### Video Transcript

In this video, weβre gonna take a look at a few vector addition and subtraction
questions that you might attempt after youβve had a go at the more basic introductory ones. In
this question, weβve got triangle ππ΄π΅ and weβre told that the point π is on the side π΄π΅ and that the
distance π΄π is a third of the distance from π΄ to π΅.

Now we got to express vector π΄π΅ in terms of
π and π, and weβve also got to express the vector ππ in terms of π and π. Now one of these is
much easier than the other, but letβs have a look at the question.

Now if we think about that
side π΄π΅, if π΄π is a third of the way from π΄ to π΅, then I can split that into three equal sized
chunks. Now the first one from π΄ to π, thatβs vector π΄π, thatβs repeated on the other two because
if itβs third the distance, then if I do the same thing three times, that will make the distance π΄π΅.

So here weβve laid the same vector in to in three times, and thatβs something that we will come back
and use later on in the question. So the f- the first part of the question, express π΄π΅ in terms of π and π.

How am I gonna get from point π΄ to point π΅? Well I could go in a straight line from π΄
to π΅, but that doesnβt give me an answer in terms of vectors π and vector π. So the other way I can
do it is go from π΄ to π and then from π to π΅. And in going from π΄ to π, Iβm going on
backtracking along an π vector; Iβm going the opposite direction of an π vector; thatβs
negative π.

And in going from π to π΅, thatβs just going a positive way along the π vector, so
thatβs just π. So the vector π΄π΅ can be expressed as the negative of an π vector plus the
positive of a π vector. Well there we go; thatβs the first part done.

Now for part b,
Iβve got to find out ππ. So Iβve got- Iβve got this journey here from π to π, and Iβve got to express that
in terms of π and π. Now the direct line from π to π doesnβt include any information about
vectors π΄ and π΅, so Iβve got to try and think about how can I get from π to π via other routes
which do have π and π in them.

So I could use this route from π to π΄ and then from π΄ to π. And we
know that π to π΄ is just simply the-the- this vector here π and from π΄ to π is a third of the way from
π΄ to π΅, and we just worked out what π΄ to π΅ was.

So letβs start writing that down. So ππ is
equal to π plus a third π΄π΅ and we know that π΄π΅ is negative π plus π, so ππ is π plus
a third of negative π plus π. And multiplying out the brackets, that means that ππ is
equal to π minus a third of π plus a third of π, so our answer is that vector ππ is equal to
two-thirds of vector π plus one-third of vector π.

Now in our next question, weβve got
a quadrilateral π΄π΅πΆπ· which is a parallelogram. πΈ and πΉ are points on the diagonal from π· to π΅, and
the distance from π· to πΈ is the same as the distance from πΈ to πΉ, which is the same as the
distance from πΉ to π΅.

Weβre told that π· to πΆ is a vector called π₯ and πΆ to π΅ is a vector called
π¦. What weβre gonna do is show that π΄πΉπΆπΈ is also a parallelogram. So looking on that diagonal
there, Iβve just shown that those three links are the same, and Iβve drawn in the shape π΄πΉπΆπΈ, and
we- basically to prove that thatβs a parallelogram, a parallelogram has got two pairs of
parallel sides.

So weβve got to prove that the vector from π΄ to πΉ is parallel to the vector from πΈ
to πΆ and the vector from π΄ to πΈ is parallel to the vector from πΉ to πΆ. If we can do that, we
know itβs a parallelogram.

Now because we know π΄π΅πΆπ· is a parallelogram, we know that π΄π΅ is
parallel to π·πΆ, so vector π΄π΅ is equal to vector π·πΆ, which is equal to vector π₯. So I can mark
that in. And likewise, the vector from π· to π΄ must be parallel to the vector from πΆ to π΅. In fact,
they must be equal because itβs a parallelogram, and that is equal to π¦. So letβs try and fill in some
of the other missing vectors on this diagram.

Well from π· to π΅, there are two ways we can go
direct from π· to π΅ or I can go from π· to πΆ and then from πΆ to π΅. And doing that, I can express
that in terms of vectors π₯ and π¦. So in going from π· to π΅, we can see that that is the same as
π₯ plus π¦. So in going from π· to πΈ, thatβs a third of the way from π· to π΅, and πΈ to
πΉ is the same and πΉ to π΅ is the same.

So all of those vectors are equal to a third of π₯
plus π¦, so we can add those to the diagram as well. Now our diagram is getting a little
bit cozy, but we can at least work out what some of the other vectors that we want to know π΄πΉ
and πΈπΆ and πΉπΆ and π΄πΈ and so on. So to go from π΄ to πΉ, Iβve got two ways. Iβve got lots of ways that I
can go. And one of them would be to go from π΄ to π΅ and then from π΅ to πΉ.

And I can see that
going from π΄ to π΅ is the same as vector π₯, and from π΅ to πΉ itβs the negative of going π₯ plus π¦, a
third of π₯ plus π¦. So I can write that down. Now, I can actually work out what they are. π₯ plus the
negative a third of π₯ is two-thirds of π₯, and then Iβve got- take away a third of π¦.

Now I need
to work out πΈ to πΆ and letβs hope we get the same vector. So from πΈ to πΆ, I can go back from πΈ
down to π· and then from π· across to πΆ. And this time going from πΈ to π· is the negative cause
weβre going backwards along that vector of a third π₯ plus π¦. And going from π· to πΆ, weβre just
moving along vector π₯. So thatβs minus a third π₯ plus π₯ is two-thirds π₯ and then weβve
got minus a third π¦.

So we can see that vector π΄πΉ, this side here, is equal to vector
πΈπΆ which is this side here. So those lengths of the same side and theyβre parallel; theyβre the same
length and theyβre parallel. Now if theyβre the same length and theyβre parallel, that means weβve
taken the same vector and just picked it up and moved it to a different place.

Now letβs look
at the other sides, π΄πΈ and πΉπΆ. And to get from π΄ to πΈ, I could go direct but that doesnβt tell me
anything about vectors π₯ and π¦, or I could go back the e- from e- back along the π¦ the negative of the π¦ and then
up from π· to πΈ; thatβs positive a third π₯ plus π¦. So the negative of π¦ plus a third of π¦ is
minus two-thirds π¦, and weβre adding a third of π₯ to that. So vector π΄πΈ is equivalent to a
third π₯ minus two-thirds of π¦.

Now letβs do the same for πΉ to πΆ. And I think the easiest way
to get from πΉ to πΆ apart from doing directly πΉ to πΆ is to go from πΉ up to π΅, which is
a third of π₯ plus π¦, and then come down from π΅ to πΆ, the negative of a π¦. And when I do
that, I find that Iβve got the same answer, a third π₯ minus two-thirds π¦.

So this
vector here and this vector here are both exactly the same vectors. That means theyβre the same
length and theyβre parallel. So just written that down in my solution, π΄πΈ is equal to πΉπΆ so theyβre
the same length and theyβre parallel.

So Iβm just gonna write down the definition of a
parallelogram; itβs a quadrilateral with two pairs of parallel edges. Now we gather the
information weβve got and see if we can show that that is a parallelogram. Now we saw that
vector π΄πΉ was equal to vector πΈπΆ, so that must mean that these two sides are parallel. And
likewise sides π΄πΈ and πΉπΆ are parallel because vector π΄πΈ is equal to vector πΉπΆ.

We can also show
that the shape π΄πΉπΆπΈ has got four sides. There we are, one two three four. Therefore, we meet all of
the criteria. And we can just write our conclusion: π΄πΉπΆπΈ is a parallelogram. So for our last
question then, weβve got parallelogram π΄π΅πΆπ· and weβve got a point πΈ which lies on the diagonal π΄πΆ,
so that the length of π΄πΈ is two-thirds of the length of π΄πΆ.

Weβve also got the vector from π· to
πΆ is three π₯ and the vector from π· to π΄ is called three π¦, and πΉ is the midpoint of π΅πΆ, so this
distance is the same as this distance. Weβve gotta prove that π·πΈπΉ, so this line going from π·
through πΈ to πΉ, is a straight line.

So our approach here is to find an equivalent vector for π·πΈ
and to find an equivalent vector for π·πΉ and prove that one is a simple multiple of the other.
And if one is a multiple of the other, then we know that theyβre parallel because the π₯ and π¦
are in the same proportions and also weβll know that π· is common to both of them. So not only
are they parallel, but theyβve got a point in common, so that will mean that they make a straight
line.

So thatβs gonna be our general approach. Letβs have a look at what weβre gonna do
first. So since there- since π΄π΅πΆπ· is a parallelogram, we can just fill in those two things. We
know that π΄π· and π΅πΆ are parallel, so πΆ to π΅ is three π¦, and we know that π΄π΅ and π·πΆ are parallel, so
π΄ to π΅ is the vector three π₯.

Now also because weβve got that line side π΅πΆ as three π¦, we know
that πΉ is the midpoint of those, we can fill in the vectors for πΆ to πΉ and for πΉ to π΅. Theyβre
simply half of three π¦.

Now the question told us that π΄πΈ is two-thirds of π΄πΆ, so if we can find
a vector for π΄ to πΆ in terms of π₯s and π¦s, then we can work out what the vector from π΄ to πΈ
is and weβve got piles in every direction through our parallelogram then. Now to get from π΄ to
πΆ, we could do the direct route, but we could also do this route here; we could come the wrong way
down a three π¦ vector and then we could go forward through a three π₯ vector. So thatβs negative
three π¦ plus three π₯.

And as we said, π΄πΈ is two-thirds of π΄πΆ, so itβs gonna be equal to two-thirds
of negative three π¦ plus three π₯. And we can add that to our diagram then. π΄πΈ is two π₯
take away two π¦. So that just leaves us with πΈπΆ to fill out and we know that πΈπΆ is the
difference between π΄πΆ and π΄πΈ, so thatβs π΄πΆ take away π΄πΈ.

And we saw up here that π΄πΆ was three
π₯ take away three π¦, so three π₯ take away three π¦ take away two π₯ take away two π¦ gives us three π₯
take away two π₯ and negative three π¦ add two π¦. So πΈπΆ comes out to be π₯ minus π¦, so we can
add that on here as well.

Right with all that information added to our diagram, now we can go
about the real business of trying to prove that π·πΈπΉ is a straight line. So how do we get from π·
to πΈ? Well we could come along here from π· to πΆ and then up to πΈ from there, so we can go from π·
to πΆ and then from πΆ to πΈ. And vector π·πΆ is just three π₯ and vector πΆ to πΈ; well we know that πΈ to
πΆ was π₯ minus π¦, so πΆ to πΈ is the opposite of that, the opposite direction is negative of π₯
minus π¦.

We have to be careful with our negative signs here. But when we work all that out, weβve
got π·πΈ is equal to two π₯ plus π¦. So letβs work out now a vector for π· to πΉ. And I think the
easiest way to get from π· to πΉ is to go from π· to πΆ and then from πΆ up to πΉ. And from π· to πΆ, that
was a vector of three π₯, and from πΆ to πΉ thatβs three over two π¦.

So Iβm just gonna factorize
that out; take the three out as a common factor. And that shows us that π·πΉ is equal three lots
of π₯ plus a half π¦. Now letβs go back to π·πΈ. And I can see that if I factorize the two out there,
Iβve got two lots of π₯ plus a half π¦. So one of these is two lots of π₯ plus a half π¦, and the other
one is three lots of π₯ plus a half π¦. So one is simply a multiple of the other. So π·πΉ is equal
to three over two lots of π·πΈ. If I multiply π·πΈ by three over two, Iβll get exactly π·πΉ.

Now that
proves that theyβre parallel. So one vector being a simple scale a multiple of another shows that
they are parallel, and point π· is common to both lines π·πΈ and π·πΉ. So not only are they parallel,
but they have one point at least one point in the same line as well.

So this situation where
weβve got π·πΉ and π·πΈ being parallel and theyβve both got a point in common, we say that those two
vectors are collinear. In other words, they make a straight line. So we can say that π·πΈπΉ is a
straight line.