## Wednesday, 9 June 2010

### More on Vectors in the HS Curriculum

Several colleagues took me to task privately, and unjustly I think (he whined), for suggesting that more stuff (vector topics) should be added to the (already overstuffed) curriculum. My idea was neither to replace the traditional y=f(x) approach commonly used, nor to introduce multiple chapters in vectors throughout the curriculum strand. Instead, I think the integration of a few “15 minute vector asides” at a number of places each year would have the potential to greatly enrich a student’s ability to do those big ideas in math, generalize, synthesize and specialize.

Distance is an abstract concept that is introduced as early as fifth or sixth grade for the number line (maybe earlier) and by grades seven or eight for the coordinate plane. Ok, I’m ready to let the fifth graders stay with Pt A –Pt B for distance on the number line. For the coordinate plane, the usual approach is to resort to a memorized “distance formula” that few of them ever realize is the Pythagorean theorem. Even fewer naturally extend that to three-space, although I admit that would seem pretty trivial. If at about this moment we gave a very brief introduction to vectors The “vector” from Point A to Point B is (B-A). Kids could use vectors to translate points on a two-or three-d grid and immediately realize the relationship between the vectors [3,5] and [-3,-5] . I would think within a single days lesson, most students could write the vectors between points in two-space and three space (how often do grade 6-7-8 students see a three dimensional point, I wonder.). Then the square root of the dot product is an easy way to define distance, and it is immediately defined for ALL dimensions. By day two, students can be finding distances between points in the room located with coordinates in feet or meters and an origin established at one corner of the room. The same students would, I suspect, immediately see slope as a “vector” relationship (although there will always be kids who get hung up on the order, x over y / y over x, as they do now).

One of the places I notice a need for at least a “vector translation” understanding of a line segment shows up in the frequency with which Alg I and II students are asked to find the midpoint of a segment (there is probably even a “midpoint formula” in the book) but are almost never asked to find the point 2/3 of the way from A to B. One leads to a view of math as using rules, and another as using ideas. My goal is to move away from the former and towards the latter.

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In a similar way, many students are taught to find the centroid of a triangle, usually by a memorized relationship. The student who can understand and apply the use of a vector approach to lines will be able to see that any point on the plane determined by triangle ABC can be represented as a linear combination P= ra + sb + tc with r+s+t=1 . Again, many students are more at ease with representing this linear combination as P=a + s(b-a) + t( c-a) since it can be seen visually as a translation from the origin to point A, then motions parallel to the two sides meeting at A to locate the point.
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Somewhere in the Alg I/Geometry sequence, they come to that point where they are given three consecutive points on a parallelogram and asked to find the fourth point. I cannot imagine many kids being successful at this who do not take a more-or-less vector approach to the motion along the segments of the parallelogram. But once they see a vector approach to “moving” a point; and if they were familiar with 3-d coordinates, how long would it take to give them four non-coplanar points representing adjacent vertices of a parallelepiped and have them find the other four points.

If we look at two sides of a triangle as vectors, call them a and b, and think of the third side c as the vector a-b, then the square of the length of c is c . c. But replacing c with a-b we get (a-b) . (a-b) and distributing we get a . a + b . b – 2 a . b and keeping in mind that a . b = |a| |b| cos(theta) we have the law of cosines . The distributive nature of the dot product gives the law of cosines as a simple application of the distance formula. And if a and b are perpendicular, their product is zero, and so the Pythagorean theorem also pops out.

A geometry student with an understanding of the dot product can show easily that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals. If we represent the two sides at Point A as vectors a and b, then the diagonal AC is a+b, and the diagonal BD is b-a. The dot product of w=a+b with itself gives |a+b|2= |a|2 + |b|2 + 2(a.b).
The same approach with v=a-b gives |a-b|2= |a|2 + |b|2 - 2(a.b), and summing the two we get the result that |v|2+|w|2 = |a|2 + |b|2+|a|2 + |b|2 and substituting in a1 and b1 in one pair gives the final result. Understanding the distributive law removes the necessity to use the memorized law of cosines. (I would love to have a student respond that they didn’t remember the law of cosines, but given a moment they could reproduce it.

As a sidebar, I would add that like my recent post about Almost Pythagorean relations this one qualifies as both also Pythagorean, and also as one of those special cases where two wrongs make a right...ie. for sides a, b, c, and d of a parallelogram with diagonals d1 and d2 it is NOT true that a2+b2=d12 and likewise for c2+d2=d22, but when both equations are added together, the result is true.

A geometry course rich with vector topics would seem to make three-dimensions a much deeper part of the course. It would also, as I hope to show in subsequent posts, allow us to easily extend some of the things we do in two-space.

Dave L. Renfro said...

"... the frequency with which Alg I and II students are asked to find the midpoint of a segment (there is probably even a “midpoint formula” in the book) but are almost never asked to find the point 2/3 of the way from A to B."

It's interesting that you bring this up now! The last session I had with a certain algebra II student I'm tutoring involved working with the midpoint and distance formulas. He's a reasonably good 10th grade student and he's taking a pretty stiff honors algebra II course, but his parents think some outside work with a tutor will be worth their money (and I'm pretty cheap, at \$20 an hour). In particular, they want me to supplement, extend, enrich, etc. what he covers in class. So naturally, when we met and he told me that the midpoint formula was one of the things covered in his class last week, I brought up a topic you can find in most any old analytic geometry text (see , a google-books search for "ratio" AND "point dividing a line segment"), namely that of a point dividing a line segment in a given ratio.

 http://tinyurl.com/22qowkq

Here's a higher-dimensional result you and your blog's readers may find interesting. For each positive epsilon > 0 and for each positive integer N, there exists a positive integer n such that at least N many n-dimensional balls of diameter (1 - epsilon) can be placed without overlap in an n-dimensional cube with edge length 1. See the solution to P.167 on pp. 605-606 of Canadian Mathematical Bulletin Vol. 14, 1971, which is freely available on the internet (at least in the U.S.) at . The solution is actually not all that difficult.

 http://tinyurl.com/2664k6o

Pat's Blog said...

Thanks Dave, Unfortunatly my experience is that many teachers never think to generalize the question (and some don't know how to do what you showed this student). I find that when I show students they find notation like 3/4 a + 1/4 b to be confusing (and often reverse the proportionals). I really prefer a+t(b-a), and it allows you to find not just a point between Pts A and B, but also points outside the segment by use of values of t>1. And they seem to find it pretty natural that to find points the other way, they go b+t(a-b).

Dave L. Renfro said...

I didn't spend much time in working with the concept (dividing a line segment in a given ratio), but rather I wanted to show how the proof his text gave for the midpoint formula easily leads to the more general situation (the congruent triangle set-up relaxes to a similar triangle set-up). I also showed the power of geometry (synthetic geometry, that is) by setting up the xy-coordinate equations for finding the midpoint, which is

dist(P,Q) = dist(Q,R) = (1/2)*dist(P,R)

for finding the midpoint Q of two given points P and R. [If you omit the right hand equals, you'll have one equation in two variables (the 2 coordinates of Q), which leads to finding the perpendicular bisector of segment PR.]

Now I'm not all that keen on geometric methods (mainly because I'm awful at geometry), but the contrast in difficulty is enough to make even a geometry-phobe like myself take notice!