Alright. 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 three D. So before we had a line, hello line. There you are. And it's just generic line. So we have to think about what did we need to describe a line in space? What 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 and 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, on this plane? Alright, 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. Like hold it in 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 or normal vector in my mind, I always think about this vector, sort of stabilizing the rotation. It points up if you will from the plane. And that point in normal vector uniquely define the plane. That point a normal vector uniquely defined 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 plane. We have uniquely described the plane. So that vector is extremely important. The one that is orthogonal to the plane. It's called the normal vector. So it's defined 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 to sort of motivate or 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. Doesn't matter the size. It could be a 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 a plane and given another point is easy to find points with the direction vector PQ. I want all. 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 P Q. Since we're starting a 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. It's head minus tail. So remember this is like q minus P and that's all still equal to zero. So a friendly reminder when the dot product is equal to zero. The vectors are orthogonal to each other. So then I say, well that's n Q and not Q. This is the position as the number remembers the dot product scalar product. So n dot Q. So Q consists of three values. We could 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 P. Q and n. An equation plane. Any point that satisfy these two things? Any one of these equations will live on a plane. This is one way to describe this 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. Ex not why not? Or Z not? 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 there's some generic point on the plane x, y and z. Okay, so given some other generic point, you can think of it call Q if you want. It doesn't quite matter whatever your other point is we're only these variables rome. So p is given the normal is given, I want that the 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 not y minus. Why not? And then z minus Z not. I want that dot product equal zero. Let's put it into an actual equation. This gives that a x minus x not plus b y minus y not plus c z minus z not 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, right? The variables are not used anymore. So we have ax not really minus ax not minus by not and then minus cz not equal 0. These constants are usually just all lumped together as one big constant, which we call d just keep the pattern going. And it's pretty common also to use 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 the line. If I cross out the C. Z part, I hold my thumb and ignore it just covered over it. Then I get the equation of a line. So we did in fact generalized equation into equation of a plane and the normal vector can always be either used to create this equation or read 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 called points. We assume they're non linear. So P. Q. And R. And they all have their numbers associated to it. You can imagine 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 co linear. You can imagine if I did draw these on a single line, then there'll be many many planes that contain them. But three points that are not linear. 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's just kind of get lost in the arithmetic. So what I want to do for a minute to sort of zoom out, see the world from space and just go through the process. So first we'll 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 PQQ 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 is orthogonal to both? Let me draw a picture here, hopefully, this gives away what I'm after. If I draw the vector that is 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 there's a cross PR cross product is a process. We set up 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 a minute how to use P, 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 Ax, By, Cz 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 A, B, C, and D, this will actually be numbers, x, y, and z will be variables. It fills up the whole slide just to go through the steps, if I were to put numbers in it, they always used to do an example first. But it just got so hairy with the arithmetic that would just kind of get the formula now stare at this. Let this sink in positivity 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 products gives orthogonal vectors, so we're going to use that vectors, and then we will go through and solve this. We will practice this with numbers, but right now just absorb, go through and understand the recipe. A 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 they're normal vectors are parallel, let's think about this for a 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 on 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 they're 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 vectors 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 in one, I have some other normal vector to the second plane, call it end two. 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 data equals arc cosine is 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, we'll do another definition formula will say the distance D from some fixed point x note, y note, and z note to a plane. So a plane will be defined by its equation, we'll say Ax plus By plus Cz plus D is 0 is given by D equals the absolute value of ax1 plus by1 plus cz1 plus 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 is 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 of a plane. So a, b, and c are given, I take the point itself and 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? 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 at 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 it's called like IP. And perpendicular to the vector negative 2, 1, 5 perpendicular to the vector. So what does that mean perpendicular to 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'll 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 to the equation of a plane, Ax plus By plus Cz plus D is 0. Your A, B, And C, they always come from the normal vector, so we can start plugging in here minus 2x plus 1y, which is y plus 5z plus 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 negative 2 times 6 plus 3 plus 5 times 2 plus D is 0, this will allow me of course to solve for D. When you do that, you get minus 12 plus 3 plus 10 plus D is 0, so minus 12 plus, it's like 13 minus 12 which is one plus D is 0, so D equals minus 1. Put it all together, you get Ax plus By plus Cz minus 1 equals 0. That is our equation of the plane. You can also set of course equal to positive 1 it does not matter. So put it all together we have minus 2x plus y plus 5z minus 1 equals zero. That is our equation of the plane through the point 6, 3 and 2. And perpendicular to the vector minus 2, 1, 5. Quick check shows that minus 2, 1, 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 a plane to the point 1 negative 1, 1 and contains the line x equals to y equals 3z. So friendly reminder, this line is presented in symmetric form. This is a symmetric equation of a line and the point 1 negative 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 always like to draw a picture when I can, it's three dimensional. So it's good 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 a 1 negative 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, okay? So let's describe this line different way. Maybe we can write it perhaps a little easier, remember symmetric equations when you saw for the parameter. So like x equals t, y equals 2t, 3z equals t. Everything is equal to t. So if I want to describe this line and perhaps the more traditional way I have that x equals t. So this would just be t and y equals would be t over 2 and then z would be t over 3. So I can put this back and perhaps an 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, the process for finding three points on a plane. Already have point P. I have to find two other points on the line. Well, how do you find two points on the line? Let's pick values of the parameter, let's pick t equals zero. And how about t equals 1? Why not? When you pick t equals zero you get l of zero equals 0, 0, 0. That's a nice little point there. So we'll pick the origin. And if you pick l is 1 is t is 1 l of 1 that just becomes one half and one third. Now, a little word of advice here if you're doing these things, especially if you're working without a calculator, one half and one third 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 clear denominators. So now I get 6, 3 and 2. And that's a little bit nicer numbers. 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. Friendly reminder, this is head minus tail. So this is 0, 0, 0 minus 1 minus 1, 1. And that's the vector zero minus 1,zero minus minus 1 is positive 1 and then minus 1 again. And if I want to find that' PQ, if I want to find the other vector from P to R. Again, it doesn't matter which one you pick, they all work out to be the same. We do head minus tail. So it's 6, 3 and 2 minus 1 minus 1 and then 1. And I get my other vector connecting those two points on the plane. So 6 minus 1 is 5, 3 minus minus 1 is 3 plus 1, that of course is 4 and then 2 minus 1 carry the pi divided by 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 1, 1, minus 1. 5, 4 and 1. All right, here we go. Take i cover up that column and then do your x pattern. So you get 1 minus minus 4, that's 1 plus 4, that's 5. 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, plus 5 is 4. And then we have our k. Cover up less column. We have minus 4 minus 5 and that's minus 9. So our normal vector in this particular case here, take it out of i, j, k notation is 5, minus 4, minus 9. Remember we just did a lot of arithmetic 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 negative 1 and 5, that's negative 5 minus 4. So it's negative 9 and then plus 9 is zero, so it's fantastic. I have a dot product of zero between these two. If I do it for the second one, I have 25 minus 16, that's positive 9 and then minus 9 is zero. 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, [INAUDIBLE] let's pick this to be our point we're going to use. So we'll call this to be our point, we're going to use the equation. So I can just set up the equation ax plus by plus cz and then d is going to be zero because you plug everything in. So you get 5x plus minus 4y plus minus 9z would be easier. So this is the equation of that plane, goes through the point 1 negative 1, 1. And check to make sure it works. And it contains the line x equals 2y plus 3z. And again how would you check that? You can replace every c and x leave it as x obviously, you put as 2y you can solve for these things, substitute and go and check. So a little more complicated example science that we went through the recipe there with actual numbers to get this one to work. Let's do another one. Find the equation of a plane that passes through the line of intersection of the planes, x minus z equals 1. And the other plane, y plus 2z equals 3and is parallel to x plus y minus 2z equals 1. A lot of information there. Other enough when there's a lot of information they're actually handing 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 that. So first are parallel to the line. Well, hello, parallel to some plane. That means that this is handing me the normal vector. The normal vector can be read off the scalar equation of the 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 negative 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 good 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 plane? So let's see, I have to use the other part of the given here. So find the equation plane that passes through the line of intersection. So I guess I should find the line of intersection of these two planes. Let's see if we can do that. So these two plane, so, what do I do when I want to find intersection things? I use the two equations. So let's play with these equations for a minute. I have X minus Z is one and Y plus 2Z is three. So if you look at these you say well wait a minute, everything is X. There's something missing in both of them which means that I can describe Z in terms of X. So this says that Z is equal to X minus one and I can also describe Y in terms of Z. So I have that Y is equal to three minus 2Z but Z was X minus one as well. So that's three minus two parentheses X minus one and I get three minus 2X plus two or five minus 2X. So I've described Z in terms of X. Y in terms of X. So it seems like if I let X be the parameter, then I'm basically done. So I can let, so we let X equals T. And you have L(T) to be my X is going to be T, my Y term is five minus 2T. And then my Z is T minus one. Remember these parameters are dummy variables. So once you write each variable in terms of one single variable you have three variables, right? In terms of one you can replace that variable with the parameter and I get a very nice equation of the 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? Take your favorite value of the parameter and just plug it in, T equals zero, tends to be a nice one. Just to make the arithmetic easy. And this gives us the point corresponding to zero, five and negative one. 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 all right the one just so you see it, it's so the normal vector is 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 minus X note) plus B (Y minus Y note) plus C (Z minus Z note) equal zero. 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 one, remember what do they come from? A, B and C, are your coordinates or components. That's what I'm looking for, components of the normal vector. So we have one for A, we have one for B, Y minus five and then C was minus two. So minus 2Z minus one is E plus one and that's all equal to zero. 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 one times X minus zero is just X. And then one times Y minus five is just Y minus five and then minus 2Z minus two is zero. We can combine the constants and we get Y plus five minus 2Z is equal to move the constants over equal to seven. If you write it as minus seven equals zero, it does not matter. But now we have the equation of the plane to stare at this, convince yourself we do have equation plane. Passes through the line of intersection of two planes and it is parallel this line. So a lot of pieces going into finding the normal or a given point. But once you do all that work, once you have all that you are all set to go to find the equation of. All right, great job on these examples. Hope you had fun sort of visualizing and playing around 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 the mathematical career. So, great job in this video. I will see you next time