8. Vectors
Direction of a Vector
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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 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, well, what I can recognize is the SOHCAHTOA memory tool, which tells us that the sine of our angle is equal to opposite over hypotenuse. Now the opposite side of this triangle is y and the hypotenuse or long 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 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 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 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 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 pi 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 PI 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. Well, to find the y component, that'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 pi over 3 radians. Now the sine of 2 pi over 3, on 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 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 x y 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 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 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.