Direction of a Vector - Video Tutorials & Practice Problems

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1

concept

Finding Direction of a Vector

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Hey, everyone and welcome back. So up to this point, we spent a lot of time talking about vectors and you may recall one of the first things that we learned about vectors is that they have a direction. Now, in this video, we're finally going to be learning about how we can calculate the direction of a vector. Now, this might sound a bit complicated or a bit scary, but don't worry about it because it turns out that calculating the direction of a vector is very similar to finding the missing angle of a right triangle, which is something we've already learned about. And this is a skill that is very important to have in this course. So let's just go ahead and get right into things. So let's say we have this vector here which we'll call vector V. Now recall that the direction of a vector is the angle that the vector makes with the x axis. So we need to figure out what this angle is right here. Now, if we had the X component as well as the Y component of our vector, could you think of a clever way that we could figure out what this angle is. Well, you need to recall a memory tool that we learned for right triangles, which is Sokoto Sokoto teaches us that tangent is opposite over adjacent. So if we had the opposite and adjacent side of a right triangle, we could calculate the angle using this equation right here. Now, this might seem a little bit out of the blue but notice something. The vectors we have here form a right triangle. So we can use this right triangle to figure out what this angle is. So if the tangent of our angle is opposite over adjacent, that means that the tangent of our angle here, which would be the direction of our vector would simply be the Y component divided by the X component. So if I wanted to find our angle theta which would be the direction of this vector, I just need to take the inverse tangent of this fraction right here. So we have the inverse tangent of the Y component divided by the X component. Now, I can see here that the Y component is 123 units up. And I can see that the X component is 1234 units to the right. So we'll have the inverse tangent of three divided by four which on a calculator comes out to approximately 37 degrees. This is not the exact result but it is a very close approximation to the answer. So this would be the direction of our vector. So as you can see, it's pretty straightforward if we recall this memory tool for right triangles. But it turns out there are going to be some problems you come across that don't give us as directive an answer where we can just plug it into our calculator and to make sure that we know how to solve these examples. Well, let's try them. So here we're told drop the vector and find the direction of each vector expresses a positive number from the X axis. So let's go ahead and start with example a where we have the vector two negative one. So what I first need to do is draw this vector well, on the X axis, this would be here at two and on the Y axis negative one is down there. So our vector is going to look something like this. Now, what I need to do is figure out the direction of this vector and I'm going to first calculate this angle right in here. Now we're called to find this angle. We should take the inverse tangent of the Y component divided by the X component. So we're going to have the Y component which is negative one divided by the X component, which is two. And if you type the inverse tangent of negative 1/2, you should get approximately negative 27 degrees. So this right here is the missing angle. Now, you might think that this would be the solution to the problem. But it turns out it's actually not because we need to express our answer as a positive number from the positive x axis. So how do we find this angle? Well, we need to start on our positive X axis and we need to go counterclockwise all the way around until we reach this vector. Now, I can see that this is close to a full 360 degree rotation. So our angle is going to be a full 360 degree rotation minus this 27 degree angle that we just calculated. So we have 360 minus 27 degrees which comes out to 333 degrees. So this right here would be the direction of our vector and the solution to this problem to really make sure we understand how to do these situations where we don't have our vector just somewhere here in the first quadrant. Let's go ahead and try another example. So in example, b we're asked to find the vector negative three negative three to sketch it and to figure out its direction. Now, what I'm first going to do is draw this vector. So I can see on the X axis negative three is right there on the y axis negative three is down here. So our vector is going to look something like this. Now, what I'm going to do is calculate this angle to find the direction and to find this angle. Well, this angle is going to be the inverse tangent of the Y component divided by the X component. Now I can see that the Y component is going to be negative three. I can see that the X component is also negative three and negative three divided by negative three is just going to give you positive one. So we have the inverse tangent of positive one which is equal to 45 degrees. So this right here is this angle. And if we want to find the total direction of our vector, we need to start on the positive X axis and go counterclockwise until we reach our vector. Well, what would this be? Well, that's a 90 degree angle that would be 100 80 degrees total. And then, so we take 180 degrees and add it to this angle. We calculated that would be the total direction. So we have 100 80 degrees this angle which is 45 degrees. So 100 and 80 plus 45 should give you an angle of 225 degrees. So this right here is the total direction of our vector and the solution to this problem. So that is how you can find the direction of vectors even if you are given a vector that is not in the first quadrant like we had in this first example. So this is the strategy. Hope you found this video helpful. Thanks. For watching.

2

Problem

Problem

Find the direction of the following vector: $u ⃗=⟨-10,10⟩$.

A

45°

B

135°

C

−135°

D

315°

3

Problem

Problem

Find the direction of the following vector: $u⃗=\langle\frac{5\surd3}{3},5\rangle$.

A

60°

B

0.030°

C

30°

D

0.010°

4

example

Finding Direction of a Vector Example 1

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4m

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Welcome back everyone. Let's give this problem a try. So in this example, we're told if vector A is equal to 22, vector B is equal to two negative three, then vector C is equal to two A minus three B, calculate the magnitude and direction of C. So what we're really trying to do first is figure out what this vector C is and that will allow us to find both the magnitude and the direction of that vector. So let's see how we can do this. Well, I see that C is equal to two A minus three B. And first going to figure out what two A is, well, that's going to be two times the vector we're given here A. So we're going to have two, multiplied by the vector 22. Now I can distribute this two into the, these brackets here. So we're going to have two times two, which is four and then we'll have two times two again, which is also four. So this is what we get for vector two A. Now for vector three B, you can see that we have vector three or we have three times the vector that we have right there, which is two negative three. So we're going to go ahead and take this three and distribute it into these brackets. We're gonna have three times two, which is six and then three times negative three, which is negative nine. So that's gonna be vector three B. Now, what I can do is find vector C by subtracting these two vectors we calculated. So we're going to have 44 and then that's going to be minus the vector here, which is six negative nine. And the reason we're doing this is because notice how this is two A and that's three B. So we're just subtracting those to get vector C. Now, what I'm going to do is to track the individual components here. So we're going to have four minus six, which is negative two. And then we're going to have four minus negative nine, which is the same thing as four plus nine, which is 13. So this is what vector C turns out to be. So now that we have vector C, we can figure out what the magnitude and direction is to find the magnitude. We need to use this equation. The magnitude of C is going to be equal to the square root of the X component squared plus the Y component squared. We've discussed this equation before it's just a rearranged version of the Pythagorean theorem. So what I can do is plug in the numbers I can see that our X component is negative two and that's gonna be squared, they can see that our Y component is 13. So we have the square root of negative two squared plus 13 squared and negative two squared turns out to be positive four and 13 squared turns out to be 169 and four plus 100 69 is 100 and 73. So our vector C turns out to be 100 the square root of 100 73. And this actually can't simplify down any further. So that means that our vector magnitude C is going to be the square root of 100 73. And that's the solution for the magnitude of C. Now, what we need to do from here is figure out what our direction is and define our direction. Well, I'm first going to think about graphically what this is going to look like. So if we have some kind of graph that I'll draw over here, and we'll do a pretty simple graph where we have our X axis and our Y axis, I can see that vector C is negative 213, I can see that the X component is negative. So we're somewhere back here and I can see that our Y component is positive. So we're somewhere up there, meaning our vector is going to look something like that. And our first step is going to be to calculate what this the first step of finding the direction is going to be to calculate what this angle is here with the X axis. And then from there, we can find the total angle if we start from the X axis in the first quadrant and go all the way around. Now, the equation defined this angle is an equation we've discussed before. The tangent of the is equal to the Y component divided by the X component. Well, I can see up here that the Y component is 13 and the X component is negative two. So what are angle of becoming is the inverse tangent of 13 divided by negative two. If you go ahead and plug this into a calculator, you should get that this is approximately equal to 81.3 degrees and make sure your calculator is in degree mode when you plug in these numbers here. So if you, you should get about 81.3 degrees for this angle. And since this angle is approximately equal to 81.3 degrees, we can actually in a pretty straightforward way, figure out what this total angle is to find the total angle. Well, we can see that's a 90 degree angle and going all the way around will give us 100 and 80 degrees. And so finding this total angle, we can take 100 and 80 degrees the travel all the way around to here and we can subtract off this 81.3 degrees, we just calculated and 81.3 degrees. If you subtract that from 100 80 degrees, you should end up with 98.7 degrees. So 98.7 degrees is the angle, meaning that is going to be the direction of our vector. So that's how you can find the magnitude and direction of vector C when we needed to first figure out what vector C was through these operations. So hope you found this video helpful. Thanks for watching.

5

concept

Finding Components from Direction and Magnitude

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4m

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Welcome back everyone. So you may recall in previous videos, we've talked about how to find both the magnitude and direction of a vector if we are given the components of that vector. So the X component and the Y component. Well, what we're gonna be talking about in this video is how you can actually go backwards, how you can find the X and Y components of the vector using the magnitude and direction. Now, this might all sound like it's super bizarre and out of the blue. But it turns out this actually is a skill that you're going to need to have both in this course, as well as future math and science courses that you'll likely take. So without further ado let's get right into things because I think you'll find this process is actually pretty intuitive. So let's say we have this vector right here which has a magnitude of 10 and a direction of 53 degrees. If we want to calculate the X and Y components, we can just think of these vectors like a right triangle. And the right, using this right triangle logic, we can use the sine and cosine trig functions to figure this out. So let's go ahead and say we have this right triangle right here. In this case, the hypotenuse is going to be equivalent to the magnitude of the vector which we can see is 10 and the angle is going to be equivalent to the direction which is 53 degrees. Finding the X and Y components would simply be finding the missing sides of this right triangle. Now let's first see if we can start by finding why to find the Y component. Well, what I can recognize is the soo a memory tool which tells us that the sign of our angle is equal to opposite over hypotenuse. Now, the opposite side of this triangle is Y and the hypotenuse or along side of the triangle is 10. And then our angle would simply be the 53 degrees which is the direction of our vector. Now rearranging this equation, I can get that Y is equal to 10 times the s of 53 degrees. Now 53 degrees is not a nice number that we have on the unit circle. So what we're going to need to do is use a calculator to approximate this. The s of 53 degrees on a calculator is approximately equal to 0.8. So we'll have 10 times 0.8 which is equal to eight. So that means that our Y value is eight, which is also going to be the Y component of this vector. So as you can see, solving for the missing components of a vector is just a big trigonometry problem. We just need to find the missing sides. And you can use the same logic for finding the X component of this vector. So notice for finding the Y component, we took 10 which is the magnitude of our vector and we multiplied it by the sign of our angle which was 53 degrees. If you want to find the X component, you can take the magnitude of your vector and multiply it by the cosine of the angle. And you figure this all out using these trigonometric functions. So let's go ahead and see if we can calculate the X component. The X component is gonna be the magnitude of our vector which is 10 multiplied by the cosine of the angle that we see here, which is 53 degrees. Now 10 times the cosine of 53 on a calculator comes out approximately equal to six. So that means that our X component is six and our Y component is eight. And that's how you can find the missing components of a vector. Now to really make sure that we have this down, let's actually try another example where we have to find these missing components. Now, in this example, we're told if vector V has a magnitude of five and the direction is two pi over three radiance, calculate the X and Y components. Now we already learned that there's a pretty straightforward equation which allows us to find each of these components. So let's go ahead and use these equations up here. So the X component is going to be the magnitude of our vector, which I can see is five multiplied by the cosine of two pi over three radiant. Now it turns out two pi over three radiant is actually a value that shows up on the unit circle. And if you look up this value, it should be about negative one half. So we'll have five times negative one half, which will give me negative 5/2. So this right here is the X component of our vector. But now let's see if we can find the Y component, we'll find the Y component that's going to be the magnitude of our vector, which is five multiplied by the sine of our angles. We have the sine of two pi over three radiance. Now, the sine of two pi over three on a unit circle, that should be the square root of 3/2. So we're going to have five times the square root of 3/2. Now again, I can move this five to the numerator, which means that we'll have five square root of three over two. And that is the Y component of our vector. So this is how you can find the X and Y components using these equations that we learned about now, you may notice that one of our components turned out negative. So what does this really mean? Well, what this means is that we have a vector that is in the second quadrant. So if you imagine that we have this XY graph since this number is negative, that means that the X component is gonna be somewhere in the negative direction and the Y component is going to be somewhere in the positive direction since we got a positive result. So our vector would look something kind of like this and that's our vector that has a magnitude of five in a direction of two pi over three. And using this strategy, we were able to find the X component which we see is negative 5/2 and the Y component which is five square root 3/2. So this is how you can find the components of a vector using this trigonometry. So I hope you found this video helpful. Thanks for watching.

6

Problem

Problem

If a vector has magnitude $|v ⃗ |=13$ and direction $θ=157.38°$, find the vector’s horizontal and vertical components.

A

$v_{x}=5$ and $v_{y}=-12$

B

$v_{x}=-12$ and $v_{y}=5$

C

$v_{x}=12$ and $v_{y}=-5$

D

$v_{x}=-12$ and $v_{y}=-5$

7

Problem

Problem

If a vector has magnitude $|v ⃗ |=5$ and direction $\theta=\frac{7\pi}{4}$, find the vector’s horizontal and vertical components.