Lecture Note
University
Massachusetts Institute of TechnologyCourse
Multivariable CalculusPages
5
Academic year
2022
Sporkz
Views
32
Geometry proofs using coordinate free vectors ● Euclidean geometry is a branch of geometry that deals with the theory of the relationship between points, lines and planes. It is one of the three classicalbranches of geometry, together with topology and trigonometry. ● The median of a triangle is the line segment that joins the midpoints of the three sides. In other words, it is a line segment that connects two midpoints of a triangle.The midpoints are located at the intersection of two sides and the included angleopposite to them. The median can be found in any triangle and its location dependson the lengths of its sides. If all three sides are equal, then the median is equidistantfrom all three vertices. This is the recitation on vectors. Throughout all topics, vectors will play a crucial role insolving equations. Here is one problem to work with in a slightly different context than we will be working inlater. The context is Euclidean geometry. You may have seen this problem before, but maybe you sis not try to solve it with vectors.Let's take a look at how it is done. Please show that the three medians of a triangle intersect at a point. The point is 2/3 of theway from each vertex. Now some terminology. A median is the segment that connects one vertex to the midpointof the opposite side in a triangle. So here, point M is exactly halfway between points B and C. The center of any triangle lieson a line that bisects the triangle and passes through its center; this line is called amedian. Thus, every triangle has three medians: one from each vertex connected to themidpoint of the opposite side. And you are asked to show that these three medians all intersect in the same point andalso divide the median into two pieces, one twice as large as the other. So this istwo-thirds of the median, and this is one-third of the median. Think about it. Try to include vectors.
To start, let us rephrase the question a little bit. We'll rephrase it to an equivalent questionthat's a little bit more clear. So to say this another way, the problem asks us to show that for each median AM whereM is the midpoint of side BC, there exists a point on AM that divides it into a 2:1 ratio. Thus, the point that is 2/3 of the distance from the vertex to the midpoint of the oppositeside. Let's say there is a point P, where AP is twice PM. And same way, there is some point Qthat is 2/3 of the way from B to the midpoint of this side. And there is some point 2/3 of theway from C to the midpoint of this side. Thus the solution of such a problem is equivalent to showing that the three points arereally one and the same point, that they are all in the same place. To determine whether three objects are in the same position, we can compare the positionvectors of those three points. If those three points all have the same position vector, thenthey're all in exactly the same position. To solve this problem, we need an origin. It happens that for this case it does not matterwhere the origin is. Instead of drawing an origin, we simply call it O. We will set up a vector coordinate systemwith the origin at point O. Now we need to look at the vector from O to P in terms of the vectors connecting O to A,B, and C. These are the vectors that determine the vertices of the triangle. We will now relate the location of P to the locations of A and M. To do so, we first need todescribe how P is related to both A and M. And we know the relationship among A, B, and C. So let's first find the position vector of Min terms of A, B, and C. We can use the midpoint M of the segment BC to determine P's position vector. You saw this before. The vector position of M is exactly the average of the vectors from Band C. The position vector is 1/2 the sum of OB and OC.
It is easy to express the position vector of a segment's midpoint in terms of the positionvectors of the endpoints. You simply add the position vectors of the points and divide by 2. This fact is equivalent to the geometric fact that the diagonals of a parallelogram bisecteach other. Thus, vector M is the position vector of P. Next, we must determine what the positionvector of P is. To solve this problem, we need to get from the origin to point P. Well, we can see that weneed to go from the origin to point A first. The vector OP is equal to the vector OA plus 2/3 times the vector AM. Because we go two thirds of the way from A to M in order to get from A to P. We chose the point P to be the segment dividing AM into a 2 to 1 ratio. So AP is 2/3 of AM. OK. Now we can find AM. As we know, the position vector of A is OA. We also know that M's position vector is OM. Thus, AM is the difference between the OA and OM vectors. OM minus OA. Another way to say this is that if you add OA to both sides, you have that OA plus AMequals OM. In other words, you can go from O to M by first going from O to A, and thengoing from A to M. We can write AM in terms of OM and OA by subtracting OA from both sides of theequation. And we also have an expression for OM here in terms of OB and OC. This means that wecan get an expression for AM in terms of OA, OB, and OC. So, let's do that. If we substitute this expression for AM into the original equation, we getthat OM is equal to 1/2 OB plus 1/2 OC. Then we can subtract OA.
The result of putting these two equations together is AM. I need to find the value of OP by substituting AM into the expression for OP. So OP is equal to OA plus 2/3 of AM, which equals 1/2 OB plus 1/2 OC minus OA. Good. Now, to multiply this 2/3 in, we can distribute the scalar multiplication across the addition.Then, we can rearrange. We will have two terms involving the OA and we can combine them. So we see that it isplus OA minus 2/3OA. So, we can set this equal. And then, we have 2/3 times 1/2 half OB. So that's going to beequal plus 1/3 OB plus 1/3 OC. So we have a simple formula for the position vector of P, that vector OP, in terms of theposition vectors of A, B, and C. So, in particular, P is a special point because it is 1/3 of the sum of OA, OB and OC. Now, to finish the proof, we must be sure that the point that trisects the other medians lieson this same line. How do we prove that? We could go back to my triangle, but it will bethe same thing again. Let’s call this midpoint N. Now it also has a name. Nice. We could make Q the point thatlies 2/3 of the way from B to N. We could then write the position vector of N in terms of OA,OB and OC. We should hope that by the end we'll have learned enough to be able to create anexpression equal to the one found earlier. In fact, it works. Thus, there is a clever shortcut to seeing this formula as symmetric in A, B, and C. So, if you re-label the points (1,2) as (A), (B) and (C), this expression for the positionvector does not change.
Rather than going through the process described previously, you can also say that in orderto look at Q rather than P, you need only switch B and A. If the roles of A and B are interchanged, then in the resulting formula you just have tointerchange the roles of A and B. However, this does not alter the value of this expression. Thus by symmetry, the point youget will be the same. If you do not like that argument, go through the computation again in the case of othermedians. In either case, the points that trisect the three medians all have position vector 1/3OA plus1/3OB plus 1/2OC, but we can see that these points are the same point. Thus, we have shown that the points that trisect the three medians divide them into 2 to 1ratios from the vertex to the midpoint of the opposite side. All three points have the same position vector, so they are in fact the same point. That’s it, we wanted to show that there is a point that trisects all three of those medians. So, we have shown that the three points that trisect a given segment are actually the samepoint. This can be interpreted as a restatement of the original conclusion.
Geometry Proofs using Coordinate Free Vectors
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