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Perpendicular plane equation help

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data

    perpendicular

    --------------------------------------------------------------------------------

    a)Find the eqution of the plane that contains points (2,-1,4) , (3,2,-1)
    and perpendicular to the plane x+y+2z=3

    b)Find the equation of the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1


    2. Relevant equations

    The vector normal to the plane is (1,1,2). You know it from the coefficients in front of x,y and z.
    i know that this vector will be part of the perpendicular plane since it is the normal vector. Then i want the vector linking points (2,-1,4) , (3,2,-1) to be in the perpendicular plane as well.
    how to Find this vector and compute the cross product with the normal vector (1,1,2). This will give you the normal vector of the perpendicular plane (say (a,b,c)).




    3. The attempt at a solution
    as above in b
     
  2. jcsd
  3. Dec 5, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: equation

    The vector normal to which plane? You are told that the plane you are seeking is perpendicular to x+ y+ 2z= 3 and (1,1,2) is perpendicular to x+ y+ 2z= 3. It is parallel to the plane you are seeking. If you take one of the points you are given, say (2, -1, 4) and add that vector you get (2+1, -1+ 1, 4+ 2)= (3, 0, 6) as a third point in the plane. Can you find the plane that contains the three points (2, -1, 4), (3, 2, -1), and (3, 0, 6)?

    Equivalently, since (2, -1, 4) and (3, 2, -1) are in the plane, the vector between them, (2-3, -2-2, 4-(-1))= (-1, -4, 5) is parallel to the plane. You now know that the vectors (-1, -4, 5) and (1, 1, 3) are both parallel to the plane and so their cross product is perpendicular to it.
     
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