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Intersection of planes and lines in space

  1. Dec 9, 2003 #1
    Could anyone help me summarize or if anyone knows good tricks in solving problems of lines intersecting with planes, etc, in 2d or 3d, the concept is same, but just want others opinion on what its basic idea is. i am able to do problems but i dont really understand them, i have to go back in my book and look at examples before i m able to do the problems. So if anyone has any tips plz tell me. thnx
  2. jcsd
  3. Dec 10, 2003 #2


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    Planes in 3d space are normally given by one linear equation in 3 unknowns: x, y, z. 3 variables, one equation connecting them: 3-1= 2
    "degrees of freedom" or two dimensions

    Lines in 3d space are normally given by three linear parametric equations for x, y, and z in a single parameter. 4 variables, 3 equations connecting them: 4- 3= 1 "degree of freedom" or 1 dimension.

    The way you solve for the intersection of planes or lines is to solve the equations as "simultaneous equations".

    Given three planes means you are given three equations for the 3 unknowns. As long as the equations are independent (i.e. two of the planes are not parallel) you can solve for the single point (x,y,z values) where the planes intersect.

    If you are given only two planes, you cannot solve for all three unknowns, but you could solve for two of them in terms of the third.
    If, for example you can solve for y(x) and z(x) as functions of x, you can write that as the parametric equations x= t, y= y(t), z= z(t) of the line where the two planes intersect.

    If you are given a line and a plane, then you are given one equation in x, y, z and three equations in x, y, z, and t (the parameter). That is, you have 4 equations you can solve for the four unknowns (although t has no "geometric significance" and so is irrelevant).
  4. Dec 10, 2003 #3
    also i had another problem, one of my homework questions was asking to find the reflection of a given point and a vector equation. the reflection of a point using the vector equation of a line as the axis.
    i m confused about how to find that.
    it looks something like this
    P(x,y,z) and r=(a,b,c)+t(A,B,C)
  5. Dec 11, 2003 #4


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    I think you would do better if you would post specific questions rather than general ones.

    If you really are given only that the line is given by
    r= (a,b,c)+ t(A,B,C) and p= (p,q,r) (I've changed from (x,y,z) because I want to reserve that for an "unknown" point), then:

    The line passes through the point (a,b,c) and has tangent vector (A,B,C). Geometrically, to find the point "symmetric" to (p,q,r) about the line, draw the perpendicular to the line from (p,q,r), and extend it beyond the line the same distance.

    Algebraically, to find the perpendicular, let (x,y,z) be the point on the given line such that the line from (x,y,z) to (p,q,r) is perependicualr to the given line. The vector from (p,q,r) to (x,y,z) is (x-p,y-q,z-r). Since it is perpendicular to (A,B,C), their dot product is 0: A(x-p)+ B(y-q)+C(z-r)= 0.
    Saying that (x,y,z) is on the line r= (a,b,c)+ t(A,B,C) means that, for some t, x= a+ tA, y= b+ tB, and z= c+ tC so that x-p= a-p+ tA, y- q= b-q+ tB, and z- r= c-r+ tC. The dot product becomes
    A(a-p+ tA)+ B(b-q+ tB)+ C(c-r+ tC)= 0 or
    A2t+ A(a-p)+ B2t+ B(b-q)+ C2t+C(c-r)= 0.
    Then (A2+B2+C2)t= A(p-a)+B(q-b)+C(r-c) so
    t= (A(p-a)+B(q-b)+C(r-c))/(A2+B2+C2).

    You can use that t to find the point (x,y,z) and then the equation of the line through (x,y,z) and (p,q,r) as well as the distance betwen them. That's the information you need to calculate the symmetric point to (p,q,r).
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