### Video Transcript

In this video, our topic is the
equation of a plane intercept and parametric forms. Weβll learn here how to write the
equation of a plane in these two forms. And weβll also see how they relate
to other ways of expressing a planeβs equation.

As we get started, letβs remind
ourselves of other ways weβve learned of writing out the equation of a plane. If we have a plane existing in
three-dimensional space and we know a vector thatβs normal to that plane as well as
a point that lies in it, then we can define a vector β here weβve called it π« zero
β traveling from the origin of our coordinate frame to our known point. And we can define a vector β we
called it π« β going from the origin of our frame to an arbitrary point π in our
plane so that if we subtract π« zero from π«, we get a vector thatβs entirely in
this plane. Therefore, the dot product of this
vector and our normal vector π§ must be zero.

Equivalently, we can write that π§
dot π« is equal to π§ dot π« zero. And this is what is known as the
vector form of the equation of a plane. Now, if we say that our known point
π zero has coordinates π₯ zero, π¦ zero, and π§ zero, that means we can write the
vector π« zero as a vector with those components. And so carrying out the two dot
products in the vector form of a planeβs equation, we get this result. And if we group together the terms
that have π, π, and π in common, then we arrive at what is known as the scalar
form of a planeβs equation.

And then to generate the general
form of the equation of a plane, we can come back to this line in our
calculation. And what weβll do is group together
everything on the right-hand side, and weβll call that negative π. If we then add π to both sides, we
arrive at this expression, which is the general form of a planeβs equation.

Knowing all this, weβre now ready
to look into two more ways of writing out the equation of a plane. As weβve seen, one is called the
intercept form and the other the parametric form. The idea behind the intercept form
is this. In general, a two-dimensional plane
in three-dimensional space will intersect the π₯-, π¦-, and π§-axes of our
coordinate frame. After all, even though weβve
represented our plane here as a four-sided object, we know that really all of these
sides extend infinitely far outward. Except in very exceptional cases
then, this plane will at some point intersect all three of our axes.

To see that more clearly, say that
we change our perspective so that now our plane is oriented to our axes like
this. If we were to mark in the points
where our plane intersects the π₯-, π¦-, and π§-axes, they might look like this. Say further that we know these
points of intersection. Along the π₯-axis, for example, our
point has coordinates capital π΄, zero, zero. Going further, on the π¦-axis,
weβll say this intersection occurs a distance capital π΅ from the origin and then a
distance of capital πΆ along the π§-axis.

Now, knowing these three points, we
can insert them one by one into the general form of the equation of our plane. For example, inserting the
intersection point along the π₯-axis into our equation, we would get π times
capital π΄ plus π times zero plus π times zero plus π equals zero. We see that this simplifies to π
times capital π΄ plus π equals zero, which then tells us if we solve for lowercase
π that it equals negative π over capital π΄.

What we can do is store that result
off to the side and next use the intersection point along the π¦-axis in the same
way. Substituting this point into our
general form, we get π times zero plus π times capital π΅ plus π times zero plus
π equals zero. This simplifies to π times capital
π΅ plus π equals zero, which tells us that π is equal to negative π over capital
π΅. Weβll record this result too. And finally, weβll substitute in
the intersection point along the π§-axis. π times zero plus π times zero
plus π times capital πΆ plus π equals zero. So π times capital πΆ plus π is
zero or π equals negative π over capital πΆ.

Now that we have expressions for
π, π, and π in terms of the points of intersection of our plane, we can return
once more to our general form and substitute these values for π, π, and π into
this equation. That gives us this equation. And as we look at the left-hand
side, we notice that our factor of π is common to all four terms. So hereβs what we can do. If we subtract π from both sides
and then divide both sides by negative π, notice that on the right-hand side, weβll
get positive one. And on the left-hand side, all
instances of the variable π cancel out.

Since we were dividing by negative
π, we also lose our minus signs. And so we end up with this
expression. This is the intercept form of the
equation of our plane, where the plane intersects the π₯-, π¦-, and π§-axes at the
points weβve identified.

This brings us to the last form of
the equation of a plane weβll talk about. As our start point, letβs go back
to our plane and letβs say that we know some point on it, not the points weβve
identified, but some other one somewhere on the planeβs surface. To completely define this plane, to
be able to access any point in the plane, we want to be able to start at this point
we know to be in the plane and then move out any distance in any direction thatβs
also in the plane. Letβs say that this known point is
π zero.

And in order to access all points
on the plane from π zero, what we need are two vectors β weβll call them π― one and
π― two β that lie in the plane and are not parallel. The idea here is that if each one
of these two vectors is multiplied by a scale factor β weβll call them π‘ one and π‘
two, respectively β then by letting π‘ one and π‘ two range over all possible
negative and positive values, we allow this expression to sweep through all the
points in our plane.

To say this another way, letβs
return to our sketch where we have π zero and these three points of
intersection. Since our intersection points all
lie in the plane, by definition, any vector going from one of these points to
another will also be in the plane. What we do then is move out on a
vector starting at our origin until we reach the known point in the plane π
zero. Mathematically, we can represent
that as a vector π« zero. If we then add to that value in the
plane all possible multiples of our nonparallel vectors π― one and π― two, then we
will have comprehensively covered all points in our plane. Weβll have defined it.

In this equation, the values π‘ one
and π‘ two are known as our parameters. They are the values we can vary in
order to sweep through or scan over all the points in our plane. Now, because weβre looking at an
equation for a three-dimensional vector, really there are three separate equations
involved here. If we write π« as a vector with
components π₯, π¦, and π§; our zero as a vector with components π₯ zero, π¦ zero,
and π§ zero; and likewise π― one and π― two in terms of their components, then we
can write out separate equations for π₯, π¦, and π§ like this. The parameters π‘ one and π‘ two
appear in all three equations. And this then is the parametric
form of the equation of a plane.

So we now know five different ways
to write out a planeβs equation. Letβs now get some practice with
these forms through a few example exercises.

Find, in parametric form, the
equation of the plane that passes through the point π΄ one, two, one and the two
vectors π one equals one, negative one, two and π two equals two, negative one,
one.

Okay, we have here all these
possible answers for the parametric form of the equation of our plane. And as weβve seen, this plane
passes through this point π΄ and contains these two vectors π one and π two. Visually then, our plane might look
like this. In writing the equation of this
plane, the idea is that we start at our known point and then we move out from that
point in multiples of the vectors π one and π two. We could say then that a vector
describing our planeβs entire surface β weβll call it π« β is equal to a vector to
our known point on the plane plus one parameter that varies across all possible
numbers multiplied by our one vector π one added to another parameter that also
varies across all possible numbers multiplied by the vector π two.

Even though it may seem weβre
looking at just one equation here, actually, there are three involved: one for the
π₯-dimension, one for the π¦, and one for the π§. To see that, we can replace this
vector π« with its components. And now we see that, for example,
π₯ is equal to one plus π‘ one times one plus π‘ two times two. Then, likewise, π¦ is equal to two
plus π‘ one times negative one plus π‘ two times negative one. And then, lastly, we also have an
equation for the π§-component of our vector. π§ equals one plus two times π‘ one
plus π‘ two. This is the parametric form of the
equation of our plane. And if we look through our answer
options, we see it matches up with option (A). The equation of our plane in
parametric form is π₯ equals one plus π‘ one plus two times π‘ two, π¦ equals two
minus π‘ one minus π‘ two, π§ equals one plus two π‘ one plus π‘ two.

Now letβs look at an example where
we solve for the parametric form of a planeβs equation using three points.

Find the parametric form of the
equation of the plane that passes through the points π΄ one, five, one; π΅ three,
four, three; and πΆ two, three, four.

Okay, so we have this plane that
contains these three points π΄, π΅, and πΆ. And we want to figure out which of
these five options gives us the parametric form of the equation of the plane. We can recall that to define the
parametric form of a planeβs equation, we need to know one point on the plane and
two vectors that lie in it. Here we have three points. And we can actually use these
points to define two coplanar vectors. For example, if we subtract point
π΄ from point π΅, we get this vector in pink. And likewise, if we subtract π΄
from πΆ, we get this vector.

Substituting in the coordinates of
these points, for π΅ minus π΄, three minus one is two, four minus five is negative
one, and three minus one is two. And then, for πΆ minus π΄, two
minus one is one, three minus five is negative two, and four minus one is three. And what weβre saying is that these
coordinates are actually the components of two vectors in our plane. Weβll call them π― one and π―
two.

And now note that we have a point β
weβve picked point π΄ β as well as two vectors that lie in this plane. We can now recall the most general
way to write the parametric form of a planeβs equation. Itβs in terms of a point in the
plane that we have a vector going to and two vectors, π― one and π― two, that lie in
the plane, each of which is multiplied by its own parameter.

Applying this equation to our
scenario, we can write that the vector π« with components π₯, π¦, π§ is equal to a
vector to our known point, one, five, one, plus a parameter π‘ one times our first
vector π― one plus a second parameter π‘ two times π― two. Note that from this expression, we
can get equations for π₯ and π¦ and π§. For example, π₯ is equal to one
plus two times π‘ one plus one times π‘ two. π¦ is equal to five minus one times
π‘ one minus two times π‘ two. And then π§ equals one plus two
times π‘ one plus three times π‘ two. And these equations altogether are
the parametric form of the equation of our plane. Looking over our answer options, we
see a match with choice (E).

Now, itβs true that there are many
equivalent ways to express the parametric form of a planeβs equation. For example, we mightβve picked a
different point, say π΅ or πΆ, rather than point π΄. And we mightβve solved for
different coplanar vectors than these two here. Nonetheless, we still wouldnβt have
chosen options (A), (B), (C), or (D) because none of those options use point π΄, π΅,
or πΆ as the point that lies in the plane.

Our final answer then is choice
(E). π₯ equals one plus two times π‘ one
plus π‘ two, π¦ equals five minus π‘ one minus two π‘ two, and π§ equals one plus
two π‘ one plus three π‘ two.

Letβs now look at an example where
we start with the parametric form of a planeβs equation and convert to another
form.

Find the general equation of the
plane π₯ equals four plus seven π‘ one plus four π‘ two, π¦ equals negative three
minus four π‘ two, π§ equals one plus three π‘ one.

Alright, so in this exercise, weβre
given the equation of a plane in parametric form. That means we have three equations
for π₯, π¦, and π§. And theyβre written in terms of two
parameters, π‘ one and π‘ two. We want to convert from the
parametric form to the general form of the equation of this plane. And we can remember that the
general or Cartesian form of a planeβs equation is given by ππ₯ plus ππ¦ plus ππ§
plus π equaling zero. So the question is, how do we
express these equations this way?

We can start by remembering that
the parametric form of a planeβs equation involves one point in the plane and two
vectors that also lie in it. We can write that mathematically
like this. This equation means that we can get
to any point in our plane by starting at a known point in the plane and then moving
out from that in these two different directions defined by these vectors π― one and
π― two, where these vectors critically are varied by these parameters π‘ one and π‘
two.

Now, these three given equations
for π₯, π¦, and π§ fit into this general parametric form. What we mean is that, for example,
π₯ is equal to four β that would be π₯ zero in this form β plus seven times π‘ one β
so π― one π₯ would be seven β plus four times π‘ two β so π― two π₯ would be
four. Carrying this out for π₯, π¦, and
π§ altogether, we get this result, which tells us that according to these parametric
equations, our plane contains the point four, negative three, one and the vectors
seven, zero, three and four, negative four, zero.

Now, at this point, letβs recall
that the general form of a plane is based on defining a plane in terms of a vector
normal to it and one point on it, while on the other hand the parametric form comes
from defining a plane in terms of a point on it and two vectors that lie in it. To go from parametric to general
form then, we need to take these two coplanar vectors and combine them in some way
to get a vector normal to the plane.

We can do exactly this by taking
the cross product of our two coplanar vectors. Weβve called them π― one and π―
two. Note that the components of π― one
are seven, zero, three and those of π― two are four, negative four, zero. And so those are the values we use
in the second two rows of our three-by-three matrix. Calculating this determinant, we
get π’ times zero minus negative 12 minus π£ times zero minus 12 plus π€ times
negative 28 minus zero. Thatβs equal to 12π’ plus 12π£
minus 28π€ or, written another way, 12, 12, negative 28.

Notice that if we divide all the
components by four in this vector, then we will get a vector that is reduced but
that is still normal to the plane of interest. For simplicityβs sake then, weβll
say that this is our normal vector π§.

Okay, so now weβve got a vector
thatβs normal to our plane. And we also have a point that lies
in it: four, negative three, one. We can use this information to
write the equation of our plane as our normal vector dotted with a vector to a
general point in the plane being equal to our normal vector dotted with a vector to
a point we know to lie in the plane. Carrying out those dot products
gives us three π₯ plus three π¦ minus seven π§ being equal to 12 minus nine minus
seven, which is equal to negative four.

With our equation written this way,
we can see weβre very close to the general form of a plane. As a last step, letβs add positive
four to both sides of this equation, leaving us with this result. So the general form of the equation
of our plane is three π₯ plus three π¦ minus seven π§ plus four equals zero.

Letβs finish up this lesson now by
summarizing a few key points. We started this lesson reminding
ourselves that the equation of a plane can be written in vector, scalar, and general
forms. We then saw that we can also use
whatβs called intercept form, where we use the π₯-, π¦-, and π§-coordinates where
our plane intersects the π₯-, π¦-, and π§-axes to define the plane. And lastly, we learned itβs
possible to write the equation of a plane in whatβs called parametric form. This involves using two vectors
that lie in the plane as well as a vector to a point in the plane to generate a set
of equations describing the π₯-, π¦-, and π§-coordinates of all the points in the
plane.