 Hi, I'm Zor. Welcome to Unizor Education. We are talking about vector product of two vectors in a three-dimensional space, and we have actually spent a lot of time analyzing the properties of the vector product, including like commutative and anti-commutative, distributive property, multiplication by null vector, et cetera. What's actually interesting is that we were trying to analyze all these properties from usually from geometrical standpoint. At the same time, during the proof of distributive law for vector products, which is this one, at the very end of the proof, I have introduced a representation of the participating vectors, A and B, in this particular case, in their coordinate form. What's interesting is to express the whole vector product in the coordinate form. Right now, we were always based on representation in geometrical form that if you multiply the lengths of two vectors by sign of the angle between them, you will get the lengths of the vector product. As far as the direction, it was defined again geometrically as the one which is perpendicular to both of them, perpendicular to the plane actually, which is defined by these two vectors. Now, we will try to do it algebraically using the coordinate representation of vectors A and B. Now, to do this, we will use the properties of the vector product, which we have already discussed and proved actually. One of the property is the distributive law, as I was just saying. Another property which we will be using is, if you have coordinate system and you have unit vectors on each coordinate i, j, and k, that's how they usually named. These are unit vectors along each axis of the coordinate system. Then, what's interesting is that i vector product by j equals to k, j vector product with k equals to i, and k vector product i equals to j. Now, why is that? Well, we already discussed it, but let me just repeat. It's obviously because the length of them is equal to one. The sign of the angle between them is 90 degree, which means the angle is 90 degree, so the sign is one. And k, for instance, is exactly the one which is perpendicular to both i and j, especially the direction is important. If you go from i to j, then the rule of the corks correct to points upwards, or if you wish, if you look from the point and point of the k vector, you will see that direction from i to j is positive counterclockwise. So, these are properties as well as the general distributive law, which we will be using to derive the representation of the vector product in coordinate form. Now, so let's say we have two vectors, vector u, which has coordinate representation u1, u2, u3, and vector v, which is v1, v2, v3. Now, vector w would be their vector product, but we would like to know is how w1, w2, w3 expressed in terms of u1, u1, u3, and v1, v2. So that's the purpose of this very small theoretical research. So, we will use, as I was saying, distributive law and the multiplication of the unit vectors. Now, what does it mean in terms of representation of the vector u in terms of u1, u2, u3 if we will be using the unit vectors along the coordinates? Now, obviously, vector u can be represented as u1i plus u2j plus u3k. Now, this is multiplication of the vector by a constant. u1, u2, and u3 are real numbers. Now, again, we did talk about this before, but let me just repeat again. So these are coordinate system. So you have a point with coordinates u1, u2, u3, which is the end point of this vector. So that's what it means that the vector u has these coordinates. It means it starts from the 0 and ends at this particular point. Now, if you will drop the perpendicular from this point to xy plane and from this point, which is a projection, this is a, let's put it b. So what do we know about these segments? We know that this segment is y coordinate, which is u2. This segment is x coordinate, which is u1. And this segment, the vertical one, it's u3. Now, let me connect this and this. So you will see how it looks. So this is a point in the space, three-dimensional space. This is the projection of the xy area. And this is a projection up onto the z. Let's put a point c. So oc is equal to u3, same as ad. By the way, because it's a parallelogram, if you look from this, it actually it's a rectangle. If you will take a look on this plane itself, oc ad plane, it just looks like a parallelogram because we look from the side, right? Same thing as this. We need the points d and e. Now, odbe looks like a parallelogram, but actually it's a rectangle. If you look from the top. So it's obvious that the vector u3, the vector u is equal to sum of the vectors oc, oc vector, plus ob. Sum of these two vectors is equal to this. At the same time, ob is equal to sum of od plus oe. Ob is equal to sum of these two, because of the rule of the parallelogram. So therefore, we can see that our vector u, which is oa, is equal to oc vector plus od and plus oe. And each one of them is equal to correspondingly. Oc is equal to u3 times k plus od is actually u1 times i plus and oe is equal to j multiplied by u2. So that's the representation, which I have just written on the top. So let's put it aside. That's simple case. And now, let's use this particular representation. And v correspondingly is v1 i plus v2 j plus v3 k. We will use this representation of each of the vectors to derive their vector product. Now, if you multiply this by this as the vector product, you can use the distributive law. So u times v is equal to, well, let me first do it with one. u product multiplied by v1 i plus v2 j plus v3 k. And now, since my distributive law works in this particular case, actually it was proven for two vectors summed together. It's a plus b times c. So a times b plus c. But at the same time, obviously it's true for the three vectors which we are adding together. Because first you can apply it to 2 and then to the third one. So that's easy. So let me just open it up. It's u1 times v1 i plus u times v2 j plus u times v3 k equals. Now, first of all, I can take v1 outside of the vector product. That's just another law which we were using. We proved it actually. It's associative law of the vector product relative to the multiplication by constant. So v1 goes out. And instead of u, I will use its representation. So it would be u1 i plus u2 j plus u3 k times i plus v2 goes out. And u times j instead of u, I will put exactly the same thing times j plus v3 u times k. And instead of u, I will use this same thing. Now I will use again my distributive law like a plus b times c. And in this case, it's three vectors which are multiplied by the force. So I can do it one by one. And what I will also use immediately here, the following. You see, i times i. You know that i times i as any vector multiplied by itself should give the zero result as a vector product. Why? Because the sign of the angle between one vector and another, since they are the same, the angle is zero, so the sign is zero. So something like i times i would never exist actually. Now, j times i, well, I have to really write it down here so we do not have any problems with this. So let me again write these equalities. i times j equals k. j times k equals i. And k times i equals to j. Now, if we reverse the sequence, you know that the vector product is anti-commutative. So it would be like i times j is k, but j times i would be minus k. So I will use this property. So let's see. So this one times this would be zero because it's i and i. Now, this is j times i. j times i is minus k. And the product is u2v1. OK, u2v1 minus k. So let me just put minus in front of it. Now, u3v1 would be ki, which is j was a plus. So u3v1j was a plus. Here, i times j would be k, right? So it's plus u1v2k. Now, j and j would give zero. So let's skip this one. k and j, k and j would be minus i. So we'll have minus u3v2i. And the third one is i times k. i times k is minus j. So it would be minus u1v3j. Now, j times k is i. So it's plus u2v3i. And k times k would be zero. 2, 3, 4, 5, 6. All right, so what do we have here? So let's collect i first. u2v3 minus u3v2i. Now, j. j is u3v1. So it's plus u3v1 minus u1v3j plus k and k, u1v2. u1v2 minus u2v1k. So that's the final representation of the vector product in coordinates. Now, if the vector product as a vector is the sum of these three vectors, each one of them is the unit vector multiplied by something. So what is this something? This something is the corresponding coordinate, right? So if i is the unit vector along x coordinate, then this in parenthesis is the x coordinate, right? So basically, what we can say is that w, which is u times v in coordinate forms, is represented as u2v3 minus u3v2 comma, u3v1 minus u1v3 comma, and u1v2 minus u2v1 close parenthesis. That's three coordinates of our vector product. Well, that's it for today. I do suggest you to do exactly the same kind of derivation just yourself without looking at my notes, et cetera. And you should really come up with exactly the same result. It would be a nice kind of exercise for you. And remember always to ask yourself why you're doing this or that particular simplification, for instance. Like, I was using these formulas for multiplication, vector multiplication of the unit vectors. I was using distributive law. What else? I was using associative law of multiplication of vectors and the constant. I was using the result of the multiplication of the vector by itself. Vector multiplication is 0 because the sign of the angle between the vectors is equal to 0. So all these properties, I would like you just mentally, just acknowledge to yourself that you are doing this because that's extremely important. Don't do it mechanically. It should be really based on certain laws. Well, that's it for this very short lecture. Thank you very much and good luck.