WEBVTT

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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 today

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is the intersection of
two parametrized lines.

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So we have two lines here.

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L_1, given with
the parametrization

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in terms of the variable t.

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And L_2, also given
with the parametrization

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in terms of the variable t.

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So the first question
that I want us to answer

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is do these lines intersect?

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And if so, then we want to find
out where do they intersect.

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So why don't you pause the
video and work on this.

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And we can check
back in a moment

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and we'll see how I solved it.

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OK, welcome back.

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Let's get started.

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So we have these
two lines in space.

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Before we start doing
any computations,

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I find it useful
to draw a picture.

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So let's see what's going on.

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OK.

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So we have these two lines.

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We can just find some
common points on the lines.

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So, well, if we put
in t equals 0 here,

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then it looks like we
get the point 2 comma 1.

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OK.

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And now if we plug in, let's
say, t is minus 1 here,

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then we get-- if we plug in t
is minus 1 here, we get x is 3

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and y is 0.

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So there's our line L_1.

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And now let's see, L_2.

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If we plug in t equals 0,
we get 2-- 1, 2, 3-- 4.

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And if we plug in, let's
say, t equals minus 1 again,

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then we get 1 and 2.

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OK?

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So there is L_2.

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And indeed, it does look
like they intersect.

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We could have probably
guessed that they intersect

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by looking back over here at
the formulas for L_1 and L_2,

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because we see that the
sort of direction that this

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is moving in, we can
take derivatives in t.

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And we see that L_2
is, this line is moving

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in the direction 1 comma 2.

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And L_1 is moving in the
direction minus 1 comma 1.

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And so those directions
are not parallel.

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And so we know that
the only way the two

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lines could fail to intersect
is if they're parallel.

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So actually, even
without drawing this,

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we could have guessed that
these lines do intersect.

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So now we know that
these lines intersect.

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And in fact, it even
looks, you know,

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it kind of looks like they
intersect-- from our sketch--

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at the point-- it
looks like-- 1 comma 2.

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It seems to be the
point of intersection.

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But, you know, we got a little
bit lucky with our sketch here.

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So let's see if
that's actually true.

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Let's see if we can verify
this in the general way

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that we discussed in lecture.

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So, now there is one place
where we have to be careful.

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We have two lines here, and
we parametrized both of these

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with the variable t.

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But we need to keep
in mind that t is

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what's called a dummy variable.

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It doesn't have any geometric
meaning to the problem.

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And in particular, what I
want to caution you about

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is if we just start
solving these two

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equations algebraically
as they're

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given to us with the
variable t, the problem

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that we could run into
is that, you know,

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we're sort of
moving-- as we vary

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t-- we're moving along this
line, and as we vary t again,

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we're moving along this line.

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And you see we're
moving at the same time.

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And so that's really
solving a different problem.

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That's not asking about when
did these lines intersect,

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but that would be
asking about when

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do two particles on these
lines collide, which

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is a harder problem.

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So instead, what
we need to do is

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we need to give a change of
variables for the line L_2.

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So what I want to do is I'm
going to write L_2-- I'm just

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going to write the
same equations,

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but I'm going to introduce
a new variable u.

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So x is 2 plus u,
and y is 4 plus 2u.

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OK.

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So now once we've done that, to
find the point of intersection,

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well, the point
of intersection is

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going to precisely
be a point on L_2

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where the x-coordinate
and the y-coordinate

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agree with another point
on L_1 with the same

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x- and y-coordinate.

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So that is, we
have the-- what we

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want to do is we want to set
the x-coordinate for L_2.

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We want to set this equal to
the x-coordinate for L_1, which

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is 2 minus t.

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So this was for L_1.

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And similarly here, we want to
set the y-coordinate for L_2

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equal to the
y-coordinate for L_1.

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So now if you think
about it, if we

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can solve this system of
equations, then what we've done

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is we've simultaneously
found a point which

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is on L_1 and on L_2.

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And that's our goal.

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So that will be a
point of intersection.

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OK, so now we just
have this system

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of two linear equations
and two variables,

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and we just need to solve it.

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Now, we could do, you
know, the-- in general,

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with an equation like this, we
might try to add or subtract

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the equations.

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But this one is
so simple, that I

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see that the top equation
is just the same thing as t

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equaling to minus u.

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That's what the top equation
says if we cancel the 2's.

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And so if we plug that
into the next equation,

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then we get 4 plus
2u equals 1 minus u.

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And so then we can
solve this, and we get,

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so it looks like 3 equals
minus 3u, which tells us

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that u equals minus
1, and then that

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tells us that t equals plus 1.

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OK?

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So we found our
parameters t and u.

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And we're not quite done yet.

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What we need to do is we need to
go back to our parametrization.

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So let me go back over to our
original parametrization here,

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and we have L_1 was 2
minus t and 1 plus t.

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And over here, we
found that t equals 1

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was the value that we're after.

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So that tells us that x is
1-- 2 minus 1-- and y is 2.

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Excuse me.

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I wrote that in a funny way.

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x is 1 and y is 2.

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Now, just as a reality check.

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We also found that
if we solved for L_2,

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we wanted the variable u
to be equal to minus 1.

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So we had 2 plus
u, and 4 plus 2u.

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And so let's see
what happens when we

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plug in u equals minus 1 here.

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We again get x equals
2 plus minus 1 is 1.

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And y equals 4
plus minus 2 is 2.

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So we just double check that
this is a point of intersection

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of both lines.

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And I'll leave it at that.