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Even easier.
Forget the Plane intersection lines, forget the average vector.
Since we know, that the plain touches the the sphere, witch has a known radius,
we make a parallel copy of each plane, so that they go right trough the center of the sphere.
Since we have 3 planes, we can solve for the coordinates.
Hesse Cartesian:
E:x = (-2x-2y+4z)/sqrt(24) = 0.5 = radius of sphere
F:x = (-x+y-z) /sqrt(3) = 0.5 = radius of sphere
G:x = (-4y-12z) /sqrt(160) = 0.5 = radius of sphereNow solve those equations for x, y and z:
solve([(-2x-2y+4z)/sqrt(24)=0.5, (-x+y-z)/sqrt(3)=0.5, (-4y-12z)/sqrt(160)=0.5], [x,y,z])
The numeric solution is:
centerpoint of sphere
= ((1/18)*(-5*sqrt(3)-4*sqrt(6)-sqrt(10)) | (1/6)*(sqrt(3)-sqrt(6)-sqrt(10)) | (1/18)*(-sqrt(3)+sqrt(6)-2*sqrt(10))
≈ (-1.2011 | -0.64662 | -0.31151)
Edit: I've corrected the mistake of putting sqrt(24) instead of sqrt(160) in G.
@garnesh: Got it.
@Bob: You're welcome. I see you start out with lines, which I do with planes.
Regarding your first and second reply:
Since it is an irregular triangle a line trough the origin, touching the sphere
might not be perpendicular to a line representing a base side.
Or do you think otherwise ?
I tried to put it mathematically:
Is every line, representing a base side and touching the sphere,
perpendicular to the line going through the origin and touching the sphere,
which it crosses, for all irregular triangular pyramids ?
It might work with regular pyramids, tough,
but how would we get to irregular ones from there ?
Following my approach would look like this:
Fist I rename my planes real quick
(starting with E cause German "Ebene", which means plain): AB=E, BC=F, CA=G:
Vectors: Hesse Cartesian:
E:x = r( 2|-3|1)+s(-2|-3|1) E:x = (-2x-2y+4z)/sqrt(24) = 0
F:x = r( 2| 0|1)+s( 0| 2|1) F:x = (-x+y-z) /sqrt(3) = 0
G:x = r(-1| 0|1)+s( 0| 1|1) G:x = (-4y-12z) /sqrt(24) = 0Plane intersection lines:
EF:x=r(1|3|2)
EG:x=r(1|-3/5|-1/5)
FG:x=r(1|3/4|-1/4)
Their average vector: V:x=r(1|3.15|1.55)
It should go right through the center of the sphere, right ?
Cause it's lovely out there, I will continue later.
@garnesh: Thank you for your patience.
@Bob: Beside the two mathematics modules,
we had one statistics module, one physics module
and many chemistry and of coarse biology modules, too.
The statistics module, held in the physics lecture hall,
was a mandatory primer for the physics internship.
So it felt more attached to physics.
I also participated in a math intro, aimed at physics students,
took a c++ lecture as a special module
and watched some regular chemistry lectures.
But after completing math I discontinued.
It is not that I do not like the subject of biology,
but i had private reasons.
Yes, many of my fellow students
did not particularly like math and physics.
Regarding referers:
The browser usually tells the server from which page it comes,
when it goes to a new page.
Especially between domains,
this can be a privacy issue, so I deactivated this.
Your forum however wants it for posting.
Since i have not had issues elsewhere, this was unusual to me.
I mentioned it in case it could be fixed somehow.
For now I temporary activate referers while posting.
hello @ganesh, i've read the rule and thought i would follow it.
please send me details about your complain.
should i for instance type "i have" instead of i've ?
is there a more detailed nettiquette ?
hello @Bob, thank you for helping. you got my definition right.
as requested,
i tried to choose some numeric values for a test:
AB:x=r(2|-3|1)+s(-2|-3|1)
BC:x=t(2|0|1)+u(0|2|1)
CA:x=v(-1|0|1)+w(0|1|1)
(would we need them in the hesse normal form ?)
r_sphere=.5
the torus would lie flat on BC,
so BC and the the torus have a circle
with r=R_torus_core in common.
but it touches AB and CA just in a single point each.
r_torus_rounding=.5
R_torus_core=1.5
this is for rounding edges of 3d-models using hull functions
(imagine a hulled cube with spheres or torusses in the edges):
you have a irregular, triangular pyramid, which is upside down.
the tip is in the origin.
now you knock the base out and put a ball in.
where is its center going to be ?
you do the same with a torus.
that's basically it.
given are 3 plains, cutting through the origin
and the radios of the sphere
the torus lies flat on the second plane
and the adjacent planes can also form a "v", if looked at from above,
which makes the torus stick to the side of the pyramid.
given are 3 plains, cutting through the origin
and the 2 radiuses of the torus
my approach, at first with the sphere, would probably be:
cutting the three planes,
going a equal distance on the cut vectors,
taking the average.
as a vector, this would point to the center of the sphere, right ?
if i could, with the normal vector of a plain,
span a rectangular triangle, with the center of the sphere
in the upper edge, i could solve it with trigonometry, right ?
before i go through it, are there more straight forward approaches ?
with the torus it gets more complicated
and i probably would have difficulties i solving that.
does somebody have a hint for me ?
hello folks,
the last exams i passed, have been about
analysis and linear algebra for biologists
in a german university.
of coarse i have not kept everything,
but it might be a rough guide about
how one could approach me best.
almost sadly, there have not been any more math modules in the subject.
right now i mainly do hobbyist stuff like 3d-printing with what i have learned.
lowercase is ok here ?
hmm, the site says bad referer header, if i send none.
do i have to have them activated ?