EPT
Math problem, please help :(
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stafu
Australia1196 Posts
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DeathByMonkeys
United States742 Posts
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Tex
United States126 Posts
Proooojections. By the looks of it, yeah, H is the projection of C onto the plane that is shared by A, B, and H. | ||
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DeathByMonkeys
United States742 Posts
Then you use the same method for finding the distance from A to B. From there you have everything you need, and can simply use the Scalar Projection formula: comp of BC onto AB = ((A vector) (dot product) (B vector)) / (magnitude of A vector) | ||
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stafu
Australia1196 Posts
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Antimage
Canada1293 Posts
Excuse me for being vague, I'm an accounting student and haven't done real math for 3+ years =/ | ||
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stafu
Australia1196 Posts
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MER
Bulgaria125 Posts
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Tex
United States126 Posts
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EtherealDeath
United States8366 Posts
A most trivial case is where A = (0,0,0), B = (1,0,0), C = (0.5,1,1). There can be no explicit solution for H in terms of A,B,C since in this case there is no H closest to C, because we can arbitrarily change the angle between the xy axis and the plane to have the plane approach C. Could be full of shit though, at first I thought it was a trivial projection problem :/ back to foods. | ||
Chill
Calgary25969 Posts
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DeathByMonkeys
United States742 Posts
On April 21 2010 05:59 stafu wrote: Thanks DeathByMonkeys, now to see if I can get my head around this hehe. Life saver. Actually I messed up. You don't need to know the distance between AB and BC, just the vectors. So disregard the distance formula part. AB will simply be = (Bx - Ax, By - Ay, Bz - Az) BC will simply be = (Cx - Bx, Cy - By, Cz - Bz) Now you can use the projection formula I provided earlier where you project BC onto AB. So... comp of BC onto AB = ((AB vector) (dot product) (BC vector)) / (magnitude of AB vector) This resultant scalar is the distance from B to H. | ||
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Oracle
Canada411 Posts
On April 21 2010 06:08 Chill wrote: Just using common sense I don't see any reason C can't be 0.00000000000000000000000001 units away from H. Seems like a retarded question. The "Shortest distance to point C" statements implies that H and C are perpendicular. C can be epsilon units away from H but that doesn't mean anything. | ||
]343[
United States10328 Posts
On April 21 2010 06:08 Chill wrote: Just using common sense I don't see any reason C can't be 0.00000000000000000000000001 units away from H. Seems like a retarded question. they're asking what the projection of C onto AB is, for general A, B, C which is ... dot product specifically, the length of (C-A) projected onto (B-A) is [(C-A) . (B-A)] so we multiply this by unit vector in direction of (B-A) and then add A | ||
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EtherealDeath
United States8366 Posts
On April 21 2010 06:17 ]343[ wrote: they're asking what the projection of C onto AB is, for general A, B, C which is ... dot product specifically, the length of (C-A) projected onto (B-A) is [(C-A) . (B-A)] so we multiply this by unit vector in direction of (B-A) and then add A That's not what they ask for. If you pick H along the projection of C onto AB, then A,B,C,H will be on the same plane, which they mentioned C is NOT on the same plane as A,B,H. | ||
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Severedevil
United States4832 Posts
The unknown point H is located at the shortest distance from Point C. The shortest distance from point C to what? If it's the shortest distance from point C to line AB, this is trivial as a lot of people have pointed out. Project C onto AB and you have H. (As has also been pointed out, all four points would be coplanar.) If it's the shortest distance from point C to the plane containing AB, well... that's a problem, because the plane isn't specified in the problem. | ||
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QuickStriker
United States3694 Posts
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Chill
Calgary25969 Posts
On April 21 2010 06:17 ]343[ wrote: they're asking what the projection of C onto AB is, for general A, B, C which is ... dot product specifically, the length of (C-A) projected onto (B-A) is [(C-A) . (B-A)] so we multiply this by unit vector in direction of (B-A) and then add A I guess I don't understand but I'd like to have it explained for my own sake. ABH is a triangle (or line) in a plane. C is not in that plane but is the closest distance to H possible. To me, the closest distance possible is like right beside it but not in the other plane so it would be 0.00000000000000000_1 units perpendicular from the plane ABC lies in. I feel like there's a line missing or I'm misunderstanding the question entirely. | ||
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Kentor
United States5784 Posts
![[image loading]](http://img34.imageshack.us/img34/9052/wtfzb.png) | ||
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stafu
Australia1196 Posts
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stafu
Australia1196 Posts
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Cambium
United States16368 Posts
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stafu
Australia1196 Posts
On April 21 2010 06:42 Cambium wrote: CH is perpendicular to the plane ABH. Yep, as a necessity from H being the closest point on the plane from C, CH must be perpendicular/normal vector to the plane. | ||
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Chill
Calgary25969 Posts
On April 21 2010 06:16 Koltz wrote: The "Shortest distance to point C" statements implies that H and C are perpendicular. C can be epsilon units away from H but that doesn't mean anything. Well, it doesn't mean anything I guess except that C and H are basically the same point. It seems like a silly question to go through all this work when the points are essentially overlapping. | ||
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Chill
Calgary25969 Posts
Thanks for helping me get it I was imagining it as some plane ABH and then move C as close as possible. | ||
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DeathByMonkeys
United States742 Posts
On April 21 2010 06:44 Chill wrote: Oh, H is the variable one not C? That changes it LOL Thanks for helping me get it I was imagining it as some plane ABH and then move C as close as possible.Haha I finally understood what you were asking, but it looks like you understand now. C is an arbitrary point and then H is placed at the distance closest to C while still being restricted in the ABH plane. | ||
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EtherealDeath
United States8366 Posts
On April 21 2010 06:44 Chill wrote: Oh, H is the variable one not C? That changes it LOL Thanks for helping me get it I was imagining it as some plane ABH and then move C as close as possible.But that's a symmetric problem no? Pretend we have ABH and move C as close as possible. Then swap C with H and we have the solution to the given problem. | ||
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EtherealDeath
United States8366 Posts
On April 21 2010 06:46 DeathByMonkeys wrote: Haha I finally understood what you were asking, but it looks like you understand now. C is an arbitrary point and then H is placed at the distance closest to C while still being restricted in the ABH plane. But I thought H's position is not yet determined, therefore the ABH plane is not a restriction when choosing H's location, since wherever we decide to place it, it will still be in the ABH plane? | ||
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Kentor
United States5784 Posts
On April 21 2010 06:48 EtherealDeath wrote: But I thought H's position is not yet determined, therefore the ABH plane is not a restriction when choosing H's location, since wherever we decide to place it, it will still be in the ABH plane? There's three restrictions to H, 1 is that it must be in the ABH plane. 2 is it is the closest distance to C. 3. it is not the same point as C I guess to solve this you use vectors AC and BC, both must be perpendicular to vector CH so their dot products must be 0. (xc-xa, yc-ya, zc-za) . (xc-xh, yc-yh, zc-zh) = 0 (xc-xb, yc-yb, zc-zb) . (xc-xh, yc-yh, zc-zh) = 0 actually... i forgot all about plane geometry :p | ||
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Chill
Calgary25969 Posts
On April 21 2010 06:46 EtherealDeath wrote: But that's a symmetric problem no? Pretend we have ABH and move C as close as possible. Then swap C with H and we have the solution to the given problem. No, because C isn't constrained to the plane. So no matter where H is I can more C to overlap it, whereas I have to keep H constrained to the plane. God I feel so dumb lol. | ||
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EtherealDeath
United States8366 Posts
On April 21 2010 06:49 Kentor wrote: There's two restrictions to H, 1 is that it must be in the ABH plane. 2 is it is the closest distance to C. Yes but (1) is not really a restriction while choosing the location of H, since there are only two fixed points given in the plane that are given in the problem. Thus given any choice of x,y,z in H = (x,y,z), H will still be in ABH. Therefore (2) is the only effective restriction while choosing the location of H. That makes this problem seem a bit strange to solve, unless I am missing something. | ||
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EtherealDeath
United States8366 Posts
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DeathByMonkeys
United States742 Posts
Based on Kentor's illustration we may have to use the knowledge that HB (cross) HA gives us the HC vector. | ||
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Kentor
United States5784 Posts
On April 21 2010 06:53 EtherealDeath wrote: Yes but (1) is not really a restriction while choosing the location of H, since there are only two fixed points given in the plane that are given in the problem. Thus given any choice of x,y,z in H = (x,y,z), H will still be in ABH. Therefore (2) is the only effective restriction while choosing the location of H. That makes this problem seem a bit strange to solve, unless I am missing something. actually C is also a given point. so you can't just pick any H otherwise CH might not be perpendicular to ABH. | ||
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EtherealDeath
United States8366 Posts
On April 21 2010 07:07 Kentor wrote: both are restrictions. C is a given point. if (1) were not a restriction, then C = H. Well the problem did explicitly say that C != H | ||
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Kentor
United States5784 Posts
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EtherealDeath
United States8366 Posts
 Yea I guess the plane must be given, or you could flip your solution about C and get another solution in certain cases. | ||
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BrTarolg
United Kingdom3574 Posts
mod(AB) indicates the modulus of a vector The key to solving this problem is finding the vector AH in terms of A, B and C only. This can be done by using dot products, as ((AB)/mod(AB)).(AC) will give you the projection of C on the AB vector, in terms of a scalar Using the unit vector in the direction AB, and multiplying it by the scalar (AB).(AC) gives you the vector AH. Thus the answer to the problem is simply A + this, or as given: (AB/mod(AB)) * (((AB)/mod(AB)).(AC)) + A = H | ||
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hypercube
Hungary2735 Posts
On April 21 2010 06:00 MER wrote: The point H is the projection of C on the plane. But your plane is not defined because you have only 2 points on it. so you can't solve the problem. There are infinite planes that go through the points A and B. This and Chill's answer seems to be correct. If you want to get fancy you can solve this: Assume that the plane where you're looking for point H is given. For example you're given the angle between this plane and the plane ABC. Let's call this angle phi. Give the locus of H as phi goes from 0 to 360 (i.e 0<phi<360, because the problem says A,B,C and H are not on the same plane). You can also give the function phi->H The plane where you're looking for H is not given. Any solution for H would include a free parameter independent from the coordinates of A, B and C. There are various ways to parametrize your solution set. "Artificially" introducing phi is probably not the best one either. A more natural way is to give H as the solution to an underdetermined system of linear equations. | ||
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Capook
United States122 Posts
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? | ||
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Kentor
United States5784 Posts
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. | ||
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BrTarolg
United Kingdom3574 Posts
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Maenander
Germany4926 Posts
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BrTarolg
United Kingdom3574 Posts
If not, it would be appropriate to provide a mathematical proof to your professor showing why this isn't possible | ||
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EtherealDeath
United States8366 Posts
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. | ||
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hypercube
Hungary2735 Posts
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. | ||
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Maenander
Germany4926 Posts
On April 21 2010 07:23 EtherealDeath wrote: 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. | ||
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BrTarolg
United Kingdom3574 Posts
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. | ||
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HeavOnEarth
United States7087 Posts
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gogogadgetflow
United States2583 Posts
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Hypnosis
United States2061 Posts
<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 | ||
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sigma_x
Australia285 Posts
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. | ||
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Hot_Bid
Braavos36372 Posts
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