DAVID JORDAN: Hello, and welcome back to recitation.
The problem I'd like to work with you right now is we have
a line which goes through two points that are given to us
explicitly, and we have a plane which is given to us by
an equation.
And what we want to know is where does this line intersect
this plane?
And so one thing I would suggest to get started is we
need to give a parametrization of our line to get started.
OK, so why don't you work on that, pause the tape, and
we'll come back in a moment and work it out together.
OK, welcome back.
Let's get started.
So let's start off by drawing a cartoon of
what's going on here.
So we have this plane sitting in space.
And we have some line kind of going through space.
So maybe it's like this.
And there's this single point of intersection.
So even from the cartoon, we can kind of, sort of see two
things which are going on.
Which is that we would expect a point of intersection, or we
would expect exactly one, if we choose kind of a generic
line and generic plane.
In order for there to be no points of intersection, we
would have to have a line which was parallel to the
plane, which is very unlikely.
And then otherwise, we expect exactly just one point of
intersection.
So I want to break this sort of into two components.
So we have an equation for the plane.
And when I see an equation describing a plane, I think of
that as a sort of test for membership.
We can plug in a point-- x, y, z--
to the equation, and we can ask, does this point make the
equation true, or doesn't it?
And if it does, then that point x, y, z is in the plane,
and otherwise it's not in the plane.
For the line, what we're going to need to do in a second is
we're going to need to come up with a parametrization.
And a parametrization is a different kind of thing than
an equation describing a line.
A parametrization, rather than being a test for membership,
is really a way of listing all the points on the line.
So when we give a parametrization--
in a second-- then we're going to be able to list all the
points in the line.
And then we're going to be able to plug our list into the
equation for the plane, and find out which point on our
list is actually in the plane.
Which one satisfies the membership equation.
So why don't we get started first with
parametrizing the line.
So the general kind of picture here is we have a point P1 in
space, and we have another point P2 in space, and we want
to parametrize the line which goes between them.
And there's actually a very simple way to do this.
What we do is we want to take our original point P1, and we
want to add a variable t times the vector P2 minus P1, which
connects them.
So that's this one here.
OK.
So this is a reasonable thing to do, because if we plug in t
equals 0, then we just get P1.
And if we plug in t equals 1, then we get P1
plus P2 minus P1.
We just get P2.
So this line definitely goes through those two points, and
that's all that we really need.
So in our specific problem here, we have P1--
we can take to be the first point--
0, minus 1, 1.
And then we have t times--
so we have 2 minus 0 is--
2--
3 minus a negative 1 is--
4--
and 3 minus 1 is--
2.
And this is the vector connecting those.
And so we can write, we can just combine these two and we
get 2t, 4t minus 1, and 2t plus 1.
OK?
So this here is a parametrization of the line.
So as we vary t-- now, walking back over to our picture-- as
we vary t, we're going to just be listing all the
points on the line.
And we're going to ask, for which point are we actually
contained in the plane?
So let's go over to the board over here and solve that.
So what we want to know is does this point satisfy the
equation for the plane?
And our plane was given to us by the equation 2x plus y
minus z equals 1.
So x on our line is 2t.
So we have 2 times 2t, plus--
y is--
4t minus 1, minus--
z is--
2t plus 1.
And all this is meant to equal 1.
OK.
So if we expand this out, we get 4t plus another 4t minus
2t and we get--
minus 1 minus another 1--
so we get minus 2 equals 1.
So altogether we get 6t equals 3, so that tells
us that t is 1/2.
OK?
And finally, to get our answer, we need to go back
over to our parametrization of the line and
plug in t equals 1/2.
So coming back over here, plugging in t
equals 1/2, we get 1--
2 minus 1 is--
1--
and 1 plus 1 is--
2.
OK, so we get the point of intersection 1, 1, 2.
So that was quite a few steps, so let's review what we did.
So to begin with, we needed to understand that the equation
for a plane is a test for membership.
It's not a list of all the points in the plane, it's a
test for membership.
The parametrization of the line, on the other hand, is a
way of listing all of the points on the line.
And so if our goal is to find which particular point on the
line is contained in the plane, then we need to
parametrize our line, and then we need to plug in our
parametrization to our equation for the plane, and
then solve for the value of t which makes it true.
Finding that t then we've, that's is equivalent to
finding a point on our line.
And I think I'll leave it at that.