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DAVID JORDAN: Hello, and
welcome back to recitation.

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The problem I'd like to work
with you right now is we

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have a line which goes
through two points that

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are given to us
explicitly, and we

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have a plane which is
given to us by an equation.

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And what we want to know
is where does this line

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intersect this plane?

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And so one thing I would
suggest to get started

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is we need to give a
parametrization of our line

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to get started.

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OK, so why don't you work
on that, pause the tape,

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and we'll come back in a moment
and work it out together.

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OK, welcome back.

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Let's get started.

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So let's start off by
drawing a cartoon of what's

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going on here.

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So we have this plane
sitting in space.

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And we have some line kind
of going through space.

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So maybe it's like this.

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And there's this single
point of intersection.

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So even from the cartoon,
we can kind of, sort of

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see two things
which are going on.

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Which is that we would expect
a point of intersection,

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or we would expect
exactly one, if we

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choose kind of a generic
line and generic plane.

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In order for there to be
no points of intersection,

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we would have to
have a line which

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was parallel to the plane,
which is very unlikely.

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And then otherwise,
we expect exactly

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just one point of intersection.

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So I want to break this
sort of into two components.

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So we have an equation
for the plane.

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And when I see an equation
describing a plane,

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I think of that as a sort
of test for membership.

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We can plug in a point
(x, y, z) to the equation,

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and we can ask, does this
point make the equation

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true, or doesn't it?

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And if it does, then that point
(x, y, z) is in the plane,

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and otherwise it's
not in the plane.

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For the line, what we're going
to need to do in a second is

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we're going to need to come
up with a parametrization.

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And a parametrization
is a different kind

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of thing than an equation
describing a line.

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A parametrization, rather than
being a test for membership,

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is really a way of listing
all the points on the line.

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So when we give a
parametrization-- in a second--

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then we're going
to be able to list

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all the points in the line.

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And then we're going
to be able to plug

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our list into the
equation for the plane,

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and find out which point on our
list is actually in the plane.

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Which one satisfies the
membership equation.

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So why don't we
get started first

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with parametrizing the line.

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So the general kind
of picture here

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is we have a point P_1 in space,
and we have another point P_2

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in space, and we want to
parametrize the line which

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goes between them.

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And there's actually a
very simple way to do this.

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What we do is we want to
take our original point P_1,

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and we want to add a variable t
times the vector P_2 minus P_1

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which connects them.

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So that's this one here.

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OK.

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So this is a reasonable thing
to do, because if we plug in t

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equals 0, then we just get P_1.

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And if we plug in
t equals 1, then

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we get P_1 plus P_2 minus
P_1; we just get P_2.

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So this line definitely goes
through those two points,

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and that's all that
we really need.

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So in our specific
problem here, we

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have-- P_1 we can take to be
the first point, 0, minus 1, 1.

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And then we have t times--
so we have 2 minus 0 is 2-- 3

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minus a negative 1 is
4-- and 3 minus 1 is 2.

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And this is the vector
connecting those.

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And so we can write, we
can just combine these two

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and we get 2t, 4t
minus 1, and 2t plus 1.

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OK?

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So this here is a
parametrization of the line.

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So as we vary t-- now, walking
back over to our picture--

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as we vary t,
we're going to just

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be listing all the
points on the line.

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And we're going to
ask, for which point

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are we actually
contained in the plane?

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So let's go over to the board
over here and solve that.

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So what we want to know is does
this point satisfy the equation

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for the plane?

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And our plane was given to us by
the equation 2x plus y minus z

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equals 1.

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So x on our line is 2t.

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So we have 2 times 2t,
plus-- y is 4t minus 1,

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minus-- z is 2t plus 1.

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And all this is
meant to equal 1.

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OK.

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So if we expand this out, we
get 4t plus another 4t minus 2t

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and we get minus 1 minus
another 1-- so we get minus 2--

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equals 1.

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So altogether we
get 6t equals 3,

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so that tells us that t is 1/2.

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OK?

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And finally, to
get our answer, we

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need to go back over to our
parametrization of the line

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and plug in t equals 1/2.

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So coming back over here,
plugging in t equals 1/2,

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we get 1-- 2 minus 1 is
1-- and 1 plus 1 is 2.

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OK, so we get the point of
intersection, (1, 1, 2).

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So that was quite a few steps,
so let's review what we did.

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So to begin with, we
needed to understand

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that the equation for a plane
is a test for membership.

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It's not a list of all
the points in the plane,

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it's a test for membership.

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The parametrization of the
line, on the other hand,

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is a way of listing all
of the points on the line.

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And so if our goal
is to find which

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particular point on the line
is contained in the plane,

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then we need to
parametrize our line,

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and then we need to plug
in our parametrization

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to our equation for
the plane, and then

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solve for the value of
t which makes it true.

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Finding that t then we've--
that's is equivalent to finding

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a point on our line.

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And I think I'll
leave it at that.