 Hi, I'm Zor. Welcome to Unisor Education. I would like to present you a couple of problems related to vector arithmetic, like addition, multiplication, the constant, etc. Now, these problems are quite trivial, but I would like to use this opportunity to basically tell about certain standard symbolics which are used in the vector arithmetic. So, what I will be using for vectors are usually lowercase Latin letters with arrow or just a bar on the top, depending on, and if this is a vector, then in its tuple representation, I will try to use the same letters as coordinates with indices. This is a three-dimensional tuple representation of vector B. Now, another thing is lengths of the vector. Now, the lengths, I will use double bars. That's the lengths. By the way, sometimes the lengths is called a norm. Norm of the vector. Now, this is the term which is borrowed from a little more abstract concept of vector spaces. So, norm or lengths, just exactly the same, so I will use it. Obviously, in this particular case, I can always say that this is a three-dimensional length, so a norm of the vector B expressed as an algebraic expression of its tuple representation. In this case, it's three-dimensional coordinate system. All right. So, that's all about different symbols which are used. Now, let's just do the problems. Okay. The first problem. Let's consider a vector B with tuple representation V1, V2. Let's say it's two-dimensional. Actually, if it's three-dimensional, it would be exactly the same. Now, let's consider the vector which is the result of multiplication of this particular vector V by some real number K. Now, what happens with coordinates? Well, the theorem states that coordinates are also multiplied. So, multiplication of the vector by a constant results in the tuple representation, multiplication by this constant, every component of the tuple representation. So, every coordinate of the vector. Now, let's just consider this as a geometric problem. Okay. This is my coordinate system and this is my vector. So, these are V1 and V2 coordinates of this vector. Now, if I multiply it by, well, let's say in this case, a positive constant first, what happens? Well, direction is the same of the result of the multiplication by the constant, but the length is increasing by that particular number. So, let's say K is something like 2, for instance. It doesn't really matter. So, this line goes twice as long. Now, what are the coordinates of this next vector? So, if this is V, this is KV. So, from 0 to A, from 0 to A is V and from 0 to B is K times V. Now, it's quite obvious that triangles A1, A2, B1, B2. Triangles OAA1 and ODD1 are similar, obviously, right? Because we know that the triangles are right triangles, right? Because this is right angle. Now, this angle is retained. It's common. It's shared. Because when we multiply the vector by some number, we retain direction. Direction is an angle. So, the angle is exactly the same. It means another angle is also the same. So, two right triangles with the same angles, they are obviously similar, which means that everything is proportional. Now, since the proportion of the hypotenuses is K, it's exactly the same proportion of Kegitus OAA1 relates to OAD1 as OA to OB. So, this is exactly the same. Same thing with Kegitus AA1 relative to GB1. So, the same coefficient of proportion analogy exists for the length of the vector itself, which is hypotenuse, and on each Kegitus, which are coordinates of the top of representation. So, that's why it's true. Now, for a negative K, situation is absolutely the same, because all it does, it just reverse the picture towards the other direction. So, if this is K, this would be multiplication. I mean, if this is V vector, then this would be a multiplication by negative. And since direction is changed on the opposite, these angles are still the same, which means that, again, the triangles are similar in the proportion analogy between the lengths of the hypotenuses would be exactly the same as the coefficient of proportionality between the Kegitus. And obviously, if this is positive, then this is negative. And that would signify that the same sign of K is the coefficient of proportionality between the Kegitus and Kegitus of these triangles. So, positive or negative K, everything is exactly the same. It's all similarity of triangles. Now, what happens? OK, so we have actually found that the multiplication of the vector by constant is basically reflected in the top of representation as multiplication of each component by this particular constant. That's that. Now, how about the lengths of the vector? So if you will take the lengths of this vector, which is the result of multiplication of the vector V by K, that's the new vector. And what's its lengths? Well, we know that the lengths of the vector multiplied by a constant is supposed to be, obviously, longer by that particular constant. Now, the only thing I cannot really write this way because K might be negative. So this is a true length. So this is all the positive thing. So we cannot just have playing K. We have to play absolute value of K. So absolute value of K times norm of the vector V is the norm of the vector KV. So that's the second problem. So multiplication by a vector by any real number results in the lengths being multiplied by absolute value of that particular real number. That's the number two problem. OK, now what happens with co-ordinates of the vectors when we add them up? Let's say you have two vectors. V and W in tuple representation. And again, I'm using two dimensional case because it's easier to draw. Basically, it's the same for any dimension. So what happens with tuple representation of the sum of these two vectors? OK, let's just write the answer. Now, the tuple representation of the sum of two vectors is sum of the corresponding co-ordinates. Now, well, this is for three dimensional case. Now, we have two dimensional case. How can we make sure that this is true? Well, let's just go again geometrically. So let's say we have vector V and this is vector W. Now, when we want to add them up, you remember the rule. We take the vector W and we attach it at the end of the vector V by a parallel shift, forming the parallelogram. So this is A, this is B, this is C. So AC vector is exactly the same as all B and this is W. Now, let's think about co-ordinates. This co-ordinate A1, this is B1, and this is C1. Now, obviously, since we took this OB and parallel shifted to AC, then the co-ordinate of the OC would be greater than co-ordinate of A by the co-ordinate of the OB, right? So the co-ordinate of OB, the x co-ordinate in this particular case, is OB1. So what I'm saying is that OC1 is equal to OA1 plus AB, sorry. OC1 is equal to OA1 plus A1C1. But A1C1, this one, it's projection of this particular vector, HC, onto the x-axis, which is exactly the same as projection of OB, right? So that's equal to plus OB1. This is the same vector as this one. So projection of this vector onto the x-axis, which is A1C1, is exactly the same as projection of this vector on the same axis, which is OA1. And this is, obviously, OA1 is projection of the vector OA on the x-axis, which is V1. And OB1 is W1, projection of the OB of the vector W on the x-axis. So OC1, this is an x-coordinate of the sum of these two vectors, right? I didn't draw the sum, but this is the sum, right? OC as a vector is equal to OA as a vector plus OB as a vector. So that's what it is with x-coordinates. Now, projection onto y-axis is exactly the same. The same logic, every single is the same. So that's why the projection on the y-axis of the sum of two vectors would be sum of its projections. So when you sum two vectors in their table representation, you just add the corresponding coordinates. But by the way, the first problem was when you multiply vector by constant, you multiply corresponding components of the table representation. Now, when you add two vectors, you correspondingly add coordinates in the table representation. So it's a very, very natural and very easy to remember rule, basically, right? All right. How about difference? OK. I'm sure you expect this, which is absolutely correct. Now, I could probably prove it in many different ways, but let's say, for a second, if v minus w equals u, what does it mean? It means v is equal to u plus w. So u has to have co-coordinates u1, u2. This is w1, w2. And v is v1, v2. So u1 plus w1 should be v1, which means u1 minus w1. Now, u2 plus w2 should be equal to v2. Therefore, v2 minus w2 is equal to u2. So that's one of the ways to do it, purely algebraically. In other ways, just to draw a picture and you will see, basically, that it's exactly the same rule. As we used for addition, it would be applicable to subtraction of two vectors without going into many details. If you have these two vectors, this is v and this is w, then v minus w is this one, right? This is u equal v minus w. Now, w plus u gives you v, right? That's right. Now, if you compare the co-coordinates of these points, the projections of these vectors on the x-axis, for instance, then obviously, if this is a and this is b, this is b1 and this is a1, then obviously, if you subtract from o, b, you subtract o, a1, you will get a1, b1, which is projection of this vector onto the x-axis, which is its first x-coordinate and the top of the representation. The same thing with y-coordinates. So either we're algebraically or geometrically we can prove it. OK. Now, from these two previous problems, I would like to prove this subtracting a vector is exactly the same as ending the vector which is opposite in direction in the same as length, which is multiplication by minus 1. Addition of this particular opposite vector. So it's like with numbers, right? 5 minus 3 is exactly the same as 5 plus minus 1 times 3, right? Which is minus 3. Now, with vectors, it's very simple actually to show. Well, one of the easiest way, for instance, OK, let's consider co-ordinate representation, top of representation. Now, what do we have in this case? Well, the representation of this is v1, v2. This is w1, w2. And the difference we already know by previous problem is v1 minus w1 and v2 minus w2. Now, what about this? This is v1 v2 plus minus 1 times w1, w2. Now, this is, as we know, multiplication by a constant results in the multiplication of each component in the top of representation. So it's exactly the same as minus w1, w2. Now, when you add these two things in the top of representation, coordinates are added together. So v1 is added to w1 and w and v2, same thing. Well, as you see, we have exactly the same result, which proves that subtracting a vector is exactly the same as edging that vector multiplied by minus 1, or if you wish, edging the opposite vector. The last problem is this one. The length of the sum of two vectors is less than or equal to lengths of the 1 plus lengths. Well, let's just consider this from the geometric perspective. We don't even need coordinates, let's frankly. Because this is the v vector. Now, this is a w vector. This is a v plus w vector. What does it say? That the length of the sum is less than sum of these two lengths, which is a triangle inequality. It's inequality, which is known for all the triangles. So geometrically, this is basically obvious. It results from the triangle inequality, which we learned before in the corresponding chapter related to geometry. But let's think about, can we do it algebraically? Not because it's better, but it might be really interesting to approach it not from the geometric perspective, but from the algebraic perspective, table representation, and expression of the lengths in terms of table components. Actually, we can do it. It's a little bit longer, but I think it has certain value just because you will familiarize yourself with manipulation with table representation. So let's say these are table representation, and I'm talking about two-dimensional case. That's easy. Now, the table representation of v plus w is v1 plus w1, v2 plus w2. Now, how is a norm or a length of the vector with coordinate x1, x2 expressed algebraically? Well, you know it's this one. That's Pythagorean theorem, and we spoke about this in the corresponding lecture about lengths of the vectors. So what's this length? Well, it's a square root of first component square, which is v1 plus w1 square, and v2 plus second component square. That's on the left. That's what we have to prove. And this one, that's the lengths of this vector in table representation. Less than or equal to lengths of the first one plus lengths of the second one. That's what we have to prove. Now, well, let's see if we can do it. If we can do it algebraically where all numbers, v1, v2, w1, w2, are just real numbers. So this is a pure algebra. Prove this particular inequality. Well, let's see if we can do it. First of all, everything is positive here. So if we will prove instead of this, if we will prove the square of this, it will be the same, right? It's equivalent. We're talking about lengths, so everything is positive here. This is positive, and this is a risk particularity of the square root. All right, so let's square both sides. This square would be everything without the radical, obviously. And I will open up the parentheses. It would be v1 squared plus 2v1, w1 plus w1 squared plus v2 squared, 2v2, w2 plus w2 squared, right? That's on the left. And we have to prove that this is less than or equal to square of this, which is square of the first one, plus 2 square root of their product, v1 square plus v2 squared, w1 squared plus w2 squared, plus square root of the second component. That's what you have to prove. OK, now, obviously, we can reduce by this, right? Everything goes out. And I will also reduce by 2, because here 2, and this is 2, and this is 2. So what's remaining is v1, w1 plus v2, w2. That's on the left. Should be less than or equal to square root of. Well, let me multiply them. v1 squared, w1 squared plus v2 squared, w1 squared plus v1, w2 squared plus v2 squared, w2 squared. That's what I have to prove. Again, let's just think about everything positive. I don't want to worry about the negative signs. Let's consider everything is positive. It's just easier to do it this way. So consider we are only dealing with vectors in the first quadrant, because everything else would be exactly similar. It's just easier for me to go without thinking about not exactly being backward compatible. But in this case, if everything is positive, then I don't have any problem with it. So let's square it again, both sides. So on the left, I will have v1 squared, w1 squared plus 2w plus 2v1v2w1w2 plus v2 squared, w2 squared. Less than or equal to this. I will not change anything. I will reduce something, I guess. This one, and this one, this one, and this one. So what remains? This, this, and this. And I will transfer this to the right by subtracting from both sides 2v1v2w1w2. So what I have is v2w1 squared squared plus v1w2 squared squared minus 2v1v2w1w2. And it should be greater than equal to 0, right? I just wrote it from right to left. So this is on the right, and this used to be on the left, but I put it also on the right with this minus sign. Now, what is this? Is this right? No, v1w1 minus v1w2 squared of this, which is this, minus 2 product, which is this, and squared of this, which is this. And this is an obvious inequality, because this is a square, so it's greater than equal to 0. Now, so from something which I have to prove by squaring a couple of times and reducing whatever is necessary, I came to an obvious thing. Now, is everything reversible? Well, almost, because as I said, in a couple of cases, I said, OK, let's just not think about negative sides. I probably should go much more rigorously if I would really want to prove it in all the general cases. I can start with this, and then gradually go backwards and reduce whatever the different cases of negative coordinates or something else. I probably could have done it a little bit more rigorously. But in any case, what I would like you to see in this particular case that I have used a purely algebraic approach to something which seems to be like a geometric problem, blanks of the vectors. In any case, it doesn't really matter that it's much longer. It's not for achieving the results. The purpose of this is not really to prove. I can prove it geometrically much simpler using the triangle inequality. But then again, how do they prove the triangle inequality? I mean, that's also a separate theme. So that's why it's always interesting to approach the problem from different sides to get the same result. So this is a mental exercise rather than something which you have to remember, anything like that. But if you think about this, everything I'm talking about here is basically your mental exercise, just to make your brain functioning. And it doesn't matter how you use it, what you use it. It's all the same. It's all to train your brain to think creatively. Well, that's it for today. These whatever six problems are very, very easy. They're basically based on definitions of whatever the concepts we were talking about. But why don't you just go through these problems again, just by yourself? They are presented on the website unison.com. If you will go to Vector's pot and the topic is vector arithmetic, that's the problems where these problems are presented to you. So thanks very much, and good luck.