Intersections Unit Assignment.docx - Intersections Unit Assignment 1 We know that vector equation of a line:⃗r =⃗ r 0 t ⃗v =� x0 y 0,2 0> t <a,b,c

Intersections Unit Assignment.docx - Intersections Unit...

This preview shows page 1 - 5 out of 16 pages.

Intersections Unit Assignment1.Weknow that vector equationof aline:r=r0+tv=¿x0, y0,20>+t<a ,b ,c>¿pointsWherer0is a position vector(¿a positiont heline)v isadirectionvector(a vector¿theline)r0=←2,3,1>¿v=¿1(2),43,21>¿<3,1,3>¿r(t)=←2,3,1>+t<3,1,3>¿r(t)=←2+3t ,3+t ,13t>¿For symmetric equation1st find parametric equation:x=−2+3t , y=3+t ,2=13tNow symmetric equation:x+23=y31=213
2.a.x13=y86=z32=t.x84=y+35=z72=sSo ,1+2gives:x=−3t+1,y=6t+8,z=−2t+3x=4s+8,y=−5s3,z=2s+7If theyhaveintersecting point3t+1=4s+8(i)6t+8=−5s3(ii)2121
2t+3=2s+7(iii)(6t+2)+6t+8=8s+165s310=3s+133z=−3→s=−13t+1=−4+83t=3→t=−1LHS:2(1)+3=5RHS:2+7=5So(iii)issatisfied at s=t=−1The pointofis:(3t+1,6t+8,2t+3)¿(4,2,5)b.Equationof line passingthroughP(1,5,2)¿¿x=axisDirectinratos of the line(xaxis)(1,010).x+11=y50=z20=t.x=t1.y=5.z=2LLzP(−1,5,2tRyx
Nowusing the point(0,5,2)aquationof line passingthroughit but¿xaxisis:x1=y50=z20=t → L1Foranother line L2take normalvectorn2=noraml¿xyplaneie(oi+o^J+^k)Now equation passingthrough(1,5,2)¿¿1.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture