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You are not logged in. #26 20110102 02:08:46
Re: Vector EquationsHi Au101 and make equations with a, b, and c by dot products AB.n = 0 and BC.n = 0 That gives two equations and three unknowns; not enough to find a single 'n', but any multiple of 'n' will also work so choose a value for, say, 'a' and then use the equations to find b and c. That will give you a solution for 'n'. (If you want, you could try another value of 'a' and generate new values for b and c. Although different, you will find this second solution is just a multiple of the first so either will do.) So now you know 'n'. You know that the transform of is position vector for a point in the new plane so substitute into r.n = p to find the constant 'p'. I've done all that and got an equation for the new plane. Now to check it. The other position vectors should also fit this solution. So far I haven't managed to acheive this! So, I think my calculations have an error but I cannot find it. I don't think the logic is faulty (I hope). Do you want to try this and post your figures. Then I can compare them with mine. Bob ps. Explanation #1 had a different line at the point marked ** Instead of transforming three position vectors, I tried transforming the two vectors lying in the first plane and assuming the result was two vectors in the second plane. That gave a completely different solution and I don't think it is correct. Does the book make the following point clear: When T transforms a plane it (i) transforms position vectors for points in the plane OR (ii) it transforms vectors that lie in the plane. Last edited by bob bundy (20110102 21:30:45) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #27 20110102 03:49:15
Re: Vector EquationsHmmmm it's a tricky one, I copied out the question exactly as it stands in the book, but if it helps I can copy out the book's example of a problem which is similar to these questions, but not quite the same, perhaps that would shed some light: Last edited by Au101 (20110103 00:36:24) #28 20110102 03:51:25
Re: Vector EquationsI don't know if that will help but that's the book's angle, I can't find any further clarification in the book I'm afraid. I'd be happy to have a go at getting my own figures, except I'm not entirely sure I understand what to do. Last edited by Au101 (20110102 05:30:48) #29 20110102 04:47:24
Re: Vector EquationsHi Au101, Last edited by bob bundy (20110102 04:51:04) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #32 20110102 21:41:23
Re: Vector Equationshi Au101,
is definitely (i), which was what I had worked with all along. Last edited by bob bundy (20110102 21:46:15) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #33 20110103 00:49:17
Re: Vector EquationsOoooh thanks ever so much for all of your tribulations Bob. I suppose that either method would suit me fine, I'm just not entirely sure I'm clear on your method, I don't suppose you could provide an example, or further instruction to clarify? (I'm sorry to request further tribulations). Otherwise the book method is perfectly fine, but I don't really see how to move from the book's finding the vector equation to finding an equation in the form Oh and yes I'd meant to say, the screen is much wider than normal, it may have been to do with my using the LaTeX text environment, which does not drop lines automatically. I tried to fix it, but it doesn't seem to have made much of a difference I'm afraid. Last edited by Au101 (20110103 00:53:34) #34 20110103 00:53:30
Re: Vector Equationshi You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #37 20110103 05:05:36
Re: Vector Equationshi Au101 form 2; the dot product form and form 3 The first form is derived from the diagram in post 20. This is the position vector for a point in the plane and are two vectors lying in the plane. In the second form is a vector that is perpendicular to all vectors lying in the plane. r and n are being 'dotted' together and the result is a constant. If you then replace 'r' by and do the dot product you get the cartesian version that is the third form. From this you can see that form 2 and form 3 are the same plane. Less obviously, so is form 1. Proof: In form 1 put lambda = 1 and mu = 1 and you get x = 3, y = 3, z = 1 put lambda = 1 and mu = 1 and you get x = 1, y = 1, z = 1 put lambda = 0 and mu = 1 and you get x = 2, y = 3, z = 2 If you try each of these sets of values in x  y + z you get 1 every time. So form 1 and form 3 have three noncollinear points in common. That's enough to prove they represent the same plane because you only need three points to define a plane (provided they are not in the same straight line). Now to the question. You know the equation of the plane before the transformation. The book method transforms points directly from this equation to get the equation after the transform. I think the algebra for this is a bit horrid and you still have to convert it into form 2. So my method was to pick three points in the plane and transform them. Let's call them A, B and C. That's just number work. *** Now get two vectors in the plane by doing AB = AO  OB etc. Any two will do because, provided they are not parallel, any two vectors will 'span' the plane ie. will enable you to reach all points in the plane. Now to get the vector that's at right angles to both these vectors. Imagine by some trick of gravity you can stand on the plane with 'up' meaning 'at right angles to' the plane. If I asked you to point a stick at right angles to the plane it would go straight up. If you draw any line on the 'ground' = 'the plane' , it would be at right angles to the stick. And it wouldn't matter how long the stick was. A stick twice as long, would still be at right angles to every line in the plane. Call the vector at right angles So do a dot product between one vector in the plane and 'n' and set it equal to zero. Do again for the second vector. You've got 2 equations with a, b and c as unknowns. Choose any 'a' to make the calculations easy. That will enable you to work out 'b' and 'c'. That gives you a possible 'n' the vector perpendicular to the plane. You might think it is cheating to choose 'a'. But if you had chosen an 'a' that was twice as big, you'd have got 'b' and 'c' twice as big as before so all that would happen is you'd get '2n' for the perpendicular vector. You can use any vector that is at right angles so the first would do! Now to get 'p'. The equation is the equation for all points, 'r', in the plane. Back at point *** we had three possible points so 'sub' in any one set of x, y and z values and you'll get 'p' Check by 'subbing' in the other values from *** to see if you get the same p. Problem done! Does that all make sense? Bob Last edited by bob bundy (20110103 05:07:51) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #38 20110103 06:35:10
Re: Vector EquationsWow what an explanation thank you so much. So we could say let Which of course gives Respectively, unless I'm very much mistaken. Now, if I have understood correctly so far we must now transform these points with T, which, I believe, gives: But at this point I don't fully understand AB = AO  OB etc. Last edited by Au101 (20110103 06:36:06) #39 20110103 07:52:15
Re: Vector Equationshi then Now do that again for BC = BO + OC Can you continue from here? Bob Last edited by bob bundy (20110103 07:53:09) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #40 20110104 01:06:35
Re: Vector EquationsWell, we only need one other vector, so we may as well do BC So we now do a dot product between one vector in the plane and 'n' and set it equal to zero. Do again for the second vector. Now if I can choose any value for a then I may as well go with 1, which gets us: Okay so let's choose: Now, at this point the most sensible thing to do would seem to be to multiply n by 10, but I happen to know that the book's answer actually requires me to multiply by 10. Now, unless I'm very much mistaken, the values of n make no difference at all, but it'd be nice to get the book's value of n, just to tie it off nicely. So we have: Which leaves me with something of a problem, because the correct answer should be 120 #41 20110104 01:34:48
Re: Vector Equationshi Au101, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #42 20110104 03:02:24
Re: Vector EquationsHi again, because that's a position vector for the original plane. You need one for the new plane so choose from Then it will work. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #44 20110104 03:48:18
Re: Vector EquationsYou do indeed. Well done! which shows what I mean about many answers being possible. Last edited by bob bundy (20110104 04:05:47) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei 