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Find the equation of the plane that satisfies the stated conditions.

  1. Feb 22, 2009 #1
    The plane that contains the line x = -2 + 3t, y = 4 + 2t, z = 3 - t and is perpendicular to the plane x - 2y + z = 5.
     
  2. jcsd
  3. Feb 22, 2009 #2
    Nevermind, I kinda just solved it myself.

    All good.
     
  4. Feb 22, 2009 #3
    Find the equation of the plane (if you dare) that is tangent to the following spheres:

    sphere 1: r=2 center P(2,2,2)

    sphere 2: r=3 center Q(3,4,5)

    Ok, I know that the plane will be parallel to the vector PQ = <1,2,3>

    For the equation of a plane I need a normal vector, and a point on the plane. I know that one of the points on the plane will be one of the points on the sphere (either sphere will do, right?).

    The equations for the spheres are:

    (x-2)^2 + (y-2)^2 + (z-2)^2=4

    and

    (x-3)^2 + (y-4)^2 + (z-5)^2=9

    What i am thinking now is that i should take the midpoint between the spheres' centers. Use that to find another vector parallel to the plane in question, and then generate a normal vector from that - is this correct thinking?

    And from there, i am unsure how to find a point common to the plane and the sphere (either sphere).

    I have drawn this out on paper and understand the geometry...
     
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