Debate Debate

Relying Exclusively of Volatile Eruptions of the Reptile Brain Mark: You should watch the debate.

Preposterously Idealized Self SErving Rational Mark: Will it be interesting?

REVERB Mark: You better be interested! It's the Presidential Debate!

PISSER Mark: I don't follow that.

REVERB Mark: You're a functioning, adult member of the society now. You live in a democracy, and that's a great privilege. It's your responsibility to take part in the governance of this nation, and you didn't even vote in the last election! Now you better turn on that debate.

PISSER Mark: Given my demographic, occupation, community, and personal interests it's not difficult to guess which candidate I support. I doubt that watching the debate will change that.

REVERB Mark: That's not the point. Even if you know who you're going to vote for, you still need to stay well informed.

PISSER Mark: Why?

REVERB Mark: Because you're more privileged than any but the tiniest fraction of human beings alive, and yet you take it all for granted. You live in a nation where you personally can have real, meaningful participation in the political process, and it's your responsibility to exercise that right.

PISSER Mark: I admit I frequently take the perks of life in America for granted, but I don't see why following politics is a responsibility, rather than a personal choice. I'd rather do something else.

REVERB Mark: What do you think politics is, then?

PISSER Mark: A hobby.

REVERB Mark: A what?

PISSER Mark: A hobby. A pastime. Like being a sports fan. You can read about sports in the newspaper, listen to experts talk about it on the radio. Then you can choose which team you like, and "support" them by buying bumper stickers and paying for admission to their games. All the same applies with politics, except the bumper sticker becomes a sign on your front lawn and the ticket to the game becomes a vote. They're both just ways for people to have something to get riled up about without doing much harm.

REVERB Mark: Not an apt analogy at all. Sports are just games. It doesn't really matter who wins the World Series. But it sure matters who wins the Presidency!

PISSER Mark: A little bit, maybe. My job prospects might be a little dimmer with one candidate in charge than the other. Their differences on social issues could affect how pissed off the people around me are. The foreign policy they enact will cause me to read different headlines in the news.

REVERB Mark: That's not just news items. All that stuff is really happening out there, to real people. It's not just all about you. To some people it makes a huge difference who's President.

PISSER Mark: But we're discussing my personal decision on whether to watch the debate.

REVERB Mark: Look, I know what you're thinking. You're thinking, "my vote is just one gumball in Candy Land. Doesn't make a bit of difference. The vote in my state always goes to the same party, and one more vote cast either way can't change the result." But you know what? One hundred thousand votes can change the result. And what if one hundred thousand people just like you are using the same logic?

PISSER Mark: Are you saying that if I go to vote it will change one hundred thousand lazy people's minds?

REVERB Mark: You know that's not what I'm saying. I'm saying you shouldn't vote because you think your particular vote is going to make all the difference. You should vote because it's a statement that you give a rat's ass about the future of this country! Not just for yourself, but for everyone you share this country with. And since the US is a leading world superpower, everyone you share this Earth with. And before you vote, you need to be informed. So you need to watch the debate.

PISSER Mark: Watching just one debate won't make me informed. Becoming a competent participant in democracy is a significant commitment.

REVERB Mark: One that every adult American ought to make.

PISSER Mark: Have you met many adult Americans? On the whole, they're not the ones I want making any important decisions. Basically, ability to choose a good Presidential candidate will undoubtedly fall on a curve like this:















So let the people who are innately interested in the whole process, and can do a good job of it, pour their efforts into politics.

REVERB Mark: All the more reason for you personally to get involved. You're way at the top of that curve. Remember high school? You don't want to common man making the decisions - then YOU better get involved and do something about it.

PISSER Mark: That's not really my point. I don't want to invest huge amounts of time and energy into becoming an "informed citizen", which by the way makes the dubious presupposition that it's possible to make a good choice from behind a computer screen, because it doesn't interest me much. Even if I did watch the debate, I don't know enough about foreign policy or economics to decide who I agree with. I'd inevitably fall back on judging demeanor and rhetoric. I could do more good by putting that effort into something I'll enjoy.

REVERB Mark: Like what? What can you do that's more important than your civic duty?

PISSER Mark: I don't know... plant a tree or something, I guess.

REVERB Mark: Oh come on. You've never planted a tree in your life.

PISSER Mark: I did, actually. I planted the free pine tree seedling they gave me third grade on Arbor Day. I think dad accidentally ran over it with the lawn mower.

REVERB Mark: Not the point. That's not what you will do with that extra time. What you will do is read stupid websites or play flash games, or something inane like that.

PISSER Mark: Or maybe I'll study for a while. Or work on that project I had planned, or...

REVERB Mark: Oh please. You've already got three tabs open and they're Pacman, SomethingAwful.com, and then something more awful than that.

PISSER Mark: sigh. Where is it streaming from?

What does a determinant have to do with a cross product? pt. 1

You're probably aware of a nifty mnemonic for calculating a cross product.



If you take a dot product of the above expressions with another vector , you'll get



What a crazy magic coincidence that vector products and determinants can be calculated the same way!

I won't say magic coincidences don't exist, but I will say that when you find one it may behoove you to look a little deeper. There is a deeper connection.

I'll hint at it today by posing a certain problem, then showing that the problem has a solution whenever either expression, the determinant or the vector product, is nonzero. So is it two different conditions that show the problem can be solved, or the same condition in different guises? We'll deal with that later.

Imagine you are given four random vectors in space, . Then you are asked to express the fourth as a linear combination of the first three. That is, find such that

I won't solve the problem. I'm just asking for a test to see if it's soluble (besides throwing water on it). Especially, I want to see if it's soluble for all possible , not just special cases.

Using A Basis

Set up a standard Cartesian basis in three dimensions, and write out the vector equation in components




which is a system of linear equations that can be written in matrix form as



(Aside: If you know how to tell to make the vertical columns the same size in those matrices, drop me a comment.)

If you've taken linear algebra, you'll recognize that the problem can be solved only if the above matrix has a non-zero determinant. That same determinant is the one we were interested in earlier, since it's also the value of

Without a Basis

Vectors and define a plane. They're two independent vectors (I'm assuming they're independent - if they're dependent, you might as well just get rid of one of them for the purposes of solving this problem, but it's clear that in general that won't work) in a two-dimensional space, so they form a basis for that space.

Break down into two parts, the part in the plane of and , which we'll call and the part perpendicular to the plane, .

and are irrelevant to the perpendicular part of the equation, so we're left with



We need to find , so we better dot with a unit vector in the perpendicular direction.



Combining the previous two equations lets us solve for .



Unless is by extraordinary chance in the plane of and (and we're interested in the general case, not special ones), we need to be finite, meaning needs to be nonzero. This condition also has built into it the requirement that and be independent, so it's a sufficient condition for the problem to have a solution.

Summary for today

The equation



can be solved for any whatsoever if



also, the same is true if



That doesn't prove the determinant and the vector product are identical, but it's hinting at why it's true. Tomorrow I'll probably post a little about what determinants are, and the geometrical interpretation of the vector product.

A X (B X C)

Any intro to physics course will teach you the identity



It's easy to prove by explicitly writing out each vector in component form, cranking through both sides of the equation, finally getting the same thing on each. But there's some insight to be gained in looking at the identity from a purely geometric point of view. Warning: the following is a bit dense, but also short. Go one sentence at a time and you should be fine.

Together, and are a basis for a plane. is perpendicular to the plane. Take that perpendicular vector product, cross it with , and you get a vector perpendicular to the perpendicular to the plane. That's back in the plane again. So we know that the double cross product is in the plane somewhere. But any vector in the plane can be written as a sum of and (that's what it means to be a basis), so we only need to find the right coefficients. Those coefficients come from requiring that we find a combination of and perpendicular to . Just dot the expression
with , and you'll get zero.

That proves both sides of the equation



point in the same direction, but doesn't prove they have the same length. It's not crazy difficult to prove the they have the same length without using a coordinate system, but I don't have a neat insightful trick for it, either.

I started by noting that you could break down into perpendicular and parallel components to the plane defined by the other two vectors, and that as far as the identity is concerned the perpendicular part of doesn't matter (plug it in and you'll see). Then, because both sides of the equation are proportional to the lengths of all three vectors, you might as well divide by their lengths to make them unit vectors. All that's left is to find the lengths of both sides of the equation in terms of trig functions of the angles between the vectors. You have tree angles between the coplanar vectors. Write one as a sum of the other two, expand out all the trig functions, and fiddle around with it a bit until you see that it's actually a trig identity. That gets the final part of the theorem in there.