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.