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.