Bignose

I'd just like to say that the uniqueness properties are important, too. Because, if there were two zeros, 0 and 0', there are a bunch of properties that start to crumble. 4 - 4 = 0 or =0'? How do you know which one to use? And, if there aren't unique answers to simple arithmetic, it's just going to get worse trying to build on it. So, it is important that there is a 0, and that there is a unique 0, same thing about 1.

Is there something you want to discuss about this? Because, otherwise, what is the point in posting some formulas that no one asked about? It's about a much use as if I'd posted a "how to add numbers to sum to 4" post and told everybody that you can add 2+2, or 1+3, or 0+4, or even -1+5 to get 4. Yay! So, if you want to discuss something about this, discuss away. However, just posting this stuff without any questions or even comments on it is a waste.

OK, I'm going to give you the benefit of the doubt, right here, right now. Please use your ideas to make one predictive, testable, verifiable statement. This will involve math -- because your predictive, testable statement will have to show why your theory predicts and explains things better than what we have now. What specifically does your theory predict more accurately than the theories we have now? Until you can answer this question directly, you are not going to find too many people receptive in the scientific community. They are not against new ideas -- but the ultimate crucible where an idea either stands up or it crumbled is whether that new idea can predict phenomena at least as well or better than the old ideas. It is as simple as that. So, show us one prediction made from your theories that is at least as good or better than the predictions the old models have.

You need to be much more explicit in defining what you mean. What exactly is a "linear plane"? I mean, I know what a line is, what linear typically means, and what a plane is, but this combination of word doesn't have any meaning I am familiar with. And, what exactly does doubling the diameter of a circle supposed to do? Besides being equal to 2d (where d is the diameter)?

There was an article recently that showed that any configuration can be solved in something like 26 moves. If a person were allowed to look at a Rubik's Cube and try to figure out that minimum (or, I guess use a computer to find the minimum), and then just perform the moves, I can easily believe that someone can do it in 10 1/2 seconds. On the other hand, I think that unless some prep work were involved, unless the cube just happened to be in a "friendly" configuration, 10.56 is awfully fast. But, "awfully fast", to me, just means rather exceptionally unlikely, but not impossible.

Unless there is a source for the angular momentum, wouldn't the fluid's own viscosity slow the swirl down. Your coffee doesn't swirl forever when you mix the sugar into it.

That first step is called reversible for a reason. That is, it assumes that it is only infinitesimal amounts of heat are being transferred, which means that that same packet of heat can be taken back out. That is the basic definition of reversible, and its definition is exceptionally important to thermodynamics. But, no real process is ever truly reversible. Because the ideal of reversibility is that an infinitesimal amount is transferred, it would take infinite time to accomplish anything finite. Some processes can be very close to reversible, but nothing is truly reversible in real life. In fact, three of the four steps of the Carnot cycle are reversible. It's the fourth step that is irreversible, and represents the efficiency loss. If your goal was just to look for thermodynamic transfers that don't lose energy, then you are looking for the concept of reversibility. A useful concept in theoretical analysis, but is a physical impossibility in the real world. ************************* A cold reservoir is needed because that is what you are going to use to cool the gas. Compression of a gas heats it up. You want to compress the gas without it gaining in temperature, because if you let is gain in temperature, you haven't done anything -- you've just gone backwards a step. The whole goal of the Carnot cycle is to use a heat differential to generate work -- if you compress the gas without keeping it at a cool temperature, you're going to use up all your work that you just gained to re-compress it.

I'm going to make a weird suggestion. I think you might like to pick up a little book called Calculus for Cats by Amdahl and Loats. I picked it up a while back, more as a joke for my fiance (whose big into cats), but I thought it was a pretty good book. It really depends on whether you can stomach lots of 'catty' examples or not. But, the information for a good years worth of calculus is in the book. Otherwise, the ... For Dummies books are usually pretty good, and definitely more geared towards self-study than the average textbook. I wouldn't put too much stock in the reviews of the textbooks, because the students who are frustrated/angry/upset with their teachers are going to take the reviews out on the books. A calculus textbook, for a university level class, is going to to be hard. Not only that, but considering how many problems and examples and just the size of the book, there are going to be mistakes in there too. It really cannot be helped. Thirdly, when a book is written for one kind of learner, anyone who isn't the same kind of learner is going to hate that books. In other words, there will always be someone who hates their particular textbook. However, you should be able to check out a few calculus books from a library, you shouldn't need to buy one yourself.

But, see, this wasn't true either. Ancient civilizations used all sorts of bases. Some used base 12, some used base 20, the Babylonians even used base 60. I'm not sure that it is completely possible to recreate the origin of mathematics in hindsight. Things that are truly obvious to us -- from many years of schooling -- wouldn't be so obvious when they weren't taught every year. There are several books on the history of mathematics, you might enjoy reading through some of them.

How is this different from the first step in the wiki article? The first step is: "Reversible isothermal expansion of the gas" And, you need a hot reservoir to heat up the gas. Also, the outside pressure has nothing to do with the calculation. The calculation is based on the amount of work the gas inside the piston does. What is works against on the outside is irrelevant.

I think that this is most of it -- to introduce a "wow" factor into the show. But, it doesn't bother me a whole lot, since I've known one person who thought like that, or at least would talk like that. Whereas you or I would say "I want to go to a bar that has Guinness on tap and isn't too smoky" he would say something like "I prefer the intersection of the set of bars that have Guinness on tap and the set of bars that don't have many cigarette smokers" He did that for as long as I knew him, so, if it was just an act, he kept it up for quite a long time. So, I don't find it unbelievable that someone like Charlie might just think like that, or it might just be the easiest way for Charlie to think about the problem. The biggest thing, to me anyway, isn't so much that the math in Numb3rs is spot on. You can pick nits in almost every single episode, if you know the material, and if you wanted to. One that stands out to me was an episode where a young man was pushed off of a bridge. And Charlie announced that the young man had to have been pushed, because if he had jumped he would have landed 10 or 12 feet in different direction. Any good mathematician knows that you can only talk in probabilities there. A correct statement would have been "it is most likely that he was pushed," because a lot of things could have happened. A large gust of wind could have kicked up right when the jumper jumped. Maybe he wanted to "go out with a bang" and leaped far off the bridge instead of just falling off. Etc. A good mathematician knows, especially when talking about one jumper (i.e. the experiment wasn't repeated numerous times) you can only talk about probabilities of what happened, and can't say anything definitively without more evidence. There are more. There was an episode near the beginning of the series where they were tracking someone or something down (I don't recall). But, Charlie comes in with a map colored different colors, roughly shaped like a bullseye with a small region in the middle, and larger regions around the middle ones and he announces, in the middle region it is 95% likely to find what the FBI as looking for is in that inner circle. But the next ring, was only 85% likely, the next one 75% likely, etc. Ummm, no, to increase the confidence that an object is in a certain region, you have to increase the size of the region. The 95% likely region would be the largest one, not the smallest one. But, I understand why they did it that way. Time was running out, and the FBI had to only a short amount of time to find what they were looking for, and if they tried to explain to the audience why 95% would have to involve searching half of L.A.'s suburbs, it wouldn't have been very interesting TV. Overall, however, I think that the show isn't too bad. Like I said, you can find errors in every episode if you look hard enough. But, it is exposing people to what math can really do. I haven't seen them ever really exaggerate beyond belief what math can do. Unlike CSI with its zoomed in and sharpened low-grade video cameras. That is an impossibility. Numb3rs stays pretty close to realism, and that's a significant improvement over what most show do. So, it's a step in the right direction.

I've seen the show a few times now (I like to have the TV or radio on while I work, so I see/hear more of these things than I probably should). I guess my reaction was mostly "meh." It's ok, but it is nothing I am going to feel bad about missing if I forget about it next week. That's pretty much how I feel about all TV today, though, so I guess take that at face value. The level of accurate physics in the show is similar to the level in Numb3rs, though just a small step down. Being a half-hour sitcom, the character interaction is definitely more of a focus, while in Numb3rs the way Charlie uses math to help solve the crime is more central to the story line. I also haven't heard any really good science-y jokes out of the show, Futurama had a few really good ones sprinkled in it. The best science-y joke out of Futurama? The episode they were at the horse track, and two horses crossed the line "in a quantum finish" as the announcer says (really, really close to the same time.) So, a white lab coated man looks through a really big microscope (which is funny in itself), and then the other guy working holds up a card that says horse #3 won. The professor, apparently holding a ticket for the other horse, stands up and yells "No fair! You changed the outcome by measuring it! " That's just the best science-y joke I've ever seen on TV. You have to know at least a little about the nature of quantum mechanics, not too much -- one of the many popularizations of QM is probably enough -- Schroedinger's cat and all. But, the scene also doesn't hand-hold and is fast. You have to have that pop-knowledge about QM to get the joke. Admittedly, I haven't been paying close attention the "The Big Bang Theory," but I haven't noticed any really good punchy science jokes to date. EDIT: in quick response to Scientia, Numb3rs is actually pretty well done. Their math is usually pretty spot on. Sometimes the amount of time they use to compute things is unrealistic, and no human being can be an expert in as many fields as Charlie is -- the guy has intimate knowledge of cutting edge research in a different field every show it seems and has brought in colleagues only a few times -- but the math they are doing and applying is pretty realistic. I think that TV is getting better about it, but, they are not at the level of good yet. Shows like CSI are mostly good. But, the funny thing about CSI, is that they are almost too good. Prosecuting attorneys are finding it harder to prosecute people these days, since juries are expecting science and forensics to connect every dot like they see on TV. Most evidence collected is relatively circumstantial and a "preponderance of evidence" presentation is not uncommon. That is, yes, each individual piece of evidence is circumstantial, but taken all together it paints a likely picture. There are the oft-pointed out many flaws in CSI, though, too. The magic cameras that can zoom and then sharpen in a fuzzy picture. But, there are good parts, too. Acquiring DNA from a wide variety of sources. Many different methods of checking for fingerprints. Etc. The biggest thing these shows have done is raise an awareness of science to the public. This is the main reason I say Hollywood has gotten better, but isn't "good" yet. At least there are some shows where science is highlighted -- it could be worse, it could be zero.

Have you read the wiki article? http://en.wikipedia.org/wiki/Carnot_cycle I thought it was pretty well done, really. It explains why the temperatures enter into the efficiency. Basically, the efficiency is the ratio of how much work can be done for a given quantity of heat that is moved from the hot reservoir to the cold reservoir. (Seems like a pretty logical choice for efficiency to me.) The work [math]\Delta W[/math]can be calculated and is equal to [math]\Delta W = (T_H - T_C)(S_B -S_A) [/math]. Same for the amount of heat put in [math]\Delta Q_H = T_H(S_B -S_A) [/math]. Putting those ratios over each other, canceling the entropy terms, and you get the efficiency just in terms of the temperatures. If you're not understanding how those quantities are computed, check out a good thermodynamics book. I recommend Smith, Van Ness, and Abbott's Introduction to Chemical Engineering Thermodynamics, though I am sure that there are other very good books that cover the Carnot cycle in detail, too.

There is one more point I'd like to make, as to why vector equations are written in vector form, and not always written out like I did in the last post. I've talked a little about this, but I'm going to supplement it with a few examples to further drive home to point. It has to do with the fact that the laws of nature do not know what coordinate system human beings choose to use to describe what is going on. In that way, the vector equations are the more general equations. For example, the conservation of mass equation for that incompressible Newtonian fluid in vector form is [math]\nabla\cdot\mathbf{v}=0[/math]. In Cartesian coordinates, what I've already written out above, this is: [math]\frac{\partial{v_x}}{\partial{x}} + \frac{\partial{v_y}}{\partial{y}} + \frac{\partial{v_z}}{\partial{z}} = 0 [/math] but in cylindrical coordinates it is [math]\frac{1}{r}\frac{\partial{(r v_r)}}{\partial{r}} + \frac{1}{r}\frac{\partial{v_{\theta}}}{\partial{\theta}} + \frac{\partial{v_z}}{\partial{z}} = 0 [/math] and in spherical coordinates the equation is: [math]\frac{1}{r^2}\frac{\partial{( r^2 v_r )}}{\partial{r}} + \frac{1}{r \sin{\phi}}\frac{\partial{ ( \sin{\phi} v_{\phi} )}}{\partial{\phi}} + \frac{1}{r \sin{\phi}}\frac{\partial{v_{\theta}}}{\partial{\theta}} = 0 [/math] Same equation, [math]\nabla\cdot\mathbf{v}=0[/math], just in three different coordinate systems. So, why use the different coordinate systems? Consider a spherically symmetric situation, where the field would only be a function of r. Then all the [math]\frac{\partial}{\partial{\theta}}[/math] and the [math]\frac{\partial}{\partial{\phi}}[/math] terms go to zero. That third equation suddenly gets a lot simpler. Now, we can describe the situation using Cartesian coordinates. Again, the phenomena is indifferent to what coordinate system human beings pick. But, as humans, we try to pick the coordinates so that the problem is as simple as possible. And, making things simpler allows a greater chance of those analytic solutions that I talked about above to exist. It also makes it easier for professors to have students solve problems for homework and exams

OK, you'll get there, then. Be patient, and please do not hesitate to ask if you have any further questions (you can start a diff thread, keep asking here, or feel free to PM me if you'd like -- I'm happy to help when I have the time). I like Severian's example of using fluids (I am much more comfortable and have studied fluids a lot more than electromagnetism). In particular, an incompressible Newtonian fluid (water is a good example) obey the mass conservation law of [math]\nabla\cdot\mathbf{v}=0[/math] where [math]\mathbf{v}[/math] is the velocity field. In words, what this means is that around a control volume, what fluid comes in must be equal to the fluid that comes out. Let's look at a 2-D example: ----------N------------ |...........................| |...........................| W........................E |...........................| |...........................| -----------S----------- Consider this crudely drawn box to be a control volume. W-E is in the x direction, and N-S is in the y direction. Let u be the velocity in the x direction, and let v be the velocity in the y direction. Δx and Δy represent the length of the face, Δx = distance from W to E, and Δy = distance from N to S. Mass in = Mass out uw*(Δy) + vs(Δx) = ue(Δy) + vn(Δx) now collect terms 0 = (ue - uw)*(Δy) + (vn - vs)*(Δx) Divide by (Δy)*(Δx) 0 = (ue - uw)/(Δx) + (vn - vs)/(Δy) Now, take the limit as both Δx and Δy do to zero. Those differences over deltas become differentials: [math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0 = \nabla\cdot\mathbf{v} [/math] Physically, What this means is that if we took a fluid flow and say somehow the outflow ue became doubled, there would have to be corresponding changes to the other outflows and inflows. That is, the outflow vn would have to decrease, or the inflows uw and vs would have to increase, or most likely, some combination of all three. Same thing with [math] \nabla\cdot\mathbf{B} = 0 [/math]. If the magnetic flux increases through one face of a control volume, then the other fluxes through the face have to change accordingly. The vector notation in this case is just the mathematical way of saying what comes in must come out. Whether that "what" is fluid flow, or magnetic flux. It also is why the different components of the vector become coupled together. From the differential equation (which, again, is just what results when we shrink the control volume to a very small, infinitesimal volume), [math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0 [/math], you see how a change in u (x-direction velocity) results in a corresponding change in v (the y-direction velocity). Same thing with the magnetic field. A change in B_x (x-direction magnetic flux) results in a corresponding change in B_y (y-direction magnetic flux). And, if the problem is three-D, the equation looks like [math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} + \frac{\partial{w}}{\partial{z}} = 0 [/math], which just says that a change in u (x-direction velocity) results in a corresponding change in v (y-direction velocity) and in w (z-direction velocity). Completely analogous for the 3-D magnetic field equation. So, the result is coupled differential equations. You are solving for each component of each vector simultaneously because when you change one component, the rest of the components will be changed too. That's why you can't have just a "set equation" for them (or, really, Maxwell's equations are that "set"). When you provide the necessary boundary conditions, the set of equations are enough to determine the field. Again, it won't be easy because of all the coupling, but, mathematically a solution does exist (proving certain conditions are met -- that's a whole different branch of mathematics, proof that solutions do exist).

Well, what you were trying to do above seems very non-valid to me. Why do you think you can set the divergence equal to the time derivative. Is there an un-curl operator? I don't think so, at least not that I am aware of. Write out the terms of the equation in Cartesian coordinates and you'll see what you get. Are you familiar with vector equations, yourdadonapogos? Do you know what I mean when I say write out the terms? Let me give you can example: I'll write out the [math]\nabla\times\mathbf{E} = -\frac{\partial{\mathbf{B}}}{\partial{t}}[/math] equation [math]\nabla\times\mathbf{E} = ( \frac{\partial{E_z}}{\partial{y}} - \frac{\partial{E_y}}{\partial{z}} )\mathbf{x} + ( \frac{\partial{E_x}}{\partial{z}} - \frac{\partial{E_z}}{\partial{x}} )\mathbf{y} + ( \frac{\partial{E_y}}{\partial{x}} - \frac{\partial{E_x}}{\partial{y}} )\mathbf{z} [/math] [math] -\frac{\partial{\mathbf{B}}}{\partial{t}} = -\frac{\partial{B_x}}{\partial{t}}\mathbf{x} - \frac{\partial{B_y}}{\partial{t}}\mathbf{y} - \frac{\partial{B_z}}{\partial{t}}\mathbf{z} [/math] Now, in vector equations, the terms in front of each base vector must be equal, so combining these two above equations, we actually get 3 separate pdes: [math]( \frac{\partial{E_z}}{\partial{y}} - \frac{\partial{E_y}}{\partial{z}} ) = -\frac{\partial{B_x}}{\partial{t}} [/math] [math]( \frac{\partial{E_x}}{\partial{z}} - \frac{\partial{E_z}}{\partial{x}} ) = -\frac{\partial{B_y}}{\partial{t}} [/math] [math]( \frac{\partial{E_y}}{\partial{x}} - \frac{\partial{E_x}}{\partial{y}} ) = -\frac{\partial{B_z}}{\partial{t}} [/math] These differential equations are coupled and need to be solve simultaneously. You can write out similar equations for the rest of Maxwell's equations by fully writing out the curl and divergence operators. The equations I wrote above are for the Cartesian coordinate system, they would be written out differently for a different coordinate system, but the idea remains the same. You will get differential equations that show the relationship between E_x, E_y, E_z, B_x, B_y, and B_z. ***************** if E only varied in time, not space, then [math]\nabla\cdot\mathbf{E} = 0 [/math] not equal to the time derivative. *************** yourdad, I don't want to seem snotty, because it is not the tone I want this to be taken in. But, have you studied a vector equation/calculus book? Because this is pretty basic stuff, and rather me typing it out piecemeal on a forum, you'll probably get a lot more out of a more organized presentation.

Well, at is basis, simultaneously means, at the same time. Obviously, this is harder to do with with pdes than say an algebraic system. Let me an example: 5x + 8y = 9 -2x + 2y = -1 You can solve these sequentially, by re-writing the second equation: x = y +1/2, plug this into the first equation, solve that first one for y, then solve for x. Or you can solve them simultaneously, by re-writing it in matrix form A*x= b where A = [ 5 8 ; -2 2], x = [x ; y] and b = [ 9 ; 1]. (I don't know how to write matrices in LaTeX, so the ; means new line. That is, A is a 2x2 matrix, and x and b are column vectors). Then compute the inverse of A,front multiply both sides of the equation by A^-1, compute A^-1*b and that is equal to x. Sometimes you can do this with pdes, often you can't. Like I said, analytic solutions are pretty rare. Computational methods can do both. They can turn the pde system and linearly approximate it as a matrix and then invert the matrix, or they can do a sequential system. The sequential system is easy to understand. Make guesses for both E & B, call them E(0) and B(0). (I'm going to stop bolding them, since it is a pain to make each one bolded.) Then, use B(0) to solve for E, call that solution E(1). Then, use that updated solution, E(1) to solve for B, and call that B(1). Then, use B(1) to solve for E(2) and so on until the solution converges. But, the idea is still the same. Because they are coupled together, you have to solve the entire system of equations together. That's what I mean by simultaneously. How did you get this? Because this doesn't seem right to me at all.

Maxwell's equations are coupled. You solve them simultaneously. This is a big reason why, in general, there won't be an analytic solution. Only in very special situations will analytic solutions be available.

Well, you are going to solve Maxwell's equations for E & B. That's the point of Maxwell's equations, that the fields E & B have to obey them. The differential equations must be supplemented with the appropriate boundary conditions to completely define a situation. J is the current density that will either be prescribed, or it may obey it's own equation and have to be solved simultaneously with Maxwell's (that is, they could be coupled). Same thing with [math]\rho[/math], it may be prescribed or have to obey it's own equation as well. Maxwell's equations are vector equations. Vector equations are great because they describe the physical situation no matter what coordinate system we choose. That is, nature does not know whether we use Cartesian, cylindrical, spherical, bipolar, or any of the other many coordinate systems to describe the situation. The equations still hold. However, the choice of coordinate system is usually made by people to make the situation easier to describe and analyze. Then, depending on what coordinate system, the appropriate divergence, curl, and gradient operators would have to be used. (see http://mathworld.wolfram.com/CylindricalCoordinates.html equation 32 for the gradient operator in cylindrical coords and compare that with equation 33 of http://mathworld.wolfram.com/SphericalCoordinates.html . In fact, the general gradient operator is equation (2) of http://mathworld.wolfram.com/Gradient.html note the use of the scale factors h_1, h_2, and h_3. Those can be used to find what the gradient of Elliptic Cylindrical Coordinates http://mathworld.wolfram.com/EllipticCylindricalCoordinates.html by using equations 6 through 12 which are the scale factors.) Is this the "set equations" you speak of? I am unsure what you mean by that term. Because Maxwell's equations are the equations for B and E.

All 4 are actually partial differential equations. You should probably bold the vector terms in there, just to keep it clear. For example, the diverence of a vector is: [math]\nabla\cdot\mathbf{E} = \frac{\partial{E_x}}{\partial{x}} + \frac{\partial{E_y}}{\partial{y}} + \frac{\partial{E_z}}{\partial{z}}[/math] where [math]E_x[/math] is the x component of the E vector. In general, there is no analytical solution to partial differential equations, that a large reason the computational techniques have gotten so much attention. However, a lot of the times you can make simplifying assumptions. Such as, dependence only on one space coordinate. This is what Klaynos was talking about, if a problem is spherically symmetric, for example, the answers will only depend on the radial direction, not either of the two angles. Then, hopefully, with the simplifications, the pde becomes a form that is easier to solve, such as a separable pde or maybe a solvable ode. I don't know of many examples in electricity/magnetism, but I could give you lots of examples from fluid mechanics, if you were interested. Also, you can consult a text on pdes for solution methods, there are many texts out there. On a broader note, a great text to learn about the vector operators is Schey's Div, Grad, Curl, and All That: An Informal Text on Vector Calculus. It is like it says fairly informal, but at the same time, good at explaining what the operators do, what they mean, how to use them, etc. Very highly recommended. Technically, you can do them over any surface if you set it up correctly. usually, the surface is chosen to make the math easier, such as a spherical surface around a point source (so, like above, only the radial dependence will show up). But, you can pick any surface -- in the derivation of the differential equations, you are taking the limit as the surface area goes to zero anyway.

Sorry, Xerxes, should have been clearer. 1/7 itself is clearly finite, but its decimal representation has an infinite number of digits. Since I was talking about the number of digits in a decimal representation, I figured that saying "infinite" in that context, and the context of the previous posts and answer to that post, it would be obvious. Apparently it wasn't. Sheesh

Ok, then I have another one, what do you do with something like [math]\frac{\pi}{4}[/math] ? That will be between 0 and 1. Also, actually the main point about 1/7 is, yes, it is a repeating decimal, but since it is infinite, which digit do you start with? This is the same issue with [math]\frac{\pi}{4}[/math]. Which digit do you start with?

I'm not sure if you are doing anything wrong, but how would you handle infinitely repeating decimal representations of fractions, like 1/3 or 1/7?

Flak, what are you specifically asking here? The wikipedia link shows how to calculate all the info you want, at least as far as I understand what you want. Are you unfamiliar with the notation and concepts used in the wikipedia article? Are you unfamiliar with vectors, and operations on vectors like dot product and cross products? Or does the wikipedia entry not show you how to calculate what you want? In that case, you need to be more specific about what you want. Please be more specific where your confusion is, so we can more directly help you. Thanks.

Yes, I am sure. Xerxes, take a piece of paper in both hands. Hold with just your thumb and one finger, and pull it so it is taut, so that it is flat like a plane would be. Where you are holding it will be the two points. Now, notice that you can rotate your hands while keeping your thumbs in place, symbolizing that there are an infinite number of planes that can go through those two points -- your thumbs. This is just like using only two points to try to define a plane -- you can keep two points fixed, but there are still an infinite number of planes that can go through them. So, which plane you you want? You have to have a third point, that is not on the same line as the first two points, then you can completely specify a single plane. Again, look at that wikipedia link I've posted twice now. Same thing Flak, if you look at the wikipedia link I posted, it tells you how to use three points to find the equation of the plane, and then how to calculate how far a point is from a given plane.