I'll Take a Linear Combo To Go

This went well.

It was one of those classes when I felt like I learned at least as much as they did, and all kinds of unexpected things happened. During class and as I wrote this post.

The idea that I wanted to get across, with as few words as possible: That under certain conditions, for every answer, not only is there is always a question to go with it, but there is only and exactly one question that goes with it.

The exact mathematical version of that, if you're into that: Given a resultant vector, and two other non-collinear vectors, under certain conditions:

1. There is always a way to linearly combine the two to get the resultant.
2. There is only one way to linearly combine  the 2 that will work out to that resultant.

This is what we did in class, on the whiteboard, which, remember, is all online, so everybody can write on and see the same board at the same time, and talk to each other while they do.

First I showed them this:

and said: "The red vector is the resultant of some number of blues and some number of greens being added. Since the red vector is the answer, what am I going to ask you to figure out?"

Someone said "The question?" Bingo. In other words, they had to figure out how many blues and greens add up to the red.

I then put them into their breakout rooms (the online equivalent to groups at tables) where they moved the blues and greens around to see how many of each would add up to the red.

Here are a couple of their results:

Everybody got that 6 blues and 4 greens were needed, which was what I expected. I wrote "6 blue + 4 green = red", and talked about how everyone got the same solution...

Audrey's first light bulb moment:

"But," said one student, "we didn't all get the same solution." She explained that really these two solutions weren't equivalent, because the paths from start to finish weren't the same. Light bulb - we're not talking the same language! I'm talking numbers & algebra, and they're talking pictures. It was a great opportunity to clarify exactly what I meant by "solution" right at the start. I meant the total number of blues and the total number of greens, not so much the sequence/path.

So I asked if they thought that there was some other NUMBER of blues and greens that would work, not just a different path but a different numerical combo. I expected and hoped they'd come back with, "Why no Mrs, there isn't any other combo." Back they went. Here was a most interesting result, which lead to

Audrey's second light bulb moment:

...in which there are more than 4 greens and 6 blues, right? TOTALLY awesome. First we traced a path that actually lead to the resultant. Now I got to connect the idea of opposite vectors to subtracting - and establish the convention that if you go one green forward then one back, you've really cancelled out one green with another. Saw a lot of "OOOHHHHH!!!" light bulbs going on everywhere today. We finally agreed that 6 blue and 4 green is the only and simplest numerical solution. Non-verbal idea 2 done!

Now that I'm blogging I can see:

I just loved how all these things came up so naturally and as a result of THEIR manipulations and thoughts - perfectly legitimate and logical thoughts, too. I would never have thought of bringing up any of these issues, which means I would have missed out on making clear the ideas that absolutely needed to be clear before we could move on. Together.

But move on we did. Time for me to mess them up again.

I kept the red, but changed the blues and the greens like so:

Now they're trying to get the exact same answer, but find a different question to go with it - were we only able to get the red the first time using those particular blues and greens that I had lovingly hand picked?

This time, I was not looking for WHAT is the question for this answer so much as IS there a question for it at all. Very quickly I had groups asking if they could change the direction of the blues or greens.

Me: *bats eyes* Well if you insist, I suppose you can make them into their opposites, but that's it.

In fact, it was only possible if the blues got opposite-ed AND you could use part of a blue or green. In fact, the previous discussion on opposites fed into this one very nicely. This lead very naturally into the motivation and meaning for non-whole scalars and negative scalars:

I casually slipped into the notation used for linear combinations, as you can see.

Now that I'm blogging I can see also:

The ease with which they could move the arrows, flip them so they were opposite-ed, and cut them into pieces cranked up the potential a few notches. If I had had them drawing, erasing, etc it would have taken way too long and been way too frustrating, not to mention the colours wouldn't have played a part. It was so much easier to say red, blue, green. Plus it's more fun for them to be manipulating things, and this is as close to observable & active learning as I can get online. Mind you, if I had had any colour-blind students, that would have been a problem.

Back to non-verbal idea 1:

Me: But is it always possible, no matter what blue and green we start with?

Some of them: Yes.

Some of them: No.

I let one group make up their own blues and greens, copy them, and voila, it was still possible. This would be the equivalent of a magician saying pick a card, any card. Now we've established that no matter what the blue and green, we can always get the red using a linear combo of them. Non-verbal idea 1 done.

Under what conditions?

OR CAN WE?!?!? Time to motivate the "under certain conditions" part. For the next part we were all together, no more break out rooms:

I asked how many "a" vectors I would need to make the "c" vector:

Astounding and wonderful to find another hole in their comprehension! Some said 4, or -4, or "You haven't taught us how to multiply and change the direction of the vector." We did some trials together, lining up 4 of them, lining up -4 of them, and eventually agreed that there was no scalar multiple of "a" that would result in "c". Even though they "knew" that collinear vectors are scalar multiples of each other, that's not the same kind of knowing as knowing that non-collinear vectors are NOT scalar multiples of each other, and never will be. I just read that back to myself. Sorry for all the negatives.

Now that I'm blogging...you get the idea:

At this point, I just felt like this lesson/activity/whatever you want to call it, was going really well. So many ideas, so many levels, so much participation, so little time! AND I had at this point stopped using colour-talk, and very sneakily slipped into the proper vocabulary: resultant, collinear, scalar multiple, linear combination. SO I was doing more talking here but I was using official language. This might have been a good time to put in a hinge question, to check that they were all still with me.... next time.

"But," I said, "what if I told you you could use vector a AND some other vector, like this?"

They got this easily, based on the manipulations they'd just done - yes it's possible. But this was another layer of knowing - we'd already established that it's always possible but now they saw that it's actually impossible without the other vector, in order to swing things back to the resultant.

Me: Will it always be possible to get c as the resultant, as long as I have two other vectors to combine linearly?

Them: Yes, Mrs, geez, we get it already.

Me: Really? You sure?

Them: *blink collectively*

Me: *smiles evilly*

Me: What if the other vector was this:

Boom! More OHHHHHH's.  Just established what those"certain conditions" are. The two have to be non-collinear, or else you can't swing back toward the resultant.

And the linear combo to go:

If I had had just a little more time in class, I would have moved into this next, but I didn't have time so I assigned a few of this type of question:

What linear combination of  <-2, 7> and  <1, 4> results in  <-1, 26> ?

Next day, the colours paid off:

Many had difficulty, not all, but many. I saw my job as helping them see the question first. Showing how this question was the same as the blue, green, and red arrow questions. In fact, it was really handy to be able to refer to the colours and make this abstract question more concrete that way:

n<-2, 7>  +  m<1, 4>  =  <-1, 26>

One student said he had tried solving these by trial and error, which at first was easy, but got harder as the examples got harder. He asked if there was a better way. Once he saw that it was a system of linear equations, he and others again went "OOHHHH!" It was an aha moment that algebra really is useful sometimes.

Now to figure out how to do this if I have a colour blind person in my class.