 In this video, I want to talk about a very important application of vectors, and that's the idea of direction, right? If we have velocity vectors that are adding together, that can give us the true direction of an airplane, of a boat, or something that's traveling that would experience some type of resistance, maybe current of a river or winds, things like that. So when it comes to direction, there's the direction we want to travel in versus the actual direction we go in. So the heading of an object is the angle which is representing the intended path of the object. So this is the place that we're pointing towards, that's our heading. And for a standard principle here, our heading will be measured clockwise with respect to due north. So for example, let's look at two headings of possible ships right here. One ship is traveling 14 miles per hour with a heading of 120. And then we'll have another one that's traveling 12 miles per hour with a heading of 240. What are the headings right here? These are going to be vectors, of course. So if we do the first one, we'll do that one in, no, sorry, that will do that one in green, 14 miles per hour, 120, like so. So with respect to north, so north is going to be pointing upward always, unless we're dwarves or something, north is going to be pointing upwards. And so with 120 degrees with respect to north going clockwise, the heading would be going in this direction. And then it's traveling 14 miles per hour. So the length of the vector will be based upon that. For the next one, if we were traveling 12 miles per hour with a heading of 240, 240, if we go clockwise from north, we get an angle that looks like this. And then the length of the vector will be 12, should be a little bit shorter than the 14 one. But this is what headings means. It means with respect to north in a clockwise manner, what's the direction we're going in. That's the intended direction, emphasis on intended. There are other forces acting on the moving object that might affect it. So although an object is headed in one direction, its actual path may be in a different direction because of those other forces. This leads to the idea of the true course of an object. The true course of an object is the angle measured again clockwise from due north to the vector representing the actual path, not the intended path, but the actual path of the object. We did an example earlier where a boat was going due east in a river that was flowing north. So the heading was eastward, but the true path was actually a northeastern direction because of the current there. There's a force pushing the boat in a different direction. So if you want, because really when it comes to headings, you have to choose your heading so that the true course is actually the direction you want to go. You have to correct because of things like wind and current. And this can be a very difficult thing for pilots and navigators, which because it involves these type of trigonometric problems. Let's look at an example of an airplane. For example, a plane is flying with an airspeed, which here airspeed means the speed of the plane relative to the air. It's traveling with an airspeed of 185 miles per hour and its heading is 120 degrees. The wind currents are currently running at a constant 32 miles per hour with a heading of 165 degrees. So we want to find the true course and the ground speed, which the ground speed is the speed of the object relative to the ground. It's going to be, of course, just the magnitude of the true course because the plane's heading is a vector. The true course is also a vector. They have magnitudes. So here airspeed represents the speed, the magnitude of the heading vector. And then the ground speed will be the magnitude, the speed of the true course vector in this problem here. So I've drawn the diagram here on the screen already, although it might be a little bit overwhelming. Let's try to unravel this a little bit. The yellow vector that you see on the screen, which is labeled V, this is the heading of our plane. So it's long because it's supposed to be 185 miles per hour. Its length is given by its speed. The direction, though, is given by 120 degrees. So 120 degrees from north would be like so. And so you see this vector right here. So then this other yellow vector labeled W represents the wind current. I've put them head to tail so that the true course will be the sum of the two vectors. We'll get to that in just a second. The wind current is going at 32 miles per hour, so much shorter. But this diagram is being drawn to scale here. This is 32 miles per hour for the wind and it has a heading. I should say it has a direction of 165 degrees as well. That is the heading. I apologize for that. And these are all with respect to north. This is what can make the headings a little bit difficult. That headings are all from the same relative point, which is north. But when we start interacting two vectors together, we form a triangle, which you can see here, but the angles of the triangle aren't the same as the angles of their headings. If the angle between the vector V and V plus W is beta, it's going to be related to these directions 120 and 165 somehow. But it might not be obvious where it is right now. And what about this angle theta? What's this angle alpha, for example? We'll get to those in just a second. So let V and W, like I said, this will be the vectors associated to the plane and to the wind, respectively, and let angles alpha, beta, theta be defined as in the diagram like so. So the first thing to notice that alpha is defined to be the supplement of this 120-degree heading of the airplane. Therefore, if alpha is supplementary to 120 there, we can find alpha by subtracting 120 from 180 degrees. This, of course, gives us that alpha is 60 degrees. All right. Why is that relevant? Well, you see that north is always pointing upward in our diagram here. And so if we think of this line right here, we see that the vector V acts like a transversal to these vertical lines representing north right here. So by the alternative angle theorem, this angle down here is also alpha because they are corresponding angles. So we're able to move the supplement of our original heading down here so we get so we're going to end up with alpha like so. So then alpha plus 165 degrees plus theta, these are going to add up to be 360 degrees because together they form one complete rotation. So this is how we're going to find theta. Theta is going to equal 360 minus 165, the heading of the wind, minus alpha, which we already found out was 60 degrees like so. This gives us that theta is equal to 135 degrees. So why is theta relevant here? Well, we need an angle interior to the triangle, right? So notice what we have right now. We know this distance, that's a side. We know this angle now, that's an angle. We know this side, that's a side there. So side angle side, this tells me that I can solve this triangle using the law of cosines. And so for the law of cosines, we're going to solve for this side right here. The law of cosines is going to tell us that the ground speed, which is the magnitude of V plus W squared, is going to equal the air speed plus the wind current minus two times the air speed times the wind current times cosine of theta in this situation here. So everything on the right-hand side we know, everything on the left-hand side, well, we don't know yet, we'll get there. So remember that the air speed was 185 miles per hour. The wind speed was 32 miles per hour. So we can minus two times 185 times 32 times cosine of 135 degrees like so. Let's try to compute this thing here. Now notice that cosine of 135 degrees, right? 135 degrees actually is, it references to 45 degrees. This is getting a little negative cosine of 45 degrees, which cosine to 45 is root two over two. So this is going to give us, when we plug that in there, 185 squared plus 32 squared. I guess I could have done those ones now. Oh, well, well, since you have a negative with an already negative there, it becomes a positive. The cosine of 45 degrees is root two over two. So that's going to cancel with this two right here. So we end up with 185 times 32 times the square root of two. That's where we're at right now. And so the rest of this, we can do this step by step by step, but I'm just going to throw all of this into the calculator. In which case we then get v plus w's magnitude, the ground speed squared is equal to approximately 43,621. So taking the square root, we see that the ground speed is going to be approximately 209 miles per hour. So it turns out the wind is actually giving the airplane a boost, right? We're getting wind behind our sails, so to speak. So that gives us the ground speed, which is one of the things we needed. But the true course is a vector, right? We needed to have magnitude, which we just found, which is 209 miles per hour, but we also need direction. Now that should be represented as a heading, right? So what's the angle with respect to north here? Well, we know what this is. We know this is 120 degrees. If we know the angle between these two vectors, we could add those together and we see that the direction, the direction that we need to find here for v plus w, that's going to be 120 degrees plus beta. That's going to be the final answer, but we didn't know what beta is. So that's where we're going to come into, that's why beta comes into play here. Now coming back to our triangle, notice here, now we know this length, we know this angle. Ooh, ooh, ooh, that's an AOS, right? We can then use the law of signs here. And this is typically what happened in the side angle side situation. You first use the law of cosines, then you can switch to the law of signs. So we have an angle opposite side there. We have another one right here, like so. So we're going to get that sine of beta over w is equal to sine of theta over v plus w, like so. And so we need to solve for beta here. Beta is going to equal sine inverse of the magnitude of the wind times sine of that angle theta there over v plus w, like so, which the winds recall was 32 miles per hour. So sine of 135 degrees, that's 135 still references 45 degrees, and the second quadrant sign is positive. So you're going to get root two over two there. And then we get this 209 in the bottom, approximate that with your calculator, and we can get that beta will be approximately 6.22 degrees, like so. So we add that to 120. So therefore, we see that the true course of the plane is going to be 126.22 degrees clockwise from due north.