# Geometric locus

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• Mar 1st 2019, 01:45 PM
hedi
Re: Geometric locus
yes'i got it by a direct computation/
Thank's
• Mar 1st 2019, 02:15 PM
Plato
Re: Geometric locus
Quote:

Originally Posted by hedi
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
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.
• Mar 1st 2019, 03:35 PM
hedi
Re: 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 PM
Walagaster
Re: Geometric locus
Quote:

Originally Posted by Plato
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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