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Equation of a plane

  1. May 31, 2004 #1


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    I'm having trouble working out the equation of a plane. Say I'm given 3 coordiantes , how would I work out the equation of the plane in the form of ax + by + cz = d?

    say (1,-1,-1), (2,1,2), (2,-2,1) for example.

    thank you in advance
  2. jcsd
  3. May 31, 2004 #2
    Let A = (1, -1, -1), B = (2, 1, 2) and C = (2, -2, 1). The vectors A - B and A - C will be parallel to the plane (of course, so will B - C, C - A, and so on). A normal vector to the plane will be given by the cross product between A - B and A - C, let's assume this vector is (n_1, n_2, n_3). The plane will then have the equation n_1x + n_2y + n_3z + d = 0, finding d is simple since you know a point on the plane (in fact, you know three).
  4. May 31, 2004 #3
    A(1, -1, -1)
    B(2, 1, 2)
    C(2, -2, 1)

    [tex]\vec{AB} = (1, 2, 3)[/tex]
    [tex]\vec{BC} = (0, -3, -1)[/tex]

    The cross product is:
    Code (Text):
        |i  j  k|
    n = |1  2  3| = -2i + 9i + 0j + j - 3k - 0k = (7, 1, -3)
        |0 -3 -1|
    (At least that's how we were taught to do it.)

    So the plane equation is: 7x + y - 3z + D = 0
    To find D, substitude the coordinates of one of the points in the equation and solve for D = -9.
    Last edited: May 31, 2004
  5. May 31, 2004 #4
    In 3D (the space with 3 dimensions) the general form of the equation is A*x +B*y +C*z +D=0. You have three points in 3D, which uniquely determine your plane.
    Since your points belong to the plane, their coordinates have to verify identically plan’s equation.
    Plug in the known values for x, y, and z; you will get a system of 4 equations with four unknowns.
    Through elimination’s you should be able to determine A, B, C and D and write the plane's equation, containing your three points in its general form.
    Another way is writing a matrix of the system.
    Plug in the known values for x, y, and z; you will get a system of 4 equations with 4 unknowns.
    You’ll get :
    First line: | x y z 1|
    2nd line: |x1 y1 z1 1 |
    3rd line | x2 y2 z2 1| = |0|
    4th line: | x3 y3 z3 1|
    This should be equal with zero(if you write it like a matrix, as I tried above). Doing the calculations(elimination or reduction) you should get the plane equation.
    If you have problems e-mail me for help.
    Here’s a site where you can easily find information regarding general math and good reference:

    Hope this helps,

  6. May 31, 2004 #5


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    thank you so much guys for your help :)
  7. May 31, 2004 #6


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    Some more questions.. regarding that plane. :(
    We also have a line going through the points (-3.2,1) and (2,1,-5)
    I worked out the equation of that line to be
    r= (-3,2,1) + t(5,-1,-6)

    I have to find the angle the line makes with the plane. The point where they intersect and the shortest distance from the line and the plane to the origin.
  8. Jun 1, 2004 #7
    I think you can just find the angle between the line and the normal vector of the plane, using the dot product. And then the angle between the line and the plane is just 90o minus that.
  9. Jun 1, 2004 #8


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    That makes sense... How do I get the normal vector of the plane?
  10. Jun 1, 2004 #9
    If the plane's equation is Ax + By + Cz + D = 0, the normal vector is (A, B, C). We used this eariler, except we found the normal vector and from there proceeded to find the plane's equation.
  11. Jun 4, 2004 #10
    The point of intersection of the line and plane can be found by solving the system of equations: Line: [tex](x, y, z) = (k, l, m)t + (a, b, c)[/tex], Plane: [tex]Ax + By + Cz + D = 0[/tex]. So:

    [tex]A(kt+a) + B(lt+b) + C(mt+c) = -D[/tex]

    Solve for t:

    [tex] t = -\frac{D + Aa + Bb + Cc}{Ak + Bl + Cm} [/tex]

    And plug it back into the equation for the line to get the point of intersection. (It's easier to do this process in each instance than it is to remember the formula.)

    For any point not on a line or plane, the shortest distance is along a line normal to the original line or plane. In particular, for the origin and the plane, the line along which the least distance is accomplished is [tex](A, B, C)t[/tex]. The closest point on the plane to the origin is the point where this line intersects the plane. Finding the point as above gives

    [tex] t = -\frac{D}{A^2 + B^2 + C^2}[/tex]

    and the point [tex](At, Bt, Ct)[/tex]. The distance from this point to the origin is then [tex]\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}[/tex].

    An alternative approach for finding the distance from the line to the origin: Let the line be [tex]\vec{r} = \vec{A}t + \vec{B}[/tex], then you need to minimize [tex]|\vec{r}|[/tex], but the square root is an increasing function, so this is the same as minimizing [tex]r^2 = A^2t^2 + 2\vec{A}\cdot\vec{B}t + B^2[/tex]. The minimum of this quadratic equation in [tex]t[/tex] is found at the vertex: [tex]t = \frac{-\vec{A}\cdot\vec{B}}{A^2}[/tex]. Using this, we find that [tex]r^2 = B^2 - \frac{(\vec{A}\cdot\vec{B})^2}{A^2}[/tex]. Since [tex]\vec{A}\cdot\vec{B} = AB \cos \theta[/tex], where [tex]\theta[/tex] is the angle between them, this simplifies to [tex]r = B\sin \theta[/tex].
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