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Planes intersect when ?

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the equation of a line of the intersecting planes
    ##3x_1-2x_2+x_3=5##
    ##2x_1+3x_2-x_3=-1##

    2. Relevant equations



    3. The attempt at a solution
    I didn't know where to start but I started at trying to find the cross product of the planes (needless to say it didn't get me and where) I got ##-x_1-5x_2+11x_3## that equation doesn't make any sense x.x I do not know what to do x.x
     
  2. jcsd
  3. Feb 19, 2013 #2

    joshmccraney

    User Avatar
    Gold Member

    first use elimination on the planes, thus add [itex](1)+(2)=5x_1+x_2=4[/itex]:
    solving gives: [itex]x_2=4-{5x_1}[/itex]
    take [itex]x_1=t[/itex] thus [itex]x_2=4-{5t}[/itex]
    now sub [itex]x_1,x_2[/itex] into [itex](1)[/itex] or [itex](2)[/itex] to find [itex]x_3[/itex] in terms of [itex]t[/itex]
    now you may write up the line in parametric form: [itex](x_1(t),x_2(t),x_3(t))[/itex]
     
  4. Feb 19, 2013 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Tenshou! :smile:
    no, your 11 should be 13, shouldn't it? :wink:

    (and i think one of your signs is wrong)

    that method should work …

    it gives you the direction that is perpendicular to both normals, and therefore it must lie in both planes …

    now you have something like x2 = Ax1 + B, x3 = Cx1 + D, where A and C are known,

    so you substitute that into the original equations, and that gives you two equations in two unknowns (B and D) …

    but i'm not sure that's any quicker than joshmccraney's :smile: method!​
     
  5. Feb 19, 2013 #4
    Thank you for your insightful answers but I think the easiest solution was IMS... Thank you so much, I mean I just did it this morning (and yes my maths were wrong.) Solved the equation of

    ##A####x##=##b##

    I was looking for the simplest way to solving it and josh, I think your way is the simplest way for solving it(thanks by the way). Although, I still did not get the equation in the book I got ##A## as a 2x3 matrix

    ##
    \left[
    \begin{array} {r r r r}
    \ 3 &\ -2 &\ 1 \\
    \ 2 &\ 3 &\ -1 \\
    \end{array}
    \right]
    ##

    Then after rref(A) [or something close to it] I got

    ##
    \left[
    \begin{array} {r r r r}
    \ -7 &\ 0 &\ 1 \\
    \ -5 &\ 1 &\ 0 \\
    \end{array}
    \right]
    ##

    I calculated by Nullity-Rank that I should have one free column left. I solved and got
    ##
    \left[
    \begin{array} {r r r r}
    \ 0 &\ 4 &\ 13 \\
    \end{array}
    \right] = x_{part}^T
    ##

    then I just found out the solution to the null space thank you for correcting my maths tiny-tin

    ##
    \left[
    \begin{array} {r r r r}
    \ -1 &\ 5 &\ 13 \\
    \end{array}
    \right] = x_{null}^T
    ##

    then I finished up by allowing ##x_{comp}=\lambda x_{null}+x_{part}; \forall \lambda \epsilon\mathbb{R}##

    Although I went though a page and a half of calculation I will remember your method josh!

    the answer in the book is ##x_1 = -k+1; x_2 = 5k-1; x_3 = 13k; \forall k \epsilon \mathbb{R}##
     
    Last edited: Feb 19, 2013
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