Direction of a Vector - Online Tutor, Practice Problems & Exam Prep

Understanding vectors involves calculating their direction and components. The direction is found using the inverse tangent of the y component divided by the x component, expressed as a positive angle from the x-axis. For example, if a vector has a magnitude of 10 and a direction of 53 degrees, the x and y components can be calculated using $x=\mathrm{magnitude}\mathrm{cosine}\left(\mathrm{angle}\right)$ and $y=\mathrm{magnitude}\mathrm{sine}\left(\mathrm{angle}\right)$. This foundational knowledge is crucial for further studies in mathematics and science.

Finding Direction of a Vector

Hey everyone, and welcome back. So up to this point, we've 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. 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. Let's say we have this vector here which we'll call vector v. 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. 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 SOHCAHTOA. SOHCAHTOA 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. 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. 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. 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. We have the inverse tangent of the y component divided by the x component. I can see here that the y component is 1,2,3 units up, and I can see that the x component is 1, 2, 3, 4 units to the right. So we'll have the inverse tangent of 3 divided by 4, which on a calculator comes out to approximately θapprox. 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 direct of 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 expressed as a positive number from the x-axis. So let's go ahead and start with example a, where we have the vector 2 negative one. So what I first need to do is draw this vector. Well, on the x-axis, this would be here at 2, 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 recall to find this angle, we see to 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 2. And if you type the inverse tangent of negative one over 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 I have 360 minus 27 degrees which comes out to θ=333°. So this right here would be the direction of our vector and the solution to this problem. Now 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 you can see on the x axis negative 3 is right there, on the y axis negative 3 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 3. I can see that the x component is also negative 3. And negative 3 divided by negative 3 is just going to give you positive 1. So we have the inverse tangent of positive 1 which is equal to θ=45°. 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 a 180 degrees total. And so we take 180 degrees and add it to this angle we calculated, that would be the total direction. So we have 180 degrees plus this angle which is 45 degrees. So 180 plus 45 should give you an angle of θ=225°. So this right here is the total direction of our vector and the solution to this problem. 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.

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

• A. $45\degree$
• B. $135\degree$
• C. $-135\degree$
• D. $315\degree$

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

• A. $60\degree$
• B. $0.030\degree$
• C. $30\degree$
• D. $0.010\degree$

Finding Direction of a Vector Example 1

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 (2−3), and vector c is equal to 2a−3b, 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 direction of that vector. So let's see how we can do this. Well, I see that c is equal to 2a−3b. I'm first going to figure out what 2a is. Well, that's going to be 2×(a). Since we're given a to be (22),
now I can distribute this 2 into these brackets here. So we're going to have 2×2 which is 4, and then we'll have 2×2 again which is also 4. So this is what we get for vector 2a. Now for vector 3b, you can see that we have vector 3 or we have 3 times the vector that we have right there which is (2−3). So we're going to go ahead and take this 3 and distribute it into these brackets we're gonna have 3×2 which is 6, and then 3×−3 which is −9. So that's going to be vector 3b. Now what I can do is find vector c by subtracting these 2 vectors we calculated. So we're going to have (44), and then that's going to be minus the vector here which is (6−9). And the reason we're doing this is because notice how this is 2a and that's 3b, so we're just subtracting those to get vector c. Now what I'm going to do is subtract the individual components here, so we're going to have 4−6 which is −2, and then we're going to have 4+9 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 2, and that's gonna be squared. They can see that our y component is 13. So we have −22+132, and −22 turns out to be 4, and 132 turns out to be 169. And 4+169 is 173. So our vector c turns out to be 173 and this actually can't simplify down any further. So that means that our vector magnitude c is going to be 173. 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 to find 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 −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 for finding the direction is gonna 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 to find this angle is an equation we've discussed before the tangent of theta 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 2. So what our angle ends up becoming is the inverse tangent of 13 divided by negative 2. 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 would give us a 180 degrees. And so finding this total angle, we can take a 180 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 180 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 let's say 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.

Finding Components from Direction and Magnitude

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, the x component and the y component. Well, what we're going to be talking about in this video is how you can actually go backwards. How you can find the x and y components of a 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 actually pretty intuitive. So let's say we have this vector right here, which has a magnitude of 10 in 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 y. To find the y component, what I can recognize is the SOHCAHTOA memory tool, which tells us that the sine of our angle is equal to the opposite over the hypotenuse. Now the opposite side of this triangle is Y and the hypotenuse or long side of the triangle is 10, and 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 sine 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 sine of 53 degrees on a calculator is approximately equal to 0.8. So we'll have 10 times 0.8, which is equal to 8. So that means that our Y value is 8, 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 sine 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 could 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 going to be the magnitude of our vector, which is 10, multiplied by the cosine of the angle, which is 53 degrees. Now 10 times the cosine of 53 on a calculator comes out approximately equal to 6. So that means that our X component is 6, and our Y component is 8. 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 5 and the direction is 2 π over 3 radians, 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 5, multiplied by the cosine of 2 π over 3 radians. Now it turns out 2 pi over 3 radians 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 5 times negative one half, which will give me negative five over 2. So this right here is the X component of our vector. But now let's see if we can find the Y component. To find the Y component, it's going to be the magnitude of our vector, which is 5, multiplied by the sine of our angle. So we'll have the sine of 2 π over 3 radians. Now the sine of 2 π over 3, on a unit circle, that should be the square root of 3 over 2. So we're going to have 5 times the square root of 3 over 2. Now, again, I can move this 5 to the numerator, which means that we'll have 5 square root of 3 over 2 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 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 going to 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 5 and a direction of 2 pi over 3. And using this strategy, we were able to find the X component, which we see is negative 5 over 2, and the Y component, which is 5 square root of 3 over 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.

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 $v_{y}=-5$

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

• A. $v_{x}=4.98$ and $v_{y}=0.479$
• B. $v_{x}=0.479$ and $v_{y}=4.98$
• C. $v_{x}=-3.54$ and $v_{y}=3.54$
• D. $v_{x}=3.54$ and $v_{y}=-3.54$

Here’s what students ask on this topic:

How do you find the direction of a vector given its components?

To find the direction of a vector given its components, you use the inverse tangent function. The direction θ is the angle the vector makes with the positive x-axis. If the vector has components (x, y), the direction is calculated as:

$\theta =tan-1\left(\frac{y}{x}\right)$

For example, if the vector components are (3, 4), then:

$\theta =tan-1\left(\frac{4}{3}\right)\approx 53.13°$

What is the formula to find the x and y components of a vector given its magnitude and direction?

To find the x and y components of a vector given its magnitude and direction, you use the following formulas:

$x=\mathrm{magnitude}\cdot cos\left(\theta \right)$

$y=\mathrm{magnitude}\cdot sin\left(\theta \right)$

For example, if a vector has a magnitude of 10 and a direction of 53 degrees, the components are:

$x=10\cdot cos(53°)\approx 6$

$y=10\cdot sin(53°)\approx 8$

How do you express the direction of a vector as a positive angle from the x-axis?

To express the direction of a vector as a positive angle from the x-axis, you need to consider the quadrant in which the vector lies. If the calculated angle θ is negative, add 360° to it. For example, if θ = -27°, the positive angle is:

$360°-27°=333°$

For vectors in other quadrants, adjust the angle accordingly by adding 180° or 270° as needed to ensure the angle is measured counterclockwise from the positive x-axis.

How do you find the components of a vector if given in radians?

To find the components of a vector given in radians, use the same formulas as for degrees but ensure the angle is in radians. For example, if a vector has a magnitude of 5 and a direction of 2π/3 radians, the components are:

$x=5\cdot cos(\frac{2}{\pi }/3)=-\frac{5}{2}$

$y=5\cdot sin(\frac{2}{\pi }/3)=\frac{5}{\surd }$

What is the significance of a negative component in a vector?

A negative component in a vector indicates the direction of the vector relative to the coordinate axes. For example, if the x component is negative, the vector points to the left of the y-axis. If the y component is negative, the vector points downward. This helps in determining the vector's position in the coordinate plane. For instance, a vector with components (-3, -3) lies in the third quadrant, indicating it points left and down from the origin.