Using Vectors to Prove CollinearityDate: 06/05/99 at 10:11:10 From: william glenn Subject: A killer vector problem I'm one of those students who always wants (and usually gets) 100% on my math assignments. I love calculus, but my other class is killing me with this problem. Three concurrent straight lines 0A, 0B, 0C are produced to D, E, and F respectively. Prove that the points of intersection of AB and DE, BC and EF, and CA and FD are collinear. At first I thought it couldn't be right, but I've drawn several diagrams, and _every time_ it works. I tried to solve it with what I know about vector analysis, but I just keep going around in circles. I call the points of intersection P, Q, and R respectively, and I think that to solve it I have to express all the points in terms of a, b, and c, but there are unknowns everywhere. P, Q and R are collinear if their vector joins are parallel and they have a point in common, right? Please help me, doctor! Thank you. Date: 06/05/99 at 12:10:07 From: Doctor Anthony Subject: Re: A killer vector problem >Three concurrent straight lines 0A, 0B, 0C are produced to D, >E, and F respectively. Prove that the points of intersection of >AB and DE, BC and EF, and CA and FD are collinear. This is Desargues's theorem. It will be more convenient to use A', B', and C' for the points D, E, and F so one can follow the proof more easily. Then if BC, B'C' meet at P; CA, C'A' meet at Q; and AB, A'B' meet at R then we must prove P, Q, and R are collinear. Use the well-known result that if P, Q, and R are collinear then the two conditions that will be satisfied are k1.p + k2.q + k3.r = 0 and k1 + k2 + k3 = 0 where p, q, and r are position vectors respectively of P, Q and R, and k1, k2, and k3 are scalar constants. Since OAA' are collinear we have (taking v as position vector of O) v + k1.a + k1'.a' = 0 1 + k1 + k1' = 0 Since OBB' are collinear we have v + k2.b + k2'.b' = 0 1 + k2 + k2' = 0 Since OCC' are collinear we have v + k3.c + k3'.c' = 0 1 + k3 + k3' = 0 From the second and third pairs of equations k2.b - k3.c k2'b' - k3'c' ------------- = -------------- k2 - k3 k2' - k3' and these must both represent the point P where BC intersects B'C' = p and so (k2-k3) x p = k2.b - k3.c Similarly (k3-k1) x q = k3.c - k1.a and (k1-k2) x r = k1.a - k2.b adding these three equations we get (k2-k3)p + (k3-k1)q + (k1-k2)r = 0 and of course (k2-k3) + (k3-k1) + (k1-k2) = 0 and this is the condition for P, Q, and R to be collinear. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 06/06/99 at 10:24:26 From: William Glenn Subject: Re: A killer vector problem To Doctor Anthony (and also maybe Doctor Floor, who seems to know a lot about Desargues and Menelaus): Dr. Anthony, thank you very much for the prompt reply. This is my first time here, so I don't know if that was lucky or this place is just very efficient! I feel a bit of a fool to say this, but I tried for hours and couldn't understand your proof entirely... I don't want to come across as ungrateful, thank you kindly for the time you took to reply. please continue... I searched for Desargues and I had no idea that this problem was actually so famous. I found all sorts of resources and diagrams, but no proofs. There were a few messages saying the proof needed to utilize the Menelaus theorem (I can understand Menelaus' theorem, but when I tried I couldn't apply it to Desargues'). There was even one message... http://mathforum.org/dr.math/problems/verhulst7.3.98.html that said that by just using the axioms on connection of points and intersection of lines, proving the theorem was impossible. This was interesting, because when our teacher gave us the problem, she said that the proof could be done with what we had learned so far. Also, that it was tedious and about six pages long. She said that when they gave the same problem in an assignment six years ago, only one student got it right; and she is a quantum physicist or something now. We have only just begun studying vectors in the "Matrices and Vectors" unit. We have not covered dot product and cross product yet, but we have done all sorts of addition and subtraction, and applications to forces, velocities, etc. >Use the well-known result that if P, Q, and R are collinear >then the two conditions that will be satisfied are > > k1.p + k2.q + k3.r = 0 >and k1 + k2 + k3 = 0 > >where p, q, and r are position vectors respectively of P, Q and >R, and k1, k2, and k3 are scalar constants. I wasn't aware of this result. Is there a proof? Could I please bother you for a little elaboration on the nature of this, because it would be nice to understand what I am doing. What determines the magnitude of each scalar constant? >>P, Q and R are collinear if their vector joins are parallel >>and they have a point in common, right? I searched for collinear results, I could not find this one - I just kept finding the determinant of a 3x3 containing the three points, with the 3rd column all ones. Apparently when that determinant is 0, the points are collinear, but I can't fathom that either. >Since OAA' are collinear we have (taking v as position vector >of O) > > v + k1.a + k1'.a' = 0 > 1 + k1 + k1' = 0 Question: Why do we take v as position vector of O here? Couldn't we just leave it as O? Also, how does v relate to 1 in the second term? >From the second and third pairs of equations > > k2.b - k3.c k2'b' - k3'c' > ------------- = -------------- > k2 - k3 k2' - k3' > >and these must both represent the point P where BC intersects >B'C' = p Algebraically, I understand what has happened here and that this is valid, but why do both these terms equal p? Thank you for your time. P.S. I think this place is great; everything here is so friendly :) Date: 06/06/99 at 11:46:22 From: Doctor Anthony Subject: Re: A killer vector problem I have given a proof of the vector condition for three points to be collinear - see further down the page. If there are other difficulties, write back. Doctor Anthony ---------------------------------------------------------------- To show that if P, Q, and R are collinear then the following pair of equations will be satisfied: k1.p + k2.q + k3.r = 0 and k1 + k2 + k3 = 0 Suppose that P, Q, and R are three collinear points with Q dividing PR in the ratio a:b With origin O we can express the position vector of Q by going from O to P then a fraction a/(a+b) along the vector PR, that is along the vector (r-p)and so we have q = p + a/(a+b) [r-p] (a+b)p + a(r-p) q = --------------- a+b (a+b)q = ap + bp + ar - ap (a+b)q = bp + ar and so bp - (a+b)q + ar = 0 or k1.p + k2.q + k3.r = 0 also k1 + k2 + k3 = b -(a+b) + a = 0 Therefore we get k1.p + k2.q + k3.r = 0 and k1 + k2 + k3 = 0 I hope the rest of the proof was clear. You might need to write down a few intermediate steps but it is all standard algebra. ---------------------------------------------------------------- >Question: Why do we take v as position vector of O here? >Couldn't we just leave it as O? To make the equation fit the pattern that I gave at the start I have taken O as a point with position vector v. It doesn't matter if you prefer to call it 0. >Also, how does v relate to 1 in the second term? If you have an equation with 0 on the righthand side you can multiply or divide by ANY factor you like to make one of the coefficients equal to 1. >>From the second and third pairs of equations >> >> k2.b - k3.c k2'b' - k3'c' >> ------------- = -------------- >> k2 - k3 k2' - k3' >> >>and these must both represent the point P where BC intersects >>B'C' = p > >Algebraically, I understand what has happened here and that >this is valid, but why do both these terms equal p? The lefthand side MUST represent some point on BC, and the righthand side MUST represent some point on B'C'. Since the two sides are equal the point must be where BC and B'C' intersect. That is the point P (position vector p) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 06/07/99 at 10:29:00 From: William. Subject: Re: A killer vector problem >I have given a proof of the vector condition for three points >to be collinear - see further down the page. If there are other >difficulties, write back. Yes that proof made sense, thank you! >I hope the rest of the proof was clear. You might need to >write down a few intermediate steps but it is all standard >algebra. Yes, that's all clear now. :) >>Question: Why do we take v as position vector of O here? >>Couldn't we just leave it as O? > >To make the equation fit the pattern that I gave at the start I >have taken O as a point with position vector v. It doesn't >matter if you prefer to call it 0. Okay, not confused here anymore. >>Also, how does v relate to 1 in the second term? > >If you have an equation with 0 on the right hand side you can >multiply or divide by ANY factor you like to make one of the >coefficients equal to 1. Yes. So, the constant in front of the null vector could just as easily be, say, seven - and that wouldn't produce different results? Dividing through by seven would change the other two constants, is that okay? It seems that using the null vector limits the freedom of the other two coefficients. Are you just allowed to decide that in this case to make it one? Couldn't the null vector coefficient validly be 0, would that mess anything up? >The left hand side MUST represent some point on BC, the right >hand side MUST represent some point on B'C'. Since the two >sides are equal the point must be where BC and B'C' intersect, >that is the point P (position vector p) In this linear combination, why must each side represent a point on BC or B'C'? It _sounds_ reasonable enough, but this is the only step which still seems trivial to me, now. Is it a problem that each side only has two variable terms? I mean, two variable points will always be collinear, just as three variable points will always be coplanar. Thanks for your time -wim Date: 06/07/99 at 11:45:46 From: Doctor Anthony Subject: Re: A killer vector problem >So, the constant in front of the null vector could just as >easily be, say, seven - and that wouldn't produce different >results? Dividing through by seven would change the other two >constants, is that okay? It seems that using the null vector >limits the freedom of the other two coefficients. Are you just >allowed to decide that in this case to make it one? Couldn't >the null vector coefficient validly be 0, would that mess >anything up? Normally you can't make one of the coefficients zero because there is no factor that will make a coefficient go to zero except zero and that would knock out all the other terms as well. >In this linear combination, why must each side represent a >point on BC or B'C'? It _sounds_ reasonable enough, but this is >the only step which still seems trivial to me, now. > >Is it a problem that each side only has two variable terms? I >mean, two variable points will always be collinear, just as >three variable points will always be coplanar. b is position vector of B and c is position vector of C. So any point on BC will be of the form: b + k(c-b) = b(1-k) + kc Note that the sum of the coefficients is 1. Provided the sum of the coefficients is 1, then ANY expression of the form p.b + q.c where p+q = 1 is a point on BC. Now look at the expression that you asked about: k2.b - k3.c k2'b' - k3'c' ------------- = -------------- k2 - k3 k2' - k3' The sum of the coefficients on the lefthand side is k2 -k3 k2-k3 ----- + ------- = ------- = 1 k2-k3 k2-k3 k2-k3 so the point in space represented by the lefthand side of the equation is a point on the line BC. Similarly, the sum of the coefficients on the righthand side is 1, so the righthand side represents a point on B'C'. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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