### Video Transcript

In this video, we’re talking about
the equation of a plane: vector, scalar, and general forms. We see then that there are several
different ways to represent a plane, which is a two-dimensional surface. In this lesson, we’ll learn these
different forms, and we’ll see how they’re related to one another.

Let’s get started by considering
this plane. Even though we’ve drawn in all four
sides, we know that, really, this plane goes infinitely far in these directions as a
two-dimensional surface. And in order to define a plane, we
need to have a frame of reference. With that frame of reference in
place, our objective is to write the equation of a plane by knowing two things about
it. First, let’s assume we have a
vector, we’ve called it 𝐧, that is normal or perpendicular to the plane. And second, let’s assume we also
know a point on the plane. We’ve called it 𝑃 zero. With these two ingredients, we
could call them, we can use vectors to develop a vector equation for this plane.

Since 𝑃 zero is a known point, we
know the three coordinates of that point. We’ll call them 𝑥 zero, 𝑦 zero,
and 𝑧 zero. And that means if we draw a vector
from the origin of our coordinate frame to 𝑃 zero, then that vector, we can call it
𝐫 zero, has components 𝑥 zero, 𝑦 zero, and 𝑧 zero. So far, then, we have these two
vectors, our normal vector and 𝐫 zero. But we can see that there’s no
particular spatial relation between them. In general, they won’t be, for
example, parallel or perpendicular.

But now let’s say that we pick an
arbitrary point elsewhere in our plane. And we’ll call this point 𝑃. Since 𝑃 is a general unknown
point, we’ll give it general coordinates 𝑥, 𝑦, and 𝑧. And to this point also, we can draw
a vector from our origin. This vector we’ll call 𝐫 with
components 𝑥, 𝑦, 𝑧. And notice what we can do now with
our two vectors 𝐫 and 𝐫 zero. If we subtract them, specifically
if we take 𝐫 zero and subtract it from 𝐫, then we’ll get a resultant vector that
looks like this. And the important thing is that
this resultant, 𝐫 minus 𝐫 zero, is in the plane. That’s because it travels in a
straight line from one point in the plane to another.

Recalling that our normal vector 𝐧
is perpendicular to the plane, it must therefore also be true that 𝐧 is
perpendicular to 𝐫 minus 𝐫 zero. And that means we can write
this. The vector 𝐧 dotted with the
vector 𝐫 minus 𝐫 zero is equal to zero. This is so because the dot product
of two perpendicular vectors is zero; the vectors don’t overlap at all.

By writing this equation, we have
defined this plane. And notice that another way of
writing this equation is 𝐧 dot 𝐫 is equal to 𝐧 dot 𝐫 zero. Written this way, we have what’s
often called the vector form of a plane equation. And so we’ve reached our goal. Given a normal vector 𝐧 and a
point 𝑃 zero on a plane, we were able to define a specific vector 𝐫 zero based on
𝑃 zero and a general vector 𝐫 to a general point on the plane 𝑃 and combine these
vectors in a way that uniquely describes this plane.

So mission accomplished. But let’s recall that in addition
to the vector form of our plane, we also want to solve for the scalar and general
forms. We’ll work on those in order. And we can start on our scalar form
by recognizing that we know the components of the three vectors in this
expression. The normal vector 𝐧 has components
𝑎, 𝑏, 𝑐, our vector 𝐫 has components 𝑥, 𝑦, 𝑧, and 𝐫 zero has components 𝑥
zero, 𝑦 zero, 𝑧 zero. And now if we compute these dot
products multiplying together the respective components of each vector and then
adding those products, we find that 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 equals 𝑎 times 𝑥 zero
plus 𝑏 times 𝑦 zero plus 𝑐 times 𝑧 zero.

And then if we take the whole
right-hand side of this expression and subtract it from both sides of the equation
and also group together terms by 𝑎, 𝑏, and 𝑐, we’ll find that 𝑎 times the
quantity 𝑥 minus 𝑥 zero plus 𝑏 times the quantity 𝑦 minus 𝑦 zero plus 𝑐 times
the quantity 𝑧 minus 𝑧 zero all is equal to zero. Written this way, we have what’s
called the scalar form of a plane equation. And note that to get this form of
our equation, we plugged in for our vectors in the vector form and then rearranged
until only scalar quantities remain. The scalar form of a plane’s
equation is sometimes also called standard or component form.

Lastly, let’s look at what’s called
the general form of a plane’s equation. To solve for that, we can actually
return to this step right here, after we took the dot product on either side of our
equation. Say we were to take the entire
right-hand side of this equation and we call it 𝑑. If we do that and then subtract 𝑑
from both sides, we can write that 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 minus 𝑑 equals
zero. And arranged this way, we have
what’s called the general form of a plane’s equation.

So now we have a vector form, a
scalar form, and a general form for the equation of our plane. And note that the general and
scalar forms both came from the vector form. This means if we can remember this
mathematical relationship and what it means, then we’ll stand a good chance of being
able to generate all three equations. And we can also note that the basic
fact, which undergirds the vector form of our equation, is that if we take the dot
product of two vectors that are perpendicular to one another, that product is
zero.

Knowing all this, let’s get some
practice with these ideas now through an example.

Find the vector form of the
equation of the plane that has normal vector 𝐧 equals 𝐢 hat plus 𝐣 hat plus 𝐤
hat and contains the point two, six, six.

Okay, so in this example, we have a
plane. We’ll say this is it. And we’re told that relative to
some set of coordinate axes, there’s a vector normal or perpendicular to the
plane. And if we write that vector in
vector form, we see that it has components of one, one, and one in the 𝑥-, 𝑦-, and
𝑧-directions, respectively. Along with this, we’re told that
the plane contains a point, we’ll call it 𝑃 zero, with coordinates two, six,
six. And knowing all this, we want to
solve for the vector form of the plane’s equation.

To write the plane’s equation that
way, we’ll want to define a vector that lies in the plane so that if we take the dot
product of that vector and the normal vector 𝐧, we get zero. Here’s how we can go about doing
that. First, let’s define a vector from
our origin to the point 𝑃 zero. We’ll call the vector 𝐫 zero. And since it comes from the origin,
it must have components two, six, six. And next, let’s do this. Let’s pick a point at random in our
plane, we’ll call it 𝑃, and we’ll say this point has coordinates 𝑥, 𝑦, 𝑧.

We don’t specify what these values
are, but nonetheless this point will help us because now we can draw a vector from
our origin to this point 𝑃, call that vector 𝐫, which we see has components 𝑥,
𝑦, 𝑧. And then if we subtract the vector
𝐫 zero from 𝐫, and this was the whole purpose of defining 𝐫 in the first place,
then we get a vector shown here, which lies in the plane, which means that this
vector is indeed perpendicular to the normal vector 𝐧. And that means if we take the dot
product of 𝐧 and 𝐫 minus 𝐫 zero, the result we’ll get is zero.

Another way to write this is 𝐧 dot
𝐫 is equal to 𝐧 dot 𝐫 zero. And we can now substitute in the
known values for the normal vector 𝐧 and the vector 𝐫 zero. We’ll do that down here, where we
see that the vector one, one, one dotted with 𝐫 is equal to one, one, one dotted
with two, six, six. We see the vector 𝐫 is made up of
the components 𝑥, 𝑦, 𝑧. But since we don’t know those more
specifically, we’ll leave this left-hand side as it is. On the right-hand side, though, we
can compute this dot product by multiplying the respective components together,
giving us two plus six plus six.

And since these numbers add up to
14, we can write a simplified form of our equation like this. This is the vector form of the
equation of our plane.

Let’s now look at another example,
where we give the equation of a plane in a different form.

Give the equation of the plane with
normal vector 10, eight, three that contains the point 10, five, five.

Okay, let’s say that this here is
our plane. And we’re told that it has a normal
vector, a vector perpendicular to the plane with components 10, eight, and
three. And along with this, the plane
contains the point, we’ve called 𝑃 zero, which has coordinates 10, five, and
five. We want to solve for the equation
of this plane. And we’ll solve for what’s called
the general or normal form of that equation.

Given a vector normal to our plane
with components 𝑎, 𝑏, and 𝑐, the general form of the plane’s equation is given by
this expression, where 𝑑 is equal to the dot product of our normal vector and a
vector called 𝐫 zero. In the case of our plane, we don’t
have an 𝐫 zero vector, but we do have a point 𝑃 zero. So if we draw a vector from the
origin of our coordinate frame to 𝑃 zero, we can call that 𝐫 zero, where its
components are the same as the coordinates of 𝑃 zero.

Returning to our general form of
the equation of a plane, we can see that to write this expression out for a
particular instance, we’ll need to know the components of a normal vector as well as
the components of this vector 𝐫 zero. We now have that information for
our plane, with 𝐧 having components 10, eight, three and 𝐫 zero having components
10, five, five. So applying this equation to our
scenario, here is what we can write. 10𝑥 plus eight 𝑦 plus three 𝑧,
where 10, eight, and three are the respective components of 𝐧, minus that normal
vector 𝐧 dotted with the vector to our point 𝑃 zero, that vector is 𝐫 zero, is
all equal to zero.

To simplify this further, let’s
take the dot product of these two vectors. In doing this, we multiply 10 by
10, eight by five, and three by five so that this dot product is equal to 100 plus
40 plus 15. And adding those values up, we get
155. This, then, is the expression that
gives us the general form for the equation of our plane.

Let’s look now at another example,
where we solve for the normal form of a plane’s equation.

Write, in normal form, the equation
of the plane one, zero, three; one, two, negative one; and six, one, six.

Okay, so let’s say that this here
is our plane defined with respect to these coordinate axes. And we’re told that our plane has
these three points, we’ve called them 𝑃 one, 𝑃 two, and 𝑃 three, in it. We’re given the coordinates of
these three points. And based on this, we want to solve
for the equation of the plane in normal form.

In order to do that, we need two
things: first, a normal vector, we can call it 𝐧, perpendicular or normal to our
plane and also a point that’s located on our plane. And we see that we actually have
three of those. Since we already have that second
bit of information, let’s look into how we would solve for a vector that’s normal to
our plane.

To start, note that we could draw
vectors from the origin of our coordinate frame to these three points. And just like the points are 𝑃
one, 𝑃 two, and 𝑃 three, we can call these vectors 𝐫 one, 𝐫 two, and 𝐫 three,
respectively. Say next that we take our vector 𝐫
two and we subtract it from the vector 𝐫 one. Graphically, that would make a
vector that lies in our plane that looks like this. And since the vector 𝐫 one has
components which are equal to the coordinates of point 𝑃 one and the vector 𝐫 two
has components that are equal to the coordinates of point 𝑃 two, we can actually
compute the components of this resultant vector.

We see it’s equal to zero, negative
two, four. So that’s one vector that lies
entirely in our plane. And we can then go about
identifying another by considering the vectors 𝐫 two and 𝐫 three this time. If we subtract 𝐫 three from 𝐫 two
on our plane, that would give us a vector that looks this way. And once again, we can use the
coordinates of our points translated to components of vectors to solve for this
vector. One minus six is negative five. Two minus one is one. Negative one minus six is negative
seven. And so we now have the components
of a second vector that also lies in our plane.

Knowing this, we can say that if we
were to take the cross product of these two vectors, then the result would be
perpendicular to them both. And since both these vectors lie in
the plane, that resultant would be normal to it. In other words, we would have a
normal vector 𝐧. And that’s good news because if we
had a normal vector, we would have all the information we need to write the equation
of our plane in normal form.

Let’s recall, then, that if we were
to take the cross product of two three-dimensional vectors, we’ve called them 𝐀 and
𝐁, then that would be equal to the determinant of this three-by-three matrix. Note that the first row of this
matrix is the 𝐢-, 𝐣-, and 𝐤-unit vectors and the next two rows are the
corresponding components of our two vectors 𝐀 and 𝐁. The cross product of two vectors
yields a vector. And as we mentioned, this resultant
is perpendicular to both the vectors we used to calculate it.

In our situation, our vectors are
𝐫 one minus 𝐫 two and 𝐫 two minus 𝐫 three. And this cross product is equal to
the determinant of this matrix, where we still have to fill in the second and third
rows. Recall that these rows are
completed by the components of the two vectors we’re crossing together. Looking at our first vector, 𝐫 one
minus 𝐫 two, we see from its components that its 𝑥-component is zero, its
𝑦-component negative two, and its 𝑧-component four.

Considering next our second vector,
𝐫 two minus 𝐫 three, this has an 𝑥-component of negative five, a 𝑦-component of
one, and a 𝑧-component of negative seven. We’re now ready to calculate this
cross product. And we’ll start with the
𝐢-component. Striking through the row and column
containing this element, we know the 𝐢-component is equal to the determinant of
this two-by-two matrix. Negative two times negative seven
is positive 14. And from that, we subtract four
times one, or four. Moving on to the 𝐣-component,
that’s equal to negative the determinant of this two-by-two matrix, where zero times
negative seven is zero. And from that, we subtract four
times negative five.

Lastly, with our 𝐤-component, this
equals the determinant of this matrix, which we see is zero times one, or zero,
minus negative two times negative five. That’s positive 10. If we add all these components
together, the result is the cross product of 𝐫 two minus 𝐫 one and 𝐫 two minus 𝐫
three. This is equal to 10𝐢 minus 20𝐣
minus 10𝐤 or, written in vector form, 10, negative 20, negative 10. So now finally, we have 𝐧, a
vector that’s normal to our plane. And as we said, this is because the
two vectors we crossed together are both in the plane. So by the property of the cross
product, their resultant must be perpendicular to them both.

Knowing 𝐧, let’s clear some space
and now start writing out the equation of our plane in normal form. We mentioned earlier that to do
this, we need a normal vector, which we now have, as well as a point in the
plane. We’ve got three of those. Let’s choose to work with point 𝑃
one. We can recall that in general, the
vector form of the equation of a plane is given by this expression. It tells us that a vector normal to
the plane dotted with a vector to a generic or general point in the plane is equal
to the dot product of that normal vector and a vector to a known point.

As we apply this relationship,
we’ll use our normal vector 𝐧 and 𝐫, we’ll say, has components 𝑥, 𝑦, and 𝑧. For the vector to a known point in
our plane, we’ll use the vector we defined, 𝐫 one. Substituting in for 𝐧 and 𝐫 one,
we get this expression. And we can now take these two dot
products. Multiplying the respective
components of these vectors together, on the left we get 10𝑥 minus 20𝑦 minus 10𝑧
and on the right we get 10 times one minus 20 times zero minus 10 times three. That adds up to negative 20.

We’re almost there. But to get the equation for our
plane in normal form, we’ll want a zero to appear on the right-hand side. So if we add positive 20 to this
side and this side of the equation, we’ll find that 10𝑥 minus 20𝑦 minus 10𝑧 plus
20 equals zero. And note that if we divide both
sides of this equation by 10, then that yields this expression. 𝑥 minus two 𝑦 minus 𝑧 plus two
equals zero. And this is our final answer for
the normal form of the equation of our plane.

Let’s now finish up our lesson by
reviewing some key points. In this lesson, we’ve seen that a
plane can be uniquely defined by a normal vector and a point on the plane. Given a normal vector with
components 𝑎, 𝑏, and 𝑐 and a point 𝑃 zero in the plane with coordinates 𝑥 zero,
𝑦 zero, and 𝑧 zero, we can define a vector 𝐫 zero from the origin of our
coordinate frame to 𝑃 zero. And choosing another point at
random on our plane and calling it 𝑃, we can define a vector to that point, calling
it 𝐫. The vector 𝐫 minus 𝐫 zero is in
the plane and perpendicular to the normal vector 𝐧 so that the equation of the
plane can be written in vector form like this, scalar form like this, or in general
form like this. These are three equivalent ways of
writing the equation of a plane.