yes'i got it by a direct computation/

Thank's

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- Mar 1st 2019, 01:45 PMhediRe: Geometric locus
yes'i got it by a direct computation/

Thank's - Mar 1st 2019, 02:15 PMPlatoRe: Geometric locus
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. - Mar 1st 2019, 03:35 PMhediRe: Geometric locus
It's okay.I also calculated the geometric locus in my way and got the same answer.

Thank's - Mar 1st 2019, 04:59 PMWalagasterRe: Geometric locus
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.