 Hi, I'm Zor. Welcome to Unisor Education. I would like to dedicate this lecture to one particular property of vector products, the distributive property. Now, the distributive law of vector product in relationship to an addition of vectors has its typical form. So, if you would like to multiply to have the vector product of some two vectors by the third one, well then you multiply the first one, the second one, and then add results. Okay, that's the typical form of the distributive law. Now, I have defined the vector product in geometrical terms, namely, if you have two vectors, then to construct their vector product, you have to go to the perpendicular to both of these. So, basically, if you have a plane which is defined by these two vectors a and b, then this is a perpendicular to the whole plane. Now, the direction is defined by a smaller angle between a and b from a to b, it's important. So, you can use this the rule of the corkscrew. So, in this particular case, the direction would be upwards, and the magnitude of this vector is equal to product of the magnitude of these two vectors times the sign of the phi angle, smaller angle between a and b. Smaller, I mean, the result is an opposite angle, but I'm talking about smaller angle less than 180 degrees. So, that's the definition. Now, from this purely geometrical definition, and again, I'm emphasizing its geometrical definition, it's not coordinate form which will be addressed later. So, using this geometrical definition, it's not really obvious that the distributive law is held. So, this lecture will be about this distributive law, and I will try to prove this particular law in geometric terms, so to speak. Okay. Now, when I was trying to prove it just for myself, it turned out to be not as trivial as other properties, like anti-commutative property that AB times, BA is minus AB, something like this, or some other properties like multiplication of the vector by itself equals to zero because the sign of the angle is equal to zero. So, this is not as trivial. Now, I came up with this particular proof, and I actually put it in the notes for this lecture. I put the hint first, so kind of encouraging you to do it yourself, with or without my hint, and then I put the whole proof into the notes. So, it turned out to be not as trivial as others, but well, that's what I came up with, and if you can come up with something simpler using purely geometric approach, I would be more than happy to put it somewhere in the website or something like this. All right. So, here is my proof. First, if you don't know how to prove the general case, you probably would like to prove something in some special, a simpler case. So, what is a simpler case which I would like to consider first? Well, the simpler case is the following. So, instead of three completely different, completely general vectors a, b, and c, I will consider the following case. This is c, and this is a, which is collinear and co-directional with c, and then somewhere there is a b. Now, what I will also do, I will draw coordinates, but not because I need the coordinates, just because it would be probably easier for you to have some space feeling about the whole thing. So, b is not in the plane x, y. It's somewhere in the space, and c is along the z-axis in my particular case. I mean, I can choose whatever the coordinate system I would like to. So, I choose z as coinciding with this c vector, and in my particular case, a is collinear with c, and same direction. That's my very particular case. Now, just to show the dimensionality of this, let me just draw the projection of the b on the x, y, z. This is on the x, right? And somewhere there is a projection on the z-axis, but I don't really need it. Now, so what's interesting about this particular case? Well, the interesting thing is, obviously, that ac is zero, no vector. Why? Because they're collinear, so the sign of the angle between them is equal to zero. Doesn't matter what the lengths of them are, all right? Okay, so what follows from here is that, let's consider now what would be the a plus b. Now, a and c coincide, right? So, let's consider the plane where a and b or c and b belong to. Now, so basically what I would like to do is, I would like to put this plane on the picture, something like this. So, let's call this point b. Well, this might be a, this might be c, and this would be say d. So, I'm talking about the plane, which is defined by vectors a and b, which is the same as the plane of a rectangle O, C, B, D. So, if you can imagine, you have a vertical plane, which is turned out in some way, so it actually contains the b vector. And when this plane cuts the x or y plane, the O, D is the result of it. Try to imagine this particular thing. Now, since a and b and c all belong to the same plane, I would like to just consider what's going on in this particular plane. Let's think about it. Since a and b belong to this plane, then there's some which I can actually draw here. This would be a plus b. It also belong to the same plane. So, if I will look at this plane from a side, I will have c vector at some angle. Now, a vector is coinciding with c. At some angle, I will have b. And here I have a plus b. So, this is just the view of this plane from the side of it. Now, if I view at this plane this way, now, where are the vectors which are a, c, b, c, and a plus b, c? I'm talking about vector product. Well, a vector product c is no vector. Now, b and c, if I will multiply them, then the vector would go this way. And a plus b vector product by c also would go this way. Why? Because all of them belong to the same plane. And the vector product should be perpendicular to all the participants. So, it's perpendicular to c. And it's perpendicular to b. And it's perpendicular to a plus b. So, all these vectors, b times c, will be perpendicular to this plane. And a plus b times c will be perpendicular to this plane. But this is the same plane, so it's the same perpendicular. So what I'm talking about is that the direction of a plus b times c, or b times c, is always this way. Now, on my three-dimensional drawing, it's somewhere here. That would be dimensional. So first of all, which I would like to establish is that a times c is no vector. bc, b times c, and a plus b times c are directed in the same direction. So if I want to establish the equality between these vectors, I have to establish equality in magnitude and the direction. So direction has been already established. And it follows from the fact that c and b and a plus b all belong to the same plane. And all these results of the multiplication should be perpendicular to this plane. And there's only one perpendicular to a plane at this particular point. Now let's talk about magnitude. So what's the magnitude of, let's say, a plus b times c? Well, it would be lengths of the a plus b times lengths of the c, times sine of this angle. Let's call it alpha. And let's call this angle b. Now, the magnitude of this vector, well, a times c is no vector, so it's bc. It would be lengths of the b times lengths of the c times sine of alpha plus beta, alpha plus beta, right? From b to c is alpha plus beta. From a plus b to c is just alpha. So this is the magnitude of the left side and this is the magnitude of the right side. Are they equal? Well, let's do it this way. Let's consider this triangle. This is our point, o. This is c. And this is, I don't remember what it is, but let's call it d. Triangle, o, c, d. This is alpha. This is beta. Now, this is parallelogram because we are adding two vectors together and the rule of the parallelogram is that their sum is the diagonal. So this angle is equal to 180 minus alpha plus beta, right? 180 minus alpha minus beta. Now, the sine of pi, which is 180, minus alpha plus beta, I put alpha plus beta in parenthesis, is equal to sine of alpha plus beta, right? That's the property of the sine. And now consider in this triangle, o, c, d, the law of sines. Now, the law of sines says that every sine, every side divided by sine of the opposite angle is equal to another side to its opposite angle and the third one, which means that length of this side, which is a plus b, length of the a plus b, divided by sine of this angle, which is the same as sine of alpha plus beta, equals to this side, which is the same as this. It's length of the b, it's parallelogram, divided by sine of alpha. So that's the law of the sines. And what do we see here? This is exactly the same as the college of these guys. Well, obviously, multiplication by the length of the c doesn't change anything. But now, b times sine of alpha plus beta, which is b times sine of alpha plus beta, equals to the lengths of a plus b times sine of a, right? So this equality is exactly the same as equality between these things. Well, provided by the sines not equal to 0, et cetera, we are all kind of considering a general case in this case. So that actually completes the proof, because we were already proven that the vector product of b and c is this type of vector directed perpendicular to the whiteboard. And the vector product of a plus b times c, also the same thing. And their magnitude is the same. Their lengths is the same. So basically, the vectors are the same. They have the same direction and the same length. So that proves that in this particular case, let's call it case a, when the vector a is collinear and has the same direction as c. We have actually proven our theorem. It was a very easy kind of a case, a very special case to the same token, when a and c are going along the same axis. Now, let's consider case b. Now, case b would be c is perpendicular to a and perpendicular to b. So this is the case when I will also draw coordinates, although I don't really need them, just to have some kind of space representation. So these are two vectors a and b. And they belong to some plane, right? So c is perpendicular to this plane. So in theory, then imagine that you have a regular Cartesian system of co-ordinate. And within the x, y plane, you have these two vectors. They are here on this plane. And c is perpendicular along the z axis. You can consider it this way, although again, I don't really need the coordinates right now. Now, what happens in this particular case? Well, let's think about it. a times c should be perpendicular to c. All vectors perpendicular to c are in this plane, right? So a times c would probably be, and also perpendicular to a, right? So a times c would be somewhere here, let's say. This is a vector product c. Now, b vector product with c, also, the result should be perpendicular to c, which means it should be in the same plane, and also perpendicular to b. That would be my b times c. Now, a plus b, again, let's put this parallelogram. This is a plus b. a plus b is in between a and b. And actually is exactly the same thing I have to tell, that the result of the multiplication of a plus b times c should also be within this plane, because it's supposed to be perpendicular to c, and probably in between here. Now, my statement is that this parallelogram is similar to this parallelogram. Why? Well, let's think about it. This side is 90 degree from this one. This side is 90 degree from this one. And this is also 90 degree of this one. So all these angles are the same, right? Now, how about the sides? Well, angles are the same. Now, a times c, and this is a plus b times c vector product. What's their length? Well, considering that the angle between all of these and c is 90 degree, you see? The side of the angle between them is always equal to 1. So the length is actually equal to product of lengths. The length of a plus b times c is the product of lengths of a plus b and c. And same thing with this. So all these lengths are proportional to original lengths, just the scaling factor is the lengths of the c, right? Because, again, the lengths of the a product c is equal to a and c. The lengths of the b times c is b times c. Again, because the sign between them is equal to 1. It's 90 degrees. They're all perpendicular to each other. And finally, the a plus b times c would be lengths of the a plus b times lengths of the c. So all lengths are scaled by a factor of c, which means every triangle, this angle is equal to this angle. This side is proportional to this side by the factor of c. And this side is proportional to this. So all the angles, everything is exactly proportional. So this is, therefore, a diagonal. Because obviously, this side would be parallel to this one because the angles would be equal. Considering these triangles are similar, the corresponding angles are equal. And that's why it's a parallelogram. So whenever you connect points a times c, a plus b times c, and b times c, you will get the parallelogram. Because all these angles are the same and all sides are proportional. Which means that since this is the parallelogram, that actually confirms that this diagonal is actually the sum of these two vectors, which is exactly what's necessary to prove. That this one is the sum of these two vectors. So this is how we prove it purely geometrically in case when we have this type of perpendicularity. Now it's time to consider the general case. And we will use these two special cases for the proof. Now in this case, I will resort to using the coordinates. And here is how. If I have three vectors, a, b, and c, I am the one who can choose any coordinates I want in this space. So I will take the point where all of them originate. And I will put my z-axis along my c-vector. Now x and y, I will put somehow, doesn't really matter how. But obviously, x, y plane should be perpendicular to c. Now my two vectors, a and b, are somewhere in the space. Doesn't matter where. So we are talking about case c, the general case. So first of all, I choose this coordinate system where one of the z-axis actually goes along the c-vector. And a and b are somewhere in the space. Now in this coordinate system, I can represent vector a as a1, a2, a3, triplet. And vector b as b1, b2, and b3, triplet. Now let me introduce unit vectors i, j, and k along each axis. So i is a unit vector along x, j is unit vector along y, and k is unit vector along z. Now obviously, I can represent, since a1, a2, and a3 are projections. So this is a1, this is a2, and this is a3. a1, a2, and vertical a3. Now obviously, I can represent the vector a as some of these vectors. Why? For obvious reasons, vector a can be represented as some of this plus this. Now this is a3 times k. Now this can be represented as this times this. This is kj, and this is the same as this equal to, this is a2j, and this is a1i. So some of these three vectors, this one, this one, and this one gives me a. Similarly, if I will represent in the coordinate form b vector, I can represent it as a combination of coordinates multiplied by corresponding unit vectors. And finally, a plus b can be represented as some of coordinates. Let's consider what is a plus b times c. I can say that this is equal to this times c. And what I will do is the following. I will call this vector b. So I will have a1 plus b1i plus b times c. Now let's think about our first property. Now the first property says, if a is collinear with c, then this is a true statement. Now I will consider this as my a, this as my b. So a would be this one, and b would be the b. Now a and c, OK, I made a small mistake, a small mistake. Now instead of this, I forgot that I put my, I will use this as b. And this separately, because I wanted it to be collinear. All right, so it's a3 plus b3, k plus b, sorry about that. All right, so now what I'm saying is that this I will consider as my a vector for the first case. And b would be my b vector. Now you see k and c, they are collinear, because I have chosen c to be along the z-axis, right? And k is a unit vector along the same z. So basically, I can apply my formula in this particular case, because my first component in the sum is exactly collinear with c. And the second is irrelevant in this particular case. So I can't open the parentheses. I can use the distributive law, because I have proven that if the first component in the case a, if the first component is collinear with c, then I can open the parentheses. The distributive law is working, right? So I can say that a plus b product c equals to a3 plus b3k times c, not times product plus b product c. Let me open even further. Now, I know that v is a combination of i and j, which means it's completely in this horizontal plane of x and y, which means that this particular vectors, both are, so v times c is equal to a1 plus b1, a1 plus b1i plus a2 plus b2j product with c. Now, both vectors, now I consider this one as a and this one as b. Now, both vectors are perpendicular to c, because they're both in the horizontal plane. They are combinations of i and j. Therefore, I can open the parentheses here as well, because it's second theorem check problem. So that would be a1 plus b1ic plus a2 plus b2jc. So what I want to say is that I have actually opened up this in the co-ordinate form. So the whole result would be this plus a3 plus b3kc. These are just numbers. These are just numbers. And because they are numbers, I can basically separate them and multiply them. The distributive law of these numbers is working. It's with vectors I wanted to prove. But these are just numbers, which means number multiplied by something, the sum of two numbers multiplied by something, it's obviously just a vector. So it's like two different constants multiplied by some vector w. Now, with multiplication by a scalar, the distributive law works without any problems. And therefore, I can see that the whole thing, I can open all these parentheses, is equal to a1 times ic plus a plus b1. Or actually, I will group them. First, I will put all a's, and then I will put all j's, all b's, plus a3k times c. It's not really time. It's product plus b1i product c plus b2j product c plus b3k product c. So that's the result of all these multiplications. What if I will multiply separately a by c and b by c? Now, a has this form. Now, if you follow the logic, it would be exactly the same thing. First, I will separate the a3k component. So a times c would be a3k plus some vector a1i plus a2j times c. Now, I will use the first theorem, which I have proved. Since k and c are collinear, I can open using the distributive law here. It would be this plus this multiplied by c. But now we'll use the second one, because i and j are perpendicular to c. Then I can see that this is a1i c plus a2jc. So I will have this. Now, with b, I will have the corresponding thing with b. And if I will add them together, I will get exactly the same thing as on the left side here. So using these co-ordinates, which I have completely artificially created, I just chose the z axis along the c and then some perpendicular x and y. Using this and the representation of any vector in the co-ordinate form as the combination of unit vectors with the co-ordinates as multipliers. Using this, I basically have reduced my very general problem with two different cases, basically. Each one of them I have already proved. So that actually concludes this general case. And I should comment probably that this is not an easy proof, at least for me. And maybe there is something easier than that. But in any case, I would like you to concentrate on this problem, how to prove the distributive law for the vector product relative to the addition of the vectors. I would like you to concentrate on this particular topic a little bit more. Read whatever the notes are. Listen again to the lecture. It's very important. It's a great exercise for the brain for how to build this logical conclusion one after another. Because as you know, this is not some skill which you will need in your real life practice. This is just the training of your mind how to logically approach the problems like this. So I first specifically considered some very particular cases. And then the general case I have reduced to these ones. How did they come up with this? Well, difficult to say. But probably I was trying first to do some general case and then realized it's too general. And let me just simplify it. So that's probably somehow how my mind was working. But it's very individual. So don't count on some similarity or anything like this. So anyway, I would like to point you again to this lecture, to the notes for this lecture. Do it again. And you will see how logical will really fill up your mind. That's it for today. Thanks very much. And good luck.