Linear Algebra Lines and Planes

Welcome back to educator.com, welcome back to linear algebra.0000

The last lesson we continued to talk about linear transformations. Today we are going to take a brief respite from that, talk about something a little bit more practical, something that you have seen before.0003

But we are just going to discuss lines and planes, before we actually launch into the study of the structure of linear mappings.0009

When we get into vector spaces next, but... So some of this that we do today will be familiar, perhaps some of it will be different, and maybe some of the techniques will be a little new.0021

In any case let's just dive right in, so again we are going to talk about lines and planes, okay...0030

... Let's talk about lines in R₂, we know that...0039

... We know that AX + BY + C is the equation of a line, you are often used to seeing this thing on the other side.0053

It doesn't really matter where you put it, believe it or not, it's actually better to put it this way, to have this 0 over here, that way.0060

All of the constants and all of the variable are on one side, and the 0's over on this side, because this idea of homogenous system is going to be very important for us, because remember we discussed homogeneous systems and the conditions under which a homogeneous system has a solution, where the determinant of the particular matrix...0066

... Coefficient matrix is equal to 0 and things like that, so it's often best to write it this way, and it's more consistent when you move on to plane and equations of things called hyper planes in N space, which are just the analogs of lines and planes and the spaces that you know, R₂ and R₃.0085

Okay, well if I have some point P, is XY...0103

And I have actually this is P₁, so X₁Y₁ and if I have a point P₂, which is X₂Y₂.0110

Well, if these two points are on that line, then they satisfy the following, basically just put the X₁Y₁ in for X and Y, so you get AX₁...0123

... + BY₁ + C = 0, actually I don’t need that C₁, because C is constant.0136

And I have A times X₂ + B times Y₂ + XC = 0. So these two points satisfy if they are on that line, they satisfy this relation, right, well let's write all three, let's write the generic version, and these one on top of the other.0148

We have AX + BY + C = 0...0168

... AX₁ + BY₁ + C = 0, AX₂ + BY₂ + C = 0.0179

And now take a look at this, this is a system. We have three equations, we have two unknowns, X and Y, we have some coefficients, we have, let me highlight these in red.0192

We have A, we have B, C, A, B, C, A, B, C...0205

... And then we have the X, the X₁, the X₂, we have the Y₂, the ₁ and the Y, we can actually write this as a matrix times a vector.0214

We were looking... Of course, here are the A, the B and the C, so these X's and Y's are actually the things that become the coefficients, and the coefficient in front of the C is the 1.0227

Let me rewrite this in matrix form as, let me actually do it this way...0237

... X, Y, 1, that's this, this and the coefficient here is 1, X₁ Y₁, 1, X₂Y₂, 1 this is our matrix.0245

And then multiplied by the A, the B and the C, so the A, B and C are the three numbers that we are actually looking for, and that's equal to, well (0, 0, 0), again it's 0 on the right.0259

This is the homogeneous system, this is the homogeneous system, and this is the matrix representation, so given two points and this equation that we know, we can set up this homogeneous system and now we know how to solve this.0271

As it turns out, so these are numbers, these are actual number, these X and Y, they stay as variables, that's why they have no subscript, so because this is a 3 by 3, it's an N by N, and it's a homogenous system.0286

We know that it has a solution, a non-trivial solution, if the determinant of this matrix is equal to 0, that's one of our theorems from some lessons past.0299

Let's go ahead and set up this determinant, now ket me actually write that down specifically, so the determinant, of, in this case let's call them matrix A = 0.0310

It implies that there exists a solution...0325

... An other words ABC can be found, they represent actual numbers, so the symbol for determinant is that straight line X, Y,1, X₁, Y₁, 1, X₂, Y₂, 1.0330

We want the determinant to be equal to 0, so let's expand the determinant and come up with some equation, conditions on X and Y such that this is satisfied.0348

Okay, so let's go ahead and expand, we are going to expand of course because X and Y are variable, we are going to expand along that row.0359

Okay, so I take my, let me actually do this in blue here, so I am going to knock that out, so what I end up with is X times...0368

... Let me see...0386

... You know what let's do this a little bit differently, let's actually stop here, and since we are dealing with X and y, when we actually do some numbers in here, then we will do the expansion.0394

Again you are going to expand along the first row, but this is what's going on here, so when you have two points X₁, Y₁, X₂ Y₂.0404

You can set up this matrix, solve for the determinant and you will actually get your equation of your particular line which is...0412

The original line that we were looking for, so let's actually do an example...0421

Let's take the point, actually let me move forward one, let's take the point P₁ as -1, 3, and let's take point P₂ as 4 and 6.0427

Okay, and now let's set up our...0444

... Determinant is before, we have X, Y and 1, so the variable stays the variable, and of ‘course we want this determinant 2 = 0, we take the first point, it's going to be -1, and 3, and 1, this stays 1 and the second point is 4 and 6, so we put 4 and 6.0449

And now we can go ahead and expand this, and I actually see what the equation turns out to be, so again I am going to expand along the first row, so when I do that, I end up with X times...0469

This determinant (3, -6), okay and I move to the next one, which is going to be a minus, you remember +, -, +, -,+, -, we alternate signs.0481

It's going to be - that entry, that I am crossing out, times -1, times 1...0495

... -1 times 4...0504

... + 1 times -1 times 6 is -6, -...0509

... 12, and we end up with 3 -6 is -3X, here we end up with -5 times of -Y.0520

And this is going to be + 5Y, and then this is going to be -6 -12 is -18, and again this determinant is equal to 0.0533

This is the equation of our line, now yes, can you write this as -3X + 5Y = 18, you can't.0546

Can you write this as 3X - 5Y = -18, yes you can, it’s a personal choice, I personally like to see everything in the same form that I did the mathematics.0558

I don't believe in simplification, simply for the sake of making something aesthetically more pleasing.0569

As it turns out, the more you see if the mathematics, even though it might look more complicated, more symbolism, it actually is more clear, that means there is more on the tale that you see.0578

Again it's a personal choice, I will go ahead and leave it like that, okay, now let's talk about lines in R₃...0588

... We just talked about lines in the lane, now we are going to talk about lines in actual space, slightly more complicated, but not too bad, okay so recall...0601

... That vectors...0617

... And points...0621

... are just different representations of the same thing...0628

... Notations of the same thing, sometimes we think of a point in the space as the point, the coordinate XYZ, sometimes we think it is a vector from the origin to that point.0637

We can write the point let's say P(0), which is some XYZ, okay it's equivalent to, well off, to symbols, the vector P(0), okay.0649

Okay, now we will let...0662

U be a vector in R₃, so this is the set symbolism, U is a member of R₃ and we will let u, will represented by its component form, point form U₁, U₂, U₃.0667

Vector form, point form, okay, the equation of a line in space, equation...0690

... Of a line in space...0700

... Is...0704

... N, we were talking about points in space, we were talking about vectors, I am going to give you the vector representation, then I will go ahead and talk about the breakdown.0708

Is some point is a reference point + some scalar times a vector in a given direction, okay T in R, and I will explain what all of this means in a second, so P(0) is a reference point.0719

It is some point on the particular line; I have to pick some point to start off with, U the vector...0741

... Is the direction from that reference point, either this way or that way...0755

... And again some of this will make sense when we actually do a problem...0765

... Let me draw a picture here, of our coordinate system, hour right handed coordinate system, and we have X, we have Y, we have Z, I am not going to label the, I hope you don't mind.0771

This is that, and move on to red, so i am going to put point P(0) over here, that's my reference point, now my vector T, let's just say T, I am sorry U is in that direction.0785

I want to, so I have a point P, the line through point P in the direction of U, well it's one direction, that's another direction and that's what's happening.0801

This point P also happens to be a vector...0816

... My reference point, my reference vector, and then + U can be any vector of any length, but T is that scalar that you are multiplying by to make it bigger or smaller, so starting a T, I can go this way, that way by changing T and having it run through all real numbers...0823

... Moves in this direction, or if I multiply it in the negative direction, and go this way, that way, that way, so this is actually how it's represented.0846

Now, these are, this is a vector representation, well as you know a vector represents just points, so...0856

... Let me see if we can't...0865

... Well if I am going to hold off on the parametric representation for a second, but I would like you to just sort of see what this is, so the equation for line is this equation right here, it says that any point, any point can be gotten to, by starting at a reference point here.0870

And moving in the direction of U, depending on what this is, either forward or backward to get the entire line, that's what's happening...0888

... Okay, so let's see, let's write, let me go back to blue here, let's write this again, X vector is equal to some E, + T times U, okay.0898

Well this can be represented as , well X₁, X₂, X₃, again points and vectors are the same thing, if you want to actually expand it, this is just a short hand notation, P(0), why hadn't we called P(0).0916

Let’s just say...0932

... I don't know let's call it as Z₁, now let's do W₁, W₂, W, the symbolism doesn't matter.0939

W₁, W₂, W₃ + T times U₁, U₂, U₃, so this is actually equivalent to three equations, notice it's the same as...0953

... X₁ is equal to W₁ + TU₁, X₂ equals W₂ + T times U₂.0969

And X₃ equals W₃ + T times U₃, okay, vector form...0980

... Component form, I broke it up, these are called the parametric equations...0991

... The component parametric equations, this is also called the parametric equation, and it's called the parametric equation...1004

... because you have a parameter, your parameter is T, that's something that varies, and it expresses relationship between some point and another point based on some parameter.1014

That's what's changing, you are not actually expressing a direct relationship between the two points, you are expressing it using a third parameter, or a parameter.1027

In this case a third entity if you will...1036

... Okay...1040

... Okay, so let's do an example...1045

... Example will be, let's find parametric equations, so find...1052

... Parametric equations...1064

... For the line, through the point 3...1071

... And now let me make this a little bit clear here...1078

... (3, 2, -1) I am not sure if that's much clear in the direction of the vector (2, -3 and 4).1084

We want the line to pass through this point and we want to move in the direction of that point, both directions positive and negative, well you can just read it right off.1100

Again the equation of line says you just take your reference point and you add the some parameter t times the vector U. Well, in component form it's like this.1109

We just take X₁, W₁, TU₁, okay...1120

... We want to pass through this point, so our X value of the point that we want is equal to3, that's the point that it passes through in the direction of that vector, + T times 2.1126

I will write it this way, just to be consistent you could write two T's, it's not a problem, but I will just, I will just keep it consistent, because here I put the T actually before the vector.1145

X₂ is equal to 2 - T times -3.1153

And X₃ is equal to -1 + T times 4, or if you prefer something a little bit more normal, X = 3 + 2T.1162

Y = 2 + 3T - (-), you get a +, and Z = -1 + 4T.1181

These equations, they give you the X, the Y and the Z coordinate of all the points that are on this line, that pass through this point and are in the direction of this vector.1193

Okay...1206

... Okay, let's do another example here, let's see, find, let me move on to blue...1212

... Find parametric equations...1228

... for the line through the point P(0), which is (2, 3, -5)...1237

... And P(1), which is (3, -2, 7) okay, well so we have a point, so we can pick either point as far as out reference is concerned, because we are given two of them.1248

We need a direction vector, well the direction vector is, well if you have a point P(0), and a point P(1), that's usually a pretty good direction vector, so just take this one - that one, that's how we get a vector in that direction, right.1263

What we want to take is the direction vector, P(0), P(1) and I have, so I have chosen 1 as my head and 0 my tail of my, I can do it in either way, I just happen to have made a choice, so that's going to equal to, let's see(3, -2)...1281

... -2, -3 , 7, -(-5)...1303

... That's how we get a vector, we just take the, the arrow, the head - the tail, so we end up with (1, -5 and 12), so this is my direction vector.1312

And now I can just read it right off, X, it's okay I will do it here, once again X is equal to P₀ + t times my direction vector, which is now that one.1328

I wills say, I will call this x, Y, and Z, I want parametric equations, P(0), 2, 3, -5, 2, 3, -5...1343

... here is my direction vector + 1T - 5T + 12T, through this point and through this point I have a direction vector, I have chosen one of the points as mu reference, I have chosen the 2, 3, -5.1358

These are the parametric equations for a line passing through those two points...1382

... Okay, and again i could have chosen the other point, it doesn’t really matter as long as there is a point on that line.1390

Okay, let's do something here, make some room, let me rewrite...1397

... We have X = some X of 0, + TU₁, Y is equal to Y₀ + TU₂.1409

And Z is equal to Z₀ + TU₃, that's all we have done here, this is just the generalized version, now T is the same, T is always going to be the same here, 5, 6, -6, radical 2, 18, 40.1424

Well because this is same, I can solve each of these equations for T, so I end up with the following...1442

... T = X - X₀ over U₁, it's equal to Y - Y, 0 over U₂, equals Z - Z₀ over U₃.1456

This version of it, when I actually solve for the parameter, and express the relationship this way, this is called the symmetric form of line...1469

... We have the vector representation, which is this, we have the parametric, well this is a parametric vector representation, this is a parametric component representation, it just breaks this one up, so we actually see everything.1481

And then we can rearrange this, and this is called the symmetric form of the line, again it's just different ways of representing something for different possible techniques.1494

Certain problems require one thing, require one form or, over another, makes it easier, excuse me...1504

... We will leave you with that, okay let's move on to, so we did lines in R₂ and then w moved to three space, we did lines in R₃, now we are going to move on to planes, in R₃...1518

... Okay...1538

... A plane in R₃...1543

... Is obtained...1551

... By specifying a point in the plane...1558

... And a vector perpendicular to it...1571

... It will make sense in a minute when we draw the picture...1577

... This vector is called...1583

... A normal vector, so any vector that's normal to a plane is perpendicular to every point in that plane, that's what normal actually means.1591

Okay let's draw a picture of this, this is important, so let's draw our right hand coordinate system, and I am going to pick three points, one on this axis, one on this axis, one on this axis, so that we actually can have a good visual of a plane.1602

We have a plane passing through these three points, in the X, Y, Z coordinate system, now if I take a...1615

... Point there, and if I have a vector, okay and let's say I have A, so this is a plane here, and let's say I draw another...1627

... Point there, so I will call this one P(0), now I will just call this one P, well the perpendicular and this is N, that's our normal vector, and this is what's happening.1642

As it turns out, you can take a point in the plain, and as long as you specify a vector, that's actually normal to it, as it turns out, you have defined every single point in that plane, and we will talk about it in just a minute.1655

Let me represent these as component, with their components, so the vector N is N₁, N₂, N₃...1670

... P₀ is let's call it X₀, Y₀, Z₀, and point P, we will just call it XYZ, so we want a way to represent all of the points of that plane, so we are looking for ultimately an equation in X Y and Z, some relationship that exist among these values.1683

Okay, so let's take this vector right here, because we know that two points in a plane, they represent a vector, if I just take the head of the vector - the tail, so let me also specify P₀, P.1705

That vector from here to P, so this not a P₀, excuse me, that's just a regular P...1721

... Well that equals, well we take this one - that one, so X - X is 0...1731

... Y - y is 0, and Z - Z is 0, so now I have all of my components, okay, well this vector made of the two points in the plane, is perpendicular to this vector right.1739

Right, and what do we know about two vectors that are perpendicular to each, they are orthogonal to each other, orthogonally means that their dot product is 0, so let's set up that equation and see what happens.1754

My vector, P, 0, P, which is this vector right here, dotted with N is equal to 0, that's the definition, I am looking for normal vector.1766

Now let's go ahead and solve this for product based on the components, well that's equal to...1782

... This times that + this times that, + this times that = 0, so I get N₁ times X - X is 0 + N₂ times Y - Y is 0...1790

... + N₃ times Z - Z₀ = 0...1810

... This is my equation for a plane, now I can expand this out, I can do N₁ times X, N, you know -N times X is 0, you are going to get numbers, you are going to get an equation in XYZ, precisely because you have variables X, Y and Z.1821

Remember these things with subscripts, they are actual numbers, and these are going to be actual numbers, it's the X, Y and Z, the things without subscripts that are your variables.1837

Now this right here, this is the equation of a plane...1847

... equation of a plane...1853

... In R₃ and again all just got it by taking two points or one point actually in specifying the vector that's normal to it, because if it's normal to that one point, well it's going to be normal to every other point in the plane, okay let's do an example.1858

Find the equation of the plane passing through...1877

... (2, 4, 3) and perpendicular to...1889

... Yeah it's fine, perpendicular to (5, -2 and 3), so we want to find the equation of the plane that passes through this point and it's perpendicular to this vector, and again any point in the plane, if one point is perpendicular to that vector, every point in that plane is perpendicular to that vector.1903

That's the whole idea behind choosing a normal vector, well let's just multiply it out, based on the equation that we just had on the previous slide, we take 5 times...1921

... X - 2 + (-2) times Y - 4...1935

... + 3 times Z - 3 = 0, 5 times X - 2 -...1947

... two times Y - 4, + 3 times Z - 3 is equal to 0...1965

... we can stop there, that's fine, this is actually a perfectly valid equation, I like this because I like to see again, I like to see the dot product, that the multiplication that I am doing, one component times another, + one component times another.1978

If you want you can multiply this out, it's not a problem, so if you end up with something like 5X - 10 - 2Y + 8 + 3Z - 9 is equal to 0.1992

You end up with 5X - 2Y + 3Z + 1 = 0.2009

This is perfectly valid, and it's the one that you are accustomed to seeing of ‘course, it's a personal choice, I would like this one, because I would like to see everything that I am doing.2019

I prefer a longer expression, more complicated expression, as supposed to things being hidden, I see what's going on, I can see the normal vector, the components of normal vector.2029

I can see my point (2, 4, 3), I can see them, it's hidden in here, I have multiplied it out, so simplification makes things look prettier, doesn't necessarily simplify the material.2039

Okay...2052

... Lets do one final example...2055

... Find an equation of the plane passing through the three points, so this time, passing through three points, well there are couple of ways to do this.2061

Let's do it the way we did for R₂, so our three points are going to be (2, -2, 1)...2075

(-1, 0, 3) and (5, -3) and 4.2086

Okay, if it passes through, so we know that the equation of a line or the equation of a plane is this, AX + BY + Z, AX + BY + CZ + D = 0, this is the generic equation for a plane.2099

For these three points to be on this plane, they actually have to satisfy this, so I can just put all of these in here, and then I can, along with that equation, so I could do (1, 2, 3) for these three points, plus my original generic equations.2123

I end up with four equations and three unknowns, just like I did for the first time around, I end up with a 1, 2, 3, 4, 1, 2, 3, 4, I end up with the 4 by 4 linear, 4 by 4 matrix.2139

Well a 4 by 4 matrix, it has a solution, if the determinant of that matrix, because it's homogeneous, they are all equal to 0, is equal to 0, therefore I can set up the following.2152

X, Y Z 1, it's that coefficient, that coefficient, that coefficient and...2166

That coefficient, well, that coefficient, the one what I am looking for is A, B. C and D, those are my variables, and then I take the point (2, -2, 1).2174

2, -2, 1, 1, I take the other point, -1, 0, 3, 1 and 5 - 3, 4, 5 - 3 , 4, 1.2187

This determinant has to equal 0, and again I am going to expand along the first row, it actually doesn't where you expand, you are going to get the same answer.2205

Don't feel like you are stuck with the first row, it's just, that's the row where the variables happen to be, so I just sort of prefer it.2216

When you do this expansion and again it's little more complicate but you have done the expansion before, you end up with the following, A to X + 15Y - 3Z + 17 = 0.2224

In this case we do end up with this sort of simplified version, before we don't necessarily have a normal vector.2240

There you go, if you are given a point in a vector, you can go ahead and just write it out, take the dot product and get the very simple equation for a plane.2249

Or if you are given the three points separately, you can just set up this determinant, set it equal to 0 and solve that way for the particular equation.2259

Thank you for joining us for lines and planes, thank you for joining us at educator.com, and linear algebra, we will see you next time.2268