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Thread: Geometric locus

  1. #16
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    Re: Geometric locus

    yes'i got it by a direct computation/
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  2. #17
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    Re: Geometric locus

    Quote Originally Posted by hedi View Post
    there might not be a solution for any arbitrary point in the space.I need to think more about it,but anywayI will appreciate some help.
    Quote Originally Posted by Walagaster View Post
    I think you will find that in 3 space, if the lines are skew, the locus is a plane parallel to both lines and mid-way between them. Agreed
    I think that I post this before. If $\ell_1: P+t\vec{D} ~\&~\ell_2: Q+t\vec{E}$ are shew lines then there is a unique perpendicular them.
    The distance between the two lines is $d(\ell_1,\ell_2)=\dfrac{|\overrightarrow {PQ} \cdot (\vec{D} \times \vec{E})|}{\|\vec{D}\times \vec{E}\|}$
    To find the endpoints of that common perpendicular is both tricky & tedious. SEE HERE
    The vector $\vec{N}=\vec{D}\times\vec{E}$ the normal for the plane you want.
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  3. #18
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    Re: Geometric locus

    It's okay.I also calculated the geometric locus in my way and got the same answer.
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  4. #19
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    Re: Geometric locus

    Quote Originally Posted by Plato View Post
    I think that I post this before. If $\ell_1: P+t\vec{D} ~\&~\ell_2: Q+t\vec{E}$ are shew lines then there is a unique perpendicular them.
    The distance between the two lines is $d(\ell_1,\ell_2)=\dfrac{|\overrightarrow {PQ} \cdot (\vec{D} \times \vec{E})|}{\|\vec{D}\times \vec{E}\|}$
    To find the endpoints of that common perpendicular is both tricky & tedious. SEE HERE
    The vector $\vec{N}=\vec{D}\times\vec{E}$ the normal for the plane you want.
    Since the OP has solved the problem, I will point out that alternatively, if $\vec R_1 = \vec P +t\vec D$ and $\vec R_2 = \vec Q + s\vec E$ then the midpoint between them is $$
    \vec M = \frac 1 2(\vec R_1 + \vec R_2) = \frac 1 2(\vec P + \vec Q) + \frac 1 2 t\vec D + \frac 1 2 s\vec E$$which is a parametric representation of the "mid" plane if $\vec D$ and $\vec E$ are not parallel.
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