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Eq. of a plane

  1. Sep 14, 2007 #1
    I recently had a quiz, and the problem was this:

    Find an equation for the plane that lies on the point P(0,1,0) and on the line, L, with parametric equations: x = t, y= t, and z = 1 + t.

    I understand for the most part how to arrive at the answer. The main problem was how to find the normal vector that stems from the point P, and to do that you choose two other points on the plane by substituting values for "t". If I substitute 0 and 1 in for "t", we obtain the vectors <0, -1, 1> and <1, 0, 2>. Taking the cross product of the two vectors gives the normal vector. What I don't understand is if the parametric equations form a "line", and you find two additional points on that line, which subsequently form vectors from the initial point P, you could get one of two sets of vectors, vectors that are anti-parallel to one another, or vectors that are parallel to one another. If you cross those two vectors and take the magnitude of the vector obtained, you would get the area of the parallelogram that the vectors would form. But if the vectors being crossed are parallel or anti-parallel, that would give an area of zero. If I cross the two vectors above, <0, -1, 1> and <1, 0, 2>, I get the vector <2, 1, -1>, whose magnitude is certainly not zero. Now I'll try to clear things up a bit. The reason, I assume, that the area is not zero is because the points above do not lie on a line, but rather, they lie on a curve. I did a bit of 3-d graphing on my notebook paper, and it appears that P, and the points that I obtain by plugging in 0 and 1 for "t" are not co-linear, in which case the parametric equations given in the beginning of the post are the equations of a curve rather than a line. In case the reader is wondering, this relates to the problem above, because if the plane lie on a curve, rather than a line, it would have been quite obvious to find two additional points via the parametric equations, and not suspect that taking the cross product of the two vectors obtained by the two additional points on the line would give the zero vector, in which case I wouldn't have been able to form an equation for the plane. Is there something that Im missing?
    Last edited: Sep 14, 2007
  2. jcsd
  3. Sep 14, 2007 #2
    you meant the line L: x=t, y=t-1, z=1+t right? I noticed the vectors you found seemed to fit those equations.

    Also I just graphed it and it does seem to be a line.
  4. Sep 14, 2007 #3
    Then I'm confused, haha. If I have two vectors pointing in opposite directions, or two vectors pointing in the same direction, how do I cross them and not arrive at the zero vector? For instance, if I crossed <1, 0, 0> with <-1, 0, 0>, that is i X -i, that gives me the zero vector. Isn't that analogous to crossing the two vectors obtained in the problem above, seeing as they lie on a straight line?
  5. Sep 14, 2007 #4
    yes you would get 0 if you cross 2 vectors in the line.
    But you only need to get 1, and then cross it to the point P to get the normal to the plane.
  6. Sep 14, 2007 #5
    So you're saying I'd use the point P(0, 1, 0) as a vector, and find an additional point on the line, say point S, and I'd cross <0, 1, 0> with PS ?
  7. Sep 14, 2007 #6
    Hm..ok w8 your right you would need 2 extra points
    point S say <0,-1,1> and point T <1,0,2>

    you don't cross those 2 points, you cross 2 vectors

    the 2 vectors you need to cross would be:
    PS and PT
    and that way you will get the normal to where the plane should be.

    P and S will lie on PS that is part of a line. same thing with P and T. But you dont cross the points you will be crossing the vectors which will not be collinear.
  8. Sep 14, 2007 #7
    If I have a line containing points P, S, and T as stated above, all three points are on the same line. If I use P as my origin to obtain vectors PS and PT, I don't see how the vectors cannot be colinear. It would look something like this:


    And the vectors would look something like:


  9. Sep 14, 2007 #8
    no P is away from the line L.
    it would look like:
    *-empty space

    ******** \** /
    ********* \ /

    Something like that. I used this to graph the line.
  10. Sep 14, 2007 #9
    Alright, I understand now. The wording confused me and I assumed P was both on the plane, and on the line. Thanks for your help =)
  11. Sep 14, 2007 #10
    no problem, but if the point P was on the line then there would be some t such that

    which has no solution so the point P is not on the line L.
  12. Sep 14, 2007 #11
    It's making more sense by the minute, haha. Thank you.
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