Welcome back, everyone. So, in the previous video, we talked about unit vectors. We got introduced to this general topic, and we specifically discussed vectors that point in the x and y directions, which were vectors I hat and j hat, respectively. Now, what we're gonna be talking about in this video is how we can actually calculate a unit vector that points in any direction. Because it's possible that you're going to have a vector that doesn't just point to the right or up. You might have a vector that points in some other direction and we're gonna talk about how to find unit vectors that point anywhere. So without further ado, let's just jump right into an example.
So let's say we have this example where we're given vector v and vector v is equal to \( 4\mathbf{i} + 3\mathbf{j} \). We're asked to find a unit vector \(\mathbf{\hat{v}}\) that points in the same direction as v. Now, since this vector is a combination of i's and j's we can't say that our unit vector is just gonna be I or j. So to figure out what this unit vector is, it turns out there's actually a straightforward way of doing this. All you need to do is take whatever vector you're given and divide it by its magnitude, and that's going to give you the unit vector. And I think this makes logical sense, because if you take a number and divide it by itself, you would typically just get 1. And we know unit vectors have a magnitude of 1, so it makes sense that we would just need to divide this vector by its own magnitude.
So to find the magnitude of a vector, you just need to take the square root of the x component squared plus the y component squared. I see the x component is 4 and that the y component is 3. Now, it turns out the square root of \(4^2 + 3^2\) is all just going to come out to equal 5. So 5 is the magnitude of our vector v. So now that we have our magnitude we can calculate \(\mathbf{\hat{v}}\). All we need to do is take our vector which is \(4\mathbf{i} + 3\mathbf{j}\) and divide it by its own magnitude which we calculated to be 5. So I'm gonna take each of these components 4 and 3 and I'm going to divide them by 5. So we'll have \(\frac{4}{5} \mathbf{i} + \frac{3}{5} \mathbf{j}\), and this right here is the unit vector \(\mathbf{\hat{v}}\) and the solution to this example. And so what happens is if we go over here to our graph, this vector \(\mathbf{\hat{v}}\) is going to look something like this. Unit vectors always have a magnitude of 1, but it's going to point in the direction of the vector that we already have.
And what this tells us what we just calculated is that this vector specifically points \(\frac{4}{5}\) in the I direction, and it points \(\frac{3}{5}\) in the j direction, and that's what gives you the unit vector \(\mathbf{\hat{v}}\). Now something else that we might need to do is figure out or prove to ourselves that this is actually a unit vector. So let's go ahead and try this example where it asks us to find if \(\mathbf{\hat{v}}\) in this example above is actually a unit vector.
Well, what I'm going to do is calculate the magnitude of \(\mathbf{\hat{v}}\). To find the magnitude of any vector, you just need to take the square root of the x component squared plus the y component squared. Now I can see that the x component is \(\frac{4}{5}\), and I need to add this to the y component which is \(\frac{3}{5}\). We're going to have the square root, and then we have \(4^2\) which is 16 divided by \(5^2\) which is 25, that's going to be plus \(3^2\), which is 9, divided by \(5^2\), which is 25. Now notice we got the same denominator in both cases. And since we have the same denominator, we can just add the numerator straight across. Now 16+9, that's equal to 25. So we'll have the square root of \(\frac{25}{25}\). \(\frac{25}{25}\) is just gonna be equal to 1, and the square root of 1 is just 1. So notice how we got that our magnitude for \(\mathbf{\hat{v}}\) is 1, and that means this is a unit vector. So the solution we calculated up here is correct.
This is the main idea of finding unit vectors in any direction. All you need to do is take whatever vector you have and divide it by its magnitude, and then you can use this strategy of finding the magnitude of your unit vector to check to make sure you did things right. Hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.