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The topic for today is going to be equations of claims and how it relates to linear systems and matrices
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as we have seen during Tuesday's lecture.
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So let's start again with equations of claims.
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So remember, we've seen briefly that an equation for a plane is of a form AX plus BY plus CZ equals D,
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where A, B, C and D are just numbers.
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And so this expresses the condition for a point that coordinates XYZ to be in the plane.
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And so an equation of this form defines a plane.
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So let's see how that works again.
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So let's start with an example.
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Let's say that we want to find an equation for the plane with normal vector,
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let's say vector n equals, let's say, the normal vector, 1, 5, 10.
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So how do we find an equation of this plane?
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Well, remember that we can get that equation by thinking geometrically.
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So what's our thinking going to be?
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Well, we have the XYZ axis, and we have this vector n, 1, 5, 10.
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And it's supposed to be perpendicular to our plane, and our plane passes for the origin here.
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So we want to think of a plane that's perpendicular to this vector.
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Well, when is a point in that plane?
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Let's say we have a point P at coordinates XYZ.
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Well, the condition for P to be in the plane should be that we have a right angle here.
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So P is in the plane whenever OP.n is 0.
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And if we write that explicitly, well, vector OP has components XYZ and has components
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1, 5, 10.
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So that will give us x plus 5y plus 10z equals 0.
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That's the equation of our plane.
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Now, let's think about a slightly different problem.
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So let's do another problem.
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Let's try to find the equation of a plane for a point P0 with coordinates, say, 2, 1 minus
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1, and with normal vector, again, the same guy, 1, 5, 10.
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So how do we find an equation of this thing?
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Well, we are going to use the same method.
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In fact, let's think first for a second.
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So I said we have a normal vector n, and it's going to be perpendicular to both planes
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at the same time.
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So in fact, our two planes will be parallel to each other.
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The difference is, well, before we had a plane that was perpendicular to n and passing
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through the origin.
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And now we have a new plane that's going to pass, well, not through the origin, but
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through maybe this point P0, and I don't plan where it is.
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But let's say, for example, that P0 is here.
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Then I will just have to shift my plane so that instead of passing through the origin,
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it passes through this new point.
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So how are we going to do that?
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Well, now for a point P to be in our new plane, we need for vector no longer OP, but P0P
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to be perpendicular to n.
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So P is in this new plane.
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If the vector P0P is perpendicular to n,
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and now let's think for a second, what's vector P0P?
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Well, we take the coordinates of P, and we subtract those of P0.
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So that should be x minus 2, y minus 1, and z plus 1, dot product with 1510 equals 0.
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So let's expand this while we get x minus 2 plus 5 times y minus 1 plus 10 times z plus
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1 equals 0.
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Let's put the constants on the other side.
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We get x plus 5y plus 10z equals, well, here we have minus 2 becomes 2,
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minus 5 becomes 5, 10 becomes minus 10.
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I think we end up with negative 3.
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So see, the only thing that changes between these two equations is the constant term
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on the right hand side.
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The thing that I call d.
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And the other common feature is that the coefficients of x, y, and z, 1, 5, and 10,
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correspond exactly to the normal vector.
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So that's something you should remember about planes.
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These coefficients here correspond exactly to a normal vector.
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And well, this constant term here roughly measures how far, you know, when you move from,
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so if you have a plane through the origin, the right hand side will be 0.
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And if you move to a parallel plane, then this number will become something else.
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So how actually could we have found that minus 3 more quickly?
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Well, we know that the first part of the equation is like this.
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And we know something else.
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We know that the point p0 is in the plane.
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So if we plug the coordinates of p0 into this, well, x is 2 plus 5 times 1 plus 10 times
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minus 1, we get minus 3.
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So in fact, the number we should have here should be minus 3 so that p0 is a solution.
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So let me point out, so maybe I'll put a 1 here again.
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So these three numbers, 1, 5, 10, are exactly the normal vector.
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And one way that we can get this number here is by computing the value of the left hand side.
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At the point p0, we plug in the point p0 into the left hand side.
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Okay?
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Any questions about that?
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No?
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Okay.
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By the way, I mean, of course, a plane doesn't have just one equation.
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It has infinitely many equations because if I take instead say, if I multiply everything
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by 2, 2x plus 10y plus 20z equals minus 6 is also an equation for this plane.
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Okay? That's because, of course, we have normal vectors of all sizes.
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We can choose how big we make it.
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So again, just to repeat the single most important thing here, in the equation,
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the x plus by plus c is equals d, the coefficients a, b, c give us actually a normal vector to the plane.
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So that's why, in fact, what matters to us mostly is finding the normal vector.
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And in particular, if you remember last time, I explained something about how we can find normal vector to the plane
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if we know points in the plane, namely we can take the cross product of two vectors containing the plane.
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Okay. So let's just do an example to see if we completely understand what's going on.
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So let's say that I give you the vector with components 1, 2, negative 1, and I give you the plane x plus y plus 3z equals 5.
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So do you think this vector is parallel to the plane? The perpendicular to it?
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The neighbor. Okay. I'm starting to see a few votes.
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So I see that most of you are answering number 2. This vector is perpendicular to the plane.
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There are some other answers to. Well, let's try to figure it out. Okay.
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So let's do the example. Say, V is 1, 2, negative 1, and the plane is x plus y plus 3z equals 5.
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So let's just throw that plane anywhere. It doesn't really matter.
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Let's first get a normal vector out of it. Well, to get a normal vector to the plane, what I will do is I will take the coefficients of x, y, and z.
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So that's 1, 1, 3. Okay. So 1, 1, 3 is perpendicular to the plane.
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How do we get all the other vectors that perpendicular to the plane? That are perpendicular to the plane?
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Well, all the perpendicular vectors are parallel to each other. Okay. So that means that they are just obtained by maybe multiplying this guy by some number.
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Okay. 2, 2, 6, for example, would still be perpendicular to the plane.
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1, 1, 1, 1, 1, 3, 1 is also perpendicular to the plane. But now, see, these guys are not proportional to each other. So V is not perpendicular to the plane.
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So it's not perpendicular to the plane. Being perpendicular to the plane is the same as being parallel to its normal vector.
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Now, what about testing if V is instead parallel to the plane? Well, it's parallel to the plane if it's perpendicular to N.
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Well, let's check that.
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So let's try to see if it's perpendicular to N. Well, let's do V.N. That's 1, 2, negative 1.1, 1, 1, 3. If you get 1 plus 2 minus 3, that's 0. So, yes.
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So if it's perpendicular to N, it means maybe it's somewhere like that. It's actually going to be parallel to the plane.
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Okay. Any questions? Yes.
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Let's see. So if I plug the vector into the plane equation, so x, 1 plus 2 minus 3, well, the left hand side becomes 0. So it's not a solution of a plane equation.
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Okay. So I should say, okay. So there's two different things here. One is the point we've coordinates 1 to minus 1 is not in the plane.
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Okay. So what that tells us is that if I put my vector V at the origin, then the point here is not going to be in the plane.
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On the other hand, you're right. The left hand side evaluates to 0. What that means is if instead I had taken the plane x plus y plus 3 is equal 0, then it would be inside. And the plane x plus y. So this one is x plus y plus 3 is equals 5.
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x plus y plus 3 is equal 0 would be a plane parallel to it, but for the origin. So that would be another way to see that the vector is parallel to the plane. If we move the plane to a parallel plane for the origin, then the end point of the vector is in the plane.
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Okay. That's another way to convince yourselves. Any other questions?
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Okay. So let's move on. So last time we saw about matrices, we learned about matrices and linear systems. So let's try to think now about linear systems in terms of equations of planes and intersections of planes.
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And remember a linear system, it's a bunch of equations. Let's say a three by three linear system. It's three different equations. Each of the equation is the equation of a plane. So in fact, if we try to solve a system of equations, that means actually we're trying to find a point that is on several planes at the same time.
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So one more board here. So let's say that we have a three by three linear system.
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And just to take an example, doesn't really matter what I give you, but let's say I give you x plus z equals one, x plus y equals two, x plus two y plus three equals three.
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What does this mean? What does it mean to solve this? It means we want to find x, y, z, which satisfy all of these conditions. So let's just look at the first equation first. Well, first equation says our point should be on the plane, which has this equation.
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So the second equation says, oh, our point should also be on this plane. So in fact, if you just look at the first two equations, that means you have two planes.
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Soluthans. So these two equations determine for you two planes and two planes intersect in a line.
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Now, well, what happens with a third equation? That's actually going to be a third plane. So if we want to solve the first two equations, we have to be on this line.
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And if we want to solve a third one, we also need to be on another plane. And in general, three planes intersect in a point because this line of intersection.
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So, okay, so if we're planes intersecting a point, and one way to think about it is that the line with the first two points, sorry, the line with the first two planes intersect meets the third plane in a point.
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And that point is the solution to the linear system. Okay, so the line, let's say, formed by the, so this is mathematical notation for the intersection between the first two planes intersects the third plane in a point, which is going to be the solution.
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Okay, so the solution, actually, how do we find it? Well, one way is to draw pictures and try to figure out where the solution is, but that's not how we do it in practice if we are given the equations.
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So, the solution is given, sorry, so let me use matrix notation. Remember, we saw on Tuesday that the solution to AX equals B is given by X equals A inverse B.
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We get from here to here by multiplying on the left by A inverse. A inverse AX simplifies to X equals A inverse B. And once again, it's A inverse B and not B inverse.
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If you try to set up the multiplication, B inverse doesn't work. The size is not compatible. You can't multiply the other way around.
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Okay, so that's pretty good, unless it doesn't work that way. Okay, and what could go wrong? Well, let's say that our first two planes do intersect nicely in the line, but let's think about the third plane.
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If the third plane does not intersect that line nicely in a point, maybe it's actually parallel to that line. Okay, so let's try to think about this question for second.
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Oh, sorry, this is too far from the left.
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Let's say that the set of solutions to a free by free linear system is not just one point. So we don't have a unique solution that we can get this way.
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What could happen? What do you think could happen?
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I see answers number three and five seem to be dominating. There's also a bit of answer number one.
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And in fact, these are pretty good answers. I see even some of you figured out that you can answer one and three at the same time.
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I just have to see somebody with free hands answer all three numbers at the same time. Okay, so indeed we'll see very soon that we could have either no solution, a line or a plane.
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So our answer is well, two points, two solutions. We will see is not actually a possibility because if you have two different solutions, then the entire line for these two points is also going to be made of solutions.
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So the headron is just there to, you know, I'm using you. It's not actually a good answer to that question. It's not very likely that you will get a hit or an out of intersecting planes.
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A plane is in it possible and I don't know is still okay for a few more minutes, but we're going to get to the bottom of this and then we will know.
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So let's try to figure out what can happen. So let me go back to my picture. And so I had my first two planes, if you determine a line.
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And now I have my third plane and maybe my third plane is actually parallel to the line. But it doesn't pass for it.
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Then I'm afraid that there's no solutions because to solve the system of equations, I need to be in the first two planes.
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So that means I need to be in that vertical line. Well, that line was supposed to be a head, but I guess it doesn't really show up as a head.
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And it also needs to be in the third plane, but the line and the plane are parallel to each other. There's just no place where they intersect. So there's no way to solve all the equations.
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On the other hand, the other thing that could happen is that actually the line is contained in the plane.
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And then, well, it means any point on that line will actually solve automatically the third equation.
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So if you try solving a system that looks like this by hand, what you will notice, if you do substitution, elimination, and so on, what you will notice is that after you've dealt with two of the equations, the third one, here, would actually turn out to be the same as what you got out of the first two.
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It doesn't give you any additional information. It's as if you had only two equations.
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The previous case would be when actually the third equation contradicts something that you can get out of the first two. For example, maybe out of the first two, you got x plus z equals one. And the third equation is x plus z equals two.
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Well, it can't be one and two at the same time. Maybe another way to say it is that this picture is one where you can get out of the equations.
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You get a number equals a different number that's impossible. And that picture is one where out of the equations, you get zero equals zero, which is certainly true, but isn't very useful equation.
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So you can't actually finish solving.
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Let me write that down. So unless the third plane is parallel to the line where p1 and p2 intersect.
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So then there's two subcases. If the line of intersections of p1 and p2 is actually contained in p3 in the third plane, then we have infinitely many solutions.
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Namely, any point on the line will automatically solve the third equation.
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Now, the other possibility, the other subcase is if the line,
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the intersection of p1 and p2 is parallel to p3 and not contained in it,
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then we get no solutions.
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So if the third line hits the third plane in a point, then that's going to be our solution.
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If that line instead is parallel to the third plane, well, if it's parallel and outside of it, then we have no solution.
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So going back to our list of possibilities, that means, let's see what can happen.
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There's no solution that happens when the line where the first two planes intersect is parallel to the third one.
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If the line of intersection of the first two planes actually has two points that are in the third plane, well, then that means the entire line must also be in the third plane.
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So if you have two solutions, then you have more than two actually.
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What about a plane? Well, that's a case that I didn't explain because I've been assuming that p1 and p2 are different planes and they intersect in the line.
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But in fact, well, they could be parallel in which case we already have no solution to the first two equations.
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Or we could be the same plane. And now if a third plane is also the same plane, if all three planes are the same plane, well, then you have a plane of solutions.
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If I give you three times the same equation, that is a linear system. It's not a very interesting one, but it is a linear system.
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Okay, and I don't know is no longer a solution either. Any questions? Yes?
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Okay, so that's a very good question. The question is what is the geometric significance of an equation like x plus y plus z or say this one equals to one, two, three or something else?
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Well, if the equation is x plus y plus z equals zero, it means that our plane is passing through the origin.
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And then if we change the constant, it means we move to a parallel plane.
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So the first guess that you might have is that this number on the right hand side is the distance between the origin and the plane.
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It tells us how far from the origin we are. That is not quite true.
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In fact, that would be true if the coefficients here form the unit vector. Then this guy would just be the distance to the origin.
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Otherwise, you have to actually scale by the length of this normal vector. And I think there's a problem in the notes that will show you exactly how this works.
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Okay, but you should think of it roughly as how much have we moved the plane away from the origin?
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That's the meaning of this last term, d, that's in the right hand side of the equation.
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Okay.
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So let's try to think about what exactly this case, you know, how do we detect now in which situation we are?
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Because that's all very nice in the picture.
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But, you know, it's difficult to draw planes. In fact, when I draw these pictures, I'm always very careful not to actually try to pretend to draw an actual plane given by an equation.
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Or when I do that, it's blatantly false. It's difficult to draw a plane correctly.
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So instead, let's try to think about it in terms of matrices. In particular, what's wrong with this? You know, why can't we always say the solution is x equals a inverse b?
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Well, so the point is actually you cannot always invert the matrix.
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Okay. So recall, we've seen this formula that a inverse is one over determinant of a times the adjoint matrix.
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And, well, we've learned how to compute this thing. Remember, we had to take minors, then flip some signs, and then transpose.
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Well, that step we can always do. We can always do these calculations. But, at the end, we have to divide by the determinant.
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Well, that's fine if a determinant is not zero, but if a determinant is zero, then certainly we cannot do that.
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So, in fact, what I didn't mention last time is that the matrix is invertible, that means it has an inverse. Exactly when it's determinant is not zero.
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Okay, so that's something we should remember.
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If a determinant is not zero, then we can use our method to find the inverse, and then we can solve using this method. If not, then not.
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Yes?
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That's correct. So, what I'm saying is we can invert the matrix a if a determinant is not zero.
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And, if you look again at the method that we saw last time, so first we had to compute this adjoint matrix, and these operations we can always do.
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If we are given a free by free matrix, we can always compute the adjoint.
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And, then the last step to find the inverse was to divide by the determinant. And, that we can only do if a determinant is not zero.
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So, if we have a matrix, which determinant is not zero, then we know how to find the inverse. If a determinant is zero, then of course this method doesn't work.
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But, I'm saying, actually, even more, there isn't an inverse at all. It's not just that our method fails. I cannot take the inverse of a matrix with determinant zero.
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And, geometrically, well, precisely the situation where the determinant is not zero is exactly this nice, usual situation where the three planes intersect in a point.
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While the situation where the determinant is zero is this situation here where the line determined by the first two planes is parallel to the third plane.
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So, let me emphasize this again. And, let's see again what happens.
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Okay, so let's start with an easier case. So, it's called the case of a homogeneous system. It's called homogeneous because it's the situation where the equations are invariant and are scaling.
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So, homogeneous system is one where the right hand side is zero. There's no B. If you want the constant terms here are all zero. 0, 0, 0.
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Okay, so this one is not homogeneous. So, let's see what happens there. Well, so what that means, you know, so let's take an example.
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Instead of this guy, we could take x plus z equals zero, x plus y equals zero, and x plus two y plus phase z also equals zero.
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Well, can we solve these equations? Yeah, I think actually you already know a very simple solution to these equations.
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Yeah, you can just take x, y, and z all to be zero. So, there's always an obvious solution, namely 0, 0, 0.
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And in mathematical jump, this is called the trivial solution. Okay, so there's always this trivial solution. And what's the geometric interpretation? Well, having zero here means that all three planes pass for the origin.
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So, certainly the origin is always a solution.
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Because the flip planes pass for the origin. Okay, so now there's two subcases.
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One case is if a determinant of a matrix A is non-zero. Well, that means that we can invert A.
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So, if we can invert A, then that means we can solve the system by multiplying by a inverse. If we multiply by a inverse, well, we'll get x equals a inverse times zero, which is zero.
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Okay, and that's the only solution. Because, well, you know, if Ax is zero, then let's multiply by a inverse, we get a inverse Ax that's x equals a inverse zero, that's zero.
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So, we get x equals zero. We've solved it. Okay, no other solution. So, to go back to this picture of that, we will enjoy its viscues.
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Now, the other case, if a determinant of A equals zero, then it means that actually this doesn't quite work.
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So, let's see. What does it mean that the determinant of A is zero? Well, remember, the entries in A, they are the coefficients in the equations.
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But now, the coefficients in the equations, they are exactly the normal vectors to the planes. So, that's the same thing as saying by the determinant of the three normal vectors to our three planes is zero.
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So, that means that N1, N2, and N3 are actually in the same plane. It's called co-planar. Okay, these three vectors are co-planar.
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So, let's see what happens. So, I claim it will correspond to this situation here. Let's draw the normal vectors to these three planes.
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Well, it's not very easy to see. But, okay, so I've tried to draw the normal vectors to my planes. Well, they're all in the direction that's perpendicular to the line of intersection. They're all in the same plane.
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So, if I try to form a box in a parallel pipette with these three normal vectors, well, I will get something that's completely flat and has no volume, has volume zero.
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Okay, so, the parallel pipette has volume zero.
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And the fact that the normal vectors are co-planar tells us that, in fact, let me start a new blackboard.
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Okay, so now, let's say that our normal vectors N1, N2, and N3 are all in the same plane. And let's think about the direction that's perpendicular to N1, N2, and N3 at the same time. Okay, I claim that will be the line of intersection.
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Okay, so, let's see.
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Okay, so let me try to draw the texture again. So, we have three planes.
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Okay, now you see why I prepared the picture in advance. It's easier.
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Okay, so, let me draw it beforehand. And I said, the normal vectors are all in the same plane. So, what else do I know? I know that all these planes pass through the origin.
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And the origin is somewhere in the intersection of the three planes. Now, I said, actually, the normal vectors to my three planes.
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Well, that's kind of hard to draw a bit. Now, all, actually, co-planar. Okay, so, N1, N2, and N3 determine a plane. Well, now, if I look at the line, for the origin, that's perpendicular to N1, N2, and N3.
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So, perpendicular to this red plane here. That's a terrible picture. It's supposed to be in all the planes. Okay, you can see that that's on the side planes.
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And why is that? Well, that's because my line, it's perpendicular to the normal vectors. So, it's parallel to the planes. It's parallel to all the planes. Now, why is it in the planes instead of parallel to them? Well, that's because my line goes for the origin, and the origin is on the plane.
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So, certainly, my line has to be contained in the planes, not parallel to them. Okay, so, the line, for the origin, and perpendicular to the plane of N1, N2, and 3,
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is parallel to all three planes. And because the planes go for the origin, it's contained in the. Okay, so what happens here is I have, in fact, infinitely many solutions.
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And how do I find these solutions? Well, if I want to find something that's perpendicular to N1, N2, and N3. Well, if I just want to be perpendicular to N1, N2, I can take their cross-product, exactly.
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So, for example, if I do N1 cross N2 is perpendicular to N1 and N2, and also to N3, because N3 is in the same plane as N1 and N2. So, if you're perpendicular to N1 and N2, you're also perpendicular to N3.
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It's automatic. So, it's an entrepreneurial solution. This vector goes along the line of intersections.
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Okay, so that's the case of homogeneous systems. And then let's finish with the other case. So, the general case.
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So, that means if we look at a system A x equals B. Well, with B now anything, there's two cases. If the determinant of A is not zero, then there's a unique solution.
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So, namely x equals A inverse B. If the determinant of A is zero, then it means actually we have the situation with claims that are all parallel to the same line.
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And then we have either no solution or infinitely many solutions. It cannot be a single solution. Now, whether you have no solutions or infinitely many solutions,
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well, we haven't actually developed the tools to answer that. But if you try solving the system by hand, by elimination, you will see that you end up maybe with something that says zero equals zero, and you have infinitely many solutions.
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You know, actually if you can find one solution, then you know that there's infinitely many. On the other hand, if you end up with something that's a contradiction, like 1 equals 2, then you know there's no solutions.
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Okay, so that's the end for today. And tomorrow we will learn about parametric equations for lines and curves.