### Video Transcript

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.