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Planes physics question

  1. Aug 4, 2005 #1
    I have been working on these two problems for a while now and I can't seem to come up with the right procedure to take them on:

    1) Find the plane through the origin and parallel to the plane 2x - y + 3z = 1

    * I thought this could be solved by taking the vector of the parallel plane <2, -1, 3> and cross multiplying with the origin, but that didn't get me anywhere.

    2) Find the plane that passes through (1, 2, 3) and contains the line x=3t, y=1+t, z=2-t

    * The same problem occured in this case.

    I know how to find a plane when given a point and a normal (perpendicular) vector... what would be of great help is if anyone could help me understand how to find the equation of planes under different circumstances (like the two problems above). Thanks a bunch =)
    Last edited: Aug 4, 2005
  2. jcsd
  3. Aug 4, 2005 #2


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    There are always multiple ways of solving these kind of problems, but the following seems straightforward.

    1) If you put in x=z=0 in the equation of the plane, you'll find the point where the plane crosses the y-axis. Now you need to translate the plane over some distance parallel to y to make this zero. (Do you know how translations work in general?)

    2) You need three noncollinear points to be able to construct a plane. The equation of the line will give you 2, the given point in the plane is a third.
  4. Aug 4, 2005 #3
    Thanks for the quick reply.
    But no, I do not know how do a translation of a plane... could you explain :smile: ?
  5. Aug 4, 2005 #4


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    Just like you translate the graph of a function. The graph of f(x-c) is translated over a distance c wrt the graph of f(x).
  6. Aug 4, 2005 #5
    i understood the first problem:

    the normal vector to the plane would be <2, -1, 3> and the equation of the plane would be the dot product of the <2, -1, 3> *<x+0, y+0, z+0>=0
    and the answer would be 2x - y + 3z = 0

    for the second problem i am given two points:
    (1, 2, 3) and (0, 1, 2)
    * could I just plug in any number for t (say, 1) to get a third point?
    Last edited: Aug 4, 2005
  7. Aug 4, 2005 #6


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    Actually, you are only "given" one point. You clearly got the point (0, 1, 2) by putting t= 0 in the equation of the line. Since all points of the line are in the plane, choosing any other value of t (t=1 would be convenient) will give you a third point in the plane.
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