EPT
Math problem, please help :( - Page 2
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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. | ||
Chill
Calgary25991 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
Calgary25991 Posts
Thanks for helping me get it I was imagining it as some plane ABH and then move C as close as possible. | ||
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
Calgary25991 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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