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A plane intersecting another plane

  1. Sep 10, 2003 #1
    Here is the question:

    Find an equation of the plane that passes through the line of intersection of the planes x-z = 1 and y +2z =3 and is perpendicular to the plane x + y -2z =1.

    Good luck!
     
  2. jcsd
  3. Sep 10, 2003 #2

    HallsofIvy

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    Homework? (Doesn't seem like the kind of problem one does for recreation.)

    What do you mean "the plane"? There are an infinite number of planes that satisfy the condition- take any one plane that does and move it "parallel to itself" along the line of intersection of the first two planes. In fact, since we are requiring that the plane is perpendicular to a plane rather than a line- there are an infinite number of planes perpendicular to a given plane.

    The first thing you should do is find the line of intersection of the two planes: x-z = 1 and y +2z =3. We can write a parametric equation for the line by taking z= t, the parameter, and solving the other two equations for x and y in terms of z: x= z+1 and y= 3- 2z so the parametric equations are x= t+1, y= 3- 2t, z= t.

    Given that the plane is perpendicular to x+ y- 2z= 1, it's normal vector is perpendicular to the normal vetor of that plane, <1, 1, -2>.
    In other words, the equation of the plane must be of the form
    a(x-x0)+ b(y-y0)+ c(z-z0)= 0 where a+b-2c= 0 so that if b and c are any two numbers then a= 2c- b gives a plane perpendicular to x+ y+ 2z= 1.

    To guarantee that the plane passes through the given line, we can make (x0,y0,z0) a point on the line. Take t= 0 so that (1, 3, 0) is a point on the line. Take c= 1, b= 1 so that a= 2(1)-1= 1. Then the plane 1(x-1)+ 1(y-3)+ 1(z-0)= x-1+ y- 3+ z= 0 or x+ y+ z= 4.
    x+ y+ z= 4 crosses the line of intersection of the two planes (at (1, 3, 0)) and is perpendicular to the plane x+ y+ 2z= 1 (it's normal vector is <1, 1, 2> and <1, 1, 2>*<1, 1, -2>= 1+ 1- 4= 0).

    But my point is that that is one of many such planes. Taking b, c, and t to be any real numbers, the plane
    (2c-b)(x-(t+1))+ b(y- (3-2t))+ c(z- t)= 0 passes through the line of intersection and is perpendicular to the given plane.
     
  4. Sep 10, 2003 #3
    Yeah it was a homework. Nobody quite knew what to make of it, and the GSA (graduate student assistant) spent all night working on it and solved it using a system of 8 equations and 8 unknowns.

    I thought of it as the two planes that intersect. They form a line L. Then this plane that we are searching for, P1 must contain L. And the plane P1 is perpendicular to, P2, must contain all the normal vectors (n2) of P1.

    so,

    n2dot(r-r0) would be the P1 plane.

    with r representing (x,y,z) and r0 being some points on L
     
    Last edited: Sep 10, 2003
  5. Sep 11, 2003 #4
    You'd have to give your GSA 100% in effort! An 8x8 matrix is pretty big, unless of course you had a computer.
     
  6. Sep 11, 2003 #5

    HallsofIvy

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    Okay, the original post said " passes through the line of intersection" which I interpreted to mean that the line and plane simply intersected (in one point). Now it is clear that you intend the line to LIE IN the plane.

    Fortunately, I gave the general formula for any plane, containing at least one point of the line of intersection and perpendicular to the given plane:
    (2c-b)(x-(t+1))+ b(y- (3-2t))+ c(z- t)= 0

    Since we want the entire line to be in the plane, we can take any point on that line to be our "base" point. In particular take t=0 so that the point is (1, 3, 0) and the equation of the plane is
    (2c-b)(x-1)+ b(y-3)+ cz= 0.

    The entire line being in the plane also requires that (2, 1, 1)
    (t= 1) be in the plane so we must have
    (2c-b)(2-1)+ b(1-3)+ c(1)= 2c-b- 2b+ c= 3(c-b)= 0.

    Two points determine a line so that there is no way to determine b and c further. If we take c= b, then 2c-b= b and the equation of the plane is b(x-1)+ b(y-3)+ bz= 0 or bx+ by+ bz= 4b. As long as b is not 0 (which wouldn't give a plane) we can divide by b to get

    x+ y+ z= 4 as the equation of the plane.

    Notice that

    (1+t)+ (3- 2t)+ t= (1+3)+ (t- 2t+ t)= 4 for all t so the entire line is in the plane.

    also, the normal vector to the plane is <1, 1, 1> and
    <1,1,1>*<1,1,-2>= 1+ 1- 2= 0 so this plane is normal to x+ y- 2z= 3.

    The desired plane is given by x+ y+ z= 4.
     
  7. Sep 11, 2003 #6
    Thank you. I suspected you didn't need to do a system of 8 equations like the GSA did. I will show him your solution (giving you credit of course) and he can realize that it was simple.



    *as a note, I was doing the same thing you did, but I now see I made a silly error. I made an addition error so that I got a plane of x+2z =4
     
  8. Sep 11, 2003 #7

    HallsofIvy

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    Hey, those GSA's are always showin' off!
     
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