 Hi. I'm Zor. Welcome to a new Zor education. I would like to continue talking about scholar product of two vectors in the context of the problems actually. What I can't present it to you in the last two lectures, which were purely theoretical, was the definition of the scholar product of two vectors. In particular, if you have representation as a couple in a coordinate form, you have these two vectors. Then the scholar product of these two vectors was derived in the following way as this formula. Again, I derived basically this formula from certain common principles or rules which scholar product of two vectors would be great to have. We invented these rules and they seem to be quite natural. As a result of these rules, we derived this formula in the coordinate table representation of the vectors. I would like as a problem actually to go backward. What if this formula is given to us? How can I demonstrate that all these reasonable and natural rules which we have established at the very beginning are actually held, are satisfied? I will go through the same set of rules and I will try to prove that this formula actually satisfies these rules. Even if we go probably more classical way of presenting this mathematical concept. Let's define the scholar product in this formula. Now let's check these properties. The properties will be the rules which we were talking about. Let's go one by one with the same set of rules which I was using to derive the formula. Now let's actually derive the rules from the formula. The first rule was independence on certain transformation of the coordinate space. The transformation which I was talking about was the one which did not change the matrix of the space. Which means if the segment had certain lengths in the old coordinate system then it would have exactly the same lengths in the new one. So the length is not changing. And the relative position of the vectors like an angle between these two vectors is not changed. So basically I'm talking about rotation of the coordinate system. Also the reflection relative to one of the axis is also of this type. It does not change the length and it does not change the angle between vectors. Now obviously I'm not talking about any kind of stretching or shrinking the coordinate system. So the unit lengths remains exactly the same. So actually that's the most important property which we are talking about. Okay so let's talk about rotation. That seems to be like a natural way of changing the coordinate system. And let's check that this particular formula is stable so to speak towards the rotation. So after we rotate the coordinates A1, A2, B1 and B2 will change but the formula will result in exactly the same number. Now to do this, to accomplish this, to demonstrate that the scalar product is actually invariant, which means it doesn't really change, invariant to rotation. Let's just examine how the rotation actually changes the coordinates. So let's say you had one particular coordinate system, x, y, and then you had a point somewhere which had certain coordinates. So the length of this is x, the length of this is y. Now let's say we change the coordinate system, we rotate it, and the new coordinate system will be this. This will be my new abscissa, which I will call u, and this will be my new coordinate system, coordinate axis which is let's say v. Now the coordinate of the same point, which used to be x, y, now will be u. So this segment is u and this segment is v. I would like to express the lengths of these two segments u and v in terms of x and y, the coordinates in the old system, and an angle of the rotation. So this angle is let's say 5, or I think I used alpha in my notes to this lecture. So let's just use alpha. So alpha is an angle of rotation. We have rotated this coordinate system this way. So the coordinates are changing. Let's try to express the new coordinates u and y in terms of the old x and v, x and y, and angle of. Now what should we do? Well, let's put some letters so we will know what we're talking about. Okay, so b and c are projections of the r point a onto the old axis. Now g and e, okay, we know that this is y. So this point is e, and we know that this is x. It's clear. So let me just do a couple of equations to derive the values of u and v in terms of x and y and angle of alpha. All right, so let's think about it. I think I can say that ab, this piece, is equal to, okay, we need a letter, let's say m here, am plus mb. Right? Now what is ab? Ab is the coordinate of the a in the old system, right? So it's y. What is am? Well, let's consider a triangle amd. Am is a hypotenuse. Ad is a calculus adjacent to the angle amad and angle is obviously equals to alpha because there are two mutually perpendicular lines. So in terms of calculus ad, which is an ordinate in the new system, which is v, I have to divide it by cosine of this angle to get the am, right? So it's v over cosine of alpha. Okay, now what is mb? mb is a calculus in the triangle omb. Ob is another calculus. I know that ob is equal to, this is an abscissa of a in the old system, which is x. And this is also the same angle alpha, right? There are verticals. So mb is equal to x divided by tangent. No, multiply by tangent, sorry, x multiplied by tangent alpha, right? Now if I multiply it by cosine, I will have y cosine alpha here. Tangent is sine over cosine. So the cosine will multiply and we will have only sine. And I will resolve it for v. So v is equal to minus x sine alpha plus y cosine alpha. That's my first equation. That's how v is expressed in terms of x and y, basically. So v we have already expressed. Now let's talk about u, this one. Okay, here's what I suggest. Let's just continue this line to point n, let's say. Now I know that na is equal to nc plus ca. Na is equal to nc plus ca, right? Now what is na? Well consider triangle nae, this one. This is again angle alpha. Ae is abscissa of a in new coordinates, which is u. So the na is a hypotenuse in this triangle so it can be expressed with ae divided by cosine, which is u divided by cosine alpha. That's what na is, right? And it's equal to nc. What's nc? nc is a casetus in nc0. C0 is another casetus and this angle again is alpha. So c0 is equal to a ordinate in the old system, which is y. So it's y times tangent alpha. And finally ca, ca is x-coordinate from which I multiply by cosine. u is equal to x cosine alpha plus y tangent times cosine dot sine. So that's my transformation of coordinates. Now my question is if instead of xy I will use uv in this formulation and substitute using this particular transformation law, each coordinate, will I have the same value? Well let's just check it out. I should have, right? So if I replace x-coordinate and y-coordinate with u-coordinate and v-coordinate, so let's talk about new coordinates of the vector a. Vector a would be, so a1 is x, a2 is y. So my new coordinate is a1 cosine alpha plus a2 sine alpha. That's my abscissa in the new coordinate and the ordinate, am I right? Yeah, I think I'm right. And the ordinate is minus a1 sine alpha plus a2 cosine alpha. So that's how my vector a1 a2 looks in the new coordinates which are rotated from the old one by angle alpha. Now similarly with b1 cosine alpha plus b2 sine alpha comma minus b1 sine alpha plus b2 cosine alpha. Okay? Now my a times b scalar product would then be equal to 2. Now the first coordinate times the first coordinate plus second times second. Okay, so we have four different members because this is the sum of two, this is the sum of two. So it's a times a1 b1 cosine square alpha. Now this times this plus a1 b1 cosine alpha sine alpha a1 a2 b1 sine alpha cosine alpha and plus a2 b2 sine square alpha. Now I have to add to this the result of multiplication of these two. So it's plus a1 b1 sine square alpha minus a1 b2 sine alpha cosine alpha minus a2 b1 cosine alpha sine alpha plus a2 b2 cosine square alpha. So that's my result, right? Now a1 b1, actually I made a mistake. This is b2. a1 b1 a1 b2, right it's b2. I know it should reduce that's why here yes a1 b2 cosine sine a1 b2 sine cosine. So these two are out. One is plus another is minus. Now this a2 b1 sine cosine a2 b1 cosine sine plus and minus out. Now a1 b1 cosine square and a1 b1 sine square if I summarize them together a1 b1 goes outside of the parenthesis and cosine square plus sine square is equal to one. So these two sum up to a1 b1. Similarly this one and this one a2 b2 goes outside of the parenthesis and in the parenthesis I have cosine square plus sine square which is also one. As you see I got exactly the same as before which proves that rotation of the coordinates changes the coordinates but it does not change the scalar product. So scalar product is invariant that's very important. Invariant relative to rotation and other transformations of the coordinates which do not change the matrix lengths and angles of the system. Now my proof actually heavily dependent on some particular drawing where my point is in the first quadrant and my angle of rotation is just smaller than the angle to the point etc. Now you obviously understand that for every other kind of a variation of this picture what if my point is not in the first quadrant but let's say the third quadrant. I will also be able to do exactly the same type of manipulations with a couple of triangles and expressing one through another. It's always exactly the same so I don't even bother with all these different cases. This is just an illustrative example actually that the system works and it would work exactly the same way in any other usual position of the points and vectors. So this was kind of a more difficult part of the proof that all the rules we were talking about are actually preserved because the rest of them are really trivial. So what was the second rule? Second rule was this. Multiplication by null vector is supposed to give the result equal to null, zero actually. So this is easy because what is zero vector, null vector? Well that's the vector which has zero ranks which means the coordinates are it's just a point. It's a vector which is reduced to one particular point which has no direction and the length is equal to zero. Well if you use this formula where b1 is equal to zero and b2 is equal to zero obviously you will get zero. So the second rule is trivial. Now the third rule is that unit vector multiplied by itself should give the result of one. Well we are talking about unit vector multiplied by itself which means that a1, a2 is exactly the same as b1, b2. So the formula would be in this case, so the a would be let's say aa and b would be aa and what do I know? I know that the length of this vector aa is equal to one which means what? Which means a square plus a square is equal to one, correct? Because what is the length of the vector? This is the k-partness and the subsistence originate are two categories, right? So by Pythagorean theorem the length of the vector square is equal to some of the squares of the subsistence and originate. So if this length is equal to one then this should be the equality which I'm actually talking about. But now what is the scalar product of a and b in this case? Well it's a times a, right? Plus a times a which is exactly a square plus a square I know is equal to one, right? So that's how we prove this one. That's easy to. Next is associated law relative to multiplication by a constant. So k times vector a scalar product by b is equal to k times scalar product a times b. But this is again very trivial because what is the multiplication of the vector by constant in quotient representation? Obviously you remember that this is k1, kA1, kA2. That's the quotients. Now if you multiply this by the vector b which is this you will have kA1 times b1 plus kA2 times b2. Now on the other hand what is ab? ab is a1b1 plus a2b2 and if you multiply it by k you get exactly the same thing as this. So that's trivial proof too. And the last is distributive law for analysis just by c equals ac plus b. Where c is c1, c2. Now how to prove that? Again very easily. First of all we know what is the sum of two vectors in cohesion form. This is sum of coordinates right? So that's this vector. Now if you multiply it by c which is c1, c2 you will get c1 times a1 plus b1 plus c2 times a2 plus b2. Now what is this? Well ac is a1c1 plus a2c2. What is bc? It's b1c1 plus b2c2. Now this is exactly the same as this because you can open the parenthesis. These are just real numbers right? So amount numbers and distributive law is working. So again c1a1 which is this one, c1b1 which is this one, c2a2 which is this and c2b2 which is this. It's exactly the same. So that proves actually that starting from the formula like this for Scholar product we can actually derive all the rules which we were talking about in the very beginning from which in turn we derived our formula. So they are very much connected to each other. It's a natural kind of way of expressing these rules in the formula like this. Now the next lecture and the problems actually which I will present will be dedicated to exactly the same rules but we would like to start from the geometric representation of the Scholar product as the product of the lengths and the cosine of an angle between them and again derive all these rules but that will be the next lecture. Meanwhile I do recommend you to go to Unisor.com and go to this particular problem one. This is called problem one in the Scholar product topic and try to do exactly the same as I did especially the first problem related to rotation of the coordinates because it's not really trivial. Anyway that's it for today. Thanks very much and good luck.