GL(s,R)

September 12, 2012

Projecting onto Projections

Filed under: High Effort/Low Payoff Ideas — Adam Glesser @ 1:53 am
Tags: , , ,

The first time I saw the expression \int_C \mathbf{F} \cdot \mathbf{n}\ d\mathbf{r}, I thought, “Why should that dot product be in there. By the time I saw \iint_S \mathbf{F} \cdot\ d\mathbf{S}, I resigned myself to the fact that there was always a dot product in these seemingly random integrals. At some point, I decided that the dot products are in there to turn vectors (or vector fields) into scalar functions—which is something we know how to integrate. More recently, I’ve decided that the purpose of these dot products is to capture the projection of one vector on the other.

For example, if I apply a force \mathbf{F} to an object, then the work done by that force in moving the object a certain distance in a given direction (denote this shift by \mathbf{v}) is \mathbf{F} \cdot \mathbf{v}. If the force is not constant over some curve parametrized by \mathbf{r}(t) (a \leq t \leq b), then we compute the work by evaluating the integral \int_a^b \mathbf{F} \cdot \mathbf{r}'(t)\ dt since, at any given point, our \mathbf{v} from above is just the tangent vector to the curve at that point, i.e., \mathbf{r}'(t).

If you understand multivariable calculus, then you are probably laughing at me. “Duh. Why did it take you so long to figure that out?”

Here is my answer: We (or maybe just I) improperly motivate the dot product. This semester, I’m using Stewart for Multivariable Calculus*. He introduces vectors in a way that seems fairly standard for math texts.

Definition: The dot product of \langle x_1, \ldots, x_n \rangle and \langle y_1, \ldots, y_n \rangle is x_1y_1 + \cdots + x_ny_n.

The great thing about this definition is that it is bloody easy to compute and understand.

Theorem: If \mathbf{a} and \mathbf{b} be vectors with angle \theta between them, then \mathbf{a} \cdot \mathbf{b} = \mid\mid \mathbf{a} \mid\mid\ \mid \mid\mathbf{b}\mid\mid \cos(\theta).

The beauty here is that you can use the dot product to help compute angles and it is immediately obvious that the dot product of orthogonal vectors is 0.

*This wasn’t my choice, but rather the choice of my department. Oh, did I mention I got a new job? Indeed I finally gave up on east coast living and moved back to California. I am now in the mathematics department at California State University Fullerton.

I’ve heard that in physics textbooks, they switch the order of the above, i.e., they define the dot product via the cosine formula and then prove the above definition as a theorem. As a mathematician, I always went with the first definition. Now, I am not so sure. What follows is the introduction to the dot product I plan to give to my students (until I come up with something better, anyway*).

*In the comments, please do set me straight about the real purpose of the dot product or how you think it best to introduce it in this context.

Let’s start with two vectors, joined by their tales.

I am interested in how far \mathbf{b} extends along \mathbf{a}, so I drop a line perpendicular to \mathbf{a} from the end of \mathbf{b}.

At this point, I’m already confused by what would happen if I had tried to see how far \mathbf{a} goes along \mathbf{b}, but I decide that I could simply extend \mathbf{b} and at least draw the following picture:

Awesome, I have a couple of right triangles. And, heck, since they are right triangles that share the angle (let’s call it \theta) between \mathbf{a} and \mathbf{b}, they are similar triangles. Let’s give some names to the important sides.

The comment about similar triangles implies that \dfrac{h}{||\mathbf{b}||} = \dfrac{k}{||\mathbf{a}||}. Ugh, let’s clear denominators to get h||\mathbf{a}|| = k||\mathbf{b}||. On the other hand, \cos(\theta) = \dfrac{h}{||\mathbf{b}||}, and so if we multiply by ||\mathbf{a}||\ ||\mathbf{b}||, we get

||\mathbf{a}||\ ||\mathbf{b}||\cos(\theta) = h||\mathbf{a}||

The moral is that this important quantity—h||\mathbf{a}|| = k||\mathbf{b}||—coming from projecting the vectors onto each other, has a very simple reformulation as ||\mathbf{a}||\ ||\mathbf{b}||\cos(\theta) which only relies on knowing the original vectors and the angle between them. Since this projection property is so important to us physically, we give a short name to this expression: \mathbf{a} \cdot \mathbf{b}, and call it the dot product of \mathbf{a} and \mathbf{b}.

If \mathbf{b} is orthogonal to \mathbf{a}, then the projection should be 0, which of course it is since the cosine of 90^\circ is 0.

At this point one can go about proving that the dot product is obtained directly from the components, i.e., without knowing the angle between them. Of course,  there is still the issue of when \theta is obtuse, and it will probably be helpful to cover that case as well. Geometrically it will look a bit different, but the algebra and trig will be almost the same*.

*You do get to use the fact that the cosine of an angle equals the cosine of the supplementary angle!

There is nothing really new here, but I think the ordering is important. Their first impression of the dot product should convey the purpose of the dot product, not just the easiest algorithm for computing it. As it stands, the projection of a vector onto another vector gets a a somewhat token reference at the end of the dot product chapter. As ubiquitous as the idea is throughout the end of the class, it deserves its time in the sun.

3 Comments »

  1. I liked this article for helping understand what dot product means. I used parts of it when I taught it this week. It worked well. http://betterexplained.com/articles/vector-calculus-understanding-the-dot-product/

    Comment by Kate Nowak — September 14, 2012 @ 8:10 am | Reply

    • That article is fantastic! There are several more on there I have to read now. Thanks, Kate.

      Comment by Adam Glesser — September 16, 2012 @ 1:08 am | Reply

  2. I am a student research assistant at Montana Tech of the University of Montana. Technology has created exciting ways to connect with others and form professional learning networks. As a part of an active member of a social media community made up of teachers, I wanted to contact you to ask you to participate in a study our research group is conducting.

    Research shows that face-to-face professional networks provide much needed professional and personal support to teachers. You and the community you belong to are providing these types of support using social media. We are interested in learning more about your experiences using social media to connect with other teachers and your opinions about online professional networks.

    The purpose of our study is to learn how professional learning networks created through social media are similar or different than face-to-face networks and what you feel are advantages of using social media to connect with other teachers. Our hope is that the results of this study will inform how professional networks for teachers are designed in the future. If you are interested in participating, please send an email to me at teacherblogPLN@gmail.com. I will send you a link to a short online survey and will set up time for a short skype interview.

    If you have any questions you would like to ask about the study, please do not hesitate to contact me.

    Sincerely,

    Kaitlyn Rudy
    Research Assistant
    Department of Mathematical Sciences
    Montana Tech of the University of Montana

    Comment by Kaitlyn Rudy — July 18, 2014 @ 10:27 am | Reply


RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

The Rubric Theme. Blog at WordPress.com.

Follow

Get every new post delivered to your Inbox.

%d bloggers like this: