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Equation of a plane containing two lines

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the plane that contains the lines given by:

    (x-1)/-2 = y - 4 = z and (x-2)/-3 = (y-1)/4 = (z-2)/-1

    3. The attempt at a solution

    Not too sure about how to go about it. I believe these are the parametric equations for the lines:

    x = 1-2t w/ normal vector = (-2,0,0)
    y=4
    z=0

    x=2-3t w/ normal vector = (-3, 4, -1)
    y=1+4t
    z=2-t

    If I take the cross product of those normal vectors, I get (0, -2, -8). What do I do from here or am I going in the wrong direction?
     
  2. jcsd
  3. Jan 31, 2010 #2

    Mark44

    Staff: Mentor

    Yes, you are going in the wrong direction. You don't want normal vectors for the two lines -- each line has an infinite number of normal vectors. What you want instead is a vector that has the same direction as the line.

    You don't need to convert the symmetric equations for your lines into parametric equations - you can read off a point on each line and a vector parallel to the line from the equations.

    For example, in the symmetric equations for L1, a point on the line is (1, 4, 0), and a vector that has the same direction as the line (is parallel to the line) is <-2, 1, 1>.

    Use the same process to find the direction of the second line. When you get that vector, cross it with <-2, 1, 1>. That will give you a vector that is perpendicular to both given vectors, and hence both given lines. For the time being, let's call it <A, B, C>.

    If you know a point (x0, y0, z0) on a plane and its normal, <A, B, C>, you can write the equation of the plane as A(x - x0) + B(y - y0) + C(z - z0) = 0.
     
  4. Jan 31, 2010 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't have "the" normal vector to a line; it has many normals. But you do have a direction vector to a line, which is what you are calculating, although the first one is wrong. If you write your first line in the symmetric form:

    [tex]\frac {x-2}{-2} = \frac{y-4}{1} =\frac{z}{1}[/tex]

    you will see that its direction vector is D = <-2,1,1>, not <-2,0,0>.

    What you get when you cross them is a vector perpendicular to both of them, so perpendicular to the intended plane. With the normal vector and a point on one line, you can calculate the equation of the plane through that point containing that line. With luck, but it doesn't have to, it may also contain the other line; you have to check that.
     
  5. Jan 31, 2010 #4
    Thanks guys, worked it out and the answer agrees with the back of the book :)
     
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