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Two planes intersecting

  1. Jul 8, 2004 #1
    This is in my multivariable course. How would I find the equation of the line where two planes meet?

    I worked out how to find the equation of a plane perpendicular to that line going through a point, Z, (in two planes, Ax + By + Cz +D = 0 and Ex + Fy + Gz + H = 0, take the cross product of (A,B,C) and (E,F,G) and plug the resulting vector into the point-normal form of a plane with the point being Z), but how do I figure out how to find the line itself and how would I find a plane parallel to the line (going through a certain point, Q)? Thanks a lot, guys!

    It seems to me, that for the second (plane parallel to a line), I'd need the first (eqn of a line) first. Then I'd find two points on the line and using those three points, I'd find the equation of a plane.

    Also, I just thought of something: how would I go about finding the equation of a plane from two vectors. Should I cross them, and then enter either vector as the point in the point-normal form of a plane (i.e. if the vector a is (a1,a2,a3) then could I make (a1,a2,a3) my point?)? Or do I need two vectors and a point not on the vectors?
     
    Last edited: Jul 8, 2004
  2. jcsd
  3. Jul 9, 2004 #2

    AKG

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    It's kind of hard to tell what your actual questions are. Anyways, to get the line of intersection of two planes, simply take the cross product of the normal vectors of both planes to get the direction vector of the line, and then find a point that exists on both planes (and hence exists on the line).
     
  4. Jul 9, 2004 #3
    To find the line between two intersecting planes, you may just solve the linear system of equations of the two planes.

    For example, I would like to find the line between the two intersecting planes x+y+z=0
    -x+y+z=2. Then I try to solve

    x+y+z=0
    -x+y+z=2

    =>

    x+y+z=0
    y+z=2

    Put z=t, the solution would be (x, y, z) = (-2, 2-t, t), and that would be the line required.
     
  5. Jul 9, 2004 #4
    Thanks, guys! That answered one of them. The next questions are:

    "how would I find the equation of a plane parallel to a line? with one other known point"

    My guess is to find two points on the line, and then using the three points find the equation of the plane.

    "how would I go about finding the equation of a plane from two vectors."

    My guess would be to cross them, and then find any point on one of the vectors (the endpoint of one of them, I guess) and then use that in point-normal form.
     
  6. Jul 9, 2004 #5
    I'm not sure what you mean here, but I think you are on the right track. Let c be the location vector of the known point on the plane. Let a and b be two location vector describing two points on the line. Now a - b and c - b are vectors that lie on the plane which is parallel to the line. Taking the cross product gives you a vector normal to this plane which you can then use to find its equation.

    I don't know what you mean here. What do these two vectors represent? What significance do they have for finding the equation of a plane?
     
  7. Jul 9, 2004 #6
    Ok, let me be more clear:

    How would I find the equation of a plane parallel to a line and a point not on that line?

    And

    How would I find the equation of the plane that two vectors are on?
     
  8. Jul 9, 2004 #7
    Where is the point then? On the plane? Or somewhere else? See my previous response if it's on the plane.
    Take the cross product and use that do derive your equation.
     
  9. Jul 9, 2004 #8
    And I use the cross of the two vectors along with one of the endpoints of the vector to get the equation of the plane, eh?

    Thanks e( , you are very helpful. Thanks a lot, bro.
     
  10. Jul 13, 2004 #9

    HallsofIvy

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    There are an infinite number of planes "parallel to a given line and containing a given point". Construct a line parallel to the given line and passing through the given point. Any plane containing that line satisfies your conditions.

    (Well, except for the one plane that also contains the given line!)
     
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