Fixed Ratio of Distances

[gudi]

When is the locus of a point, moving so as to retain a fixed ratio of
distances between a fixed point and a fixed straight line in R^3,
also another straight line?

Best Regards and happy New MMXI !
Narasimham

[Avni Pllana]

Hi Narasimham,

I think the problem has to do with the Apollonian Circles Theorem. Let A be the given fixed point and L the given straight line. Let B be the intersection point of L with the plane E through the point A and orthogonal to L. Then if a point P moves on E such that PA/PB=1, the locus of P is a straight line, it is the perpendicular bisector of the segment AB.

Best regards,
Avni

[gudi]

But B or variable point B1 which is the foot of the perpendicular
dropped from P should not be a fixed point. i.e., when P is moving
away from BA, B should also move along L, PB or PB1 should always be
perpendicular to L.

Best Regards,
Narasimham

[Avni Pllana]

Hi Narasimham,

I think that in cases other than that described previously, the locus of P is not a straight line.

Best regards,
Avni

[gudi]

Hi Avni,

B is not a given point ,but is on a given line L. So there cannot
exist such a locus, may be?

Regards,
Narasimham

[Avni Pllana]

Hi Narasimham,

Let A=[0, 1, 0] be the fixed point and let the z-axis be the fixed line. Let r be the fixed ratio of the distance of a point P from z-axis over the distance of P from A. The ratio r takes values between 0 and 1. Then the line through B=[0,r/(1+r), 0], and direction vector a=[r, 0, sqrt(1-r^2)], is the locus of points P that have the fixed ratio r of distances from z-axis and point A. Now we can write

P = B+l*a = [0,r/(1+r), 0] +l*[r, 0, sqrt(1-r^2)], or

P = [ l*r, r/(1+r), l*sqrt(1-r^2)] ...(1)

The distance of P from z-axis is

d1 = sqrt(l^2*r^2 + r^2/(1+r)^2) ...(2)

The distance of P from A is

d2 = sqrt(l^2 + 1/(1+r)^2) ...(3)

>From (2) and (3) we have

d1/d2 = r .

Best regards,
Avni

[Philippe 92]

Hi,

You wrote :
>> Then the line ... is the locus of points P that have the fixed

>> ratio r of distances from z-axis and point A.

???

Such a locus in R^3 is obviously a *surface*, not any line,
straight or not.

Let M(x,y,z) a point in R^3
distance from M to line L x=y=0 (z axis) is sqrt(x^2 + y^2)
distance from M to A = (0,1,0) is sqrt(x^2 + (y-1)^2 + z^2)
Let H the perpendicular projection of M on z axis.
The searched locus is MA/MH = 1/r = e

so that the locus is the *surface* with equation :
x^2 + (y-1)^2 + z^2 = e^2(x^2 + y^2)

That is z^2 = (e^2 - 1)(y^2 + x^2) + 2y - 1

Consider the intersection of this surface with a plane z = k
parallel to xOy plane.
That is (y^2 + x^2) + 2y/(e^2 - 1) = (k^2+1)/(e^2 - 1)

If e != 1 it is the equation of a circle
The center of this circle is x = 0, |y| = 1/sqrt(|e^2 - 1|)
the sign of y depending on e > 1 or e < 1, but this center doesn't
depend on z, hence this center is on a straight line parallel to Oz
When z = 0, this circle is on the xOy plane and is the well known
Appolonius circle, locus of points M in the plane with MA/MO = e
Let I the center of this Appolonius circle.

*************************************************
* If e != 1, the locus is a revolution surface, *
* the axis being the parallel to Oz in I *
*************************************************

Intersect now this surface with the yOz plane, that is in this yOz
plane the locus of points with MA/MH = e, H being the perpendicular
projection of M on axis Oz.

It is well known that this curve is a conic section with
excentricity e.
A is the focus and Oz axis the directrix.
That can also be seen at once by writing the equation as
z^2 - (e^2 - 1)y^2 - 2y = (e^2 - 1)x^2 - 1
with x = 0 (plane yOz) it is z^2 - (e^2 - 1)y^2 - 2y = - 1
if e > 1 it is an hyperbola
if e = 1 it is a parabola
if e < 1 it is an ellipse

*******************************************************
* if e = 1 the locus is a parabolic cylinder *
* if e > 1 it is a revolution one sheeted hyperboloid *
* if e < 1 a revolution ellipsoid *
*******************************************************

Now for straight lines **included** in this locus.
If e = 1 the cylinder generators are infinitely many straight lines,
parallel to Ox axis.
If e > 1 there are also two families of infinitely many straight
lines (well known property of the one sheeted hyperboloid :
http://en.wikipedia.org/wiki/Hyperboloid "doubly ruled")
if e < 1 the ellipsoid contains no straight line.

In the case of hyperboloid, the straight lines may be obtained by
intersecting the surface with a tangent plane at vertex of the
generating hyperbola. That is a plane perpendicular to Oy and
tangent to the Apollonius circle.
These lines are then y = k, z^2 = (e^2 - 1)x^2
It can be constructed as follows :
let D, D' the intersections of the Apollonius circle in xOy with Ox
OD^2 = OD'^2 = 1/(e^2 - 1) = x^2/z^2
that is the lines are parallel to KD and KD', with K = (0,0,1)

Best Regards, and Happy New Year.

[gudi]

Hi Philippe,

Yes,had the same results; after the full surface doubly ruled
hyperboloid of one sheet is obtained the straight line generators/
rulings are selected.The skewed asymptote generators for case e > 1are
the solution.

I believe even the simple extension of equal distances ( e = 1) in 3D
applicable as a parabolic cylinder as a generalization is unknown to
many. For e ->1, at z = 0 sections the Apollonius circles tend to the
parabola.

Although an elementary, I regard the study as a fundamental extension
of conic sections. We can use the same concept and words directrix,
focus for the locus in R^3 here as well as in planar conic sections.
Apollonius circles, one sheet hyperboloid asymptotes, parabolic
cylinder all nicely appear here together.

What still beats me though is, why are not all asymptotes in the
rulings family not considered the answer? Afterall, they are all on
the surface of revolution (y^2 + x^2) + 2y /(e^2 - 1) = (k^2+1) /(e^2
- 1), satisfying MA/MH = e.That is the motivation of this new post.

I shall shortly upload Mathematica visualization and a sketch of the
hyperbolic paraboloid ( z/h = x y (a b) ) spanning between the
asymptotes (inside the hypar, normal to it) made much before posting
this.Their edges can be also regarded as the required locus.

Hope all enjoyed the topic. So familiar, but interesting
generalization.

Best Regards,
Narasimham

[Walter Whiteley]

Just as an aside - perhaps known to those who play with such loci, 

the locus of points equal distant to two straight lines in 3-space is a hyperbolic paraboloid.  

This surface does appear in some surprising settings.  For example portions are the boundaries between modules of a pig's liver or rats live - as the boundaries are formed for blood flow into modules - at equal distance from two drain holes (roughly straight line segments).  

There is an article about this in the google book on 

Multidisciplinary approaches to visual representations ... Volume 2

Walter Whiteley

York University

[gudi]

On Jan 5, 7:11�pm, Walter Whiteley <white...@mathstat.yorku.ca> wrote:
> Just as an aside - perhaps known to those who play with such loci,
> the locus of points equal distant to two straight lines in 3-space is �
> a hyperbolic paraboloid
.

Yes, and so, could you or anyone please confirm this:

Hypar parameterization is

(x,y,z) = ( u + k v, u - k v, u*v/c )

So that u = a, u = 3a are 'bisected' by u =2a,

( Equal R3 embedded straight euclidean lengths, not arcs along
surface)

and u = a, u = 4a are 'trisected' by u =2a and u= 3a etc.?

Best Regards,
Narasimham

> Walter Whiteley
> York University

[gudi]

Asymptotic lines extracted from One sheet hyperboloid surface and
hyperbolic paraboloid surfaces.

The two pictures are drawn to scale ( Mathematica and Ansys structural
analysis package).
They depict a case when ratio of distances = eccentricity = 2 ( > 1
case).
They are hyperbolic parallels.
For eccentricity >1, < 1 Apollonius circles would obviously appear on
main centre section,
from ellipsoid surfaces.

http://i51.tinypic.com/2628ieo.png
http://i55.tinypic.com/nfinoo.jpg

Invariant Cross Ratio = 1 is also involved for point P. Both in the
plane as Apollonius circles
and out of this plane shown in the black/white line sketch.

With Best Regards,
Narasimham

[max]

thanks for the solution