 Hi, I'm Zor. Welcome to Unizor Education. Today we continue talking about certain less standard problems, in this case in geometry. This particular lecture is part of the course which is called Mass Plus and Problems. It's a continuation of the course which was called Mass for Teens. All the courses presented on Unizor.com. This is a totally free website, no ads, no strings attached. And there are a few courses there like Mass for Teens. This one, Mass Plus and Problems. There is a Physics for Teens course. There is a course called Relativity for All. So, again, everything is absolutely free. Use it as well. For all those curious and seekers for knowledge. Okay, so, Geometry 04. So, this is the fourth lecture of Geometry in this Mass Plus and Problems course. Today we will talk about 3G, three-dimensional space in Cartesian coordinates. And we will talk about planes and vectors which are normal, which is perpendicular to these planes. So, it's basically like this. If you have a vector which is normal to a plane, then what's the equation of the plane in Cartesian coordinates? And vice versa. If you know the plane, how to find the vector which is normal to this plane, perpendicular to this plane at certain point. Okay, so let's start from the simplest problem. Let's say you have certain vector in three-dimensional space. This is point P with coordinates A, B and C. What I'm interested is an equation of the plane which goes through the origin of the coordinates. So, some kind of plane like this, which is perpendicular to this vector. Very, very simple problem. How can I solve this? Okay, let's assume that we have some point Q, X, Y and Z on the plane. It's supposed to basically satisfy the equation of the plane which we are looking for. Now, consider these two vectors. This is O. Vector OP, which is ABC, and vector OQ, which is XYZ. So, point O is origin of coordinates, and I was saying that the plane should go through this point. And point Q is chosen on the plane itself with coordinates XYZ. So, the whole vector OQ belongs to this plane, obviously, because both beginning and the end of this vector belongs to this plane. Now, vector OP is by assumption supposed to be perpendicular to this plane equation of which we want to know. Now, the plane is perpendicular to vector, which means every line on the plane is perpendicular to this particular vector OP, including this line. Including this vector OQ. Now, let's use the vector algebra. What do we know about perpendicular vectors? We know that their scalar product in Cartesian coordinates is equal to zero. So, which can be written as A times X plus B times Y plus C times Z is equal to zero. This is OP scalar product OQ in Cartesian coordinates, as we know, from vector algebra. But look, this is what we're looking for. This is an equation. Every point XYZ on the plane must satisfy this particular equation. So, we found this is equation. If you have this vector A, B, and C, and you know that this is vector perpendicular to the plane which goes through the origin of coordinates, then this is an equation of the plane. That's it. Problem solved. Next problem. Next problem we will do slightly differently. It's exactly the same problem, but we don't want this plane to be going through origin of coordinates. Instead, I would like this plane to go through the far end of this vector OP. So, that's where the plane is. So, from origin of coordinates, this vector goes to point ABC. So, ABC belongs to this plane, the equation of which we want to know. Okay. How to do this? We will do very similarly. We will choose point Q here. This is point Q. This is point Q. Now, what we know is that vector OP is perpendicular to PQ. OP is perpendicular to PQ. What's PQ? Well, PQ vector is OP minus OP. This vector is a difference between this and this. Or, if you wish, OP plus PQ gives you the OQ. So, OQ is, as usually, this is a point x, y, z. We don't know equation, but we're just choosing any point x, y, z. So, let's just express these two in coordinates form. OQ is x, y, z. OP is ABC. So, the difference between them is x minus a, y minus b, z, z minus c. This is Cartesian representation of this vector, of vector PQ. Now, OP is ABC. So, Cartesian, so, scalar product, these two, of these two vectors is what? a times x minus a plus b times y minus b plus c minus z minus c equals to zero. Okay, we've got this equation. We can just express it slightly differently, which is ax plus by plus cz equals to x square b square. So, it will be ax square plus b square plus c square. So, these are all with a minus. So, I put it to the right. So, this is an equation of the plane, which goes through the end of this vector OP, which has coordinates ABC. Now, point ABC is supposed to be part of this plane, which means it's supposed to satisfy this equation. If we use ABC as x, y, z, well, let's see if it satisfies. If x equals a and y equals b and z equals c, we will get exactly a square plus b square plus c square equals to a square plus b square plus c square, which means point ABC satisfies our equation. Just to check, just to check. But basically, this is equation. So, if you have a vector ABC, then this is equation of the plane which goes through its far end and perpendicular to this. Next problem. Next problem will be in reverse order. In these two problems, we have the normal vector to a plane and we want it to find out the equation of the plane. Now, we'll do vice versa. Let's say we want to find the vector which is perpendicular to a plane. So, let's start with a simpler case. What is our simpler case? So, again, we have a system of coordinates. We have a certain plane and let's start with a plane which goes through zero and it has an equation. ax plus by plus cz equals to zero. Well, basically, this is the problem which we have kind of solved. All we have to do is reverse our original logic. So, let's just say that we have to find out what is the coordinates of point p. Let's say coordinates are p, q and r, such that op is perpendicular to this plane. Now, the problem which we used, the first problem was what? So, if this is given coordinates of the point and then we are using this vector op as a normal vector, perpendicular vector to a plane which goes through origin of coordinates, then its equation is supposed to be like this. r times z equals to zero. Now, we have an equation this, but, hey, what we have to do is we just put p is equal to a, q is equal to b and r is equal to c. So, if this equation is given, then using this particular line, this particular point p, we will get exactly the vector op about which we have already proven that the equation should be this. So, we'll just reverse all the logic basically. So, the first problem was knowing this point, find these coefficients. Now, it's just completely in reverse, but everything is reversible, everything is completely symmetrical. Knowing this, we can find this equals to a, b, c. So, if you have this equation which is an equation of the plane which goes through the origin of coordinates, then the coefficients a, b, and c define basically the normal vector. Well, in this particular case, we can say that not only these coefficients, but any proportional to these coefficients like p is equal to k, a, q is equal to k, b, and r is equal to k, c, where k is any coefficient not equal to zero. It will also satisfy this equation, because this vector is perpendicular to the plane and shorter vector or longer vector, they're all normal or perpendicular to the plane. And so, this is kind of a general expression of all vectors which are going through zero to some point and it's perpendicular to the plane. So, that's it, that's easy. Again, just use these coefficients and they will give you the vector. It's very, very simple thing from vector to get to the plane which goes through zero or from plane to the vector. All you have to do is just reuse the same coefficients. Okay, and the last problem is slightly different. What if our plane is not actually going through the origin of coordinates, but it's somewhere else, somewhere else. Now, general equation would be ax plus by plus cz plus dz is equal to zero. So, a, b, c and d can be any coefficients and that basically goes as an equation of any kind of a plane in the three-dimensional space. Now, given this particular plane, I would like to find the vector which goes from zero to some point on the plane perpendicularly to the plane. And we will use exactly the same logic as before, but again in reverse order. So, let's say this is point p with coordinates p, q and r. I choose any point q with coordinates x, y, z and vector p, q supposed to be perpendicular to o, p. Now, p, o, p vector is unknown p, q and r. Vector o, q is x, y and z, which satisfy this equation. And I know that they are perpendicular to each other. But again, no, sorry, perpendicular to each other would be p, q. So, p, q would be x minus p, y minus q, z minus r. And vector p, q is perpendicular to o, p, p, q, they are perpendicular to each other because p, q completely within the plane and o, p is supposed to be perpendicular. Okay. How can I use again this thing the same way? Scalar product should be equal to zero, right? So, well, Scalar product of this, this is p times x minus p plus q, y minus q plus r, z minus r equals to zero. So, our equation supposed to be p, x plus q, y plus r, z minus p square plus q square plus r square equals to zero, right? p square with a minus, q square with a minus, r square with a minus, p, x, q, y, z. Okay, this is our equation. Now, let's compare it with this equation. Well, it's very, very similar. The only problem is if we will let p is equal to a, and q is equal to b, and r is equal to c, this part would be the same. But this is not necessarily d. However, we can always stretch or shrink the vector to get this thing to be equal to d. So, let's just assume that p is not equal to just a, but equal to some coefficient times a. q is equal to some coefficient, the same coefficient times b, and r is equal to the same coefficient times c. It will not change this thing, basically, because it's zero. But now, what we can say is if we will substitute these p, q, and r into our equation, we should have that k, a, x, what our equation is, plus b, sorry, instead of this, q is k, b, k, b, y, plus k, c, z, minus square of these, which is, well, it's a square plus b square plus c square times k square, right? This is equal to zero. So, find k from this, because this now is equal to d, from which k is equal to what? From which k is equal to, now, a, x, plus b, y, plus c, z, it's equal to, so, this is supposed to be k d. So, minus k, if it's equal to minus k d, everything will be fine, right? So, k should be equal to, if I will cancel this, minus d divided by a square plus b square plus c square, right? So, if I will use this coefficient k and define p, q, and r using this coefficient times a, b, and c, everything should be fine. Well, let's just check it out. So, first of all, we can check if point p, q, r belongs to the plane. So, if we will put p equals this, q equals this, r equals this, where k is equal to minus d divided by a square plus b square plus c square, then we will have the point p, q, r, which lies on the plane, which means it's supposed to satisfy this equation. Well, let's just check. So, if my point p, q, and r is substituted instead of x, y, z, we will have a times p, which is a, which is minus d a divided by a square plus b square plus c square plus by plus b, and then instead of y, we should put q, which is k b, which is, again, minus d b divided by the same a square plus b square plus c square plus c minus d c divided by this thing, and plus d. Is it equal to 0? Well, actually, yes, because if you will d take outside, you will have minus a square minus b square minus c square, and divided by a square plus b square plus c square, so you will have minus d and plus d will be equal to 0. So, we have proven that the point p belongs to the plane, and the fact that it's perpendicular, so the vector op is perpendicular to the plane basically follows from the fact that these are proportional coefficients, and we know that abc is always perpendicular, which means k, a, kb, and kc will also be perpendicular. So, vector op, p belongs to the plane, and op vector is perpendicular to that plane, and the plane is given by this general equation. So, again, from equation, we came with the vector, which goes from 0 to the point on this plane, and perpendicular to the plane. So, we basically did both sides of this thing, from vector to the plane, or from plane to the vector, and we have chosen two different kinds of planes, planes which go through the origin of coordinates. It's just a little bit easier piece, or general plane, which is somewhere in the space. Well, that's it. Now, every lecture on the website Unisor.com has the textual part, so to speak, which basically is written like a textbook. So, I suggest you to read the description of this lecture in the corresponding text. So, to go to both lecture and the text, you should start from Unisor.com, choose the course, math, plots, and problems. Then the submenu would be geometry in this case, and this lecture is geometry 0, 4, and the next menu you will see. Not only the notes are on this website, in other cases, well, not in these cases, because we are actually talking about problems, but in the main course, which is basically prerequisite for this called math for teens, there are many exams, which you can take as many times as you want, to basically to check yourself. So, I do recommend you to go to the website and read notes, and if you have any problems with whatever I was just talking about, like vectors, scalar product, etc., go to the prerequisite course, math for teens, where everything is explained in corresponding chapters of the course. That's it for today. Thank you very much, and good luck.