# C 35 4 tt 213 s tbr yz 15 6 ts c 354 tt cz 13 ter

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C-3,5,4)tt.(2/1,3);S,tBR(×,Y,Z)=(1,5,6)ts.C-35,4).tt.CZ,1,3);§tERvectorequationi.Parametricequationarez=1-3st2ty=St55t4E2-Gt4st3f
April19ZawcrkhcnIntersectionzoizAssignment3b)Planedoesnotpassthroughtheorigini.equationinformaxtbytcz=1pointwhereplaneinterceptaxisfor(4,2)=(0,0)x=la(y,-2)=@,o)→x=taforCK,z)to,o)→Y=lbCx,2)-40,0)y=Ifor(x,y)=§0)2=1Cex,y)=P,o)2=2Since±=tb=E=kti.Searchedequationxtytz-K
April19ZawcrkhcnIntersectionzoizAssignmentua)vectorequationvectora=[-2-2,7-6,5-3]=EU,-13,2]Vectorbe[2-1,6-4,3C-I)]=[1,2/4]Normalvectorn=axb=I4I3k2=[-56/18,5]I24Equationofplaneis-56Got2)+184+77+54--57=0-56×+18.y+5-2=11line2=5+2£,4=1-t,2=15+4tSubformulasforK,yand2intoequationforPandSoweto-56×+18-1+52=11-56(5+24+184t)+545+44=11-280-112£+18-18£t75t2ot=11-Hot-198=0T=tsf=s±Intersectionpoint2=5+2(E)y=tC→⇒=¥zestu.c¥)=
April19ZawcrkhcnIntersectionzoizhb)assignmentline,=¥=2=15uplaneI=56×+184+5211=0u=(2,-1,4)n=(-56,18/5)Sina=12C-56)+C-1)C18)+4(5).l(J22t.DZ+42|jC56>2+182+52Sina=110(4,35)(59.3of=Sri'(0.4283)=25.363°
April193awcrKhanIntersectionzoizAssignmentuc)ConsiderXtYt2=3Paralleltoplane:Gqy,2)=(1+41-t,2)LoneLiesonplane:(z,y,7)=(Itt,It,1)(meintersectonplane:(x,y,2)=€,Ztf,5-t)Laneparalleltoplanehaedirectionvectorperpendiculartonormalvectorofplaneplanehasnormalvector[1,1,1]BothlineparalleltoplanehavedirectionnectarCl,-1,0]whrchrsperpendiculartonormalvector,sincedotproductsoForlineparalleltopkneChosemitralnotonplane(1,1,2)ForlineonplanechoseMitapointthatonplane4,1,\)Laneintersectsplaneatsinglepointifdirection