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Math Help - Vector equation for the line..

  1. #1
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    Question Vector equation for the line..

    Find the vector equation for the line of intersection of the planes and

    , ,0 8, , .


    i think this problem have something to do with distance of point and plane.but that is just my guess
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  2. #2
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    Quote Originally Posted by DMDil View Post
    Find the vector equation for the line of intersection of the planes and

    , ,0 8, , .


    i think this problem have something to do with distance of point and plane.but that is just my guess
    Solve the second for x, then subs. into the first and solve for y. I got


     x = -1 - \frac{2}{3} z,\;\;\;y = \frac{3}{2} - \frac{1}{4} z. If we let z = - 12t

    then the line is given by

    \bold{r} = \left(-1,\frac{3}{2},0 \right) \;+ <8,3, -12> t
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    Quote Originally Posted by DMDil View Post
    Find the vector equation for the line of intersection of the planes and

    , ,0 8, , .


    i think this problem have something to do with distance of point and plane.but that is just my guess
    The point (3,3,-6) is on both planes.
    The direction vector is \left\langle {3,4,3} \right\rangle  \times \left\langle {3,0,2} \right\rangle , the cross product of the normals.
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  4. #4
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    sorry guys,, i am little lost now...plz show more steps..

    i understand that i have to take cross propuct of this <3,4,3> X<3,0,2> gives me <8, 3, -12> but i don't get how find this points (-1, 3/2, 0)
    Last edited by DMDil; February 8th 2009 at 05:19 PM.
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