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Math Help - Intersection of a plane 3D Geo

  1. #1
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    Intersection of a plane 3D Geo

    Could anybody pease explain to me where I have gone wrong in the sum below the answer is ment to be t = 0.5


    The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2] and the plane defined by 3x+y -2z = -2

    using the formula: t =
    -n.s + n.po
    _________
    n.v

    I took s = [2,-1,1] v = [1,3,-2] n = [3,1 -2] and n.p0 = -2

    -n.s = [3,-1,2].[2,-1,1] = [-6,1,2]

    n.v = [3,1,-2].[1,3,-2] = [3,3,4]

    [-6,1,2] -2
    _________
    [3,3,4]

    [-8,-1,0]
    _______
    [3,3,4]

    t = [-2.6, -0.3,]
    t = [ -2. 9]
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  2. #2
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    Hello, aoibha!

    Sorry, I don't understand that formula you used.

    You didn't tell us what we're looking for.
    Is it the intersection of the line and the plane?


    The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2]
    and the plane defined by: .3x + y - 2z .= .-2

    . . . . . . . . . . . . . . . . . . . . . . . x .= . 2 + t
    The equations of the line are: . y .= .-1 + 3t
    . . . . . . . . . . . . . . . . . . . . . . . z .= . 1 - 2t


    Substitute into the equation of the plane:

    . . 3(2 + t) + (-1 + 3t) - 2(1 - 2t) .= .-2

    . . . . . . . .6 + 3t - 1 + 3t - 2 + 4t .= .-2

    . . . . . . . . . . . . . . . . . . . . . . 10t .= .-5

    . . . . . . . . . . . . . . . . . . . . . . . . t .= .-0.5


    The intersection is: .(1.5, -2.5, 2)

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  3. #3
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    Thanks! I was just trying to find t.
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  4. #4
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    Quote Originally Posted by aoibha View Post
    Could anybody pease explain to me where I have gone wrong in the sum below the answer is ment to be t = 0.5


    The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2] and the plane defined by 3x+y -2z = -2

    using the formula: t =
    -n.s + n.po
    _________
    n.v

    I took s = [2,-1,1] v = [1,3,-2] n = [3,1 -2] and n.p0 = -2

    -n.s = [3,-1,2].[2,-1,1] = [-6,1,2] <--- NO! The result is -3

    n.v = [3,1,-2].[1,3,-2] = [3,3,4] <--- NO! The result is 10

    [-6,1,2] -2
    _________
    [3,3,4]

    [-8,-1,0]
    _______
    [3,3,4]

    t = [-2.6, -0.3,]
    t = [ -2. 9]
    Your formula works fine but ...

    1. The equation of the plane is p: \vec n \cdot \vec r = -2

    The equation of the line is \vec r = \vec s + t \cdot \vec v

    Therefore the Point of intersection is determined by:

    \vec n \cdot (\vec s + t \cdot \vec v) = -2~\implies~t=\dfrac{-2-\vec n \cdot \vec s}{\vec n \cdot \vec v}

    2. And now comes the point where you have made a (I don't know an appropriate adjective) mistake: The result of a scalar product of 2 vectors is a real number. So using your values you'll get:

    t = \dfrac{-2-3}{10} = -\dfrac12
    Last edited by earboth; April 30th 2011 at 01:51 AM.
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