### Video Transcript

In this video, we’re gonna use our
vector knowledge that we’ve learnt so far to solve a real-life problem. And in fact, the problem is gonna
be about an airplane flying along and there’s some wind behind it. We’ve got its airspeed and we’ve
got to work out its groundspeed. And we’re just gonna use dot
products and we’re gonna use some vectors to represent this problem and solve
it.

Right, so here’s the problem.

An airplane is flying on a
bearing of one hundred and twenty degrees. So we start off looking north
as we have here, come round in a clockwise direction of a hundred and twenty
degrees; this is the direction that our airplane is flying in. It has an airspeed of three
hundred miles per hour. Now that means that it’s
travelling at three hundred miles per hour in relation to the air around it. Now the next sentence says that
there’s a thirty-mile an hour wind blowing from the north, directly
southwards. So we’ve got a thirty-mile an
hour wind blowing in this direction. So if it’s travelling three
hundred miles per hour in relation to the wind — relative to the air around it —
that air is travelling southwards at thirty miles an hour to start off with. So that’s kinda of an
additional component of velocity blowing in the southward direction — travelling
in that southward direction. Now we’ve gotta find the actual
bearing of the airplane and its groundspeed.

So we’re gonna represent all
this information in a vector diagram and then we’re gonna use some of our vector
knowledge that we’ve gained in order to solve that problem. Right, so here’s our
diagram. We’ve got vector 𝑎 represents
the air speed or the air velocity of the aircraft. So it’s on a bearing of a
hundred and twenty degrees as we said before. And the magnitude of vector 𝑎
is going to be three hundred because that’s the speed with which it’s
travelling. So the magnitude of vector 𝑎
is three hundred moving in that direction. Now because the wind is
travelling at thirty miles an hour in this direction, the groundspeed is gonna
have this extra component of southerly speed to it. So the groundspeed is gonna be
slightly bigger than that and it’s gonna be heading in a slightly more southerly
direction.

And we’re also trying to find
the size of this angle here; call that 𝜃. And if we add that 𝜃 to the
hundred and twenty degrees we had up here, that would give us the bearing of the
groundspeed — the actual bearing — that the-the aircraft is heading on. Now what you might see is we’ve
got two vectors and we’re trying to find the angle between them. So we’re gonna be using vector
dot products in here. And we’re also gonna be looking
at magnitudes of vectors. So we’re gonna be using some of
those skills as well.

So let’s try and fill in some
of the details about vector 𝑎. Well, it’s got an 𝑥-component,
which is in this case it’s an easterly direction on our diagram. And it’s got a 𝑦-component,
which in this case is in a southerly direction. So it’s gonna be a negative
number; it’s heading downwards in the negative 𝑦-direction. We said that the magnitude of
our vector was three hundred because it’s travelling at three hundred miles per
hour. So let’s just do a little bit
of work on this triangle that we’ve created before we work out what the
𝑥-component and the 𝑦-component are.

So I’ve just drawn in my west-,
east-, and southern and I filled in this angle here of thirty degrees because
the angle between the north and the east direction is ninety degrees, which
leaves thirty degrees for this angle here. So just looking at the cosine
of that then, we got a right-angled triangle. The cosine of thirty degrees is
the adjacent side over the hypotenuse; so that’s 𝑎𝑥 over three hundred. So we can rearrange that to
work out what 𝑎𝑥 is, simply by multiplying both sides by three hundred. So that’s 𝑎𝑥 is three hundred
cos thirty. And then cos thirty of course
is root three over two.

So when we work all that out,
the-the easterly component of 𝑎’s velocity is a hundred and fifty root three
miles per hour. Okay, let’s look at 𝑎𝑦 now —
the 𝑦-component. And sine of thirty is the
opposite side of the hypotenuse which is 𝑎𝑦 over three hundred, which again
multiply both sides by three hundred and we get three hundred sin thirty. Now sin thirty is a half, so
that equals a hundred and fifty. So the airplane’s speed if you
like in the easterly direction is a hundred and fifty root three and in the
southerly direction is a hundred and fifty miles per hour. So we’ve worked out what the
components of the 𝑎 vector are.

Now the one thing that we do
have to watch out for is that remember the-the positive 𝑦-direction would be
going in a more northerly direction. But because we’re heading
towards the south, we’ve got a negative 𝑦-component. So that a hundred and fifty
miles per hour is in the southerly direction. So on our scale, the 𝑎 vector
is gonna be this here: a hundred and fifty root three, negative one hundred and
fifty.

So moving on to the groundspeed
vector, we can see that the horizontal component is exactly the same as the
airspeed. The extra wind was only blowing
in the southerly direction; so it’s not-not affecting the horizontal
component. So the the horizontal component
— the 𝑥-component — of the groundspeed is the same as for the airspeed. Now with the southerly
component, the 𝑦-component, we’ve got this extra thirty miles an hour kicking
in. So we’ve got whatever the
airspeed was up here, but we’re adding an extra thirty miles an hour to
that. So the-the-the groundspeed
𝑦-component is gonna be a hundred and fifty same as airspeed plus the extra
thirty is a hundred and eighty miles per hour.

And again because that is in a
southerly direction, not a northerly direction, it’s the negative 𝑦-component,
so it’s gonna be minus a hundred and eighty miles an hour so that our 𝑔 vector
is a hundred and fifty root three for the 𝑥-component same as the airspeed and
negative a hundred and eighty that extra thirty miles an hour in the
𝑦-component.

So we’ve worked out our
velocity vectors then for the airspeed and groundspeed 𝑎 and 𝑔 vectors. And what we now need to do is
work out what that- the magnitude of that groundspeed is. So we’re going to try and find
the magnitude of 𝑔. And to do that remember we
square the 𝑥-component and we add that to the square of the 𝑦-component. And calculating those, we get
the square root of ninety-nine thousand nine hundred, which is equal to thirty
root one one one if you wanna be exact about it or to one decimal place three
one six point one miles per hour. So the magnitude of the
groundspeed is three hundred and sixteen point one miles per hour to one decimal
place. So that’s our groundspeed
then. Now what we need to do is work
out the size of this angle here so that we can add it to a hundred and twenty
degrees and get the size of our bearing.

And to work out the angle,
we’re going to use the dot product of the unit vectors in the direction of the
airspeed and the groundspeed. So 𝑎 over the magnitude of 𝑎
dot 𝑔 over the magnitude of 𝑔 in this case. So the magnitude of 𝑎 is the
components squared and added together and take the square root of that. We’ve already worked out the
magnitude of 𝑔; so we can now just simplify that down. And to work out these dot
products, I’m gonna take this component and multiply it by that component and
add it to this component multiplied by this component.

So that’s what that gives
us. Remember because we’re doing
dot products, we mustn’t use the-the cross-multiplication sign; we need to be
consistent and use those dots for multiplying the numbers together. So just got a common
denominator that already multiplying those numbers out and adding them together
gives us twenty-one over two root one one one. So what we’re saying is that
this cos 𝜃 here is equal to twenty-one over two root one one one. So if I do cos to the minus one
of this, it will tell me what 𝜃 is.

And to one decimal place that
gives us four point seven degrees. Now it’s just worth mentioning
at this point I haven’t been rounding these numbers as I’ve been going along;
I’ve been trying to keep them in a surd format — root format — just to keep the
accuracy in the question as long as possible. So it’s not always possible to
do that because you know some numbers don’t come out quite as easily; the
calculator can’t handle them in that format. But if they can, it’s best to
work in that format for as long as possible. Now remember in our question,
so we’ve just worked out this bit here. So we got to add that to a
hundred and twenty to work out what the actual bearing was. So that’s gonna give us an
answer of a hundred and twenty-four point seven degrees So there we have it. Our answers were three hundred
and sixteen point one miles per hour to one decimal place for the groundspeed
and a hundred and twenty-four point seven degrees to one decimal place for the
bearing.

So let’s just kinda recap some
top tips as we go. Always worth doing a diagram,
so that definitely helped us doing a diagram. We represented the airspeed and
the groundspeed as vectors, so that turned out to be quite useful. We were able to work out the
magnitude of vector 𝑔 fairly easily just using effectively the Pythagorean
theorem. We were trying to keep our
values in surd format — in this root format — as far as possible to keep the
numbers as accurate as possible as well. And we were also able to use
the-the cos 𝜃 equals the dot product of the unit vector in each direction to
work out the angle between those two vectors.

So hopefully that’s given you a bit
more insight into using vectors to solve some problems.