exam2solutions - Calculus III Math 143 Spring 2008...

This preview shows page 1 - 2 out of 4 pages.

The preview shows page 1 - 2 out of 4 pages.
Calculus IIIMath 143 Spring 2008Professor Ben RichertExam 2SolutionsProblem 1.(20pts) Consider the pointsA= (1,2,3)andB= (2,3,5), and the liner(t) =(2,2,3)+t(1,2,5).(a–5pts) Give parametric equations for the line containingAand parallel to−−AB.Solution.The line through (a, b, c) in direction(d, e, f)is(a, b, c)+t(d, e, f), giving parametric equationsx(t)=a+dty(t)=b+etz(t)=c+ftWe use the point (1,2,3) and the direction−−AB=(1,1,2)which we get by subtracting the pointBfromA. So ourline is:x(t)=1 +ty(t)=2 +tz(t)=3 + 2tsquare(b–5pts) Give the equation of a plane which is perpendicular tor(t)and containsB.Solution.The equation of a plane containing (x0, y0, z0) with normal(a, b, c)isa(xx0) +b(yy0) +c(zz0) = 0.Since the plane we are looking for is perpendicular to−−AB, its normal must be parallel to−−AB=(1,1,2), that is, wecan use(1,1,2)as its normal. With the pointB= (2,3,5), we get the plane(x2) + (y3) + 2(z5) = 0.square(c–10pts) Give the equation of the plane containingr(t)and the pointB.Solution.LetC= (2,2,3), and note thatCis a point onr(t). Then the normal to our plane will be perpendicularto both(1,2,5), the direction ofr(t), and−−CB=(0,1,2). We know that the(1,2,5) × (0,1,2)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleijk125012vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=(−1,2,1)is normal to both those directions. Thus the plane in question is(x2)2(y

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 4 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Fall
Professor
staff
Tags

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture