All right, hi everyone. Welcome to our lecture on the equation of planes in space. So we're going to talk about a plane, we're going to talk about it in space. What does that mean? We're in 3D. So before we had a line, hello line, there you are. And it's just a generic line, so we have to think about what did we need to describe a line in space? Well, we needed some point on the line and then we needed some direction vector v that was either parallel to or on the line itself. You give me a p and a v and I'll give you the equation of a line. Okay, so if I want to translate that information as to describe some plane in space, you can imagine there are lots and lots and lots of them. So let's draw some plane in space. Maybe on the floor just a little above. What do we need to do that? How can I uniquely describe all the points that lie on this plane? All right, so first off, we're going to use a point again, so some point p on the plane. But if I just tell you I'm thinking of a point, there's infinitely many planes that go through that point, right? Imagine a point on space, holding your finger and then rotate a piece of paper, rotate your hand, rotate some plane through it. So in order to talk about one particular plane, I need to tell you which way a vector is pointing off the plane. Not just any vector, a vector that is orthogonal to the plane or right angle to the plane. And that vector that is right angle to the plane is called our normal vector. In my mind, I always think about this vector sort of stabilizing the rotation. It points if you will, from the plane and that point and normal vector uniquely define the plane. Think about that for a minute. If I tell you I'm thinking of a point and I have a vector and then I want all the points that are perpendicular orthogonal to that vector. We're all thinking about the same thing. We have uniquely described the plane so that vector is extremely important, the one that is orthogonal to the plane. It's called a normal vector, so let's define that as well. So a vector that is orthogonal to every vector in the plane is a normal vector. This is going to be key. Okay, so let's try to find the equation of a plane. First, let's draw a picture, sort of motivate what we're about to do. So I have some plane in space. I have some point P that lives on the plane. Maybe I have some point Q that's also on the plane. You can always draw the vector that is between them. They will go from P to Q, and then I want this normal vector that comes off the plane. Okay, it doesn't matter the size, it could be unit normal. Those are always nice, but it could be just any vector. It doesn't quite matter which one it is. So to find the equation of plane and given another point, it's easy to find points, with the direction vector PQ, I wanted to find any generic point as a point such that the normal vector and the direction vector QP or from P to Q, they're right angles. So how do we describe right angles and vectors? We use dot products. So I'm looking to define the normal vector dot the vector QP, or let's say PQ since we're starting at P in my picture. P to Q, I want that to be zero. Okay, well, how do I find the direction vector between two points? So let's keep playing with this equation, its head minus tail. So remember, this is Q- P and that's all still equal to 0. So friendly reminder, when the dot product is equal to 0, the vectors are orthogonal to each other. So then I say, well, that's n Q. And that Q is the position as a number, remember, this is a dot product, scalar product. So n dot Q, so Q consists of three values, we do a dot product there, would be equal to n dot P. This equation again, any one of these forms really will do just fine. This is called the vector equation of the plane. This is a perfectly fine way to describe, assume you're given PQ and n, an equation of a plane. Any point that satisfy these two things, any one of these equations will live on a plane. This is one way to describe a plane. Let's write another way to do it. Let's write out the scalar equation of the plane. And so what are we going to start with? We're going to actually write out some numbers here, so I have a point P, x naught, y naught, or z naught, and have some normal vector a which is a, b and c. And I want to have a generic point x, some point x, y and z, be some other point on the plane. So we're going to let x, y and z, be some generic point on the plane, x, y and z. Okay, so given some other generic point, you can think of it called Q if you want, doesn't quite matter. Or whatever your other point is, we're going to let these variables roam. So P is given, the normal's given, I want that Maybe we'll use the first one here or the second one I should say. That the normal- so I want a, b and c dot, the difference head minus tail. So I'll just write that out as one vector x minus x, naught y minus y naught and then z minus z naught. I want that dot product equals 0. Let's put it into an actual equation. This gives a a x minus x naught plus b y minus y naught plus cz minus z naught is 0. This equation is usually rewritten if you distribute the a and you get ax plus by plus cz and then you get a whole bunch of constants. The variables are not used anymore so we have ax naught, really minus ax naught minus by naught and then minus cz naught equal 0. These constants are usually just all lumped together as one big constant which we call D just to keep the pattern going, and it's pretty common also these capital letters. Although it doesn't quite matter, so we have Ax plus By plus Cz plus some constant D0. This is the scalar equation of the plane. There's a couple things to note about this here. Whenever you're handed this equation, remember it is the generalization of the two dimensional version of a line. If I cross out the Cz part. If I hold my thumb and ignore it, just cover it over it, then I get the equation of a line. So we did in fact generalize equation relating to equation of a plane and the normal vector can always be either used to create this equation or write off this equation as the numbers that are coefficients on the variables. So A, B and C, this is called the scalar equation of the plane. One thing that we're going to do in the exercises and it's pretty common to do is we always start with a couple points. We assume they're non collinear, so P, Q and R and they all have their numbers associated to it. You can imagine it's nine numbers. I want to give you a recipe for finding the plane that contains them. So find the plane that contains three points. And the points will be given such that they're not collinear. You can imagine if I did draw these on a single line, then there would be many, many planes that contain them, but three points that are not collinear. Then there's one unique point, I'm not going to work this out with numbers at the moment because it gets kind of messy. You can imagine three points with three coordinates each. That's nine numbers, it just kind of get lost in the arithmetic. So what I want to do for a minute is sort of zoom out, see the world from space, and just go through the process. So first, I will write down the recipe and then we'll go actually cook the ingredients later. Let's pick some sort of base point and we're going to find the vectors that connect the two points. We know how to do that, this is head minus tail method. Just some easy subtraction, so we're going to find the vectors PQ, and we're going to find the vector PR. Head minus tail subtraction, we can do that. Now, the plane that contains these things is going to- right, we're going to need a normal vector to describe this plane. So how do we find a normal vector given two other vectors? How do I find a vector that's orthogonal to both? Let me draw a picture here. Hopefully this gives away what I'm after. If I draw the vector that's orthogonal to both, I need a cross product and that's cross product is going to find the normal vector. Remember that normal vector helps make the equation, it's three out of the four numbers you need in the equation. So the normal vector becomes PQ times where this is a cross, PR, cross product is a process, we set out the diagram. We go through it, but we get back some vector and then I have almost everything I need. So I have my point. So I'm going to use point P, although any point will be fine. I'm going to use my normal vector and then I plug into the equation Ax plus By plus Cz. So remember your normal vector. The numbers you get, these are A, B and C, these are coefficients on the variable plus D is 0. So I have A, B and C and I say, wait, well how do you use P? Well what we're going to do is you're going to plug in for P. This will be your x, your y and your z. So you have six of the seven variables that appear in this formula. You have AXBYCZ, you just don't have D, so you plug into the equation to solve for D. Now you have everything and last but not least, you can return the formula in full generality and now you have Ax plus By plus Cz plus D is 0, so ABC and D. These are actually B numbers. XY and Z will be variables. Again, it fills up a whole slide just to go through the steps. If I were to put numbers in it and I used to do an example first, but it just got so hairy with the arithmetic that we just kind of give the formula now. Stare at this, let this sink in. Pause the video if you want, go through the steps and it should make sense how you're constructing the vectors. Constructing the normal vector, of course cross product gives orthogonal vectors. So we're going to use that vectors, and then we would go through and solve this. We will practice with numbers, but right now just absorbed go through and understand that the recipe. Couple other definitions that I want to point out as we talk about planes. The first ones were going to say that two planes are parallel. If their normal vectors are parallel, let's think about this for minute. How do we know when two planes don't touch? What does it mean to have two parallel planes like the roof and the floor? The ceiling in the floor? Well, we want to describe things as normal vector turns out to be pretty important. If I have some normal vector one and some other normal vector and two that are right angles, then the planes will never touch, if these two vectors point in the same direction. So we'll say two planes are parallel if their normal vectors are in fact parallel. Okay, so what happens if you don't have parallel planes? You have some planes that actually do touch. Think of it like a book of some sort. So what does it mean when two planes intersect? Well, one thing we're after is the angle that they form, so we can find this. This is like a formula, I guess definition combo here. So the angle between two non parallel planes, is the angle formed between their normal vectors. Once again normal vector is playing a major role in the description of planes. Let's think about this for a second. I have some normal vector to the first plane, call it n1. I have some other normal vector to the second plane, call it n2. And the angle that these vectors form is exactly the same as the angles formed by the plane. Well, how do we find the angle between two vectors? Hey, we have a formula for that. That's Theta equals arc cosine, it's coming from the dot product formula. The alternate version of the two normals dot each other, and then divided by the product of the magnitude of each one. Remember, this formula comes from the dot product. So a simple formula to just work through and just plug and chug and get your numbers. But the idea of this, the visualization this is really important and I want you to see how we're using these formulas from prior to get new information. Last but not least, do another definition formula will say the distance D from some fixed point X not, y not and Z not to a plane. So plan we defined by its equation will say ax+by+cz +d= 0, is given by D equals the absolute value of ax1+by1+cz1+d divided by the square root of a squared plus b squared plus c squared. Important thing to note here that we have an absolute value. The reason for absolute value of course that distance can't be negative, so I need the top to be positive. I wrote ones a subscript instead of zeros on the point. So let me just clarify that. So I take a, b, and c. Remember that these are given, I know the equation will plane. So a, b,c are given. I take the point itself when I plug all that in and I also plug in d. So I plug in all the numbers I have, and I take the absolute value and I have my numerator. And then I take the square root of a, b, c squared. Friendly reminder, what is this? Is just the magnitude of the normal vector to the plane. So once again this is coming in as well. Just a little picture to help motivate this. If I have a plane, and some point I want to find the distance to it. Be careful when we find distance, right, there's lots of ways to sort of travel from the point to the plane. When we say the distance, it's always implied that we want the minimal distance. What's the shortest path? The straight line that connects the point to the plane, that will always be defined as the distance? Let's do a couple examples to work through and some numbers and actually see these equations in use here. So the first one up is find the equation of the plane through a certain point. So let's call like IP. And perpendicular to the vector (-2,1,5). Perpendicular to the vector, so what does that mean? Perpendicular to the vector hey, wait a minute. That is the normal easy questions are always nice and easy. They give you exactly the information that you need. So I have my normal and I have my point as we get into more complicated examples, it will be a little harder to get the normal vector. Perhaps a little hard to get the point, but once you have this information well then you're ready to go. From the reminder, the equation of plane Ax+By+C Cz + D is 0. Your A, B and C they always come from the normal vector, so we can start plugging in here -2x + 1y, which is y, + 5z + D is 0. I have almost everything I need, I get to plug in my point now, the point is always x, y, z so I can say -2(6) + 3 + 5(2) + D is 0. This will allow me of course to solve for D. When you do that, you get -12 + 3 + 10 + D is 0. So -12 plus, this is like 13- 12 which is 1 + D is 0, so D equals -1. Put it all together, you get Ax + By + Cz- 1 = 0. That is our equation of the plane, you could also set of course equal to positive 1 it does not matter. So put it all together we have -2x + y + 5z- 1 = 0. That is our equation of the plane through the point (6, 3, 2) and perpendicular to the vector (-2, 1, 5). Quick check shows that -2, 1 and 5 are still the coefficients of x, y and z and if you plug in 6, 3 and 2 it will satisfy the equation. And therefore the point P is on the plane. Let's do another example. Find the equation of the plane with the point (1, -1, 1) and contains the line x = 2y = 3z. So a friendly reminder this line is presented in symmetric forms. This is a symmetric equation of a line and the point (1, -1, 1) that's our P. So our goal here is to try to find that normal vector. Once I find the normal vector, I'm sort of back in the prior case and I can go through the same process and find the equation of a line. I was like to draw a picture when I can is 3 dimensional, so it's good to visualization practice. I have a plane, I have some point P, again don't worry about drawing to scale, just get the general idea, and then I have some line that's in the plane. So the first thing we can ask ourselves is is the point on the line. So when I draw my line, am I going through this point or am I off the point? Well, how do you check? We can plug in this x which is 1 from my point equal to 2y, no of course not, so it's 1, -1, so it's not. So the line I'm drawing is just some other line L that just goes through in the plane but not actually containing the point. OK, so let's describe this line in different way, maybe we can write it, perhaps a little easier. Remember symmetric equations is when you solve for the parameter, so like x = t, y = 2t, 3z = t, everything is equal to t. So if I want to describe this line in perhaps the more traditional way I have that x = t, so this would just be t and y equals would be t/2 and then z would be t/3. So I can put this back and perhaps and easier way to do it. All right so let's think about what I need, I need a normal vector so I can sort of use the formula or process for finding three points on a plane. I already have point P, I have to find two other points on line. Well, how do you find 2 points on a line? Let's pick values of the parameter, lets pick t = 0 and how about t = 1? Why not? When you pick t = 0 you get l(0) equals <0, 0, 0>, that's a nice little point there. So pick the origin and if you pick l(1), when t is 1, l(1) that just becomes <1, 1/2, 1/3>. Now a little word of advice here, if you're doing these things, especially if you're working without a calculator 1/2 and 1/3, they're fine numbers with their fractions. They're not something I really want to deal with, so I'm going to be a little clever here and use 6 as my parameter. Why is that? Because then I just cleared denominators. So now I get <6, 3, 2> and that's a little bit nicer number. So t is 1 even though it's a feel good number for parameter kind of gives us fractions. Let's use some other points here. So now I have point Q and now I have point R. So I actually found more than I needed, I'm not going to use this other point, so now I have P, Q, and R. I have three points on a plane and remember the process, should we go through it? Let's go through it. So if I want to find P and R, so we use P as my base point. Let's go find the vector from P to Q and friendly reminder, this is head minus tail, so this is <0, 0, 0>- <1, -1, 1> and that's the vector 0- 1, 0 minus minus 1 is positive 1 and then minus 1 again. And if I want to find sets P, Q. If I want to find the other vector from P to R, again it doesn't matter which one you pick, they all got to be the same. We do head minus tail, so it's <6, 3, 2>. - 1- 1 and then 1 and I get my other vector connecting those two points in the plane. So 6- 1 is 5, three minus minus one is 3 + 1, that of course is 4 and then 2- 1. Carry the π / 13, that's also so I have these two vectors that live on the plane. How do I get a normal vector? I have to take a cross product, so let's see if we can do it over here. So I want the normal vector, so I set up i,j and k minus one 1- 1, five, 4 and 1. Here we go take I cover up that column and then do your ex pattern so you get 1 minus minus 4. Let's 1 + 4 and that's five we do, minus j, we don't forget about the minus on the j vector and then we do cover up the middle column. We have minus 1 + 5- 1 + 5 is 4 and then we have our k cover up less column we have minus 4- 5. And that's minus 9 so, our normal vector in this particular case here take it out of my jk notation is 5- 4- 9. Remember we just did a lot of rithmetic negative numbers, subtraction, very easy to make a silly mistake, so let's just do a quick check. This normal vector should be orthogonal to the two vectors we found. So we take a dot product with the first component, we have -1 and 5 this -5- 4, so it's -9 and then + 9 is 0, so it's fantastic I have a dot product of 0 between these two. If I do for the second one I have 25- 16, lets positive 9 and then minus nine is 0. So I take the dot product just a quick check after I do all this work I want to be confident that I actually have the right vector. And then last but not least since this goes through the origin I heck, let's pick this to be our point, we going to use so will call this to be our point. We going to use the equation so I can just set up the equation ax + by + cz and then d is going to be 0 cause you plug everything in so you get 5x +- 4 y+- 9z would be easier. So, this is the equation of that plane, it goes through the point(1, -1,1) and check, make sure it works and it contains the line x equals 2y + 3z. And again, how would you check that? You can replace every cnx, leave it as x service area or you put as to why you can solve for these things Substitute and go and check. So a little more complicated example science that we went through the rest of you there with actual numbers to get this one to work let's do another one. Find the equation plane that passes through the line of intersection of the planes x- z = 1 and the other plane y +2z = 3 and is perpendicular to x= y- 2z=1. Lot of information there, actually hand you a lot of information, so even though it's a mouthful to read. And it looks kind of scary at first, just feel like there's a lot of information there, so let me get. So, first of all, perpendicular to the line well, hello, perpendicular to some plain that means that this is handy me the normal vector. The normal vector can be read off the scalar equation of plane is the coefficients on the variables. So my normal vector that I'm after for the plane that they're trying to find is 1,1 And -2, well, that's really nice. That's really nice because that's the normal vector, and that's usually the thing that's difficult to find. All I need now is a point on this plane, and then I'm going to go. I have a point, I have my normal and I can find the equation that I want. So, the question is how do I get a point on this on this plane? So let's see I have to use the other part of the given here. So, find the equation playing that passes through the line of intersection. So, I guess I should find the line of intersection of these two planes and let's see if we can do that, so these two planes, so how do I want it? What do I do when I want to find intersection things? I used the two equations, so let's play with these equations for a minute I have x- z = 1 and y +2z = 3. So if you look at these you say, well, wait a minute, everything is x. There's something missing in both of ' which means that I can describe z in terms of x. So this says that z = x- 1, and I can also describe y In terms of z, so I have that y= 3- 2z, but z was z minus 1 as well, so that's =3- 2( x- 1) and I get 3-2x +2 or 5- 2x. So, I've described in terms of xy in terms of x, so it seems like if I let x be the parameter. I'm basically done so I would let x = t, and you have l of t to be, where my x is going to be t, my y term is 5- 2t, and then my z is t- 1. Remember these parameters are dummy variables. So once you write each variable in terms of one single variable, if three variables, right, in terms of one, you can replace that variable with the parameter and I get a very nice equation of line. Now remember what I'm after? I did all this work because I need some point. I need some point. We'll give an equation of a line. How do you find some point? Pick your favorite value of the parameter and just plug it in. t = 0 tends to be a nice one. Just to make the arithmetic easy. And this gives us the point corresponding to 0, 5, and -1. So we'll use that as my point P. Let's plug it all into my equation. We will write it as, so it's like I'll write the one to see it. So normal vectors always coefficient. We do x minus the point, so I'm using the form, I guess I'll write the formula first, so it's A x- x knot, + B, y- y knot, + C, z- z knot = 0. When I write it this way, I don't have to solve for the D. Remember D was the collection of all the other terms. This is perfectly fine way to do it, so it's good that you see both ways. So A is 1. Remember worth, where do A come from? A, B, and C are your coordinates or components that sort of looking for, components of the normal vector. So we have 1 for A, we have 1 for B, y- 5, and then C was- 2, so- 2, Z minus,- 1 is Z + 1, and that's all equal to 0. Please don't forget the equality sign in the equation of a plane. A lot of students do that. Let's clean this up just a little bit. So 1 times x- 0 is just x, and then 1 times y- 5 is just y- 5, and then- 2z- 2 is 0. We can combine the constants we get y + 5- 2z is equal to, we move the constants over equal to 7. If you're write as- 7 = 0, it does not matter. But now we have the equation of the plane. We'll stare at this. Convince yourself we do have equation of plane, passes through line of intersection of two planes, and it is perpendicular to this line. So lot of pieces going in to finding the normal or a given point. But once you do all that work, once you have all that, you're all set to go to find the equation. All right, great job on these examples. Hope you had fun sort of visualizing and playing around with things in 3D, and it was a lot more variables, and can be a little tricky. But you'll see that this really sets us up to do all the calculus that we want to do, all the analysis, and just found fantastic foundations to go forward in your mathematical career. So great job in this video. I will see you next time.