Friday, September 26, 2008

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.

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