Doing Math with Vectors in any Quadrant (More Trig)

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9m

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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.

2

Problem

Problem

Vector F is 65 m long, directed 30.5° below the positive x-axis. (a) Find the x-component, F_{x}. (b) Find the y-component, F_{y}.

A

F_{x}=30.6 m, F_{y}=51.6 m

B

F_{x}=56 m, F_{y}=33 m

C

F_{x}=30.6 m, F_{y}=−51.6 m

D

F_{x}=56 m, F_{y}=−33 m

3

Problem

Problem

The vector A represented is by the pair of components A_{x} = -77 cm, A_{y} = 36 cm. (a) Find the magnitude of vector A. (b) Find the absolute angle of this vector.

A

(a) A = 68 cm (b) 179.6°

B

(a) A = 85 cm (b) 179.6°

C

(a) A = 68 cm (b) 154.9°

D

(a) A = 85 cm (b) 154.9°

E

(a) A = 68 cm (b) −25.1°

F

(a) A = 85 cm (b) −25.1°

4

concept

Describing Vectors with Words (More Trig)

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6m

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Hey, guys, so many times when you're working out in your vectors problems, you'll see that problems will use different words to describe the directions of vectors. That's what we're gonna check out in this video. We'll see that there's really just a few keywords to look out for, like counterclockwise and clockwise, and then north, east, south and west. We're gonna check out those examples, Um, so let's just get to it. So what you need to about clockwise or counterclockwise angles, or sometimes abbreviated A C C. W is that these are positive angles. So wherever access you start from, as long as you're angles measured in this direction here against the clock, it's gonna be positive. So this is a counterclockwise. It's positive. So, for example, if this is 45 degrees is gonna be positive 45 degrees now. Clockwise angles, which are previewed by C. W, are negative, so these are negative angles. So for this, be vector Over here, this is a clockwise angle. It's negative. So if this was 60 degrees, it's actually negative 60 degrees Now, even though the positive or negative sign indicates just where it's going counterclockwise or clockwise, the reference angle that we use for our component equations. A co signed data. A sign data is always gonna be a positive number relative to the nearest X axis, for example, we're gonna use 45 degrees. So when we plug this into our calculators for our component equations, we're gonna plug in 45 degrees. Now, even though this is a negative 60 degrees, which all it does is tell us that it's that it's clockwise. The reference angle Theta X that we use is going to be positive 60 degrees. It doesn't matter what that negative sign out in front is. For example, let's just go ahead and do some calculations here. We're gonna draw each vector and calculate the components. So we've got 5 m at positive, 37 degrees from the negative X axis. So it's important to pay attention to the signs off all of the information that they're so positive. 37 means we're gonna be going counterclockwise like this from the negative X axis. So this is negative X. This is negative. Y we'll start here and I'm gonna go 37 degrees this way. So this is 37 degrees like this and this vector would be a equals five in this direction. So if I wanted my ex components in my white components, then I just have to use a cosine theta as long as this angle is nearest to the X axis, which it is. So I have five times the co sign of 37 and I get four. Then I have five times the sign of 37 I get three. The one thing that you do have to do is still use the rules of the quadrants. We know we're in the third quadrant, so these just pick up negative science because these components points to the left and down. Let's be wannabe. Now we have, ah, clockwise angle from the positive y axis, and it's 53 degrees eso We're gonna have this vector here be degrees clockwise from the positive y axis. So this is my positive. Why? Positive X and so this angle is gonna look like this. So we're gonna look at 53 degrees and this is gonna be a negative sign over here. So this is my B vector, which equals five. Now we just have to calculate the legs or the components over here my B y and my B X. We know that be X is just gonna be be times the coastline of data. The problem is, is that I have this negative angle here. That's relatives the Y axis. I can't use that. I have to figure out what the with the complementary angle is, which is 37 degrees. So I'm still just gonna use five times the co sign of 37 because that's the reference angle. It's positive relative to the nearest X axis, and that's just four. And it's positive because it points to the right and then my B Y is five times the sign of 37 which gives me three. It's also positive because I'm in the first quadrant. Alright, that's really all there is for that one. So the other thing that you might see is you might also see cardinal or compass directions like north, east, south and west. So you're gonna see directions described, like 30 degrees north of east, and this actually really straightforward. All you're doing here is you're gonna draw your arrow in the second direction. The second direction is the second, uh, you know, direction that they give you and you're gonna curve towards the first one here. So you're basically gonna go from here? You're gonna curve towards this. Let me just show you an example. We're gonna drive, vector and calculate just the X component for these two examples. Let's just get to it. So a equals six at 30 degrees north of east. So this is my second directions. We're gonna start basically by drawing the arrow, so we're gonna draw in the second direction. And now you're just gonna curve towards the first, which is to the north. So you're gonna curve like this 30 degrees, and then this is gonna be your arrow, so this is a equal six, and so you have to curve towards the first. All right, so now we just calculate the components. My a X component. I just need to magnitude times the reference angle, the coastline of the reference angle. So this is just gonna be six times the co sign of 30 and then this ends up being actually 5.2. So that is our components. So what about be now? We have 10 at 53 degrees west of south. So again, this is our second direction. This is our first. So we're gonna draw the arrow in the direction of the first one, which is south, and then we're gonna curve 53 degrees. So, like this towards the West. So this is 53 degrees like that, and this is our arrow, so we know B equals 10. So I want to calculate the X component over here is my B X. So I have to use be cosine theta. But remember that r theta term has to be against the X axis, and this one is actually against the why. So that's why these directions air important. So actually, this is the bad angle. I have to figure out this angle over here, which is 37 degrees. That's my reference angle. So that means that for this one, I'm gonna use 10 times the co sign off 37 degrees, and it's positive because it doesn't matter. It's always just plug in a positive number. And when you do this, you're gonna get eight. So this is our components, but we have to stick in a negative sign. You know, sometimes you'll see West conventionally is chosen to be negative. So that would be the component for that. Alright, guys, that's it for this one. Let me know if you have any questions.

5

Problem

Problem

A small helicopter travels 225 m across a city in a direction 53.1° south of east. What are the components of the helicopter's trip?