 Hi, I'm Zor. Welcome to a new Zor education. I think it's time to solve a few problems related to vector product. It's actually a scalar product as well as vector product, because there are certain problems where both products are involved. Now, before starting with solving these problems, two comments. Number one, the usual comment, please try to do these problems yourself. On unizor.com, at the topic of vector product, this is problems number four, actually. Now, another comment which I want to make is, I will be extensively using in these problems expression of the vector product in Cartesian coordinates. So, if you have two vectors, u and v, u having coordinates, u1, u2, u3, and v having v1, v3, v3, then their vector product will have these coordinates. Now, I have presented this in the lecture, which is dedicated, completely dedicated to coordinate representation. So, it's derived, it's properly proven, et cetera. So, I'll just use these formulas. They are a little bit more complex, I would say. So, I'm not sure that I will remember by heart of them, so that's why I wrote it down. But I will be using them. All right, so, problems. Problem number one. Okay, that's a very simple one. We have to prove that the square of the vector product of two vectors equals to the product of squares of their lengths minus the square of their scalar product. So, square lengths of the vector product equals to product of squares of their lengths minus the square of the scalar product. Well, it's very easy, and here is why. You know that the lengths of the vector, which is a vector product of two other vectors, equals to lengths of the first one times lengths of the second one times sine of the angle between them. Let's have absolute value of the sine. Now, how about scalar product? Well, scalar product, that's the constant, and it's equal to lengths of the one times lengths of the other times cosine of the angle between them. If you will square this one, you will have this, this, this, and this. And we don't really need the absolute value because we will put square here. And if you will square this, you will have this, this, and this. And if you will add them up, you will have this, this, and that's it. Sine squared plus cosine squared will give you one, right? And since sum of these is equal to this, then the square of the vector product is product of squares of their lengths minus square of the scalar product. So it's basically a consequence of the fact how the lengths of the vector product is expressed, how the scalar product is expressed, the theorem that sine squared plus cosine squared equals to one, very simple. Now, next problem is, it's simple, but it involves actually lots of calculations. So that's why I don't, personally, I don't like these problems. Unfortunately, they do exist, and the problem actually looks quite, you know, quite neat. It's called Jacobi's identity. Jacobi's identity. Now, it looks really neat, as I was saying. Now, to prove it, it's just very mechanical usage of the formula of how the vector product looks in the coordinates. I'm sure there are certain geometric thoughts which Jacobi had first before actually discovering this particular identity. Let me just think about this. Let's consider, A, B, and C are i, j, and k unit vectors in the Cartesian system of coordinates. What do we know? We know that i, now this is i, this is j, and this is k. You know that i times j equals k, j times k equals i, and k times i equals j. Notice the cyclicity, i, j, k, j, i, k, i, and k, i, j. So, these three characters are cyclically substituting each other. So, if you will put them into some kind of a ring, let's have it, i, j, k, this way. Then i substituted with j. You see, i, j, j substituted with k, k substituted with i. So, j substituted with k, k with i, k with i with j. So, it goes cyclical, or cyclical, whatever. Same thing, by the way, in here. A, B, and C are also cyclical here. A goes to B, goes to C. B goes to C, goes to A, and C goes to A, goes to B. So, it's all cycle. Now, if you consider, instead of A, B, and C, unit vectors along the axis, x, y, i, i, j, and k. Now, what happens in this case? Let's just think about it. First, you have i times j times k in parentheses, which means j times k is first, and the result is i. And then you add it to this i. I mean, you multiply it with i, and the result is zero, because i and i, vector product of i by i is zero, and vectors are two linear, zero angle between them. Next, instead of VCA, you have j times k times i, right? And k times i is j. So, it's j times j. So, you have again zero. And the third one is exactly the same. The third one would be k times i times j. Now, which is k times k, i times j is k, which is zero again. So, I think his thoughts came from here. But in any case, if you would like to do it in general form, then basically you have to do exactly the same. Let's just use the cohesion representation. And let's see it this way. Consider, I would probably consider only this one, and then I will use cyclicity to come up with the rest of it. Now, B times C has, so A is A1, A2, A3. And B times C is, I will use these formulas. So, it's B2 C3 minus B3 C2 comma. That's the first coordinate. The second coordinate, and it's also cyclical, by the way, after B2 goes B3, after C3 goes C1 minus, now we're changing their order, comma. Now, after B3 goes B1, and after C2 goes C1 goes C2, minus B2 C1. So, that's my B times C vector problem. And if I will multiply this time this, again, my first coordinate would be U2 V3, U2 V3. So, it's A2 B1 C2 minus A2 B2 C1 minus U3 V2, minus A3 times B3 C1. And then, since this is minus, so it will be plus A3 B1 C3. So, that's my first coordinate. Now, my second coordinate of the result, I'm talking about only this thing, my second coordinate would be U3 V1, U3 V1. So, it's A3 B2 C3 minus A3 B3 C2. By the way, you already have to notice the cyclicity. A2 A3 B1 B2 C2 C3. So, it's always 1 2 3 1 2 3 1 2 3. Again, 2 3 2 3 1 2. Now, next, next is minus U1 U3 U1 V3 with a minus sign. So, it's minus A1 B1 C2 and plus A1 B2 C1 comma. Again, after 3 goes 1, after 3 goes 1, after 1 goes 2, after 3 1, after 1 2, after 3 1. And the third coordinate would be U1 V2 U1 and V2. So, it's A1 B3 C1 minus A1 B1 C3 minus U2 V1. So, it's minus A2 B2 C3 plus A2 B3 C2. Close. So, that's my first coordinate of, I mean, that's 3 coordinates of the first element in this sum. Now, and again, you see the cyclicity 2 3 1. So, it's always 1 2 3 1 2 3 on every level. Now, how about the second component? Well, the second component is different from the first component in one very simple thing. Instead of A, I should put B instead of B, I should put C. And instead of C, I should put A, right? Again, cyclical correspondence. So, let me wipe out this. And let me write 3 components of the, but you know what, let me write just the first component of this. And then the first component of this. Because if this is, now B times C is a vector, A times the vector, again, it's vector product, it's a vector. So, it's sum of 3 vectors. And I'm saying it's supposed to be equal to a new vector. Now, what does it mean? Actually, I should put here. It means every coordinate is equal to 0, right? So, let's talk about the first coordinate only, this one. And I will have the first coordinate for B. Now, it would be what? It would be B2 C1 A2, right? So, it's BCA minus B2 C2 A1 minus B3 C3 A1 plus B3 C1 A3. That's the first component. Let me check again. A, B, B, C, C, A, A, B, B, C, C, A, A, B, B, C, C, A, A, B, B, C, C, A. Okay. Now, the third component is, instead of B, I should put C, instead of C, A, and instead of A, B, right? They're all symmetrical. So, it's C2 A1 B2 minus. C2 A2 B1 minus C3 A3 B1 plus C3 A1 B3 and second and third component. And you know what I will do? I will not prove anything for the second and the third coordinate. I'll just wipe it out and I'll do it only for the first coordinate. If my first coordinate is equal to zero, then I can say, you know what? You probably will be sure that the second and the third will be exactly the same because they're all cyclical. So, now, let's talk about the first component. Say, you will add up the first components of these three vectors. It's the x coordinate, if you wish. You should have actually zero, right? Well, let's just think about it. Do we have it or do we not? Let's check again if this seems to be right. Okay, looks like. All right, so let's just see if something will just reduce itself, if we will add out these together. A2 B1 C2 with a plus. A2 B1 and C2 with a minus. A2 B1 C2, right? So, we have this. It's reduced. It's plus. This is minus. Next, A2 B2 C1. A2 B2 C1. This one. This is minus. This is plus. A3 B3 C1. A3 B3 C1. Plus and minus. Okay. A3 B1 C3. This one. Plus and minus. B2 C2 A1. B2 C2 A1. This is reduced with this minus and plus. And finally, this is A1 B3 C3 also reduced. So, as you see, all these three elements which represent the first component, the x-coordinate of this, x-coordinate of this, and x-coordinate of this nullify each other. And again, since everything is really cyclical, you will have exactly the same with y-component, here, here, and here, and with z-component. So, that's basically the proof. It's a little bit involved calculation-wise, but it's really straightforward. There is nothing to it. Just have to be very careful and accurate, right? So, that's the proof of the Jacobi's identity. Which does look interesting, actually, right? I mean, it's three vectors and, apparently, some of them is equal to zero in the three-dimensional space. That's an interesting property. Next. Okay. Here is the problem. Consider you have this. Now, you know that among the numbers, if you have, let's say, two times x equals two times y. Now, obviously, it immediately follows that x is equal to y, right? Because both parts can be divided by two. Now, is it the same with vectors? Can you say that from this, you can reduce it by a and vectors b and c are equal to each other? Well, no. This is not an implication. From this, this doesn't fall at all. And here is why. It's a very simple explanation. Now, let's bring them both to the left. We can do that with vectors. Why? Because from both vectors, we can subtract the same vector, which is a times c, and you will have this one. So, addition and subtraction of vectors as far as you remember is commutative, distributive, associative, whatever. I mean, all the nice properties of addition you have, right? Well, distributive actually is not applicable, but associative and commutative is applicable. Now, what's also interesting is the vector multiplication is associative, right? So, it's a times b minus c equals to zero. Again, with numbers, when you have something like two times x minus y equals to zero, you immediately conclude that x minus y is equal to zero or x equals to y. With vectors, that's not the case. Why? Well, for a very simple reason. b minus c doesn't have to be actually equal to zero to have this vector product equal to zero. It can be just a collinear vector with a. So, let's say, for instance, b is equal to 2a and c is equal to a. Then b minus c would be equal to 2a minus a, and again, distributive law versus multiplication by constant work. So, it would be 2a minus a, which is equal to a. And then vector multiplication by a would give you zero. So, because the a and a are collinear and the angle between them is equal to zero. So, that's why b minus c is not necessarily equal to zero. So, from this, you do not conclude that b equals to c. But the next problem makes it actually interesting. What if not only the vector product of a and b, but also a scalar product is equal... Can I conclude, based on these two equalages that b is equal to c? And now the answer is yes. Because, let's think about this. That's what we have from the first. And that's what we have from the second. Notice that in this case, it's a null vector. In this case, it's a zero number because this is a scalar product. Now, what can you conclude from this? If a is not equal to null vector, obviously. Because then, again, even with numbers, if zero times x is equal to zero times y, even within the numbers, it doesn't really follow from here that x is equal to y. Because we cannot divide by zero. So, we always assume that a is not null vector. So, if a is not null vector, what I'm actually stating right now, that b minus c is a null vector. Well, let's consider an opposite. And we will prove that the opposite is wrong. What if b minus c is not a null vector? It's some real vector with some real direction and real lengths. Well, if it's real lengths, then the lengths of this thing is supposed to be lengths of this times lengths of this times the absolute value of the sign between the directions of a and b minus c. So, a and b minus c, the angle between them, should be equal to what? If sine is equal to zero, basically either zero when they are directed towards the same thing, or it can be 180 degree. That's what I can say, right? So, if b minus c is not a null vector, then it should be collinear with the a vector. Otherwise, the vector product would not be equal to zero. So, it's directed towards a. Now, if it's directed towards a and b minus c, and it's not a null vector, then what would be this color product? It would be lengths of the a, which is not zero. It would be lengths of the b minus c, which we assume is not zero, and the cosine of the angle between them. And what's the angle between them? Either zero or 180, right? Now, cosine of zero is one, cosine of 180 is minus one. So, in any case, this is not equal to zero in this case. So, unless b minus c has zero lengths, this would not be equal to zero. So, that's the contradiction, right? So, we can notice here that b minus c is not equal to zero, because we come to a contradiction. Therefore, b minus c is equal to zero. The lengths of the b minus c is equal to zero, and therefore the length, therefore the vector itself, if b minus c lengths is equal to zero, that means that vector b and c are exactly the same. Because if they're not the same, then the difference would not be equal to zero, right? All right. So, that's an interesting property. So, vector product by itself does not resolve this particular equation, a times b equals to a times c, it doesn't resolve to b equals c. But if you consider both vector product and the scalar product of exactly the same vectors and the same equality exists, then you can conclude that b equals to c. Next. Next is the following. It's just an identity which has no name according to my sources, but nevertheless... So, this is vector product of two vectors, which is a vector. This is vector product of vector and vector, which is a vector. Now, this is a scalar times vector, which is also a vector, and this is also a scalar times vector, which is also a vector, and difference between vectors is a vector. So, it's a vector equals to vector, and we have to prove this identity. Well, the only reasonable way to prove this identity is to use the coordinates. Otherwise, I would probably be in difficulties to prove it geometrically. So, in any case, let's try to do it with coordinates, right? So... And again, let's just use the first coordinate because the second and the third will be really a cyclical transformation. So, if the first coordinates will be the same, then the second and the third will also be the same. So, the first coordinate is... So, we know what B times C is. It's this one. And multiplication by A would be... A2... So, the first coordinate will be A2 times B3. So, the third coordinate of this, which is equal to B1 C2 minus B2 C1. So, the first coordinate is U2 B3 and the third coordinate. U3, which is A3 times B2, the second coordinate of B times C, which is this, B3 C1 minus B1 C3. Okay, so that's the first coordinate of this vector. Now, what's the first coordinate of this vector? Well, A times C is A1 C1 plus A2 C2 plus A3 C3 and that should be multiplied by vector B and the first coordinate would be B1 minus A1 B1 plus A2 B2 plus A3 B3. That's this color product. Times first coordinate of this C1. Well, let's just compare. A2 B1 C2. A2 B1 C2. Same thing, right? So, this one and this one correspond. Well, you know, it's probably easier if I will put B1 on each and every one of them and that would be probably easier. So, it's B1 here, B1 here and B1 here and then I will do the same thing with C. I will put C1 C1 C1, okay? And now I can say that this is this. A2 B1 C2. Correct. Now A2 B2 C1. A2 B2 C1 with a minus sign here and minus sign here. So, this is also good. Okay. Now A3 B3 C1. A3 B3 C1 minus and this is minus. So, that goes out. A3 B1 C3. A3 B1 C3 minus and minus plus and this is plus. So, this goes out. And finally A1 B1 C1 plus and A1 B1 C1 minus also out. So, as you see, the first co-ordinate is exactly the same in both cases. Now, the second co-ordinate would be cyclically transformed from the first co-ordinate. So, instead of A2 I will have A3 instead of B1 B2, et cetera, et cetera. So, since it's all cyclical, it will be exactly the same thing. So, I don't want to spend time on this, basically. So, that proves this particular identity. Now, the last problem which I wanted to address today is more of a geometrical quality. You have a three-dimensional space and three points. One, one, three. So, this is one, this is one. And, okay? Two minus one, five. Two minus one, something like this. Two minus one, five. The second point. So, this is A, this is B. And the third B, minus three, three, one. Minus three goes this way. Three on the Y and one. Something here. That's B. Minus three, three, one. So, you have three points in the space. I don't think I really have to draw them on the co-ordinate system. I just, you know, give the co-ordinates, basically. That's it. The question is, what's the area of the triangle which is formed by these three points? Well, it's not easy to do it in space if you don't really know anything about vector product. But you do know that the vector product is actually an area of the parallelogram which is built on the two vectors. So, if you have two vectors, A and B, and build parallelogram, then the vector product is perpendicular to both. That's true. But the magnitude of this vector is equal to the area of the parallelogram, right? Well, number one. Triangle is half of the parallelogram. So, if we are building a parallelogram on these two vectors, then we can find the area of the triangle. So, why don't we arrange these three points in such a way that they form a zero point and two end points of two vectors? So, what do we have to do for this? Well, very simply. Why don't we move the origin of the co-ordinate instead of zero, zero, zero into the point where, let's say, the first point is located, which is this one. So, if I will move my point of origin here, I'll make a transformation of co-ordinates. Well, my triangle will be exactly the same and its area will be exactly the same, but co-ordinates will be different. But now, point A would be in the origin of the co-ordinate and the vector from A to B and from A to C would be, this is actually C, right, would actually form two vectors and I can calculate their scalar product, which would be an area of the parallelogram, which they make, and then I would take half of this area, which is the triangle which I need, right? So, all I have to do right now is, number one, to make the transformation of co-ordinates from zero, zero, zero to one, one, three, and that correspondingly changes the co-ordinate of other two points. And secondly, I have to calculate the vector product using the co-ordinate form. All right, so, if I have A one, one, three, B two minus one, five and C minus three, three, one, and I will change the co-ordinate system in such a way that my zero point moves to one, one, three, which means I am shifting my x-coordinate by one, y-coordinate by one, and z-coordinate by three. So, what happens? Well, relative to the new origin, my point B will move. Well, relative to one, my point B, it used to be two when this was one. Now, if this is zero, now this is one, right? Now, here the difference would be minus two, and here the difference would be two, right? So, if I shift my z upwards by three, my co-ordinate, which used to be five, now becomes two. If I shift y, which used to be zero to one, then minus two will increase the distance, right? It will be minus two. In this case, if I move upwards, no, not upwards, x means towards the x from zero to one, but used to be minus three, now will be minus four, right? Relative to this, if one becomes zero, minus three becomes minus four. Now, this becomes two, and this becomes minus two, right? From three to one is minus two. You shift it by three upwards. So, these are new co-ordinates, right? So, I have point zero, I have point where my one vector should go from the zero towards this point, and another vector from zero to this point, and now these two vectors form a triangle which I can now calculate the area of using the vector product, because the vector product of the vectors with endpoints this and this is equal to the area of the parallelogram. So, I just have to take half of that, and that's it, right? So, what's the area of the parallelogram, or what's the vector product of the vectors with endpoints this and this? Well, again, let's just use my formula for the vector product. Now, my first coordinate would be u2 v3, which is four, minus u3 v2, which is four. So, this is w1, which is zero. So, u2 is equal to u3 v1, u3 v1. So, it's minus eight, u1 u3 v3, and it's minus. So, it's plus, plus two, equals to minus six. And my third coordinate would be u1 v2, which is two, minus u2 v1, minus u2 v1. That's it, minus eight. So, it's also minus six. So, that's my coordinates of the vector product. And the magnitude of this vector is the area of the parallelogram I'm looking for. And the magnitude is square root of w1 squared plus w2 squared plus w3 squared, which is equal to square root of 36 plus 36, which is 36 times two, which is six square root of two. So, that's the area of the parallelogram. And the area of the triangle is half of it. So, that's the answer. Well, I think it's very important if you will go through these problems yourself right now. Go to unizor.com, vector product, this particular problems are the problem number four, problems number four. Go through these problems, try to do it yourself. I think it's very important to repeat it. And basically, that's it for today. I thanks for your attention and good luck.