### Video Transcript

We already know that a vector is a
set of numbers that can be represented in a suitable space by a line segment with a
specific length and direction. Weβve also seen that a line segment
has a magnitude and direction, which basically means that we can describe it by
saying how long it is and in which direction itβs pointing. In this video, weβre gonna talk
about horizontal and vertical components of two-dimensional vectors and introduce
the π and π unit vector notation.

Any two-dimensional vector has two
components. The first represents the amount of
movement in the π₯-direction and the second, the amount of movement in the
π¦-direction. Of course, when I say movement, Iβm
just talking about differences in π₯- and π¦-coordinates in a graphical
representation of the vector by a line segment on a graph. The vector itself might be
representing something completely different, force or acceleration, for example. So in this case, the amount of
π₯-movement is this distance here. So weβre going from an
π₯-coordinate of three to an π₯-coordinate of six, so thatβs a difference of plus
three. So our π₯-component of the vector
is positive three. The π¦-component is this bit
here. The π¦-coordinate of π΄ was two,
the π¦-coordinate of π΅ was nine, so thatβs a difference of plus seven. And we can write that as π΄π΅,
vector π΄π΅, in this format here: three, seven. So the three is the π₯-component
and the seven is the π¦-component.

And if we had a three-dimensional
coordinate system with π₯- and π¦- and π§-coordinates, we would just be inserting
another number at the end here onto our vector. So we can extend this system to any
number of dimensions.

We define two special vectors, π
and π, to be positive one in the π₯-direction or positive one in the π¦-direction,
respectively. So this is π, itβs just a movement
of one in the π₯-direction. And hereβs π, a movement of one in
the π¦-direction. So remember that the π and π
vectors β are at one, zero and zero, one β can be placed anywhere on the graph. They donβt have to start at the
origin. So there we are; I placed them
somewhere else. Each vector is just describing a
particular journey. In this particular case here, for
π, weβre adding one to the π₯-coordinate and weβre leaving the π¦-coordinate as it
is. So weβre doing this journey from
here to here. In the π case, weβre not adding
anything to the π₯-coordinate, but weβre adding one to the π¦-coordinate. It describes this movement from
here up to here.

Now remember, we can also use the
rules of adding and subtracting vectors to stack up these πs and πs, to create
bigger vectors. So, for example, this vector here,
π, represents a movement of one in the π₯-direction and none in the
π¦-direction. If I increase that to this vector
here, so this whole length here, that would be two πs following on from each
other. Or, it will be a translation of two
in the π₯-direction and zero in the π¦-direction. Now if I added on to that this
vector here β which starts at the end here and then moves up not one, not two, but
three β that would be three π vectors added on together, making a vector of three
π or zero, three. So if I add two π plus three π,
that represents this green journey from the start point to the end point up
here. So two π, the π₯-component would
be two. Three π, the π¦-component would be
three. And the direct journey two π plus
three π is the green line. So this means the π₯-component was
two and the π¦-component was three. So weβve now got two different ways
of representing this green vector here. We can use it in the standard
vector notation that weβre familiar with. But weβve got this new notation
here, in terms of π and π vectors. The number of steps in the
π₯-direction is the πs and the number of steps in the π¦-direction is the πs.

So letβs sum that up in the general
case. If we start off at point πΆ here
with coordinates π₯ one, π¦ one and we end up at point π· here with coordinates π₯
two, π¦ two, then vector πΆπ· is, the π₯-component here is the difference in the
π₯-coordinates and the π¦-component here is the difference in the
π¦-coordinates. So thatβs our standard vector way
of writing it. Or, we can say that that is π₯ two
minus π₯ one lots of the π unit vector plus π¦ two minus π¦ one lots of the π unit
vector. That make some kind of sense. All weβre saying is that the
π₯-component of a vector is the coefficient of π, in this format, and the
π¦-component of the vector is the coefficient of π, in this format. So weβve just got this new notation
where we have the π vector, which is a step of one in the π₯-direction, the π
vector, which is a step of one in the π¦-direction, and we just say how many of
those weβre doing in each case to make up our vector.

Right. So now we know this new format. Letβs just have a look at a couple
of quick questions that involve using πs and πs in our questions. So weβre going to do this question
here.

Weβve got π΄π΅; itβs the vector
three π plus four π. ππ is the vector negative two π
plus three π. And weβve just got to add those two
vectors together.

So the first stage is just to take
vector π΄π΅ and add vector ππ. And all we have to do is add the
πs together first, and then add the πs together second. So three π add negative two π is
just one π, so thatβs π. And four π add positive three π
is seven π. So thereβs our answer, π plus
seven π. Simple as that, add the
π-components together, add the π-components together, look out for the negative
signs and- when youβre doing those calculations; but otherwise, thatβs a pretty
straightforward process.

So letβs just visualise that
example. So we had three π plus four
π. So weβll be going positive three π
and then positive four π, this is the ππ vector. So there we are. Weβve just laid that down at the
origin. We couldβve laid it anywhere on
the-on the graph. Now adding vectors, we just lay
them end-to-end. So what we were adding, π₯π¦, was
negative two π plus three π. So weβre effectually starting π₯
from π΅, so π₯ lays on top of π΅ and weβre going negative two π. So weβre going two in the negative
π₯-direction and weβre going positive three π, up to here.

So adding vectors is just a matter
of laying them end-to-end on the graph. So we laid π΄π΅ down, which started
here and ended here, and then we just added ππ to the end of that, laid that onto
the end. So that started from where we just
finished off and then ended up here. So the resultant vector is this
green one here. And to get from the beginning of
the green vector to the end of the green vector, we had to go positive one in the
π₯-direction. So thatβs one π, or just π. And in the π¦-direction, weβre
going up seven all the way up here. So thatβs plus seven π.

So when youβre doing these
questions, it really is just a matter of adding the π₯-components together, adding
the π¦-components together, and coming up with a simple answer. You donβt need to do all this
graphical checking. But Iβm just hoping that thatβs
giving you some extra insight into the process and why it works.

Right. Letβs take a look at one final
question then.

Weβve got vector π΄π΅, is three π
take away four π. Vector ππ is negative four π add
seven π. And weβve got to find vector π΄π΅
take away vector ππ.

So just writing that out, π΄π΅ is
three π take away four π. ππ is negative four π plus seven
π, and thatβs what weβre taking away from vector π΄π΅. So we just need to be really
careful, when weβre taking these away, about the signs here because weβve got a
negative outside the bracket, weβve got negatives and positive inside the
bracket.

So letβs just start off with the
π-components then. Iβve got three π and Iβm taking
away negative four π, so that means Iβm adding four π. So three π add four π is seven
π. And then for the π-components, Iβm
starting off with negative four π and Iβm taking away positive seven π, so f-
negative four take away another seven is negative eleven π. So thereβs our answer, seven π
take away eleven π.

And just to summarise then what
weβve learnt. π is the unit vector in the
π₯-direction one, zero and π is the unit vector in the π¦-direction zero, one. And given any vector β like π΄π΅ is
five, three β this here is the π₯-component, this here is the π¦-component. We can rewrite this as five π,
because thatβs the number of π₯s, plus three π, because thatβs the number of
π¦s. And given two vectors, π΄π΅ and
πΆπ·, with their π- and π-components like this, in this format, we can add or
subtract them just by adding or subtracting their π-components and their
π-components, separately.

So, for example, adding π΄π΅ and
πΆπ·, we can add the two and the negative two π and we can add the negative three
and the four π. Well, in this case, two plus
negative two π, thatβs zero πs, so we donβt need to bother writing zero π. And negative three plus four is
just positive one, so we end up with an answer of just π, one π. And if we want to subtract the
vectors π΄π΅ minus πΆπ·, weβve got two of the πs here take away negative two πs,
and we had negative three take away four of the πs. Then two take away negative two is
two add two, so thatβs four of the πs. And negative three take away
another four is negative seven πs. So weβre adding negative seven
π. So we probably not write plus
negative π, we just write four π take away seven π.

So hopefully, youβll be comfortable
using the π and π unit vectors just to represent the π₯- and the π¦-component of
any vectors that you come across now.