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# Defining a plane in R3 with a point and normal vector

Determining the equation for a plane in R3 using a point on the plane and a normal vector. Created by Sal Khan.

Video transcript

Let's take a little bit of a
hiatus from our more rigorous math where we're building the
mathematics of vector algebra and just think a little bit
about something that you'll probably encounter if you ever
have to have to write a three-dimensional computer
program or have to do any mathematics dealing with
three dimensions. And that's the idea of just the
equation of a plane in R3. And you know what a plane is. I mean we live in a
three-dimensional world and we see planes all around us. The surface of your monitor, of
your computer monitor is a plane, regardless of what
angle you hold it at. And I can draw here in
three dimensions. Let me do a little bit
better job than that. So let's say that that
is the x-axis. This is my y-axis. And then that's my z-axis. And we know what a
plane looks like. it looks something like that. I'm just drawing it at an
arbitrary angle and it goes off in every direction. Now, the equation of a
plane-- and you've probably seen this before. It's a linear function
of x, y and z. So it's ax plus by plus
cz is equal to d. If this is the graph on that
plane, then that means that every point on this plane, every
x, y and z on this plane satisfies this equation. Now another way, just as valid
a way to specify a point or specify a plane is to give you
an actual point on the plane. So we'll say, look, that's
the point on the plane. Let's say that this is the
point x0, y0, and z0. It could be one of the instances
of this point right here, but I'm just saying
this is another point that's on the plane. That obviously by itself isn't
going to specify the plane. You could pivot the plane
around that point in an infinite number of ways. But if you specify that point
and you specify a vector that's perpendicular to the
plane-- and I can draw that starting from here, but I can
shift a vector wherever. But let me just draw
it right there. So if I also specify a normal
vector to the plane. And I just used a word that I
haven't defined for you yet. But when I say a normal
vector, so n is a normal vector. Or which just means that it's
perpendicular to the plane. It's perpendicular to everything
on the plane. It's perpendicular to every
vector on the plane. Perpendicular to, I guess the
best way to say it, I'll just say it in very imprecise
terms-- everything on the plane. So if you have some vectors
lying on the plane, if I have some vector here, let's say
that's lying on the plane. If you imagine the plane as
a piece of cardboard, that yellow arrow I just drew-- I
would've actually drawn that on the cardboard. It's sitting on the plane. If this yellow vector, let me
call it vector a, then if this is just some arbitrary vector
sitting on the plane and this is the normal vector through
the plane, we know from our definition of vector angles that
this is perpendicular to this if and only if, n dot a--
only if the dot product of these two things
are equal to 0. And that's true for any vector
that we pick that actually lies on the plane. So let's see if we can use this
definition of a plane, if we can use the-- I'll call it
the normal or let me call it the n plus-- let me
do it this way. The n plus some x0,
y0, z0 definition. And if I can go from that to
just the standard linear equation definition, ax plus
by plus cz is equal to 0. Let's see if there's some way
based on what we already know and using this information
that we can do that. So the way to think about it,
this point, this little blue point that lies on the plane,
I can specify it by a position vector. So let me set some position
vector x0 to be equal to-- so I'm going to define x0
to be equal to the scalar x0, y0, z0. And I want to be very clear. This specifies the coordinate
that lies on the plane. This vector does not lie
entirely on the plane. The way I drew it right here,
it's starting at the origin. It's a position vector. And the way I drew it it's
behind the plane. The tip of its arrow
sits on the plane. But this vector itself,
it's not necessarily drawn on the plane. This plane might not even go
through the origin, while this vector does touch
on the origin. It just specifies some
point on the plane. Similarly, let me define
another vector. I said this was some other
arbitrary point on the plane xyz, and so you know,
this is true for any point on the plane. Let me define another vector
x, and I'm going to define that as x, y, and z. So once again, like x0, the
vector x-- let me draw it right there. This vector x does not
lie on the plane. It goes from the origin. It's a position vector
that specifies a point on the plane. So it goes from the origin
and it goes out. And you can almost view them
as if the plane was like a coffee table that this would
kind of be these vectors. Let me see if I can draw it. If this was the flat surface of
the plane that the vector x0 is going from the origin
to specify some point on the plane. And then the vector x is also
going from the origin to specify-- let me do it
in a different color. The vector x is also going from
the origin to specify some other point on the
plane right there. I just took the plane and made
it flat, so you see it right along its side. If you could imagine sitting
right at the surface of the plane and then you could see
that these guys clearly do not lie in the plane. But using these guys, I can
construct a vector that does lie on the plane. What does the vector x
minus x0 look like? Well, I just drew a little
triangle here. x minus x0-- I'll do it in this
green color. It'll look exactly like this. x
minus x0 will be this green line right here. This is x minus x0. We could view x0 plus this
vector plus x minus x0 is going to be equal to x. If I were to do it on this
graph, it's going to look like this. It's going to be like this. So let me draw it better
than that. It's going to go from that
point from x0, the point specified by x0 to the
point specified by x. And it's going to lie
along the plane. So this right here
is x minus x0. I know this drawing is getting
very dirty, but you can see that this is definitely
lying on the plane. So this vector right here must
be perpendicular to n. Perpendicular to our
normal vector. Now, if my normal vector-- let's
say my normal vector n. So this vector is perpendicular
to this guy right here. It's perpendicular to the
vector n1, n2, n3. Now using this information, how
can we get to this type of an expression, just this linear
equation of x, y and z? Well, we know that n-- let me
switch to a neutral color. We know that n-- actually, I
didn't want to do this caret here. n is not a unit vector. But let's say n, so it's
perpendicular to this. So the dot product of it,
we saw that right there. We saw it in the
previous video. The dot product of n with this
vector right here-- actually, let me draw it. You know, I already drew the
plane sideways, so I can actually draw my n vector. My n vector's going to look
something like this. It's going to be popping
straight out of the plane. And I could shift it over, but
it's always going to be in that same direction. It's going to be perpendicular
to this vector right here. So n is perpendicular
to x minus x0. Which means that their dot
product is equal to 0. Well what does x minus
x0 look like? So this is going to look
like this expression. If I write out the vectors
themselves it's the vector n1, n2, n3 being dotted with-- well,
if I take x minus x0, that's just the scalar x
minus the scalar x0. The first term subtracted. And then the scalar y
minus the scalar y0. And then the scalar z
minus the scalar z0. We know that this whole thing
has to be equal to 0 because they're perpendicular. And then if we take the dot
product here, we get n1 times x minus x0 plus n2 times y minus
y0 plus n3 times z minus z0 is equal to 0. And you might not completely
recognize it, but this is-- you'll have to do a little
algebra to clean it up-- but this is the form ax plus by
plus cz is equal to d. And actually, I think I
made a mistake here. This should be not 0,
this is equal to d. This is the general form
for a plane in R3. A plane is just a linear
surface in R3. I shouldn't have written
a 0 there. So this does take this form. And if you don't believe
me, we can do it with an actual example. So let's say we have--
I gave you my normal-- I specify a plane. If I give you a normal vector
and I tell you that normal vector is the point 1, 3, minus
2 and I say that it intersects the point, or a point
that lies on the plane-- The normal vector and the point
don't necessarily have to intersect. But let's say for a point that
lies on the plane, I have the point 1, 2 and 3. And I say give me the equation
for this plane. Well, I would say well, if I
take any other point on that plane-- so if I take any other
point on that plane, xyz and it's specified by this vector,
the vector that's defined by the difference between
these two is going to lie on the plane. This point and this point
lie on the plane, so the difference between these two
vectors, the whole vector will lie on the plane. Let me take the difference. So x minus x0 is equal to x
minus 1, y minus 2, and then z minus 3. I'm saying this will
lie on the plane. This is on the plane. And it's going to
be perpendicular to our normal vector. So if I dot my normal vector, 1,
3, minus 2, with this thing right here, with x minus 1,
y minus 2, z minus 3, I should get 0. Because this has to be
perpendicular to anything that lies on the plane. So what do we get? We get 1 times x minus
1 is x minus 1. Plus 3 times y minus 2. Just taking the dot product. Minus 2 times z minus
3 is equal to 0. Let's if we can do a little bit
of algebra here to clean this up a little bit. I get x minus 1 plus 3y
minus 6 minus 2z plus 6 is equal to 0. And let's see. Minus 6 and a plus
6 cancel out. And then I can take
this minus 1. I could add 1 to both sides and
I get x plus 3y minus 2z-- add that 1 to both sidess--
is equal to 1. And there you have it. Just by using the simple fact
that this is a point on the plane and this is a normal
vector, I was able to use the idea that this has to be normal
or its dot product with any point, with any vector that
lies on the plane, I was able to get this right here. I didn't have to go
through this whole business right here. You could have just used this
formula right here. You could have just said n1
is 1 times x minus x1. Or x0 I could call it. So x minus this 1 plus n2, 3
times y minus 2 plus minus 2 times z minus 3 is equal to 0. And then if you just did
a little bit of math. a little bit of algebra, you
would've gotten there. So hopefully find this
reasonable useful. This is actually quite useful
if you ever have to do anything that involves any
type of three-dimensional mathematics. And if you ever become a game
programmer this would be-- there's thousands of
other applications. But this is kind of a useful
byproduct of some of the formal mathematics that
we've been doing.