### 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.