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 “dist! ance 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” r! elations hip (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.

addendum Addendum to “proportion of a line segment”.
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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 w! ill 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-coplan! ar points representing adjacent vertices of a parallelepiped a! nd 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 momen! t they could reproduce it.

As a sidebar, I woul! d add th at 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.

distance midpoint formula