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Equations of Planes from Symmetric Equation of a Line

  1. Jun 7, 2015 #1
    Hello,


    Suppose I have a vector equation:

    \begin{cases}
    x=0+10t\\
    y=0+10t\\
    z=0+10t
    \end{cases}

    Which forms the symmetric equation [itex]\frac{x-0}{10}=\frac{y-0}{10}=\frac{z-0}{10}[/itex]

    Now, I know the symmetric equations can be split up so that you can form the two planes whose intersection yields the initial vector:

    [itex]\frac{x-0}{10}=\frac{y-0}{10}[/itex] and [itex]\frac{y-0}{10}=\frac{z-0}{10}[/itex]

    but I haven't been able to find any examples on how to get from the split symmetric equations of the line to two separate equations of a plane in standard form.

    Would I just cross multiply each?

    and get

    Plane 1: x = y or x - y + 0z = 0
    Plane 2: y = z or 0x + y - z = 0
     
    Last edited: Jun 7, 2015
  2. jcsd
  3. Jun 7, 2015 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is what you want. More generally, given the line with parametric equations x= At+ a, y= Bt+ b, z= Ct+ d, we can write
    [tex]t= \frac{x- a}{A}= \frac{y- b}{B}= \frac{z- c}{C}[/tex]

    Break that into the two equations
    [tex]\frac{x- a}{A}= \frac{y- b}{B}[/tex]
    and
    [tex]\frac{y- b}{B}= \frac{z- c}{C}[/tex]

    Multiply the first equation by AB to get B(x- a)= A(y- b) which is the same as Bx- Ay= aB- bA.
    Multiply the second equation by BC to get C(y- b)= B(z- c) which is the same as Cy- Bz= bC- cB.

    Those are the equations of the two planes that intersect in the original line.
     
  4. Jun 7, 2015 #3
    Thank you. Now I'm going to work in reverse on my own and see if the intersection of the two planes via cross product yields the original line.

    Appreciate your clarification on this topic.
     
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