Hey, guys. We've already seen how position and displacement work in one dimension along a flat line like on the x or y axis. But we're already talking about 2-dimensional motion now. So I want to show you in this video how position and displacement work in 2 dimensions, so along an angle like this. And here's the whole idea. If we ever have position and displacement that are 2-dimensional vectors, then they just turn into a bunch of triangles. So that means we can use all of our triangle or vector equations to jump back and forth between 2-dimensional vectors and their one-dimensional x and y components. That's the whole thing. It really is just more triangle math. So let's get started.

So let's talk about the position first. Now, the position when we talked about in one dimension was basically just an x or y value. So it was either a point along the x or the y axis. But now we're talking about 2-dimensional motions now. So it's not x or y. It's actually x and y. So it's a coordinate, x, y that cord, that basically describes where you are. Now, another way you can think about the position is that the position is a vector in 2-D. So it gets a letter r with an arrow on top of it. So it's a vector or an arrow that points from the origin to the point where you are. Let's take a look at an example. So a position is, in this diagram here, 3.6 at 33.7 degrees and then later on, our position is something else. We're going to calculate the x and y components of our positions at a and b. So we want the components of these vectors, but first, I have to draw out the position vectors in the first place. So remember, it's the origin, it's the arrow from origin to point, so that means that my vector at a is just an arrow. This is r_{a}. And then at b, this is just an arrow pointing from the origin to b, so this is r_{b}. So notice how these vectors here don't just point in the x or the y axis, they actually point at an angle, which means that we can break them down into triangles. So this turns into a triangle like this and this is my x and y components x_{a}y_{a}, this is an angle theta_{a}. The same thing happens for b. I'm going to break it down into a triangle and these are my components, x_{b}, y_{b} and the angle theta_{b}. So in this case, in this particular problem here, I want to figure out the components of these 2-dimensional vectors. So I'm just going to use all of my vector equations. I'm just going to use my decomposition and composition equations to figure out my components. And so those equations are my x and y equals r cosine theta r sine theta. That's really all there is to it. So for example, my position at a is just going to be r_{a} times the cosine of theta_{a}. So this is just going to be 3.6 times the cosine of 33.7, and you'll get 3. And if you do the same thing for y, you're just going to get 3.6 times the sine of 33.7 and you'll get 2. So basically, these are my x and y positions or my x and y coordinates. You can think about this as the legs of the triangle 32. Let's do the same thing for b. So x_{b} is just going to be r_{b} times the cosine of theta_{b}. So, this is just going to be 8.49. That's what I'm told in the problem and times the cosine of 45 degrees. These are magnitudes and directions. You'll get 6 and you do the same thing for y, you're going to do 8.49 times the sine of 45 and you'll also get 6. So my y is 6 and my x is 6. So notice how we can take any 2-dimensional vector and then we can just use vector equations to break them down into their x and y components. So that's it, just vector equations. Let's move on now.

So another thing we need to talk about is the displacement. So the displacement, the difference between the displacement and position is that whereas the position is an arrow from the origin to a point, the displacement is the shortest path from one point to another point. And so basically, you can think about this as a change in your position. So instead of the symbol r, like we use for position, it's delta r. Remember, delta always means change. So let's take a look at this example. So we're going to use the same dots a and b as before, and we're going to calculate the magnitude and direction of the displacement from a to b. So now the displacement is the shortest path between this point, point a to point b. So this is my delta r. It's the change in position. So notice how this vector also points at an angle not just in the x or y, so it also gets broken down into a triangle like this. Except now the symbols are going to be delta x and delta y. And so we're just going to use again vectors equations to figure out the magnitude and the direction. The magnitude is just the Pythagorean theorem. That's just your square root and your x squared plus y squared. And then your angle here, your direction is just the_TAG_will_inverse. So if I want to figure out the magnitude of my displacement, then I'm just going to have to use the Pythagorean theorem. So this is delta x squared plus delta y squared, and then my angle theta is going to be the tangent inverse of the absolute value of delta y over delta x. Alright? So notice how both of these equations involve me having delta x and delta y. But I actually don't know what those are. I don't know what the legs of this triangle are, so I'm going to have to go figure them out to figure out the magnitude and the direction. So how do we do that? Well, one way you can think about the displacement in the x direction is that and remember, it's just a change in your position. So the delta x here is going to be remember that my x position at a was 3. And then my x position at b, when I calculated this, was 6. And in a similar way, we had our y position over here, which was 2, and then our y position at b was equal to 6. So what happens is my delta x is really just the change in the x position. So it's x_{b} minus x_{a}. So that's 6 minus 3 and that's 3. And then similarly for the y direction, it's just going to be the difference. So 6 minus 2 is 4. So now I actually have the legs of the triangle. I know this is just 34, so I can plug in into my Pythagorean theorem. So it means the magnitude of the displacement is just going to be Pythagorean theorem, 3 squared plus 4 squared, and that's 5 meters. So this delta r is 5. And the direction theta is going to be the tangent inverse of my absolute value of 4 over 3. So if you plug this into your calculator, you are going to get, 53 degrees. Alright? So that is the magnitude and the direction. Alright, guys. That's all there is to it. Again, it's a bunch of just vectors things we've already seen before. So let's move on. Thanks for watching.