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Math Help - Vectors and Linear Algebra Question

  1. #1
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    Vectors and Linear Algebra Question

    Point P lies on the line:
    x=6+3t, y=4+8t, z=511t

    Point Q lies on the line:
    x=1312t, y=377+2t, z=1565t

    Find points P and Q when they are closest together.
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  2. #2
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    Hey mate,

    Whilst I'm sure there is a more efficient way of solving this, the following thought came to mind,

    Any point P can be expressed as P(t) = ( x1(t) , y1(t) , z1(t) )
    Similarly, Q = Q(t) =( x2(t) , y2(t) , z2(t) )
    **I will impose that the two lines do not intersect, (which can easily be shown as the evaluation of P(t) = Q(t) - however I imagine no solution for t exists).

    Then for any value of t the Distance between the points P(t), Q(t) is solved simply using,

    Dist(t) = sqrt( ( x2(t) - x1(t) )^2 + ( y2(t) - y1(t) )^2 + ( z2(t) - z1(t) )^2 )

    We observe that our original problem now involves a simply 1D minimisation problem,

    To be honest I don't have to strength to plug through the algebra, so I will attempt this in a general sense.

    Following high school mathematics, to identify any relative extrema solve,

    d/dt ( Dist(t) ) = 0

    d/dt ( Dist(t) ) = L(t)/(2*Dist(t)) = 0
    Where L(t) = 2 ( x2(t) - x1(t) ) ( d/dt (x2(t) - x1(t) ) + ....
    as x1(t) , x2(t) , ... , z2(t) are linear, thus there derivatives are merely constant values, Hence L(t) is a linear polynomial.

    Returning to our original problem,

    d/dt ( Dist(t) ) = L(t)/(2*Dist(t)) = 0

    Given our assumption that Dist(t) ~= 0 for all t, then we arrive at the simplified equation,

    d/dt ( Dist(t) ) = L(t)/(2*Dist(t)) = L(t) = 0

    Which given the nature of L(t) has a definate simple solution say t* and thus form the single point of relative extrema has been found, whilst one may state due to the nature of P(t), Q(t) that this point must reflect a relative minima we will persue onward and enforce this using the double derivative test,
    thus,
    d^2/dt^2 ( Dist(t) ) = (1/2)*( Dist(t) (d/dt ( L(t) ) - L(t) (d/dt ( Dist(t) ) ) / (Dist(t)^2)

    At t = t* L(t) = 0,

    thus,
    d^2/dt^2 ( Dist(t) ) = (1/2)*( Dist(t) (d/dt ( L(t) ) / ( Dist(t)^2 )
    = (1/2)*( d/dt( L(t) )/ Dist(t) )
    Given the nature of Dist(t) we observe Dist(t) > 0 and thus for t = t* to be a relative minimina d/dt( L(t) ) at t = t* must be positive - which Im sure it will be.

    Once this has been achieved simply sub in the value t = t* into P(t), Q(t) and Dist(t) and you will have identified the points P, Q and the length PQ

    Hope this helps,

    David
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