1. Not finding help here? Sign up for a free 30min 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!

Finding a plane parallel to a line, knowing a point on the plane

  1. Jun 11, 2010 #1
    Hi there, I got this problem to solve:

    Given L:[tex]\begin{Bmatrix} x-y+z=1 & \mbox{ }& \\y+5z=0 & \mbox{ }& \end{matrix}[/tex], find the equation of the plane parallel to L that pass through the origin. Is it unic?

    Well, I think its unic, cause there are not two different planes parallel to a line that pass through P(0,0,0).

    I've started finding the director of L, [tex]\vec{d_L}[/tex] lets call:
    [tex]\pi:x-y+z=1[/tex] [tex]\pi':y+5z=0[/tex]

    [tex](1,-1,1)\wedge{(0,1,5)}=(-6,-5,1)=\vec{d_L}[/tex]

    Lets call the plane [tex]\gamma[/tex], and [tex]\vec{n_{\gamma}}=(x_0,y_0,z_0)[/tex] a normal vector of [tex]\gamma[/tex]

    [tex]<\vec{n_{\gamma}},\vec{d_L}>=0\Rightarrow{-6x_0-5y_0+z_0=0}[/tex]

    [tex]\gamma:-6x-5y+z=0[/tex], [tex]P(0,0,0)\in{\gamma}[/tex]

    So, that result is wrong, cause then [tex]\vec{n_{\gamma}}=d_L[/tex]

    How should I continue? I know that I should find a vector normal to L, and normal to [tex]\gamma[/tex], but I don't know how to proceed. Cause, for L I can find infinite normal vectors. Now I'm starting to think that the answer is not unique.
     
  2. jcsd
  3. Jun 11, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ??? There are, in fact, an infinite number of different planes, parallel to a given line, through the origin! Draw the unique line through the origin that is parallel to the given line. Any plane that contains that line and not the original line satisfies the conditions.

    Okay, another way to do that is to write the line as parametric equations. Using, say, z as parameter, we can write the second equation y= -5z. Then the first equation becomes x- (-5z)+ z= x+ 6z= 1 so that x= 1- 6z. That is the parametric equations are x= 1- 6t, y= -5t, z= t so the "direction vector" is (-6, -5, 1), just as you say.

    Any plane through the origin can be written in the form Ax+ By+ Cz= 0. It will be parallel to the given line as long as its normal vector, (A, B, C), is perpendicular to the direction vector, (-6, -5, 1). That is, we must have -6A- 5B+ C= 0. That gives C= 6A+ 5B so that any plane Ax+ By+ (6A+ 5B)z= 0 satisfies the conditions.

     
  4. Jun 11, 2010 #3
    Thank you very much.

    Sorry for the misspelling :P

    Bye there!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding a plane parallel to a line, knowing a point on the plane
  1. Points, lines and planes (Replies: 49)

Loading...