WEBVTT

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JOEL LEWIS: Hi.

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Welcome to recitation.

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In lecture, you started learning
about vectors.

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Now vectors are going to be
really important throughout

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the whole of this course.

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And I wanted to give you one
problem just to work with them

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in a slightly different context
than what we're going

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to do in the future.

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So this is the context of
Euclidean geometry.

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So some of you have probably
seen this problem that we're

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going to solve, but you probably
haven't seen it

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solved with vectors.

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So let's take a look at it.

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So what I'd like you to do is
show that the three medians of

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a triangle intersect at a point,
and the point is 2/3 of

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the way from each vertex.

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So let me just remind you
of some terminology.

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So in a triangle, a median is
the segment that connects one

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vertex to the midpoint
of the opposite side.

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So here, this point M is exactly
halfway between B and

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C.

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So OK.

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So every triangle has three
medians-- one from each vertex

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connected to the midpoint of the
opposite side-- and what

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I'm asking you to show is that
these three medians all

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intersect in the same point.

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And also, that this point
divides the median into two

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pieces, and the big piece
is twice as large

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as the small piece.

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So this is 2/3 of the median,
and this is 1/3 of the median.

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So, why don't you take a few
minutes, work that out-- try

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and do it using vectors
as much as possible--

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pause the video, come
back, and we

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can work on it together.

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So hopefully you had some luck
working on this problem.

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Let's get started on it.

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So to start, I actually want
to rephrase the question a

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little bit.

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And I'll rephrase it to an
equivalent question that's a

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little bit more clear about how
we want to get started.

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So another way to say this
problem is that it's

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asking us to show--

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so for each median, say this
median AM here, where M is the

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midpoint of side BC--

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there exists a point on the
median that divides it into a

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2:1 ratio, so the point that's
2/3 from the vertex to the

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midpoint of the opposite side.

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So for example, you know,
there's a point--

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so, let's call it P,
say, at first--

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so there's a point P such
that AP is twice PM.

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OK?

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And similarly, there's some
point-- maybe called Q--

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that's 2/3 of the way from B to
the midpoint of this side.

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And there's some point that's
2/3 of the way from C to the

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midpoint of this side.

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And so an equivalent formulation
of the question is

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to show that these
three points are

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really the same point.

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That they're all in
the same place.

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So one way we can do that is
that we can compare the

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position vectors of those
three points.

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And if those three points all
have the same position vector,

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then they're all in exactly
the same position.

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So in order to do that
we need some origin.

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And it happens that for this
problem, it doesn't matter

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where the origin is, and so
I'm not going to draw an

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origin, but I'm going
to call it O.

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So we're going to set up a
vector coordinate system with

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origin O. And now I want to look
at what the vector from O

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to P is in terms of the vectors
connecting O to A, B,

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and C. Right?

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Those are the vectors
that determine the

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vertices of the triangle.

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And so I want to relate the
location of P to the locations

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of A, B, and C.

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So the first thing
to do is that--

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well, in order to talk about
where P is, I know how P is

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related to A and M and I know
how M is related to B and C.

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So let's first figure out what
the position vector of M is in

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terms of the position vectors
of A, B, and C, and then we

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can use that to figure out
the position vector of P.

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So M is the midpoint
of the segment BC.

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So I think we saw
this in lecture.

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What this means is that the
position vector OM is exactly

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the average of the position
vectors of B and C. It's 1/2

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of the quantity OB plus OC.

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All right?

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So it's easy to express the
position vector of the

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midpoint of a segment in
terms of the position

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vectors of the endpoints.

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You just add the position
vectors of the endpoints and

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divide by 2.

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So if you like, this is
equivalent to the geometric

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fact that the diagonals of a

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parallelogram bisect each other.

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So that's the position
vector of M.

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Now we have to figure out what
the position vector of P is.

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So in order to do this we can
note, that in order to get

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from the origin to point P,
well, what we have to go from

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the origin-- wherever it is--
to A, and then we have to go

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from A 2/3 of the way
to M. All right?

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So the vector OP is equal to OA
plus 2/3 of the vector AM.

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Right?

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Because we go 2/3 of the way
from A to M in order to get

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from A to P. This is because
we've chosen P to be the point

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that divides segment AM into
a 2:1 ratio so that

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AP is 2/3 of AM.

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OK.

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So good.

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So now we need the vector AM.

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Well, we know what the position
vector of A is.

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It's just OA.

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And we also know what the
position vector of M is.

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It's OM.

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So that means that AM is
just the difference

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of those two vectors.

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It's going to be OM minus OA.

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Another way to say this is that
if you add OA to both

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sides, you have that OA
plus AM equals OM.

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In other words, to go from O to
M, first you can go from O

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to A, and then go from
A to M. All right.

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And I've just subtracted OA
onto the other side here.

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So we can write AM in
terms of OM and OA.

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And we also, we have an
expression for OM here in

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terms of OB and OC.

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So that means we can get an
expression for AM in terms of

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OA, OB, and OC.

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So let's do that.

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So that's just by substituting
from here into here.

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So if I do that, I get that AM
is equal to-- so OM is 1/2 OB

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plus 1/2 OC, and now
I just subtract OA.

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All right.

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So that's what AM is, putting
these two equations together.

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I get that that's AM.

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And so now I need to figure
out what OP is.

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So for OP, I just need to
substitute in this new

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expression that I've
got for AM.

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So I have OP is equal to, well
it's equal to OA plus 2/3 of

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what I've written just
right above--

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2/3 of AM.

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So that's 1/2 OB plus
1/2 OC minus OA.

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OK.

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And so now you can multiply this
2/3 in-- you know, just

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distribute the scalar
multiplication across the

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addition there-- and then
we can rearrange.

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We'll have two terms involving
OA and we can combine them.

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So we'll see we have a
plus OA minus 2/3 OA.

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So that's going to be
equal to 1/3 OA.

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And then we have, OK so
2/3 times 1/2 OB.

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So that's plus 1/3
OB plus 1/3 OC.

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So this gives us a simple
formula for the position

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vector of P--

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that vector OP--

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in terms of the position vectors
of A, B, and C. So in

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particular, it's actually
because P is the special

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point, it's 1/3 of their sum.

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Of the sum OA plus OB plus OC.

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OK, so that's where P is.

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Now to finish the problem, I
just have to show that this is

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the same location as
the point that

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trisects the other medians.

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So how would I do that?

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Well, I could go back to my
triangle and I could do

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exactly the same thing.

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So I could-- maybe I'll give
this point a name, also.

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I'll call this midpoint
N, say.

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So I could let Q be the point
that lies 2/3 of the way from

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B to N. And then I could write
down the position vector of N

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in terms of OA, OB, and OC.

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And then I can use that to write
down the position vector

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of Q in terms of OA, OB, and
OC, and I'll get some

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expression.

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And what will happen
at the end--

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I hope if I'm lucky--

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that expression will be equal
to this expression that I

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found over here.

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OK?

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So you can go through and do
that, and if you do that, what

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you'll find is that
in fact it works.

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So there's actually a sort
of clever, shorter

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way of seeing that.

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Which is that this formula is
symmetric in A, B, and C.

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So that means if I just relabel
the points A, B, and

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C, this expression
for the position

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vector doesn't change.

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So rather than going through
that process that I just

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described, you can also say,
well, in order to look at, say

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Q, instead of P, what I need to
do is I just need to switch

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B and A. I need to do exactly
the same thing but the roles

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of A and B are interchanged.

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Well, if the roles of A and B
are interchanged, then in the

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resulting formula, I just have
to interchange the roles of A

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and B, but that won't change the
value of this expression.

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So by symmetry, the point
I get really is

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going to be the same.

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If you don't like that argument,
I invite you to go

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through this computation
again in the case

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of the other medians.

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In either case, what you'll find
is that the points that

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trisect the three medians all
have position vector 1/3 OA

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plus 1/3 OB plus 1/3
OC, but that means

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they're the same point.

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So what we've shown then, is
that the points that trisect

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the three medians-- that
trisect-- that divide them

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into 2:1 ratios from the vertex
to the midpoint of the

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opposite side, that those three
points all have the same

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position vector.

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So in fact, they're the same
point, and that's what we

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wanted to show, right?

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We wanted to show that there's
one point that trisects all

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three of those medians.

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So we've shown that the three
points that trisect them are

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actually the same.

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So that's the same conclusion,
phrased differently.

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So I think I'll end there.