1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find Parametric Equation for Line Parallel to Two Planes

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Find parametric equations for the line which passes through the point (1; 2; 3)
    and is parallel to both of the planes 3x + y + 5z = 4 and z = 1 -2x.

    I have seen the result for this problem, but it's different than mine. I'm not sure, what I'm doing wrong. Please, help.

    3. The attempt at a solution
    If the line is parallel to two planes, then the line must be orthogonal to both planes' normals.
    First, I notice that the planes' normal are:
    n1=(3,1,5) and n2=(2,0,1)

    Then, the direction vector of the line is p=<a,b,c>
    and since, the vector and the normals are orthogonal.
    *Dot Products*
    p.n1=0
    p.n2=0

    I get :
    3a+b+5c=0
    2a+0+c=0
    Using determinants, I find that the direction of the line is: 1i+7j-2k
    {x-x_{1}} over {1} ={y-y_{1}} over {7} ={z-z_{1}} over {-2} =t
    or
    (x-1)=(y+2)/7 =(z-3)/-2 =t ...where the values of the point passed are plugged in on the numerator.
    So, my parametric equation yields:
    x=t+1
    y=7t-2
    z=-2t+3
     
  2. jcsd
  3. Jan 30, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi knowLittle! welcome to pf! :smile:
    looks ok (apart from the -2 in the y) …

    maybe the book is using a different parameter (eg t+1)

    what answer does the book give?
     
  4. Jan 30, 2012 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And you might notice that a quick way to get a direction vector along the line of intersection is to take the cross product of the normals.
     
  5. Jan 30, 2012 #4
    This is the book's answer. It's very short and unclear.

    1a: Take cross products of the displacement vectors to get a normal to plane P:
    Cross [{0, 1, 2} - {1, 0, -1} , {1, 2, 3} - {0, 1, 2}]
    {-2, 4, -2}
    That gives an equation of the plane from the normal vector and a point:
    -2x + 4(y - 1) - 2 (z - 2) ==0
    -2 x + 4 (-1 + y) - 2 (-2 + z)==0

    Also, I don't see any problem in the substraction of "2". It's the result of the determinant.
     
  6. Jan 30, 2012 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    I'm confused … that seems to be a different question :confused:
     
  7. Jan 30, 2012 #6
    I don't understand the cross product shown on the book.
    My cross product of the normals give: 1i+ 7j-2k , where normal1=<3,1,5> && normal2=,2,0,1>
     
  8. Jan 30, 2012 #7
    I think that you are right. The pages, where uploaded incorrectly by the staff. Wow.
    Thank you all.
     
  9. Jan 30, 2012 #8
    I think that you are right. The pages were uploaded incorrectly by the staff. Wow.
    Thank you all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook