 Okay, so in this short video I'm just going to look at four more examples of the cross-product for practice. And here they are. Okay, so here's the first one. We want the first element of this cross-product. So we ignore the first elements of the two source vectors. We do the falling diagonal three times zero. That's zero. And we subtract the rising diagonal seven times minus one. That is minus seven. So we're subtracting minus seven. That means we'll get plus seven. So the first element here is going to be a seven. Okay, so now we want the second element. That means we ignore the second element of the two source vectors. We do, however, the rising diagonal first. Seven, two, is fourteen minus one times zero is zero. So that's fourteen. So the second one was the rising diagonal first, if you follow me. And then finally to get the third component, we ignore the third component of the source vectors, and we do the falling diagonal. One times minus one is minus one minus three, two, six. So that is minus seven. Okay, so there's our solution, seven, fourteen, minus seven. But is that correct, or have we made a slip? It's a good time to check the older dot product trick. So if we call this A cross B equals C, then we should find that if we do the dot product of one of the input vectors, say B, with C, then it should be zero. Let's check that. Seven, two, is fourteen. Minus one times fourteen is minus fourteen. Zero times minus seven is zero. So that's fourteen minus fourteen. It's correct. Let's do the other one. It's harder. So one times seven is seven. Three times fourteen is forty-two. So that's forty-nine in total. And then the final term here, seven, sevens are forty-nine. But that was with a minus number. So we've got, in fact, forty-nine minus forty-nine is zero. So another one of those dot products is correctly zero. So what we found out is that A dot C and B dot C are both equal to zero, as they must be. So we're now very confident that we have the right cross product there. Let's do another one. Okay, so we're going to want the first element. So we ignore the first element of the two source vectors. And we do eight threes, eight threes are twenty-four minus two. Two times twenty-two. So that's twenty-two. Let's do the next element. So we ignore the middle elements and we do the rising diagonal. Four twos are eight minus eight. That's just going to be zero. And then finally, we ignore the bottom elements and we do the falling diagonal minus the rising diagonal. One minus twelve is minus eleven. So there's our solution, twenty-two, zero minus eleven. We notice we could take eleven out of that as a common factor that would make the next stage very easy. But let's just do it the hard way and do the dot product. So four times twenty-two is eighty-eight. One times zero is zero and minus eighty-eight actually pretty easy to confirm that zero. Let's do the other one. One times twenty-two is twenty-two. Three times zero. And again, two times minus eleven again zero. So that's fine. That one's past its checks as well. On to the third one. Okay, so this time I think I might take a common factor out just to show us doing that because I see that this twenty-five, five, fifteen chap is going to lead to some pretty big numbers. But maybe I don't need to do that. I can just take the common factor of five out of the first vector, we're calling it vector A. So that's just five one minus three. And then I go ahead and write vector B which can't be simplified as just one three minus two. We'll do this cross product, excuse me, we'll do this cross product and then we'll put the factor of five in at the end. That's fine to do it that way round. Okay, so let's go ahead and write that out. There's our factor of five and here's our cross product. So the first element of our cross product, we ignore the first elements of the two source factors. We do the falling diagonal. That gives us a minus two. We subtract the rising diagonal. That's a minus nine. So that's minus two plus nine. That's going to give us a seven. And now the middle element, we ignore the middle elements on the two source factors. We do the rising diagonal this time. That gives us minus three. We subtract the falling diagonal. That gives us minus ten, which means we're going to have to add on ten. So that's minus three plus ten. It's another seven. Okay, and then finally, the third element, we ignore the third elements on the source factor. We do the falling diagonal. That's five threes of fifteen and we subtract the rising diagonal one. That's going to give us another 14. So in fact, a really simple vector here because we could take out a factor of seven if we want to. But let's check those dot products. Do it before or after we take out the factor of seven. It's pretty easy. That's going to be four times seven minus and minus two times 14. Yes, that goes to zero. Let's do this one. Just quickly. Thirty-five and another seven is forty-two. But minus three times fourteen is exactly minus forty-two. So that one is also satisfied. We've passed our checks. That looks pretty good. We can leave it like this or if we want, we can take out that factor of seven and do thirty-five times one, one, two. Very simple. Very nice vector there. Okay, let's come here. Now come down to the bottom and look at the final one. We notice it's actually the cross product of a vector with itself. It's the same vector here. So what are we going to get? Well, we can just easily enough work it out. We ignore the first two elements and we do two times minus four and minus four times two. So it's something minus itself. That's just going to give us a zero, obviously. And let's keep going if we ignore the middle terms and do the rising diagonal minus the falling diagonal. Again, threes and minus fours, the same product. So something minus itself, zero, and it's going to be the same for the final element. So the cross product of a vector with itself is always going to be the zero vector. Now it's important not to write that just as the scalar zero because it is a different object. It's the vector zero. It's a set of, in three-dimensional space, three zeros. That's what we get when we cross a vector with itself. Of course, this is going to trivially satisfy our condition on the a dot c is equal to zero and b dot c is equal to zero. That's clear. And so I think that's a nice set of four examples done quite quickly there. They're not too bad, are they? So that's the end of the video.