Algebra

Cross product introduction | Vectors and spaces | Linear Algebra | Khan Academy

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We learned a lot about the scalar work. But when I first mentioned it, that this is just one kind of multiplication of vectors, and the other kind is the vector work you probably know from his course in vector higher mathematics or from your physics course. Vector work. But the vector product is much more limited from the scalar work. It is useful, but much more limited. The scalar product is defined in each dimension. This is defined for every two vectors that are in Rn. You can take the scalar product of vectors, which have two components. You can take the scalar product of vectors that have millions of components. The vector product is defined only in R3. And the other main difference is that the scalar product – and we’ll see that in a moment, after I define the scalar product, I have not yet defined the scalar product. The scalar product gives us a scalar quantity. You take the scalar product of two vectors and you only get a number. But in the vector work you will see that we will get another vector. And the vector we get is going to be a vector, which is perpendicular to the two vectors, on which we take the vector product. After I excited you with anticipation, let me define this. He’s probably seen it once or twice before in his mathematical career. Let’s say I have a vector a. It must be in R3, so there are only three components: a1, a2 and a3. And I’m going to multiply that vector by vector b, which has three components: b1, b2 and b3. and the vector multiplied by b is defined as the third vector. And now this is going to look a little weird and hard to remember, because that is the definition. But I’ll show you how to think about it, when my vectors are written in this form. If you look at the physics playlist, I have a lot of videos for the vector work and I show you how I think for the vector product, when I have it in the form i, j, k. But when I have it that way, you think about it this way – the first term above will be another vector in R3, so there will be 1, 2, 3 members. For the first term, you ignore these top two terms of this vector and then you look at the bottom two members and you take a2 * b3 – a3 * b2. And I made some videos about the determinants, although I haven’t done them formally in this linear algebra playlist. But if you remember, finding the members co-coefficients when determining the determinant, or if you just take the determinant for a 2×2 matrix, this may seem very familiar. This first article here is the determinant of – if you get rid of the first row of these two here, you take a2 * b3 – a3 * b2. This is a2 * b3 – a3 * b2. I hope it was pretty easy. So as not to make your life more complicated, when you do the second, middle row, when you do that here, you multiply this by vector. You may want to take a1 * b3 – a3 * b1. And that would be logical, because we did that up here. But for the middle row you do the opposite. You take a3 * b1 – a1 * b3. Or you can look at this as a negative version of that, which you would do in the way of logic. You would take a1b3 – a3b1. Now we will take a3b1 – a1b3. And that was only for this middle order. And for the bottom row we multiply this vector again or we ignore it. We take a1 * b2, just like we did with the first row. Or minus a2b1. This is not easy to remember. That is why we must create this system, as I told you. But this may seem rather strange and long. Let me give you a few examples to help you understand the definition for the vector product in R3. Let’s say I have the vector – let’s say I multiply by vector … I have a vector (1; -7; 1). And I will multiply this vector by a vector (5; 2; 4). This will be equal to the third vector. Let me make room to do the calculations. We just ignore the first components of these vectors and we take -7 * 4 – 1 * 2. And this is just a simple multiplication. I do not take the scalar work. These are just ordinary numbers. We ignore the middle members here and do the opposite. We take 1 * 5 – 1 * 4. Remember, you may want to take 1 * 4 – 1 * 5, because that’s what we did with the first article. But the middle member is the opposite. Finally, you ignore the third article here and do as you did with the first article. You start from the top left. 1 * 2 – (-7) … You put this in parentheses. – (-7 * 5). And that will be equal to – let’s see. What do we get? -7 * 4. I don’t want to make the mistake of carelessness. That’s -28 – 2. For the first member we have -30. That’s 5 – 4. 5 – 4 is just 1. And after 2 – (-35). 2 – (-35). This is 2 + 35. That’s 37. Here. I hope you understand at least the mechanics of the vector product. Then you say to yourself: “Okay, I can find the vector product of two things. But what is this good for? What does this do for me? ” And the answer is that this third vector here and depending on whether I stick to the abstract case or do this case with numbers, it is perpendicular to the two vectors on which we take the vector product. This vector here is rectangular to a and b. Which is pretty nice. If we just think about the last video where we talked about normal vectors in a plane, we can determine plane through two vectors. Let’s say we have a vector a here and then I have the vector b. Let me make a vector b here. These define a plane in R3. Let him define the plane. All linear combinations of these two – this is a plane in R3. You can look at it as if they could form subspace in R3. This forms a plane. If you multiply a vector by b, you get a third vector, which is at right angles to these two. and vector multiplied by b will look like it pops out of the page. It will be at right angles to both of them and it will look like this. This vector here is a vector multiplied by b. And you can tell there are a lot of vectors, which are at right angles. Obviously the length of this vector – I didn’t specify it here, but it may look like it’s popping straight up or it could look like it was popping down, like this. This will also be at a first angle to a and b. And a vector multiplied by b is defined as follows: you can find the direction visually using something called the right-hand rule. And you use your right hand. Let’s see if I can draw a right hand. You point your index finger in the direction of a. If your index finger is in the direction of a, then point your middle finger in the direction of b – in this case, my middle finger will look like this. My middle finger will look like this. And I don’t point my other fingers anywhere. Then my thumb will be in the direction of a vector multiplied by b. You can see this here. My thumb is in the direction of a vector multiplied by b. And assuming you look like me anatomically, then you will get the same result. Let me draw everything. This is a vector a. Vector b moves in this direction. I hope your thumb isn’t hanging down here. You know that a is vector multiplied by b in this example will point upwards and is at right angles to both. To make you happy, this vector definitely is at right angles to – or that thing definitely is at right angles to these two. Let’s take a look at this and see for sure. What does it mean to be at right angles? What is the definition of “right angle” in our context? Vectors at right angles. If a and b are at right angles, it means that a scalar multiplied by b is equal to 0. Remember the difference between at right angles and is perpendicular to that at right angles also applies to vectors 0. These can also be vectors 0. Notice I’m not saying any of these here must be non-zero. In a moment we will talk about the angle between the vectors and you will have to assume that they are non-zero. But if you just find a vector work, there is no reason why these numbers cannot be 0. But let me show you that a vector multiplied by b is definitely will be at right angles to both a and b. I think that can satisfy you. Let me copy a vector multiplied by b here. I don’t want to rewrite it. Okay. To put it. I copied other things with it. Let me find the scalar product of this, with only vector a, which was just a1, a2, a3. What does a scalar work look like? This article on this – this is a1, let me free up some space. This is a1 * a2b3 – a1 for that. Minus a1 * a3b2. Then you have a plus for that. That is, plus a2 * a3. Plus a2 * a3 * b1. And then minus a2a1b3. And I will continue down here. Plus a3 * a1b2 – a3 * a2b1. And I just took the scalar product of those two things. I just took each of these. That was equal to these two members. This was equivalent to the next two members, these two. And then that was equal to those two members. And if these are really at right angles, then this must be equal to 0. Let’s see if that’s the case. I have a1a2b3, here is +, and then I subtract the same thing. It’s a1a2b3, but it’s simple -. I can shorten it with this. Let’s see what else we have? We have a – a1a3b2. We have + a1a3b2, so these two will be shortened. I think you see where this is going. You have + a2a3b1, and then you have -a2a3b1. These two will also be shortened. I just showed you that this is at right angles to a. Let me show you that this is at right angles to b. Let me make another version of the vector product of two vectors. I’ll take it down a bit. Let me go back. And let me multiply this by the vector b. b1, b2 and b3. I will do it here to have a little space. b1 on this whole thing here is b1a2b3 – b1 on this. Minus b1a3b2. Let me change the colors. And then b2 on that will be b2 – this will be a plus. It’s just one expression, I just write it in a few lines. This is not a vector. Remember when you find the scalar product of two things, you get a scalar quantity. + b2 on this. b2a3b1 – b2a1b3. Finally we have b3 on this. + b3a1b2 – b3a2b1. If these things are definitely at right angles, then this thing must be equal to 0. Let’s see if that’s the case. We have b1a2b3. b1 and b3. b1a2b3, this is positive and this is negative. You have b3a2b1. This is shortened. Here you have -b1a3b2. You have b1 and b2. This is -b1a3b2. It’s the same thing, but positive. We just change the order of multiplication. But these two are the same member. They are just opposite to each other, so they are shortened. Finally you have b2a1b3. This is negative. And then you have the positive version of the same thing. These two are shortened. You can see that this is also equal to 0. I hope you showed that vector here is definitely at right angles to both a and b. And that’s because it was created that way. This is the definition. You can do some calculations without me explaining this definition you can find this definition on your own. But apparently it is designed to have some interesting properties. We will talk about this in the next few videos. I hope this was helpful to you.

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