 Hi, I'm Zor. Welcome to Yizor Education. I will continue talking about scalar product of two vectors in two and three-dimensional cases. Now, the previous lecture about scalar product, the most important result of the previous lecture was that if you have two vectors A, let's say, considered in two-dimensional case and vector B with coordinate representation A1, A2, and B1, B2, then their scalar product looks like this. So that's the most important result of the previous lecture. I also mentioned that scalar product should not really depend on the coordinates. It should depend on physical characteristics of vectors in their mutual position, their lengths, and the angle between them. Well, but looking at the formula, you see that it looks like it depends on the coordinates. Well, it remains to be proven that if I change the coordinates in some nice way, so the matrix of the space is not changed, like lengths is preserved and the angle between the vectors is preserved, like a rotation for instance. Then the formula though it does depend on the coordinate, but the result of this calculation with different coordinates will be exactly the same. So that remains to be shown and probably I will have some problems related to this. Today I would like to concentrate on different expression of the scalar product, not in the coordinate form, but in the form of these physical characteristics, which I was talking about, which are unchangeable, so to speak, like lengths and angle between vectors. Well, let me give you the final result, which I'm going to prove, lengths of the first vector times lengths of the second and the cosine of the angle between them. So that's something which is a final result which I will prove today. So if I will, and I hope I will, then it's reasonable to, it's actually quite obvious here that no matter how I change coordinates, if this is a geometric interpretation of the scalar product, so no matter how I change the coordinates, the result should be exactly the same as this one, as long as coordinates are changed in a nice fashion, in orthogonal fashion. So the lengths remains the same, so we don't really scale the coordinate system and the angle between the vectors remain the same. So that's my final result and I'm going to prove it in two cases, three-dimensional case and three-dimensional case. By the way, I did not mention it in the three-dimensional case. There is a similar formula. If you have a third coordinate, A3 and B3, then the result would be here that would be in three-dimensional case, right? So I didn't mention it in the lecture, but I decided to put it into the notes of the previous lecture. So if you're interested, it's exactly the same basically derivation with A3 and B3 as an two-dimensional case. But in the case of a geometric interpretation, there is a significant complication in a three-dimensional case, so I will probably spend a little bit more time on this. Okay, now let's go to a two-dimensional case in geometric interpretation. All right, so this is easier because I can draw it on the board and everything is kind of obvious here. Let's say this is my vector A and this is my vector B. Now, these are coordinates, and I also would like to mention this is angle alpha and this is angle beta. All right, now, I can express coordinates in terms of lengths and angles, well, very easily. Let's say this length is A. Length of the vector A is A and length of the vector B is B. So this is A, this is B. If you draw the perpendicular here, here, here, and here, basically you can express the coordinates in case of A, for instance. This is A1 and this is A2. For a B, this is B1 and this is B2. So I can use the hypotenuse of this right triangle to find A1 equals to A, the lengths of the hypogenuse times cosine of alpha and A2 is equal to A cosine of beta. B1 is equal to B cosine of beta and B2 is equal to B cosine, sorry, sine and this is alpha. Okay, now this is correct, right. Alpha is for A vector, so it's A cosine and A sine of alpha. Beta is the angle of the B vector, so it's B cosine and B sine. Okay, fine, now, since I have this expression, I can substitute this into my scalar product and what will I have? Well, scalar product of A times B is equal to A1 B1, it's A times B times the cosine and cosine. So A, B, cosine alpha, cosine beta. And A2 B2 is A, B, sine alpha, sine beta. Well, if the angle between alpha and beta, between A and B, is phi, phi is equal to beta minus alpha, or if you wish beta is equal to alpha plus phi. Now, in this case, I can substitute this beta for these and what I will have is AB, I will factor out AB. Cosine alpha, cosine alpha plus phi plus sine alpha, sine alpha plus phi, we'll use square break. So that's my final expression, now this is what I'm thinking I would like to get, this is capital A, this is capital D, and this is a cosine of phi. This doesn't look like this one, right? So A and B I do have, but the cosine of phi is not exactly this, but maybe it will be exactly like that if I will transform it slightly. Let's just think about it. Now, cosine of sum of two is equal to cosine alpha, cosine alpha, cosine beta minus sine alpha, sine, sorry, not beta, phi. So that's the cosine of alpha plus beta, alpha plus phi, cosine cosine minus sine sine, now plus sine alpha times sine alpha, cosine phi, right? That's the formula, sine cosine plus cosine sine. Now, let's just multiply, what do we have? AB, cosine square alpha, cosine phi minus, cosine alpha sine alpha and sine phi plus sine square cosine, sine phi plus sine square cosine, plus cosine phi plus sine cosine and sine phi, sine alpha cosine alpha and sine phi equals. Well, first of all, this and this are the same with the opposite sines, right? Here, I can factor out cosine phi and what do I have cosine square plus sine square? Now, cosine square plus sine square is one, so I get AB and cosine phi, which is exactly what I wanted to get. So as you see, no matter how I position the coordinates, I can always express these coordinates in terms of lengths of the vectors and some angles, but after the manipulation, this particular formula basically gets rid of all these angles except one, angle between the vectors, which is fine. So this is a final formula for the same scalar product but in a geometric fashion. Well, actually, geometric plus trigonometric, but that's okay. So we are using geometric characteristics of vector or if you want physical characteristics, whatever, to express this scalar product and that's very important. It depends only on the vectors, their lengths and their position in space rather than how we choose the coordinates. Another thing which I wanted to present to you today is exactly the same problem in three-dimensional space. Well, in three-dimensional space it will be a little bit more involved as far as the calculations are concerned, but the theory is exactly the same and the formula will be exactly the same. It's still lengths times lengths times the cosine of an angle between these two vectors in three-dimensional space. So let's go for it and see if I will not make any mistakes on the way. In any case, I put all these calculations in the notes for this lecture. So basically, it's really straightforward. I'll try not to lose my way around these three-dimensional coordinates. So it's three-dimensional coordinates like this. Okay, we will do exactly the same. So first of all, I know that the scalar product is expressed as this formula using the coordinates. So I will choose certain physical or geometric characteristics of the two vectors. Obviously, there are lengths and some angles. I need more angles in this particular case because it's three-dimensional space. And I will use these physical characteristics of the vectors to express all the coordinates. Put it in the formula and see if whatever is necessary will be just reduced and that will have only the product of these two lengths and the cosine of the angle between them. And they will, hopefully. So first of all, how can I use this geometry of the three-dimensional space to express my coordinates in terms of lengths of the vectors? Okay, here it is. Here is the picture. Now, this is three-dimensional space. So we have x, y, and z coordinates. Now, x, y imagine it to be horizontal and z vertical up, all right? Now, if I have a vector, here is my vector a. Now, let's project it down onto the horizontal x, y space. And this would be the projection of this point. So this is a1, a2, a3 coordinates. Now, the projection here would have coordinates, obviously, a1, a2, 0, right? Now, this would be a1, this would be a2. Now, projection onto the z will give me a3. And this point, obviously, will be 0, 0, a3. Let's have two angles. The angle number one would be from the x-axis to the projection on the x-y plane. And let's call it alpha one. This is a horizontal rotation from the x-axis to the projection of the a vector onto the plane, on the x-y plane. Now, the second angle would be vertical, alpha two between this projection and the real vector. So I will use these two angles. Now, remember, in a two-dimensional case, I only needed one angle from x-axis to my vector. Now, in this case, I need two. One from x-axis to the projection of the vector on the x-y plane. And then another angle vertically up. This is called azimuth vector, and this is called altitude vector. So azimuth is within the horizontal plane, and altitude is within the vertical plane. So if the length of the vector a original length is a. Now, this is the projection, which means this is the right angle. So a is a hypotenuse. So the projection, length of the projection is equal to a times cosine alpha. a times cosine alpha, alpha two. So I have this one. Now, this is the right angle. So this is the hypotenuse. So a one is equal to a one is equal to this times, cosine of alpha one. Hypotenuse, this is the casualties. The angle between them is alpha one. So hypotenuse times the cosine of the alpha one. So this is how my a one coordinate is expressed. Now, a two is a cosine alpha two. It's the same thing, but now I need this calculus, which is sine of alpha one. And finally, a three is equal to, it's this, which is the same as this. So it's opposite casualties with hypotenuse a and alpha two is angle. So it's a sine alpha two. Obviously with b, I will have exactly the same thing. I will have b and angles would be beta one and beta two. That's how I will use it, right? So I don't need this anymore because everything else is just trigonometric manipulation. Okay, so my scalar product would look like a one times b one, which is a b cosine alpha two cosine alpha one, cosine beta two and cosine beta one plus a b. Now a two is cosine alpha two sine alpha one. Sine alpha one and same thing would be for the vector b. And the third coordinate, the product of the third coordinates would be a b and two sines. I wanted to prove that this is a times b times cosine of the angle between them. So this big expression a b is obviously is factored out. Cosine cosine cosine cosine plus this plus altogether they actually constitute a cosine of phi where phi is an angle between these two vectors. Question is why? Okay, that's what I'm going to prove. I'm going to prove that this sum of these very complicated things is actually a cosine of the angle between these two vectors. And here is how. Let me go back to the drawing. Let's say this is vector a and this is vector b. Let's consider this triangle and I'm going to use, now this is the angle phi between them. I'm going to use the law of cosines for this particular triangle. I know the length of this is b, this is a and the length of this is something which I can obviously calculate based on knowing the coordinates. I know the coordinates of a, it's these. Coordinance of b are similar with the letter b and angle beta. So I can use just the formula between two points, distance between two points knowing their coordinates. Remember the formula is the distance squared is equal to a one minus b one square plus a two minus b two square plus a three minus b three square. That's the formula for distance in Euclidean space between points a one, a two, a three and b one, b two, b three. So I will use this as the third coordinate and I also know that d square is supposed to be a square plus b square minus two a b cosine phi. Now this is the law of cosines for this particular triangle. So equating these two things, I can see what will be cosine phi in terms of all these cosines of alpha one and alpha two and science, et cetera, et cetera. So the equality between these two should give me the expression of cosine phi in terms of other angles, alpha one, alpha two, beta one and beta two. So that's what I'm going to do. I'm going to calculate this and compare it with this and that's how I will find cosine phi. And I will actually find that the cosine phi is equal to exactly this thing. I know this, right? So let's try a one minus b one square. Okay. It's a one square which is a square cosine square alpha two and cosine square alpha one minus two a one times b one, a one times b one with the a b cosine alpha two, cosine alpha one, cosine beta two and cosine beta one. And plus b one square which is b square cosine square b beta two and cosine square beta one. That's the first plus. Now second one is a two minus b two square which is a two square minus two a two b two plus b two square, right? So a two square is cosine square alpha, sine square alpha one minus two a b cosine alpha two, sine alpha one, cosine beta two, sine beta one and plus b two square, b square cosine square b two, sine square b one and the third one, a three square which is this minus two a b sine alpha two sine beta two and plus b two square, b three square which is b square sine square alpha, sorry beta. Now, let's think about what this is. I don't need this anymore. Now, how can we simplify it? Well, very easy actually. Look at this. A, b, c, c, c, c, c, c, c, c, c, c, c, c, c, c, c, c, look at this, a two square cosine square alpha two times cosine square of alpha one plus sine square. So that's a square cosine square alpha two, right? Now, if I will add this, a square cosine square and a square sine square would be just a square, right? So that's the result of this. Similarly, this would give me b square cosine square beta two times cosine square plus sine square, which is one, right? Now, if I will add this one to this, I will have b square cosine plus b square sine square, it will be just b square, okay? And what do I have left? Minus, minus two a b times all this thing. This, this, and this. Cosine alpha two, cosine alpha one, cosine beta two, cosine beta one. Well, let me open the parenthesis here. Now I will have only pluses. Cosine alpha two, sine alpha one, cosine beta two, sine beta one plus, and the last one is sine alpha two, sine beta two, that's what I will have. That's the result of, that's the result of this. Now, let's compare it with this one, a square plus b square, a b square plus b square. Minus two a b minus two a b, cosine five, and this is I have in square brackets. So, whatever I have in the square brackets is actually a cosine five, right? And look at this, it's exactly the same expression. Factor out a b will have exactly the same thing, cosine alpha two, cosine alpha one, cosine beta, and cosine beta one. Here, cosine sine, cosine sine, and sine sine. So it's exactly the same thing. So this expression can be replaced here as a b times cosine five, and that's the end of it basically. So as you see, in a three-dimensional case, situation is exactly the same. I have product of the lengths of two vectors and the cosine of the end of it. Again, it depends only on geometrical characteristics of the vectors, not on their coordinates. Well, that was actually everything I wanted to tell today. I wanted to present the case of geometric interpretation of the scalar product, and it is the product of their lengths times the cosine of the end of between them, independent on the coordinates. So as long as the coordinates are Euclidean, the vectors in space are positioned somehow, vector, the vector, and the angle between them. So no matter where is the coordinate system, how we move this couple of vectors around without changing their lengths and without changing the angle between them, their scalar product will be exactly the same. And that's the beauty of this particular thing. And if you remember, when we started the previous lecture, I started with certain rules, how it would be nice to have this particular scalar product adhere to certain properties, certain rules, as I was calling them. And one of the rules was independence on the coordinates. So it depends only on the internal characteristics or inner characteristics. Maybe that's why sometimes this scalar product is called inner product. Okay, that's it for today. I do recommend you to go through notes for this lecture. They are on Unisor.com, and just to refresh that thing. It would be even better if you can just completely do it by yourself. Just have a piece of paper, draw some drawings, and try to derive, especially the three-dimensional case, which is a little bit more involved, and it needs certain spatial view. I do recommend you to do it just by yourself. That would be great, of course. So that's it for today. Thank you very much, and good luck.