 So now let's consider an aerospace example of vector addition in two dimensions. In this case, we're going to consider the situation of a plane that is traveling at 400 miles per hour and is flying northeast with a heading of 40 degrees. There is a wind of 50 miles per hour from the south. And if there's no correction made for the wind, what are the bearing and the ground speed of the plane? So let's clarify a little bit by drawing. I've already created a little bit of a sketch of a plan view of an airplane over here. And let's identify some of the information we have. The first thing we have here is a value of vector magnitude of 400 miles per hour. This is something we're going to call the air speed. And that air speed is given a direction. That's what makes it a vector, a direction heading of 40 degrees. Now typically in geography and therefore in aerospace and other associated fields, we're going to define a lot of our directions from north. And in this case, we're going to go 40 degrees from north. And our magnitude of our vector, our air speed, is 400 miles per hour. And again, we're going to call that our air speed. And I'll code it in blue here with my ink. Our other piece of information is that we have a wind of 50 miles per hour from the south. From the south means it's heading due north. So I'm going to go ahead and code that, my wind speed. And notice typically when we measure air speed, the air speed is measured in the air that we're moving in. Well, if the air itself is then moving, then our two velocities are going to need to be combined in a fashion that tells us what our overall speed is. And that overall speed, we've already identified what we're looking for, is called the ground speed. And just like our air speed and our wind speed, this ground speed is a two-dimensional vector. It has a magnitude that we're trying to find, and it also has a direction that we're trying to find. Well, the name of that direction has a slightly different name, and that's going to be the bearing. In one hand, our airplane is heading in a particular direction, but it's going to actually have a motion that's some combination of its motion through the air and it's the motion of the air itself. Let me go ahead and draw the motion of the air itself. In this case, that is our wind speed, which we just established to be due north, and which has a magnitude of 50 miles per hour. So now if I want to combine those two vectors, well, we might want to draw them slightly differently here. In this case, I'm thinking about the wind speed is acting on the plane and the plane itself moving. But because these are velocity vectors, their position in space is unimportant. Remember, a zero-velocity vector doesn't represent a place. A zero-velocity vector means something that's not moving. So we can take these vectors and move them around to help us conceptually understand combining the vectors. So let me take my 400 mile per hour vector here, 400 MPH, and again, based on due north, there's my 40 degrees, and that is my heading. And we're going to combine that head to tail with my 50 mile per hour wind speed vector. And when we combine the two things together, we will end up with our ground speed vector. And that vector will have our unknown bearing as well as its own magnitude. So now let's establish some steps for working through this kind of problem, adding these two vectors. First thing I want to make sure I do whenever I'm working in some sort of vector problem is I'm going to define my basis. In other words, I'm going to think about what system I want to consider. And generally, we want to make a perpendicular system that we want to consider these vectors in. Typically, in a case like this, we already have a direction that's important, the direction that's north. So I'm going to go ahead and align my basis with north, and I'm going to define an axis there. We'll go ahead and call this my y-axis, and we'll make it positive in the north direction. And then we will also make a perpendicular axis to that. The perpendicular part piece is important, so we can do our vector addition easily. I'll call that my x-axis, and I will make that positive to the right, or positive to the east. So that's the first step is make sure you define your basis. Sometimes, actually very often as we'll see in later problems, you can actually take your basis and rotate it, and by rotating it conveniently, you can actually make the math even easier with an appropriate choice of basis. In this case, again, we'll choose north to be a positive for our y-dimension, and east to be positive for our x-dimension. So now that we've done that, our second step is to take the vectors that we know that are currently described in a magnitude and direction format. We're going to take them and we're going to determine components. In other words, we're going to consider each of the vectors that we know in terms not of their magnitude and direction, but in terms of component pieces aligned with the basis that we just created. So let's consider our first vector here, our airspeed vector. If I take that airspeed vector, I can draw it, and I can create a right triangle associated with it, where the sides of the right triangle are aligned with my basis that I just created, and I can label any angles that I happen to know, and go ahead and label the magnitude that I happen to know. And I'm going to go ahead and create names for the pieces that I want to determine. I have velocities. This is my 400 mile per hour as a velocity and airspeed velocity. Maybe I'll call that Va. And I'm going to go ahead and call this Va y, the velocity, the component of my airspeed velocity in the y direction, and Va x, the component of my airspeed velocity in the x direction. Now is where I need to use my trigonometry from trigonometry class. You probably remember the mnemonic Saucatoa, a series of letters that tell you about the relationships between the sines, cosines and tangent of an angle in a triangle, and the appropriate sides of the triangle. Well in this case, I want to determine what Va x and Va y are, so I can recognize some of these relationships. Let's start with the sine. The sine of this angle, the sine of the 40 degrees that I've identified here, is going to be the opposite side. In this case, there's the opposite side, Va x, over the hypotenuse, which is 400 miles per hour. So now I have a relationship that only has one variable in it. In this case, Va x is equal to, well let's see here, Va x is equal to what? Let's actually step back a second and recognize the sine of the angle. Sine of 40 degrees is equal to the opposite side, Va x, over the hypotenuse, the 400 miles per hour. Now I do a little bit of an algebraic solution, multiply both sides by the 400 miles per hour, and I determine that Va x is equal to 400 miles per hour times the sine of 40 degrees. Now when I do that calculation, and I make sure that I have the right settings on my calculator so it makes sense, I get a value of 257 miles per hour. I can do a similar thing now, recognizing one of my other pieces. In that case I use the sine, let's go ahead and look at the cosine of the angle. In this case, the cosine of my 40 degree angle is going to relate two different sides of the triangle. Perhaps I'm going to relate my adjacent side, here's the side that's adjacent to the 40 degree angle, to the hypotenuse, the 400 miles per hour. So I could write the similar thing here, cosine of 40 degrees is equal to Va y. I'm going to skip this particular writing and go ahead and do the algebra and recognize that Va y is equal to the 400 miles per hour times the cosine of 40 degrees. And when I solve that particular equation or plug in the appropriate values, I get 306 miles per hour. Now one small warning here is typically when you work with angles, when you first learn about angles in trigonometry, is you almost always measure those angles from the horizontal value in a counterclockwise direction. Because we were working in the other direction, things are a little bit different. And typically what often happens is the triangles you work with, that your y component or your vertical component almost always goes along with the sine of the angle. It didn't happen in this case, the main reason why is because the angle we knew was not measured counterclockwise from the horizontal, but rather clockwise from the vertical. So it's important to solve these problems to think about where your angles actually are, whatever angle you choose, and where the components actually are that you're looking for. So now I've found these components of my airspeed. I'll go ahead and underline them both in blue, the components of the airspeed. And I want to do a quick check and sort of look here. Notice Va y is a longer component, it's a larger value than the Va x. And if I look at it here, I didn't quite draw it that way, but I stop and think, a 45 degree angle they would be equal, a 40 degree angle would make the Va x a little shorter than the Va y. So even though I didn't quite sketch it correctly, I can stop and think about it and check myself, at least it seems like these values are appropriate. Another way I could check myself is I could take these two values and use the Pythagorean theorem. a squared plus b squared should equal c squared. And if you take those two values, you should find at least within the significant figures that those two are indeed, that combination is indeed true. So now we're going to follow the same process with our other vector. In this case, our wind speed vector. Well, that one's a lot simpler. Our wind speed vector is already aligned with one of our axes in our basis. Maybe it was a smart choice of basis in this particular case. So in that case, my 50 miles per hour is simply identified as the velocity, but not the velocity of the airspeed. Let's call it the velocity of my wind speed in the y direction is 50 miles per hour. I'll underline that in pink. What is our velocity of the wind speed in the x direction? Well, there is no component in the x direction or actually saying there is no component isn't quite right. The component has a value of zero. There is no motion in the x direction which we associate with the zero velocity in the x direction. So I can go ahead and write that as zero. I'll keep the units miles per hour. Although when you have zero, the units are sort of meaningless. So that's the second step of our process. Why did we do that step? Well, the entire reason, the entire purpose of doing that step is so that we could actually then proceed to combine our vectors. So our step three is to go ahead and do the math to add our components. We recognize from our picture here where we're doing our vector addition head to tail that what we're adding are the following vectors. That our ground speed is equal to the, let me get the correct color, air speed plus the wind speed. Well, in each of these cases you see we have our air speed components and we have our wind speed components. So now to find the appropriate ground speed components, vgx and vgy, we're simply going to add the appropriate pieces. Here's my x component of 257 miles per hour for my air speed. Here's my x component of zero miles per hour for my wind speed. Similarly, my y component for my air speed plus my additional y component for my wind speed. And then I can total those two, 257 miles per hour in the x direction and 356 miles per hour in the y direction. So one thing to be aware of when we do problems like this is that this problem was actually simplified by the fact that all of our components were pointing in the same direction. Notice our x component of our air speed is pointing in our positive direction, so that would be considered to be a positive value. Our component of our air speed in the y direction was north, which we defined it as being positive, and so was our direction of the wind from the south. The only thing that isn't officially positive is the zero, but in this case it made the addition simple, we just simply added all the values. Notice in slightly more complicated problems you might have to pay attention to the direction that each of the components is pointing so you know whether to add or to subtract or to add positive and or negative values. So now that we've actually determined our components and we've added them together, now we actually have a final vector. We actually have the vector that we were looking for. However, the vector we were looking for isn't in the form that we would like to express it in. If you remember the question asked us what was the final bearing and ground speed of the plane. So we actually want to know the magnitude and direction of the final vector. To do so we're going to have to do the final step here, which is to translate our components. And by translate I don't necessarily mean move them in space in a geometric sense. We've already done some of that translation. What we mean is to take the components and turn them back mathematically into our original sort of magnitude and direction form. And again we're going to need to go back to our right triangle trigonometry in order to do so. So now I have a sketch here but let's actually sketch our ground speed. We're looking for our velocity over the ground or our ground speed. And let's actually record the components that we had. We had vgx which was 257 miles per hour in an eastward direction. And we had vgy which was 356 miles per hour in a northward direction. And because we defined our basis perpendicular we can recognize that this is a right triangle. Well now there's two pieces of information we would like to have here. Let's go ahead and solve, well let's see here. I've drawn this a particular way but I kind of was looking for the angle that came from the north. So maybe I should redefine it using this triangle instead. Notice that's the exact same triangle, same values, but it's easier to think about what angle I'm looking for here because the angle I was looking for was the angle coming from north. So let me choose to use that angle instead or at least as I think about it. So now that I'm doing that let's consider, I'm going to write this vg here inside this other one so it's outside of my triangle. Let's consider which of our relationships are trigonometric relationships we want to use here now. Well we used the cosine before, well as it turns out now is the time to use the toa, the tangent of the angle. Because that relationship is going to give me a relationship between the opposite side of the triangle and the adjacent side of the triangle and how they relate to our angle. Again the adjacent side here is the 356 and the opposite side is the 257. So now I have a relationship that says the tangent of my angle is equal to the opposite 257 miles per hour over the adjacent 356 miles per hour. As expected, because it's a ratio between two things, the units cancel out. And then we actually have a relationship between those two things. That value is .7219 and then we use our inverse tangent function making sure that we're translating it into degrees or into radians depending on what we want. I want it in degrees so I make sure my calculator is set appropriately and I find an angle value in this particular case of 35.8 degrees. So that now actually tells us where the direction that my plane is actually traveling. And this is the direction that we're calling the bearing of the plane. This is the bearing of the plane. And if I sketch that up here basically we can see that the plane is being pushed, pushed in a direction that's a little bit off of what it was originally. Even though it's pointing in this direction, it's traveling at a more northerly direction due to that southern wind. How fast is it traveling? That's the second piece of the question. So how do we calculate that? Well in this case we need to use our old friend Pythagoras. We recognize that we have a right triangle here. We can use a squared plus b squared equals c squared. So I can recognize that my vg is the c squared we're talking about here. vg squared is equal to 257 miles per hour squared plus 356 miles per hour squared. If I do the math there then I get that that's equal to 192,785 miles per hour squared. Notice the miles per hour are currently squared until I take the square root of both sides and I get a ground speed in this case of 439 miles per hour. And in this case the best I can keep is three significant figures, probably only two because of that but perhaps we can assume since it's a word problem that we might have an extra significant figure that's just not listed. So that's the second part of our question. 439 miles per hour. If I wanted to label it up here, 439 miles per hour. Notice we are going faster over the ground than we are actually going through the air because we get that extra boost from that wind that's coming from the south. However, it's not the same as if we had just added 450. That would only occur if the wind was what we call a tailwind coming directly behind us and pushing us good putting all of that 50 miles per hour in that direction. Instead because we're traveling at an angle there we have to use vector addition to solve the problem.