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Finding the line of intersection of 2 planes

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    The planes 5x-2y-2z = -1 and x-4y+2z = 25 are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is ....

    3. The attempt at a solution

    Ugh.... this is a doozie...

    So I started the problem by adding both plane equations

    5x - 2y - 2z = -1 (1)
    x - 4y + 2z = 25 (2)

    I get

    6x = 6y-24

    now here is where I run a little brain snag...

    x = y -4 (I'm not sure if this is correct... shouldn't it be x = (6y-24)/6?) anyway carry on

    so then substituting x into the second equation up there...I gets

    -3y + 2z = 29
    2z = 29 - 3y
    z = 29/2 - 3y/2

    I have, hopefully, done this correctly and got x= y - 4 and z =29/2-3y/2
    t = y
    I need to find y and the parametric equation for it

    so far i got (- 4, *, 29/2)+t(1, *,-3/2)

    please halp...
  2. jcsd
  3. Nov 15, 2008 #2


    Staff: Mentor

    x = y - 4 is equivalent to x = (6y - 24)/6.

    You haven't done enough that I can check your work, so here's another approach.
    1. Find a normal to plane 1. You can do this by inspection.
    2. Find a normal to plane 2. ditto
    3. Calculate the cross product of these normals. This will give you a vector that is perpendicular to both normals, and that will have the same direction as line you want. Let's say that the coordinates of this vector are (A, B, C)
    4. Find a point that is common to both planes, by using the equation you found, x = y - 4. In this equation, z and y are arbitrary. Let's say that the point is [tex](x_0, y_0, z_0) [/tex].
    The equation of your line will be [tex](x, y, z) = (x_0, y_0, z_0) + t(A, B, C) [/tex].
    You should be able to find values for [tex]x_0, y_0, z_0, A, B, and C[/tex].
  4. Nov 15, 2008 #3
    n1 = <5, -2, -2>
    n2 = <1, -4, 2>
    n1 x n2 = <-12,-12,-18>

    mkay... so now

    (x, y, z) = (x0, y0, z0) + t(-12, -12, -18)

    and this point
    x = y -4 is used to find (x0,y0,z0).....

    I'm stuck, yuck.
  5. Nov 15, 2008 #4


    User Avatar
    Homework Helper

    Pick any value for y (or x) and then find the corresponding value for x (or y). For example, put y=0 and find x.

    the put those values into either equations for the planes and get z. You will now have a point on the line that you want to get.

    EDIT: If you wanted to, you could have just let y=t and just find x and z in terms of t and you'd get that vector line.
    Last edited: Nov 15, 2008
  6. Nov 16, 2008 #5


    User Avatar
    Science Advisor

    Yes, and (6y- 24)/6= (6y/6)- (24/6)= y- 4

    At this point, I think I would let y itself be the parameter: y= t so x= t- 4. Now put those back into x- 4y+ 2z= 25: t-4- 4t+ 2z= 25 and solve for z. 2z= 25+ 3t+ 4= 29+ 3t so z= 29/2+ (3/2)t.
    Parametric equations for the line are x= t-4, y= t, and z= 29/2- (3/2)t.
    The vector equation would be r= (-4, 0, 29/2)+ t(1, 1, -3/2).

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