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Discrete systems totaly stuck

  1. Aug 7, 2008 #1
    i have an exam in a few days and am certain a question like this is going to pop up but i have no solutions to this question and no idea how to work it out the question is as follows

    Find the equations of the line L1 through the point with position vector (4,2,1) and parallel to the vector (-3,3,-9)
    Find the equation of the line L2 through the points with position vectors (1,-1,2) and (3,0,3)
    Show that L1 and L2 intersect and find the point of intersection, thanks in advance for your help
     
  2. jcsd
  3. Aug 7, 2008 #2

    HallsofIvy

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    If you have a test you should have already covered this! And they are not all that hard. A line, with position vector <A, B, C>, passing through point (x0, y0, z0) has parametric equations
    x= At+ x0, y= Bt+ y0, z= Ct+ z0

    If you are given two points, (x0, y0, z0) and (x1, y1, z1), then a direction vector for the line is the vector from one point to the other: <x1- x0, y1- y0, z1- z0>.

    To find where they intersect (if they do) call one parameter t, the other s, and set the x, y, z coordinates of the two points equal. That gives you 3 equations for the two unknown numbers t and s.
     
    Last edited: Aug 9, 2008
  4. Aug 8, 2008 #3
    so for the first 2 parts would my answers be ok like this or would i have to arrange it differently

    X=t(4,2,1)+(-3,3,-9)

    X=t(2,1,1,)+(3,0,3)
     
  5. Aug 9, 2008 #4
    so for the first 2 parts would my answers be ok like this or would i have to arrange it differently

    X=t(4,2,1)+(-3,3,-9)

    X=t(2,1,1,)+(3,0,3)
     
  6. Aug 9, 2008 #5

    HallsofIvy

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    Are Franky2727 and pokerfan91 clones?
     
  7. Aug 9, 2008 #6
    no its me and a friend we study together and i hadnt realised he was still logged into my PC when i posted and guessed that the reason no one had replied was as the last poster was not me
     
  8. Aug 9, 2008 #7

    Defennder

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    The first one is incorrect.
     
  9. Aug 11, 2008 #8
    what exactly is incorect about the first one?
     
  10. Aug 11, 2008 #9

    Defennder

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    Which is the parallel vector from your vector equation?
     
  11. Aug 11, 2008 #10
    -3 3 -9
     
  12. Aug 11, 2008 #11
    also how do i find the intersection? which parameters do i set as s and t the position vectors?
     
  13. Aug 11, 2008 #12

    Defennder

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    No, that's not what your vector equation of a line shows. The parallel vector is the one which is scaled by a parameter t, not the reference position vector.

    First get the above two vector equation of lines correct. Then it is apparent that where the two lines intersect, there exists some value of s and t, where s and t are the parameters of the vector equation, where the vector line equations are equal. Solve them for s,t.
     
  14. Aug 11, 2008 #13
    its clicked! right so i'll get my position vector and my paralel vector and because the vector is paralel and not acctuly on the line (however in the same direction) it needs to be that direction but to a diffrent scale (hence t) plus the position vector giving X=(-3,3,-9)t +(4,2,1) right so that along with my other answer of X=t(2,1,1)+(3,0,3) what do i need to do with S's and T's to get the right value? do you mean replace one of the t's with an S for instance the second equation and then i need a value where t=s therefore there at the same point? i hope you do because that would mean i get this! thanks for the help so far
     
  15. Aug 11, 2008 #14

    Defennder

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    As said before you have X1 and X2. These are both vectors from the origin pointing to any point on the lines of X1 and X2. Where X1 and X2 coincide, their vector equation of the lines are the same because those two values of s,t are such that they give the same intersecting coordinate. So equate the two vector line equations and find s,t
     
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