Video Transcript
In this video, we’re gonna look at
what happens when you add or subtract vectors to or from each other. We know that vectors can be
represented by line segments with a specific length and direction, and using these
will really help us to visualise vector addition and subtraction.
The addition process amounts to
laying vectors end-to-end and calculating the components of the resultant vector by
examining the difference between the coordinates of the initial point of the first
vector and the terminal point of the last vector in the chain. So let’s see an example of
that.
We’ve got 𝑎 which is the
vector three, two and 𝑏 which is the vector four, negative one. And we want to add those two
vectors together. So just quickly sketching those
out, here we’ve got vector 𝑎 which has an 𝑥-component of positive three and a
𝑦-component of positive two — so we’ve gone three in that direction, two in
that direction — and 𝑏, which has an 𝑥-component of four and a 𝑦-component of
negative one.
So if we add these two vectors,
it’s like laying them end to end and then working out how we get from the
beginning of the first vector to the end of the last vector. So to get from here to here,
we’ve taken a journey of three in the 𝑥-direction here and another four in the
𝑥-direction here. So to find the 𝑥-component of
the resultant vector, 𝑎 plus 𝑏, we simply add the 𝑥-component of 𝑎 to the
𝑥-component of 𝑏.
And now let’s visit the
𝑦-components. So to get from here to here
again, we’ve done our positive two to take us up to the top of the diagram but
then we came back one. So we’ve got to add those two
components together. So that’s two plus negative
one. And three and four makes seven,
so the resultant 𝑥-component is seven. And two add negative one is
one, so the resultant 𝑦-component is one.
So addition of vectors is just
a case of adding the 𝑥-components and adding the 𝑦-components.
Now let’s add three vectors
together.
So we’ve got vector 𝑎, which
is four minus two; 𝑏 is one, six; and 𝑐 is negative five, negative four. And we’ve got to add all three
of those vectors together.
So first let’s just sketch out
a diagram. Well here’s vector 𝑎 going
positive four in the 𝑥-direction and negative two in the 𝑦-direction. There’s 𝑏 going positive one
in the 𝑥-direction and positive six in the 𝑦-direction. And lastly we lay 𝑐 onto the
end of 𝑏, and we’ve got negative five in the 𝑥-direction and negative four in
the 𝑦-direction. And we can see that when we’ve
added 𝑎, 𝑏, and 𝑐, the initial point of the first vector 𝑎 is in exactly the
same place as the terminal point of the last vector 𝑐.
So although we’ve gone around
the houses, we started off here and we’ve ended up in exactly the same place
here. Okay let’s add those
𝑥-components together. So we’ve gone four and then
another one, and then we’ve taken away five. And in the 𝑦-direction, we
went negative two then added six and then finally took away four. And four add one take away five
is zero, and negative two add six take away four is also zero.
So the resultant vector by
doing the adding up of the components is zero, zero. And that tallies with what we
knew at the beginning: that we started off and we ended in exactly the same
position. So the resultant vector, we
haven’t moved in the 𝑥-direction we haven’t moved in the 𝑦-direction; our
initial point and our terminal point are coincident.
Having looked at some basic
addition of vectors, let’s take a quick look at repeated addition of vectors, or as
it’s more properly called multiplying vectors by scalars. So let’s start up with 𝑎 is equal
to the vector two, three. Let’s say we wanted to find out the
what the value of two 𝑎 was. Well two 𝑎 is just 𝑎 plus 𝑎. And as we’ve just seen, to add two
vectors together you just add the components together, so two plus two and three
plus three.
So yep we’re labouring the point
here a little bit. But that means, we’ve got two lots
of the 𝑥-component and two lots of the 𝑦-component, so this two here. We’ve multiplied each of the
components by two, which gives us the vector four, six. Okay let’s think about three 𝑎
now, so that’s 𝑎 plus 𝑎 plus 𝑎.
Well I’m just gonna jump straight
to the stage. When we’re multiplying the
components, because it’s three 𝑎 we’re multiplying the 𝑥-component by three and
the 𝑦-component by three, which gives us six, nine. And let’s just look at one more
example: a fractional scalar, so a quarter. So a quarter of 𝑎, we’re just
gonna multiply each component by a quarter. So that’s a quarter of two and a
quarter of three. So that gives us a half and
three-quarters as our components.
So just to summarise that, the
process of multiplying a vector by a scalar, we take the scalar and we just multiply
each component of the vector by that scalar to get our components of the resultant
vector. So let’s start off with the same
vector 𝑎. And now let’s have a look at
negative scalar multiplication, works in exactly the same way. So let’s just see a couple of
examples.
Okay we’re trying to find negative
𝑎. That just means minus one times 𝑎,
so we’re gonna take the scalar negative one and we’re gonna multiply both components
by negative one. And that gives us an answer of
negative two and negative three. And okay, let’s look at negative
two 𝑎. So this time the scalar is negative
two and we’re gonna multiply each of our 𝑥- and 𝑦-components by negative two. And that gives us an answer of
negative four, negative six.
Now it’s just worth taking a moment
to notice 𝑎 and negative 𝑎: the components have the same numbers, two- two and two
for the 𝑥-components and three and three for the 𝑦-components. But the in the negative 𝑎, the
signs are the opposite signs. In fact, what we’re doing is we’re
going exactly the opposite direction. So vector 𝑎, we’re going positive
two in the 𝑥-direction and positive three in the 𝑦-direction; that’s this journey
here. Negative 𝑎, we’re going negative
two in the 𝑥-direction and negative three in the 𝑦-direction.
So we’ve done the exact reverse
journey in the neg- in the negative direction, in the opposite direction. So positive 𝑎 and negative 𝑎 are
the same length of line, the same angle of vector, but they’re just going in the
opposite direction.
Okay let’s look at a quick example
or two of subtracting some vectors.
So we’ve got vector 𝑎 is two,
three and vector 𝑏 is four, negative one. And we wanna calculate the
resultant of vector 𝑎 take away vector 𝑏. Well there’s different ways,
you could just take the 𝑥-components of 𝑎 and subtract the 𝑥-components of 𝑏
and the 𝑦-components of 𝑎 subtract the 𝑦-component of 𝑏 to get your
resultant. But I just like to suggest that
we look at it in this slightly different way so 𝑎 take away 𝑏 is the same as
𝑎 add the negative of 𝑏.
So when we write that out,
we’ve got two plus the negative of four and we got three plus the negative of
negative one, which looks a bit on odd on the page. But there it is. So two add negative four is
negative two, and three add the negative of negative one, which is positive one,
makes four.
Now this business of converting
𝑎 minus 𝑏 into 𝑎 plus the negative of 𝑏 seems like we’re overcomplicated
things. But let’s just take a look at
it visually first and see how that pans out. So we’ve got here vector 𝑎,
which I’m going across two and up three. So if I was adding 𝑏, I’d be
going right four and down one, but I’m adding the negative of 𝑏. So adding the negative of 𝑏
takes me from the end of 𝑎 left four, so we’re going four in that
direction. And the negative of negative
one, we’re going up one in that direction.
So this idea of laying vectors
end to end, thinking of taking away 𝑏 as adding negative 𝑏, means we can just
lay down the negative vector to the end of 𝑎 and we can then look for this
resultant vector, going from the beginning of 𝑎 to the end of 𝑏 here. And when we do that, we were
going negative two in this direction and up here positive four in that
direction. So when subtracting vectors, we
can just subtract the components if you want to or we can add the negative of
the components for the second vector, which helps us to visualise the vector
diagram a bit more easily.
Let’s just take a look at one more
example. So we’re gonna start off with the
same 𝑎 and 𝑏 vectors.
But this time we’re gonna
calculate 𝑏 take away 𝑎, which is the same as 𝑏 add the negative of 𝑎, which
gives us four add negative two and negative one add negative three. So that’s a resultant vector of
two, negative four.
Well let’s look at the diagram
again. Well first of all, vector 𝑏,
we’re going positive four in the 𝑥-direction and negative one in the
𝑦-direction and then adding the negative of 𝑎, that’s negative two in the
𝑥-direction, so we’re coming back, negative two. And negative of three so we’re
gonna go down three — one, two, three — in the 𝑦-direction.
So we start off at the initial
point of 𝑏; we add the negative of 𝑎; and we end up at the terminal point of
𝑎. So our resulting journey is
this one here. And in completing that journey,
we went positive four in the 𝑥-direction but then we came back two, which gave
us an 𝑥-component of two. So this distance here is
two. And we went down one and then
we went down another three, which gave us our 𝑦-component of negative four,
which means we’ve come down here four. So this vector here is 𝑏 take
away 𝑎, 𝑏 subtract 𝑎, or 𝑏 plus negative 𝑎.
So just putting those two
results side by side, 𝑎 take away 𝑏 was negative two four, and 𝑏 take away 𝑎
was two negative four. So they were both in the same
position here, this was the position. And this was the length of the
vector, but one was going in this direction and the other was going in that
direction.
So they’re the same vector, but
in opposite directions; one vector is the negative of the other. So vector 𝑎 minus 𝑏 is the
negative of vector 𝑏 minus 𝑎. And that makes sense because
negative of 𝑏 is negative 𝑏 and the negative of negative 𝑎 is positive
𝑎. So algebraically, they mean the
same thing; and in vector format, they mean the same thing as well.
Now let’s look at one last
problem.
We’ve got a regular hexagon
𝐴𝐵𝐶𝐷𝐸𝐹 and 𝐺 is the midpoint of that and we have to express 𝐴𝐸 in terms
of vectors 𝑢 and 𝑣. So the vector 𝑣 is from 𝐺 to
𝐶 and the vector 𝑢 is from 𝐷 to 𝐶. Now because this is a regular
hexagon, we know that a number of these things are parallel. So 𝐴𝐵 and 𝐸𝐷 and 𝐹𝐺 and
𝐺𝐶 are all parallel; 𝐴𝐹 and 𝐵𝐺 and 𝐺𝐸 and 𝐶𝐷 are all parallel; and
𝐸𝐹, 𝐷𝐺, 𝐺𝐴, and 𝐶𝐵 are all parallel.
So we know for example that
vector 𝑢 runs from 𝐷 to 𝐶, or we can put in vector 𝑢 in some various
different places in there as well. So those distances are
parallel, but they’re also the same length. So they- we can pick the vector
𝑢 up and place them in each of those locations. And likewise, vector 𝑣,
running from 𝐺 to 𝐶, that will also be vector 𝑣; that will also be vector 𝑣;
and that will also be vector 𝑣.
So we got a few gaps on our
hexagon here. How would I get for example
from 𝐺 to 𝐷 along this vector here? Well I could go the
straight-line route, but that doesn’t tell me anything in terms of 𝑢 and
𝑣. So I could also go by this
other route; I could go along from 𝐺 to 𝐶, which is the vector 𝑣, and I could
go from 𝐶 to 𝐷, which is the opposite way to 𝑢, so it’s a negative 𝑢.
So vector 𝐺𝐷 as we said is
𝐺𝐶 plus 𝐶𝐷, which is 𝑣 plus the negative of 𝑢. In other words, 𝑣 take away
𝑢. So let’s draw that in on the
diagram then: 𝐺𝐷 is 𝑣 take away 𝑢. And likewise, 𝐹𝐸 is parallel
and the same length, so that is also 𝑣 minus 𝑢; A𝐺 is too; and so is
𝐵𝐶.
So when you’re trying to
summarise the journey from 𝐴 to 𝐸 in terms of 𝑢 and 𝑣, so all of these
journeys between individual points on our hexagon are already in terms of 𝑢 and
𝑣, so we just need to pick a convenient route. So let’s go along here, which
is a negative 𝑢 — it’s the opposite direction to a 𝑢 — and then down here,
which is 𝑣 minus 𝑢. We better tidy those up. So when we write that out,
we’ve got 𝑢 plus 𝑣 minus 𝑢. So when we write that out,
we’ve got a negative 𝑢 plus 𝑣 minus 𝑢. And it doesn’t matter what
route we took; however convoluted, would’ve still come up with that same answer
for 𝐴𝐸.
