 So given a set of vectors, we can try to find a vector orthogonal to all of them by setting each dot product equal to zero and trying to find a solution to the resulting system of equations. Unfortunately, this is too easy. And it would leave us nothing else to do with this lecture, which is supposed to last 50 minutes. So we can summarize all the steps into a single formula. You really don't want to use the formula. It's probably still easier to just find a solution to the system of equations, but we like formulas. So we do need to introduce some notation to make the formula usable. So while we can write vectors by listing their components, it's sometimes useful to write them as the sum of simple vectors. The sum of simple vectors is what's known as a linear combination, and the simplest vectors in three dimensions are i, which is the vector 1, 0, 0, j, 0, 1, 0, and k, 0, 0, 1. So for example, we might consider what vector is represented by 3i minus 2j plus 5k. And it's convenient when dealing with linear combinations to not subtract but instead add the additive inverse. So 3i minus 2j plus 5k. We'll rewrite that as 3i plus negative 2j plus 5k. And then we have our vectors i, j, and k. i is 1, 0, 0. j is 0, 1, 0. And k is 0, 0, 1. We'll do our scalar multiplication. We'll add the vectors. We can also try to go backwards. We can try to express a vector in terms of i, j, and k. So the first thing to notice is that only i, 1, 0, 0, has a non-zero first component, which means that the first component for must come from a multiple of i. And since we want that first component to be 4, we need 4 times 1, 0, 0. Next, since j, 0, 1, 0, has a non-zero second component, this means the second component must come from a multiple of j. And since we want our second component to be negative 3, we need negative 3 times 0, 1, 0. And finally, only k, 0, 0, 1, has a non-zero third component, which means that the third component must come from a multiple of k. And since we want that third component to be negative 7, we need negative 7 times 0, 0, 1. And so our linear combination should be 4i plus negative 3j plus negative 7k. And we can write the addition of the additive inverse as a subtraction, so we might write this as 4i minus 3j minus 7k. And this allows us to introduce the cross-product. If we find a vector orthogonal to two other vectors, we could write an equation setting each dot product equal to 0, and find a solution to the resulting system of equations. We should actually do this, but it's useful to have a formula if you can remember it. So here goes. Let x and y be two vectors with three real components. The cross-product, vector x cross vector y, will be defined as and this vector will be perpendicular to both x and y. So let's say we want to find a non-zero vector perpendicular to both 1, negative 1, 3, and 2, 5, negative 4. We can find the cross-product. And the first few times you find the cross-product, you'll probably want to have the formula right in front of you. Let's see, that's x1, x2, 3, minus 2. So that's going to be x2 times y3 minus x3 times y2. And we fill in all of the numbers. And then we do a bunch of arithmetic. And that gives us our orthogonal vector. Now, as you can guess, trying to remember the formula for the cross-product is kind of challenging. And again, you're better off writing down the system of equations and just solving it. As we saw, that's very easy to do. But if you insist on memorizing a formula, one way to remember the formula for the cross-product is to find the determinant. Of course, you have to know what the determinant is. So, well, let's not worry about that. And let's just set this up. You can think about this as a useful mnemonic device, which is more trouble than it's worth. So what we'll do is we'll set up this table. And our first row are going to be those vectors i, j, and k. And the second and third rows are the two vectors. And as we'll see, it does make a difference which order we put them. Now, to find the determinant, you should take a course in linear algebra. But in this particular case, one way to find the determinant is to double this matrix. So now I have this i, j, k. I'm going to put a second copy down. And there's a couple of diagonals here. So we have these down-right diagonals. And we have these down-left diagonals. And so we can distinguish them, these non-repeated down-right diagonals. If I multiply along them, I get i, x2, y3, j, x3, y1, k, x1, y2. And the next diagonal is a duplicate of what I already have, so I'll ignore it. And then I have a bunch of these down-left diagonals. k, x2, y1, i, x3, y2, j, x1, y3. And then this diagonal here is a duplicate, so we'll ignore it. And then we have this rule of sorrow, which incidentally only works in this one instance. In this one particular case, we can add the non-repeated down-right diagonals and subtract the non-repeated down-left diagonals. And if we do that, that gives us our formula. So let's go back to that cross-product. We'll set up our matrix i, j, k, 1, negative 1, 3, 2, 5, negative 4. We'll double it, and then we'll take those down-right diagonals, i, negative 1, negative 4, j, 3, 2, k, 1, 5. And then the down-left diagonals, k, negative 1, 2, i, 3, 5, j, 1, negative 4. And remember the down-right diagonals are added, and the down-left diagonals are subtracted. And after all the dust settles, we get our cross-product.