 Hi, I'm Zor. Welcome to InDesert Education. Talking about scalar product of the vectors. In the beginning, in the theoretical lectures, I have introduced certain rules or principles or axioms, if you wish, about the properties of the scalar product which it's supposed to satisfy. There are natural rules like independence of the rotation of the system of coordinates, so it depends only on physical characteristics of vectors, their lengths and an angle between them. Commutative, distributive law, multiplication by null vector or multiplication by unit vectors by themselves. So these rules were sufficient to come up with certain formulas for scalar product. Now, in case of let's say two dimensional case, which is a little easier, so if you have two vectors with coordinates a1, a2 and b1, b2, then in the coordinate form, their scalar product, according to these rules, must actually be expressed as the formula like this in the coordinate form or geometrically speaking, it's expressed as the lengths of one vector times the lengths of another vector times cosine of an angle between them. So we have two representations which were derived from these rules which I was talking about. Now, the previous lecture which was like problems number one was about, okay, let's take the coordinate representation of this scalar product as given and then derive all the rules which we were talking about, independence of the rotation, commutative, distributive law, etc. Now I would like to do exactly the same with geometric representation. So let's consider that we have defined our scalar product as this lengths times length times times cosine of the angle between them. Can I derive all these very useful and nice properties which we were kind of put in the beginning as a foundation? Can I derive all these from the formula? So basically what I'm trying to do is to prove the necessity and sufficiency of the expressions in coordinate form or in geometric form for satisfaction of those rules. So rules and these formulas are in one-to-one correspondence, so to speak. They are necessary and sufficient condition for each other. So this lecture which is basically a problem number two lecture is prove that all these rules which we were talking about follow from the geometric expression of the scalar product. So let's just try to prove it. Okay, the rule number one, the scalar product should not depend on transformations of the coordinate system which preserve the metric of the system, which means lengths of segments are not changed and the angle between segments is not changed either. So it's like rotation, for instance, or reflection of the coordinate system relative to one of the axes, for instance. So, well, let's just look at this formula. It depends on the lengths of the vectors and the angle between them. And we just said that we are talking about transformations of the coordinate system which do not change it. So basically from the problem itself actually, from the conditions of the problem itself, this rule actually follows because it depends only on the lengths and the angles. So no matter how we rotate the coordinate system or reflected relative to some axes, the lengths of the vectors is not changed in these transformations and the angle between them not changed either. And by the way, considering this is a cosine which is an even function, it doesn't really matter, we count an angle from A to B or from B to A because all we are doing, we are changing the sine of the angle and the change of the sine doesn't change the value of the cosine because sine is an even function. So the rule number one is satisfied basically by definition of the transformation we are talking about. Now, rule number two is this. If I multiply a vector, any vector, by a null vector, null vector is the one which has lengths of zero, then I will get zero. Well, obviously, this B is a null vector, so the lengths of the B again by definition is zero, so this is the multiplication of lengths of the A which is something by zero and by cosine of the angle which doesn't really exist and it doesn't really matter because this is zero. So this is satisfied as well. Next is another simple case. If I will have two unit vectors, well, if I have one unit vector multiplied by itself, then result is supposed to be one. Well, again, it's very easy because the length is equal to one, this and this, and the cosine of the angle, if I'm talking about multiplication by itself, it means that I have two vectors which have directed in exactly the same direction, which means that the angle between them is zero and the cosine of zero is one. So we have one, one, and one, and that's why this product is equal to one. So that's true too. Next is associative law. Okay, I have to prove that if I will multiply one of the vectors which is a component of the scalar product by a constant, then it will be the same as multiplication of this constant by scalar product of these two vectors. Well, that's very easy to see because multiplication by a constant does not change the direction of the vector. It just makes it longer or shorter. It's stretching or shrinking, right? So the length is the only thing which is changed here. So what's the length of K times vector A? Well, the length is equal to... This is by definition of the multiplication... Actually, I have to put absolute value here. This is the definition of the multiplication of the vector by a constant. We were talking about when we were defining this that the direction remains the same and multiplication by constant is changing the length. The only little detail is that if the constant is negative, we change the direction to an opposite. But again, it doesn't change the absolute value of the length which remains to be exactly the same. Now, B is what it is. So on the left, I will have... On the left, I will have absolute value of K by A by B by cosine of phi. Now, speaking about cosine of phi, if K positive, then the direction between A and B remains exactly the same. So angle is exactly the same as it was before. Now, if direction is changed to the opposite, the angle... if this... what used to be A and B, and now this is A times K for K negative, then the angle is this one. Right? So the angle is changed to the opposite. The direction is changed to the opposite, so the angle is changed to 180 if this is phi. So this is 180 minus phi. Right? So as a result, my cosine is changing from the cosine of phi to cosine of 180 degree minus phi, which is... which is minus cosine of phi, as we know from the trigonometry. So for positive constant K, the left side is equal to... basically I have to put, instead of this angle, I have to put 180 minus phi. And for positive, it will be the same phi. So let me just have two cases here. So this is for K greater or equal to zero, and for K less than zero, I have the same thing, cosine... Well, I can actually immediately write minus here and cosine of phi here. So that's what this part is. Now let's talk about this part. Well, this is always absolute length of A times length of B times cosine of phi. And we multiply by K. So for positive K, for positive K, we have exactly the same, because the absolute value of K for positive K is K. For negative K, we have this, because this is exactly the same cosine of phi, right? So negative K would be equal to minus absolute value of K. So in both cases, positive K or negative K, we have exactly the same result. So that's why the left part is equal to the right part. And obviously, if I will multiply this vector, instead of A multiply B by K, it would also be exactly the same thing for exactly the same reasons. What I mean is this. It's equal to A... Well, I don't really need this parenthesis. A times KB. That's exactly the same thing, exactly the same logic but applied to the vector B should prove it. So that's about associative law relative to multiplication by constant. Okay, and the last one is a little bit more involved, is distributive law. So distributive law is this. This is a scalar product of some of two vectors by the third one, and these are two scalar products of these two vectors. Now, when we were talking in the previous lecture about the same distributive law in coordinate form, it was actually very easy to prove because the distributive law was actually... between the vectors in the scalar product was reduced to distributive law among numbers. And geometrically, it's a little bit more involved. So let's do it differently. And again, it's not really for, like, really proving this thing. It's more of a mental exercise because it's quite interesting if we can prove exactly the same thing but using a completely different approach, a geometric approach rather than algebra of coordinates. All right, so here is the drawing. My proof will probably depend quite significantly on the drawing. However, even in differently positioned vectors, the proof would be exactly the same thing. So I'll do it in this particular case. So let's say this is my angle B, my vector B. This is my vector A. And we need C, right? So this is C. This is my three vectors which are involved in this thing. So first I have to draw A plus B. So this is a parallelogram. That's how we draw the sum of two vectors, right? You take this one, attach it to the end of this one, or draw a parallelogram. It doesn't really matter how we'll do it. And this is V equals to A plus B. Since we are talking about geometry, I need some angles, right? So let's say this is angle alpha, angle between A and B would be vector. And I will also use this angle gamma between the sum of A and B relative to A. Okay. Now, let me just wipe out everything from here. Let me know all this. And I will use that space for something. All right, so what I'm going to do is I would like to... Okay, let me just write again what I'm going to prove. To prove this, let me express the lengths of A and B in terms of the lengths of their sum and whatever the angles I have here. Now, I will use this triangle from here to here and to here and use the law of signs to basically express the lengths of all these components. Now, by the way, this is beta, so this is 180 degree minus beta, right? Now, I will use the same letters but without the bars on the top to express the lengths of these vectors, right? So I'm talking about lengths only. So V, which is this one, lengths of this vector, which is the sum of these two, divided by sine of this, V divided by sine of 180 minus beta. Now, sine of 180 minus something is the sine of that something, right? Again, we know this from trigonometry. Equals to... Now, this is A and this angle is equal to this angle, which is equal to beta minus gamma, right? So it's equal to A divided by sine of beta minus gamma and equals to B divided by... Now, this is B and this is gamma, so this is sine of gamma. So from this, we can say that A is equal to V times sine of beta minus gamma divided by sine of beta and B is equal to V times sine of gamma divided by sine of beta. Now, what do we have to prove, actually? We have to prove that V times C times cosine of alpha plus gamma, right? V is the length of this, C is the length of this and alpha plus gamma is the angle between them. We have to prove that this is equal to A C cosine... A C cosine of alpha, right? That's what this is. So, V C and cosine of alpha plus gamma is this one. V is the length of this, C is the length of this and cosine of alpha plus gamma is their angle. A C is this, A C and cosine of alpha between them. And B C is B C cosine of B C. It's alpha plus beta. That's what we have to prove. So what I'm going to do, I will substitute this and this into A and B here and I'll just try to prove this trigonometric equality. So we don't need this anymore. Again, I was saying that I would kind of depend on this particular drawing but you understand that if any other composition, there will be very similar considerations. So, if I will substitute, instead of A and B, I will substitute this. What I will do, well, first of all, I can obviously reduce by C and then V would be here and V would be here so I can reduce by V as well. So I will retain only the trigonometric functions. So cosine of alpha plus gamma equals, instead of A, I will put this. So the cosine of alpha will remain so it will be cosine of alpha times sine of beta minus gamma divided by sine beta. And here I will have cosine of alpha plus beta times sine gamma divided by sine beta. Or if I multiply by sine beta, I will get this and this would be sine of beta. So that's what I would like to prove. Now, I still remember sine and cosine of the sum of the difference between two angles so I will just open these parentheses and see what happens. Okay, sine of beta times cosine of sum. This is a cosine, cosine minus sine, sine. All right, so I will have cosine, cosine, and sine. All right, so cosine of alpha. I will put it in alphabetical order. Then sine of beta then cosine of gamma. Right? I'll give you these parentheses. Minus, because this is the minus, sine sine, right? So I will have three sines. Sine alpha, sine beta, sine gamma. That's what I have to prove that it's equal to this. All right, same thing. Cosine alpha, sine cosine minus cosine sine, right? So it's sine beta cosine gamma minus the same cosine alpha, cosine beta sine gamma. Plus this thing. Cosine cosine minus sine sine, right? So it's cosine alpha cosine beta and sine gamma minus sine alpha sine beta sine gamma. Now, if I did not make mistakes, it should all be somehow equal. Well, let's think about it. This is equal to this. That's true. Cosine sine cosine. Cosine sine cosine. This is equal to this. And this cosine cosine sine, cosine cosine sine minus and plus they nullify each other. Yeah, everything is fine. So basically this, a little bit more involved trigonometrically proves that all the rules, including this one actually, the distributive law are satisfied if we will give the definition of this color product only in geometric form. Length times length times cosine of the angle between them. So that actually completes this exercise of proving of the equivalence of giving these six rules, which I was actually starting with, and derived the coordinate and geometric representation of the vectors. Equivalence of an opposite approach. We can define the coordinate as a coordinate representation of this color product in the beginning and then we derive the rules as properties of this. Or if we define geometrically the color product of two vectors and then again we derive all the rules. So all these three logical systems, six rules, which seems to be quite a major role, definition in the coordinate form and definition in the geometric form, they're all equivalent to each other. From any one I can derive any others. I do suggest you to go to unison.com again to this particular lecture to problems too, it's called. And try to prove yourself exactly the same thing which I did here. I think it's a very useful exercise. And that probably would complete the theoretical part of this color product. There might be some other problems which I will add in the future, but basically this is the most important because it actually explains all the nuances of the definition of this color product and how the properties of this are. Well, that's it for today. Thank you very much and good luck.