|
Kentor
United States5784 Posts
Yeah seriously. Instead of saying
On April 21 2010 07:17 Capook wrote:
Given three points A,B,C, find a fourth point H not coplanar with A,B,C such that the distance between H and C is minimized.
They had to make it complicated AND with a stupid picture. But
On April 21 2010 07:17 Capook wrote:
If the restriction of not coplanar is lifted, then the answer is obvious: choose H=C, so that the distance is zero. However, the "not coplanar" eliminates this choice (as well as some other choices). But there are tons of points left and you are asked to pick the one that minimizes the distance to C. That is like asking you to find the smallest positive number. (There isn't one.)
you are forgetting that H makes a plane with AB too.
I think this is what's making it unclear
![[image loading]](http://img339.imageshack.us/img339/3048/30129237.png)
I think by this you assume that it means H is the point on the ABH plane that is the shortest away from C.
|
note that my answer assumes that ABH are collinear otherwise the question doesn't quite make sense.
|
Stupid question. Worst math thread ever!
|
Actually i'm giving this a second thought. Ill see if it is possible to come up with a general solution that doesnt assume that ABH are collinear.
If not, it would be appropriate to provide a mathematical proof to your professor showing why this isn't possible
|
On April 21 2010 07:17 Capook wrote:
As others have said, the question is stupid. Here is an equivalent reformulation that makes it obvious.
Given three points A,B,C, find a fourth point H not coplanar with A,B,C such that the distance between H and C is minimized.
(This is equivalent because C not being on the ABH plane is equivalent to H not being on the ABC plane.)
If the restriction of not coplanar is lifted, then the answer is obvious: choose H=C, so that the distance is zero. However, the "not coplanar" eliminates this choice (as well as some other choices). But there are tons of points left and you are asked to pick the one that minimizes the distance to C. That is like asking you to find the smallest positive number. (There isn't one.)
Who came up with this nonsense? In addition to having no content it is written like its author doesn't understand basic facts about euclidean geometry. For example, any three points lie on plane, so why say they do?
So.. I wasn't full of shit when I said it seems with this formulation there are arbitrarily close points and given any H you can find a closer H?
Unless of course that line in the diagram is the plane that A and B are on, since by themselves they do not determine a plane.
|
On April 21 2010 07:17 Capook wrote:
As others have said, the question is stupid. Here is an equivalent reformulation that makes it obvious.
Given three points A,B,C, find a fourth point H not coplanar with A,B,C such that the distance between H and C is minimized.
(This is equivalent because C not being on the ABH plane is equivalent to H not being on the ABC plane.)
If the restriction of not coplanar is lifted, then the answer is obvious: choose H=C, so that the distance is zero. However, the "not coplanar" eliminates this choice (as well as some other choices). But there are tons of points left and you are asked to pick the one that minimizes the distance to C. That is like asking you to find the smallest positive number. (There isn't one.)
Who came up with this nonsense? In addition to having no content it is written like its author doesn't understand basic facts about euclidean geometry. For example, any three points lie on plane, so why say they do?
Harsh but true. I still think the original intention was to ask for H for all planes through AB, except ABC.
edit: not sure what's up with ABH being colinear on the diagram. If it's a condition then the solution won't depend on the choice of the plane and H is the projection of C to AB.
|
On April 21 2010 07:23 EtherealDeath wrote:Show nested quote +On April 21 2010 07:17 Capook wrote:
As others have said, the question is stupid. Here is an equivalent reformulation that makes it obvious.
Given three points A,B,C, find a fourth point H not coplanar with A,B,C such that the distance between H and C is minimized.
(This is equivalent because C not being on the ABH plane is equivalent to H not being on the ABC plane.)
If the restriction of not coplanar is lifted, then the answer is obvious: choose H=C, so that the distance is zero. However, the "not coplanar" eliminates this choice (as well as some other choices). But there are tons of points left and you are asked to pick the one that minimizes the distance to C. That is like asking you to find the smallest positive number. (There isn't one.)
Who came up with this nonsense? In addition to having no content it is written like its author doesn't understand basic facts about euclidean geometry. For example, any three points lie on plane, so why say they do? So.. I wasn't full of shit when I said it seems with this formulation there are arbitrarily close points and given any H you can find a closer H? Unless of course that line in the diagram is the plane that A and B are on, since by themselves they do not determine a plane.
I assume the intention of the diagram was to imply that the planes ABC and ABH are perpendicular, which leads to an easy solution.
|
Ok ive generalised the answer slightly
pick ANY 3 points, A, B, C.
Now create ANY plane that passes through A and B.
HC will ALWAYS be a normal vector to this plane, except when C = H
Thus, C-H = HC
and thus HC.A = HC.H
Leaving you
A.C (known scalar) = A.H + C.H - H.H
A.C = H.(A+C) - H.H
Note the infinite number of solutions for h1, h2, h3 due to the fact that the plane passing though AB is not specified.
However you get to see the relation between h1-3 and the boundaries they imply on each other. By picking elements of the vector h you essentially define the plane ABH.
|
|
|
Give tl a textbook math problem and it will be convoluted into an infinite machine of fascist propoganda. *Bravo*
|
If you project BC onto HC considering H is directly above (on the z axis) the Point C you come up with:
<xhc-xbc, yhc-ybc, zhc>/(SQRT(xbc^2 + ybc^2)
If you had some sample points you could just plug them in and get the projection in terms of numbers and unknown H point variables.. not sure where to go from here tho. I should fuckin know this im almost done calc 3~!!
Also I just realized that once you get a projection you need to solve for the orthogonal by doing your "top" vector (in this case it would be BC) minus the projection.
so
BC - <xhc-xbc, yhc-ybc, zhc>/(SQRT(xbc^2 + ybc^2) = orthogonal
but still idk what this means other than taking the vector BC and making it lie along vector HC
|
This is how i interpret the question:
Fix points A, B and C. Let H be a point on the line defined by AB, and such that C is closest to H. ABCH do not lie on the same plane. Find H.
First note that ((x.y)/|y|^2)y is the projection of x onto y. Now, AH is the projection of AC onto AB. Therefore, AH=((AC.AB)/|AB|^2)AB. But by definition AH=OH-OA. Therefore,
OH=((AC.AB)/|AB|^2)AB + OA
I'm not clear why the requirement ABCH is not coplanar is necessary.
|
Braavos36362 Posts
I really hate homework threads but this already has 3 pages of discussion, so I guess it can stay open :/
|
|
|
|
|
|
EPT
