 In this video, we're going to see how to calculate something called the cross product of two vectors. It's also called the vector product because the output is a new vector. And we'll see how to test that the answer is correct. So here I've written A cross B is equal to C. And notice that the symbol for the cross product is just the multiplication symbol that you're familiar with from basic arithmetic. I've given the vector A a particular form, this two, three, four column vector. And similarly B is written as 456. So we're going to go ahead and find out what is the cross product of these two vectors C. Because it's a vector, we'll need to do some working for each of the three components. Now what I'm going to do is I'm going to paste up some structure to help us work through the problem. So don't worry because it's going to look like a lot. But you don't need to write all this out every time you want to do a cross product. I'm just putting it here so we can really spell out the process. Okay, so let's go ahead and work out the first component of the output vector C. Strangely enough what we're going to do is we're going to ignore the first component of vectors A and B. So I'm just going to cross those out, those aren't used. And what we're going to do is we're going to multiply a certain of the other components. What we're going to do is we're going to multiply the second component of vector A with the third component of vector B. I call that the falling diagonal because when we draw it like this we start high and then go low. And then we're going to subtract off the multiple of the rising diagonal 4 and 6 here. The last component of vector A and the middle component of vector B. So what we have here is 21, that's 7 3's are 21, minus 6 4's are 24, that's minus 3. We can go ahead now and write that in as our first element minus 3. Now let's move to the second element of the output vector C. We'll start by ignoring the second component of the two source vectors A and B. We can cross those off and again we're going to multiply some diagonals. But what's different here is we start with the rising diagonal 4 times 5. The last component of vector A times the first component of vector B, the rising diagonal 5 4's are 20 and then we subtract off the falling diagonal so 2 7's are 14 and that's going to give us 6. So we can put that in. Now let's move to the third and final component. As before, we start by noting that we will ignore the third component of the two source vectors and we're going to need some diagonals. It's the same pattern as the first falling diagonal first. So 2 times 6 and subtract which is 12 and then subtract off the rising diagonal 5 3's are 15. Alright, so that's going to be minus 3. Pop that in. We see that we have quite a simple vector here. There's a common factor of 3. Let's bring that out. 3 then minus 1, 2 minus 1. That is our vector C. That is A cross B. Notice again the pattern. It was the falling diagonal minus the rising diagonal for the first component and then the rising diagonal minus the falling diagonal for the second component and then for the third it was back to the same pattern as for the first. These look a bit like letters to me. They look a bit like a V, the middle one perhaps an N and the final one a V. I like to remember that as a little sentence which is Vols never vary. Because in my opinion Vols don't vary very much. Here's a Vol. This one doesn't vary at all because it's stuffed in a museum. However if you compare it to some other Vols which I found these on the internet I think they are all pretty much identical and it's a big difference there. For me Vols never vary. If for you they do seem to vary then think of a different way of remembering it. But the important thing is that the first thing is the falling diagonal and then subtract the rising diagonal of V shape and it alternates. How to check your cross product has been worked out correctly. This is really useful stuff. Let's give ourselves another example. We'll have 2 3 1 and then we'll have let's say 3 7 minus 1. Let's get a minus in there. And that's going to be equal to something. We'll work it out in a minute. For now I'll put X, Y, Z. Now how am I going to test once I found those X, Y and Z that I haven't made some kind of slip? I mean there's a lot of mental arithmetic. If we don't write it all out we're going to be doing a bunch of multiplications. I could easily slip up. How am I going to test that? It turns out there's a very interesting property of the vector C that we get out after the operation if we've done it correctly. That is, as I've written here, that A dot dot product with vector C is 0 and so is B. So either of the input vectors A and B dotted with the correct cross product C should give us 0. And that's great because the dot product is very easy to work out even by I as a check. Let's go ahead and do it. So I've copied it down here. We're going to want to work down our various components. Let's do the first component of C. So what do we do? We ignore the first components of A and B and we do the falling diagonal. So that's going to be 3 times minus 1 and we subtract the rising diagonal, 1 times 7. So actually let's just write that out. Normally I wouldn't bother to write all this out but let's go ahead and do it here. So it's minus 3, minus 7 and so that's going to be minus 10 as our first component. Now we work out second component. We ignore the second component on the input vectors. We do the rising diagonal, 1 times 3 and subtract the falling diagonal, 2 times minus 1. So what have we got? We've got 3 here, minus minus 2 and so that's going to give us 5. And then finally the third component, ignore the third component of the input vectors, do the falling diagonal, 2 times 7, 7, 2 is a 14, subtract the rising diagonal, 3, 3 is a 9. So we're going to have, for our final component, 14 minus 9, which is another 5. So that's quite a simple vector, has a common factor of 5 in there if we wanted to write it out that way. Now let's test that guy versus the A and B vectors to see if it passes our test or have we made a slip. So let's just be completely explicit about that. We're going to start by testing the dot product of the vector A with our hopefully correct cross product C. I'll write it out, 2, 3, 1. Dot product minus 10, 5, 5. What's that going to be equal to? Minus 20 and then 3, 5 is a 15 and then 1, 5 is 5. Aha! It does equal 0. That's correct. That's a very, very encouraging thing but for real thoroughness we're going to test the other one as well. So this is B dot C. Let's check that out. So that's 3, 7 minus 1 dotted with again minus 10, 5, 5. This time it's going to be minus 30 from 3 times minus 10 and then 7, 5 is a 35 but then minus 5 from the last 10 meant 0 again. Aha! So it has in fact passed both of our tests and we're now very confident that's correct. This is a great test to do. One word of warning though, the one thing it won't pick up is if you've done your rising and falling diagonals in exactly the wrong way round by starting with the wrong pattern. So do remember the VNV pattern and this test will check for any particular slips in your multiplications. And that's the end of the video.