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Cartesian equation of the plane

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data
    If the line "l" is given by the equations 2x-y+z=0, x+z-1=0, and if M is the point (1,3,-2), find a Cartesian equation of the plane.
    a) passing through M and l
    b) passing through M and orthogonal to l
    2. Relevant equations
    (r-r0)n=0


    3. The attempt at a solution
    I expanded the equation above, so i got rn=r0n
    then, (x,y,z)n=(1,3,-2)n.
    And i don't know how to get the normal vector from those two equations
     
    Last edited: Jan 17, 2013
  2. jcsd
  3. Jan 17, 2013 #2

    jedishrfu

    Staff: Mentor

    can you be more specific? does the plane contain the line or the point? or does it contain both?
     
  4. Jan 17, 2013 #3
    sorry, i forgot some of them.

    a) passing through M and l
    b) passing through M and orthogonal to l
     
  5. Jan 17, 2013 #4

    jedishrfu

    Staff: Mentor

    and what course? vector algebra?

    also you really need to show more work before anyone here at PF will help.
     
  6. Jan 17, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I presume you know: If <A, B, C> is a vector normal to a plane and [itex](x_0, y_0, z_0)[/itex] is a point in the plane, then the Cartesian equation of the plane is [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].

    2x- y+ z= 0, x+ z=1. Subtract the second equation from the first to eliminate z: x- y= -1 or x= y- 1. If you let y= 0, x= -1 so x+ z- 1= -1+ z- 1= 0 gives z= 2. (1, 0, 2) is a point on the given line. If you let y= 1 in x= y- 1, x= 0 so x+ z- 1= z- 1= 0 gives z= 1. (0, 1, 1) is a point on the given line. Those, together with (1, 3, -2) give you three points in the plane. <1- 1, 3- 0, -2- 2>= <0, 3, -4> is a vector in the plane. <1- 0, 3- 1, -2- 1>= <1, 2, -3> is another vector in the plane. Their cross product is normal to the plane.

    <2, 1, 1> is a vector normal to 2x- y+ z= 0. <1, 0, 1> is a vector normal to x+ z= 1. Their cross product is normal to both so normal to any plane perpendicular to the line. You have a normal vector and a point in the plane.

     
    Last edited: Jan 17, 2013
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