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Math Help - Vector ( Distane between 2 lines)

  1. #1
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    Vector ( Distane between 2 lines)

    Question: Two lines, l1 and l2, have equations r = -2i + k + t(i + j + k) and r = i + 5j + 2k + s(i + 2j + 3k) where s and t are constants. Points P and Q lies on the lines l1 and l2 respectively. If PQ is perpendicular to both l1 and l2, find the coordinates of P and Q and hence show that the length of PQ is √6 .

    This is how I solve the problem:
    l1 : r = (t - 2) i + t j + (t + 1)k
    l2 : r = (1 + s )i + (5 +2s) j + (2 + 3s) k

    Let b1 = i + j + k
    Let b2 = i + 2j + 3k

    b1 b2 = i 2j + k

    Since line b1 b2 is perpendicular to l1 and l2 , then I try to use the scalar product of (b1 b2 ) ∙ P and (b1 b2 ) ∙ Q to get the constants s and t , but I failed, could anyone help me to solve this problem? Thanks.
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Vector ( Distane between 2 lines)

    First I don't think that s and t can be constants ,because if it is, there cannot be a line l_1 and l_2

    Now to the problem,

    take the position vector of P and Q are respectively, (t_0-2)\,\mathbf{i}+t_0\,\mathbf{j}+(t+1)\,\mathbf{k} and (1+s_0)\,\mathbf{i}+(5+2s_0)\, \mathbf{j}+(2+3s_0)\, \mathbf{k}

    then

    \begin{align*}\overrightarrow{PQ}&=[(1+s_0)-(t_0-2)]\,\mathbf{i}+[(5+2s_0)-t_0]\,\mathbf{j}+[(2+3s_0)-(t_0+1)]\,\mathbf{k}\\ &=(3+s_0-t_0)\,\mathbf{i}+ (5+2s_0-t_0)\,\mathbf{j}+(1+3s_0-t_0)\,\mathbf{k}\\ \end{align*}

    take vectors parallel to l_1 and l_2 respectively \underline{b_1} and  \underline{b_2} then,

    \underline{b_1}= \mathbf{i} +\mathbf{j} + \mathbf{k}
    \underline{b_2}= \mathbf{i} + 2\,\mathbf{j} + 3\,\mathbf{k}

    So let's consider the given fact now, it says the \overrightarrow{PQ} is perpendicular to both \underline{b_1} and \underline{b_2}. This means following dot products should be equal to zero.

    \underline{b_1}.\overrightarrow{PQ}=0

    \underline{b_2}.\overrightarrow{PQ}=0

    from these two equation you'll get two equations involving t_0,s_0. Solve them and find the position vector of P and Q

    you'll eventually get  s_0=1, t_0=5
    Thanks from alexander9408 and HallsofIvy
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  3. #3
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    Re: Vector ( Distane between 2 lines)

    The vector b1 b2 does not have to be perpendicular to the vector from the origin to P. You can construct a system of three equations P + u(b1 b2) = Q where u is a real number and find s, t and u (I get s = 1, t = 5 and u = -1). Searching the web for "distance between skew lines" also returns many results, such this this PDF.
    Thanks from BAdhi, alexander9408 and HallsofIvy
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