和(-2,-4,4),(2)(-2,3,5)和(16,-24,40)5.:(2,1,3),(4,2,),,.abxabx已知求的值练习9②Ｇ为△ABC的重心OBOAOP21OCOBOAOG311OBOAOP３、①的中点为ABP③PBAP即所成的比为分ABP练习9、(1)已知P(2,-1,3)为AB中点且A(0,4,7)求B(2)已知△ABC中,A(2,0,1),B(3,5,-2),重心G(1,3,5),求顶点C坐标(3)ABCD中,A(4,1,3),B(2,-5,1),C(3,7,-5)求D坐标(4)已知A(4,1,3),B(2,-5,1),C在线段AB上,且AC:AB=1:3,求C坐标作业P42.3,4,5,6