Hesse normal form

The Hesse normal form named after Otto Hesse, is an equation used in analytic geometry, and describes a line in $\mathbb {R} ^{2}$ or a plane in Euclidean space $\mathbb {R} ^{3}$ or a hyperplane in higher dimensions.^[1]^[2] It is primarily used for calculating distances (see point-plane distance and point-line distance).

It is written in vector notation as

{\vec {r}}\cdot {\vec {n}}_{0}-d=0.\,

The dot $\cdot$ indicates the scalar product or dot product. Vector ${\vec {r}}$ points from the origin of the coordinate system, O, to any point P that lies precisely in plane or on line E. The vector ${\vec {n}}_{0}$ represents the unit normal vector of plane or line E. The distance $d\geq 0$ is the shortest distance from the origin O to the plane or line.

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What I want to do in this video is make sure that we're good at picking out what the normal vector to a plane is, if we are given the equation for a plane. So to understand that, let's just start off with some plane here. Let's just start off-- so this is a plane, I'm drawing part of it, obviously it keeps going in every direction. So let's say that is our plane. And let's say that this is a normal vector to the plane. So that is our normal vector to the plane. It's given by ai plus bj plus ck. So that is our normal vector to the plane. So it's perpendicular. It's perpendicular to every other vector that's on the plane. And let's say we have some point on the plane. We have some point. It's the point x sub p. I'll say p for plane. So it's a point on the plane. Xp yp zp. If we pick the origin. So let's say that our axes are here. So let me draw our coordinate axes. So let's say our coordinate axes look like that. This is our z-axis. This is, let's say that's a y-axis. And let's say that this is our x-axis. Let's say this is our x-axis coming out like this. This is our x-axis. You can specify this is a position vector. There is a position vector. Let me draw it like this. Then it would be behind the plane, right over there. You have a position vector. That position vector would be xpi plus ypj plus zpk. It specifies this coordinate, right here, that sits on the plane. Let me just call that something. Let me call that position vector, I don't know-- let me call that p1. So this is a point on the plane. So it's p-- it is p1 and it is equal to this. Now, we could take another point on the plane. This is a particular point of the plane. Let's say we just say, any other point on the plane, xyz. But we're saying that xyz sits on the plane. So let's say we take this point right over here, xyz. That clearly, same logic, can be specified by another position vector. We could have a position vector that looks like this. And dotted line. It's going under the plane right over here. And this position vector, I don't know, let me just call it p, instead of that particular, that P1. This would just be xi plus yj plus zk. Now, the whole reason why I did this set up is because, given some particular point that I know is on the plane, and any other xyz that is on the plane, I can find-- I can construct-- a vector that is definitely on the plane. And we've done this before, when we tried to figure out what the equations of a plane are. A vector that's definitely on the plane is going to be the difference of these two vectors. And I'll do that in blue. So if you take the yellow vector, minus the green vector. We take this position, you'll get the vector that if you view it that way, that connects this point in that point. Although you can shift the vector. But you'll get a vector that definitely lies along the plane So if you start one of these points it will definitely lie along the plane. So the vector will look like this. And it would be lying along our plane. So this vector lies along our plan. And that vector is p minus p1. This is the vector p minus p1. It's this position vector minus that position vector, gives you this one. Or another way to view it is this green position vector plus this blue vector that sits on the plane will clearly equal this yellow vector, right? Heads to tails. It clearly equals it. And the whole reason why did that is we can now take the dot product, between this blue thing and this magenta thing. And we've done it before. And they have to be equal to 0, because this lies on the plane. This is perpendicular to everything that sits on the plane and it equals 0. And so we will get the equation for the plane. But before I do that, let me make sure we know what the components of this blue vector are. So p minus p1, that's the blue vector. You're just going to subtract each of the components. So it's going to be x minus xp. It's going to be x minus xpi plus y minus ypj plus z minus zpk. And we just said, this is in the plane. And this is, this right, the normal vector is normal to the plane. You take their dot product-- it's going to be equal to zero. So n dot this vector is going to be equal to 0. But it's also equal to this a times this expression. I'll do it right over here. So these-- find some good color. So a times that, which is ax minus axp plus b times that. So that is plus by minus byp. And then-- let me make sure I have enough colors-- and then it's going to be plus that times that. So that's plus cz minus czp. And all of this is equal to 0. Now what I'm going to do is, I'm going to rewrite this. So we have all of these terms I'm looking for, right? Color. We have all of the x terms-- ax. Remember, this is any x that's on the plane, will satisfy this. So ax, by and cz. Let me leave that on the right hand side. So we have ax plus by plus cz is equal to-- and what I want to do is I'm going to subtract each of these from both sides. Another way is, I'm going to move them all over. Let me do it-- let me not do too many things. I'm going to move them over to the left hand side. So I'm going to add positive axp to both sides. That's equivalent of subtracting negative axp. So this is going to be positive axp. And then we're going to have positive byp plus-- I'll do that same green-- plus byp, and then finally plus czp. Plus czp is going to be equal to that. Now, the whole reason why did this-- and I've done this in previous videos, where we're trying to find the formula, or trying to find the equation of a plane, is now we say, hey, if you have a normal vector, and if you're given a point on the plane-- where it's in this case is xp yp zp-- we now have a very quick way of figuring out the equation. But I want to go the other way. I want you to be able to, if I were to give you a equation for plane, where I were to say, ax plus by plus cz, is equal to d. So this is the general equation for a plane. If I were to give you this, I want to be able to figure out the normal vector very quickly. So how could you do that? Well, this ax plus by plus cz is completely analogous to this part right up over here. Let me rewrite all this over here, so it becomes clear. This part is ax plus by plus cz is equal to all of this stuff on the left hand side. So let me copy and paste it. So I just essentially flipped this expression. But now you see this, all of this, this a has to be this a. This b has to be this b. This c has to be this thing. And then the d is all of this. And this is just going to be a number. This is just going to be a number, assuming you knew what the normal vector is, what your a's, b's and c's are, and you know a particular value. So this is what d is. So this is how you could get the equation for a plane. Now if I were to give you equation or plane, what is the normal vector? Well, we just saw it. The normal vector, this a corresponds to that a, this b corresponds to that b, that c corresponds to that c. The normal vector to this plane we started off with, it has the component a, b, and c. So if you're given equation for plane here, the normal vector to this plane right over here, is going to be ai plus bj plus ck. So it's a very easy thing to do. If I were to give you the equation of a plane-- let me give you a particular example. If I were to tell you that I have some plane in three dimensions-- let's say it's negative 3, although it'll work for more dimensions. Let's say I have negative 3 x plus the square root of 2 y-- let me put it this way-- minus, or let's say, plus 7 z is equal to pi. So you have this crazy-- I mean it's not crazy. It's just a plane in three dimensions. And I say what is a normal vector to this plane? You literally can just pick out these coefficients, and you say, a normal vector to this plane is negative 3i plus the square root of 2 plus 2 square root of 2 j plus 7 k. And you could ignore the d part there. And the reason why you can ignore that is that will just shift the plane, but it won't fundamentally change how the plane is tilted. So a this normal vector, will also be normal if this was e, or if this was 100, it would be normal to all of those planes, because all those planes are just shifted, but they all have the same inclination. So they would all kind of point the same direction. And so the normal vectors would point in the same direction. So hopefully you found that vaguely useful. We'll now build on this to find the distance between any point in three dimensions, and some plane. The shortest distance that we can get to that plane.

Derivation/Calculation from the normal form

Note: For simplicity, the following derivation discusses the 3D case. However, it is also applicable in 2D.

In the normal form,

({\vec {r}}-{\vec {a}})\cdot {\vec {n}}=0\,

a plane is given by a normal vector ${\vec {n}}$ as well as an arbitrary position vector ${\vec {a}}$ of a point $A\in E$ . The direction of ${\vec {n}}$ is chosen to satisfy the following inequality

{\vec {a}}\cdot {\vec {n}}\geq 0\,

By dividing the normal vector ${\vec {n}}$ by its magnitude $|{\vec {n}}|$ , we obtain the unit (or normalized) normal vector

{\vec {n}}_{0}={{\vec {n}} \over {|{\vec {n}}|}}\,

and the above equation can be rewritten as

({\vec {r}}-{\vec {a}})\cdot {\vec {n}}_{0}=0.\,

Substituting

d={\vec {a}}\cdot {\vec {n}}_{0}\geq 0\,

we obtain the Hesse normal form

{\vec {r}}\cdot {\vec {n}}_{0}-d=0.\,

In this diagram, d is the distance from the origin. Because ${\vec {r}}\cdot {\vec {n}}_{0}=d$ holds for every point in the plane, it is also true at point Q (the point where the vector from the origin meets the plane E), with ${\vec {r}}={\vec {r}}_{s}$ , per the definition of the Scalar product