Hey, guys. So up until now, all the vectors we worked with have been in quadrant 1 or the top right corner of the xy plane. But you're going to have to get really good at handling vectors in all quadrants and all sections of the xy plane. So we're going to need a couple more conceptual points and also just a little bit more trigonometry to be able to solve just about any one of these kinds of problems. Let's check it out.

One thing you need to know now is the signs of magnitudes and the components of vectors. Basically, the magnitudes of vectors, no matter which quadrant they're in, are always going to be positive. For example, I've got these 4 vectors a, b, c, and d. They're all just combinations of 3, 4, 5 triangles. But the hypotenuses, the magnitudes, are all going to be positive here. Where things get a little different is when we start breaking them down into the legs of the triangles or their components over here. So these components, which again are all combinations of 3, 4, 5 in this example, may be positive or negative depending on which direction they point in. But there's a really simple rule to follow: Positive components are always going to be ones that point up and to the right, whereas negative components are always going to point down and to the left. So let's just go through this really quickly. In quadrant 1, we had vector a. We break it up into its components and it points to the right and up. So my ax is positive, my ay is positive. For vector b, we break it down. This vector points to the left and this component points up, so this is positive and this is negative. For vector c, this component here points to the left. This component points down, so this is negative and negative. And for vector d, this component points to the right, so it's positive and this one points down, so it's negative. So again, all these things, all these components are all just combinations of 345, just in this example, but the sign will change depending on which quadrant they're in. Alright, guys. So that's it for that one. Let's keep going.

So let's get to some trigonometry. Sometimes, and this applies to any quadrant, you're going to have to find a non-reference angle. So what do I mean by that? Well, we've got this vector here in which the angle that we're given is relative to the y-axis. But the reference angle that we need is against the x-axis. So this is my reference angle theta x. Now I need this angle theta x in order to calculate components using my \(a \cos \theta\) and \(a \sin \theta\) equations. So I'm given this angle in this diagram here, which is an angle relative to the y-axis. So I'm going to call it theta y and this is bad. I can't use this inside of these equations that I have. So I need to figure out the reference angle. I need to figure out the good one. And fortunately, there's an easy way to do this, and basically, I'm going to break it down for you. We're to break down this triangle or vector into a triangle. And we can make some right angles here. We know that this makes a 90 degree angle, and we also know that this quadrant here makes a 90 degree angle. So this angle is also 90 degrees. So all right angles add up to 90 degrees and we can use this to come up with a really simple equation for theta x. Theta x and theta y all just add up and always add up to form 90 degrees. So that just means that if this is 10 degrees, my bad angle, and I need to find my good one, then this is just gonna make up the difference between 10 and 90 degrees. So \(\theta_x\) is just \(90 - 10\), which is 80 degrees. So this is 80 degrees over here. Whatever ends up being your bad angle, your good angle is just gonna be 90 minus the bad one. So, that's it for that one.

The other angles you might be asked for a couple more, so I'm going to show you. You might be asked for this alternate angle or this other interior angle of the triangle. And basically, what happens is that these two lines form parallel lines and this one is sort of diagonal. And so what happens is these two angles here are paired up. They're perfectly symmetrical. So this means this is also 10 degrees over here. So the last thing that you might be asked for, might see in problems is you might have a situation where this vector gets extended to the other side of the axis like this, and you might be asked for some of these angles over here. So let's just break it down really quickly. We're going to do the same exact thing, break it up into a triangle. This is right angles over here. And basically, this triangle just gets mirrored opposite like this. So for example, this was my skinny angle. It was the 10 degrees measured relative to the y axis, and that's exactly what's going to happen over here. This is my skinny angle. It's measured relative to y, and it's 10 degrees. And this is going to be the good angle, which is my \(80^\circ\), and notice how they form perfectly opposites of each other, and this is always against the x-axis. So that's it for that one guys. Basically, you're always just going to figure out what your good angle is before you plug them into your problems. Now the last thing we have to do, or you might have to do, is figure out something called the absolute angle in any quadrants. So let's say we have this angle or this vector, it's \(a = 5\). It's in the top left or the second quadrant. And the absolute angle is going to be the angle relative to the positive x-axis over here. So this is also where \(0^\circ\) is located. So the absolute angle is just if I extend from the positive x-axis and I go all the way until I hit the vector over here. So this is going to be my theta absolute. And sometimes you're going to have to calculate this. So let's check it out. This is different from an angle that we already know how to calculate called the reference angle. Remember that the reference angle for this vector is going to be the angle relative to the nearest x-axis. So this guy is actually my reference angle \(\theta_x\). And I get it just by using my inverse tangent or my arc tangent here. And I'm always going to plug in the positive values of the components, and it doesn't matter which quadrant I'm in. So I'm always just gonna plug in 3, for example. I've got my 3 as my y component and my 4osterone, they're always just gonna be positives. And if I do this, I'm gonna get \(37^\circ\). So \(37^\circ\) is this angle over here. But what if I wanted to find the absolute angle? Well, all we have to do is to find the absolute angle. We just have to work our way mathematically back to the positive x-axis or 0. We might have to add or subtract some angles. What do I mean by this? Well, I know that this is \(37^\circ\) and I'm trying to figure out this angle over here. One angle that I do know is I do know that a straight line from positive x to negative x is \(180^\circ\) . So if I know that this is \(180^\circ\), and I know that the my reference angle is \(37^\circ\), then that means that my absolute angle is just gonna be the difference. I'm gonna take \(180\), I'm subtract \(37\). So my absolute angle is \(143^\circ\) and this is obviously not the same thing as my reference angle of \(37^\circ\). So that's how you do that. Alright, guys. So it might seem like a lot, but we're gonna get some practice with this. So let's take a look at this example. We're gonna calculate the components and then the absolute angle for our vector over here. So we've got this vector, it's just 13 and we've got this angle \(22.6\). So let's start out with the first part, calculating the components. So we've got this vector. I'm gonna break it up into its x and y pieces. This is my Ax. This is my Ay. So I know how to calculate the components..ToTable\(Ax\) is going to be \(a \times \cos(\theta_x)\) and my \(Ay\) components is going to be \(a \times \sin(\theta_x)\). So now the question is, do I have this \(\theta_x\)? And the answer is no. So you have to be very careful here. So our angle here is relative to the Y-axis. So this is the bad angle. So this is my \(\theta_y\) . It's bad. I need to figure out what the good angle is, my reference angle. And so, all I just use is the equation that we saw up there which is that \(\theta_x\) is \(90 - 22.6\), and this is going to be \(67.4^\circ\) . So this angle, \(67.4^\circ\) is the good angle, and this is the one that I use inside of these equations over here. So we're going to use that my \(Ax\) component is going to be a \(13 \times \cos(67.4^\circ)\). And if you plug this in, you're going to get 5. But remember that this 5 here that this component points in the left direction. So it picks up a negative sign because it points to the left. And now we do the same thing for Ay. This is going to be \(13 \times \sin(67.4^\circ)\), and you're going to get 12. But again, this component also points downwards. So that means that we have to add a negative sign. This is going to be \(-12\). So these are our components here.

Alright. Let's move on to the second part. The second part says, what's the absolute angle for this vector that's given over here? Remember, the absolute angle is going to be the angle that's read measured relative to the positive x-axis. So we're going to draw this huge angle all the way until we hit this vector over here, and this guy is going to be my theta absolute. So what is that \(\theta_{absolute}\)? Well, basically, we just have to work backward and get all the way back to 0, and we can do this by adding and subtracting some angles that we know. I know the reference angle is \(67.4^\circ\). So I know that little piece there in my triangle is \(67.4^\circ\). So what we can do is we can add the \(67.4\) to the flat line that we already know is \(180^\circ\). So we know this is \(180^\circ\). So that means that my absolute angle is really just \(180 + 67.4\), and that's how we're going to get that \(\theta_{absolute}\). So my \(\theta_{absolute}\) is going to be \(180 + 67.4\) degrees, and so, therefore, the absolute angle is \(247.4^\circ\). Alright, guys. That's it for this one. Let me know if you have any questions.