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Calculus 3 problems (equations of planes and lines) 3 Space

  1. May 1, 2009 #1
    Hi, Im currently doing a practice test for my final exam coming up, im wondering anyone can double check the questions to see if i did them write, below is a picture of the questions, the answers i got are listed at the bottom,

    If you could, please post whether you agree with my answers to any of the following or if you disagree.

    28k3d47.jpg
    1. The problem statement, all variables and given/known data


    for 1 A) i got, (X-2)/-3=(Y-4)/-4=(Z-1)/2

    for 1 B) after simplifying i came up with -3x-4y+2z=-24


    For 2) i got the following,

    x= 3+5t
    y=1+2t
    z=2+5t

    for 3) after simplifying i got, 3x-2y-5z=-42
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 1, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If x= 2, y= 4, and z= 1, your equation becomes (2- 2)/(-3)= (4- 4)/(-4)= (1-1)/2 or 0= 0= 0 which is true. If x= -1, y= 0, and z= 3, your equation becomes (-1-2)/(-3)= (0-4)/(-4)= (3-1)/2 or 1= 1= 1 which is also true.

    If x= 2, y= 5, and z= 1, this becomes -3(2)- 4(5)+ 2(1)= -6-20+ 2= -24. Yes, that point is in the plane. A vector perpendicular to this plane is <-2, -4, 2> which is the direction vector for the given line.

    When t= 0 that gives (3,1,2) which is also the point given by t= 1 in the given parametric equations. Further, <5, 2, 5> is the direction vector and is the tangent vector to the given curve at t= 1.

    If x= -2, y= 3, and z= 6, your equation becomes 3(-2)- 2(3)- 5(6)= - 6- 6- 30= -42 so the point is on that plane. However, the normal vector to that plane is <3, -2, -5> while the the gradient of the given function is <3x^2, -3y^2, 2z> and at (-2, 3, 6), that is <12, -27, 12> which is NOT a multiple of <3, -2, -5> and so not in the same direction.

     
  4. May 1, 2009 #3

    Do you agree with the answers i have?
     
  5. May 2, 2009 #4

    Mark44

    Staff: Mentor

    HallsOfIvy disagrees with your answer to problem 3. Show us how you got the answer you've shown, and we'll figure it out.
     
  6. May 3, 2009 #5

    I figured the problem i did wrong, i was thinking of a tangent plane to a line instead of a tangent plane to a surface, to fix it i ended up taking the gradient of the surface function, plugging in the point, then dot producting it with (x-xi),(y-yi),(z-zi)

    i ended up with12x-27y+12z = -33


    pretty sure this is right now
     
  7. May 3, 2009 #6

    Mark44

    Staff: Mentor

    Looks fine.
     
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