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Hey, guys. So up until now, all the vectors that we worked with have all just been in the quadrant one or the top right corner of the X Y plane. But you're gonna have to get really good how to do vectors in all of these quadrants and all sections of the X Y plane. So we're gonna get a couple more conceptual points and also just a little bit more trig to be able to solve just about any one of these kinds of problems. Let's check it out. Now, One thing you need to know now is the signs of magnitudes and the components of vectors and basically the magnitudes of vectors, no matter which quadrant there in are always going to be positive. For example, I've got these four vectors A, B, C and D. They're all just combinations of 345 triangles. But the iPod news is the magnitudes are all gonna be positive here, where things get a little different 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 345 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 gonna be one that point up and to the right, whereas negative components are always gonna point down and to the left. So let's just go through this really quickly. In quadrant one, we have vector A. We break it up into its components and a point to the right and up. So my excess positive my a wise positive for vector B, we break it down, this vector points to the left or this components and this one points up So this is positive and this is negative for vector. See, this component here points to the left. This component points down, so this is negative and negative. And for D, this component points of the rights. What'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 signs will change depending on which quadrant there in alright guys. So that's it for that one. Let's keep going. So let's get to some trig so Sometimes in this applies to any quadrant, you're gonna have to find a non or you're gonna have to be worked with 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 equal. I need this angle theta X in order to calculate components using my A CO sign data and a synthetic equations. So I'm given this angle in this equation or in this diagram here, which is a Nangle relative to the Y axis. So I'm gonna call it Fate A 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 on. Basically, I'm gonna break it down for you. We're gonna break down this triangle vector into a triangle. We could make some right angles here. We know that this makes a 90 degree angle. 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 data X data X and theta. Y all just add up 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 with Theta X is just 90 minus 10 which is 80 degrees. So this is 80 degrees over here. Whatever ends up being your bad angle, you're good. Angle is just gonna be 90 minus the bad one. So that's it for that one. So the other angles you might be asked for a couple more s. I'm gonna show you. You might be asked for this alternate angle or this other interior angle, 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. Their perfectly symmetrical so 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 the victor 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 gonna 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 gonna happen over here. This is my skinny angles measured relative to why, and it's 10 degrees and this is gonna be the good angle, which is my 80 degrees and notice how they form perfectly opposites of each other. And this is always against the X access. So that's it for that one. Guys, basically, you're always just gonna figure out what you're good angle is before you plug them into your problems. Now, The last thing we have to dio is or you might have to do is figure out something called the absolute angle in any quadrants. So I'm gonna show you how to do that. So that's why we have this angle or the specter that's equals five. It's in the top left or the second quadrant, and the absolute angle is gonna be the angle relative to the positive X axis over here. So this is also where zero degrees is located. So the absolute angle is justify extend from the positive X axis. And I go all the way until I hit the vector over here. So this is gonna be my theta absolute. And sometimes you're gonna have to calculate this, so let's check it out. So 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 gonna 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 on my arc tangent here. Andi, I'm always gonna 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 three. For example, I've got my three as my wife component. And my four, they're always just gonna be positives. And if I do this, I'm gonna get 37 degrees. So 37 degrees 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 zero we might have to add or subtract some angles. What do I mean by this? Well, I know that this is 37 degrees, 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 two negative X is 180 degrees. So if I know that this is 1 80 I know that the my reference angle is 37 then that means that my absolute angle is just gonna be the difference. I'm gonna take 1 80. I'm subtract 37. So my absolute angle is 143 degrees. And this is obviously not the same thing as my reference angle of 37. So that's how you do that. All right, guys, we 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 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 ex. This is my A Why? So I know how to calculate the components. So my a X component is gonna be a times the co sign of Data X and my A Y components is gonna be a times the sign of Data X. So now the question is, do I have this Data 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 fatal. Why? It's bad. I need to figure out what the good angle is. My reference angle. And so all I just uses the equation that we saw up there, which is the data X is 90 minus 22.6, and this is gonna be 67.4 degrees. So this angle 67.4 is the good angle. And this is the one that I use inside of these equations over here. So we're gonna use that my A X component is gonna be 13 times the cosine of 67.4 degrees. And if you plug this in, you're gonna get five. But remember that this five 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 a wide. This is gonna be 13 times the sign of 67.4, and you're gonna get 12. But again, this component also points downwards. So that means that we have to add a negative sign. This is gonna be negative. 12. So these air are components here. All right, Let's move on to the second part. Second part says, what's the absolute angle for this? This, uh, this vector that's given over here, remember, the absolute angle is going to be the angle that's right, measured relative to the positive X axis. So we're gonna 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 data absolute? Well, basically, we just have to work backwards and get all the way back to zero. And we can do this by adding and subtracting some angles that we know. I know the reference angle is 67.4 degrees. So I know that little piece there in my triangle 67.4. So what weaken dio is we can add to 67.4 to the flat line that we already know is 180. So we know this is 180 degrees, so that means that my absolute angle is really just plus this 67.4. And that's how we're gonna get that data absolute. So my fate a absolute is gonna be 180 plus my 67.4 degrees. And so therefore, the absolute angle is 247 0.4 degrees. Alright, guys, that's it for this one. Let me know if you have any questions.

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