 Now we're going to look at a little bit of a different way to calculate the work using something called the dot product. Well, you remember before, if we had a constant force, we described our work equation as the magnitude of the force times the magnitude of the displacement times the cosine of the angle in between them. So this equation used the magnitudes of those two vector quantities, force and displacement. And the directions come in using the angle. There's another way for us to think about this angle down here, the angle between the force and the displacement. If I think about this force and project it down onto the displacement, I have this projection down here. And this projection is f cosine theta, because it's the adjacent side of this triangle. So that means this f cosine theta could be thought of as the parallel component of the force to the displacement. In other words, what work really cares about is the parallel parts of the force and the displacement. So what if I start with my vectors in vector equation notation? So I've got things like, rather than giving the magnitude and direction, I've given the force in my x and y components with my i hat and j hat. And the same thing for my displacement. How am I going to find the parallel components of these two things without knowing what the angle is? Well, it turns out if we look at the x component of the force and the x component of the displacement, those two things are automatically parallel to each other because they're both in the x direction. Similarly, my f y and my delta y are parallel to each other. So if I multiply not the entire force by the entire displacement, but the components in the x and the components in the y, those two things are both parallel components. Now to get the full work, I have to include both of these. So I simply take those two numbers and add them together. So I've got the x force and x displacement multiplied plus the y force and the y displacement multiplied. Now this is called a dot product, and it's one of our vector math things we can do. So what I would write this then is that the work is the dot product between the force vector and the displacement vector. And notice here, I'm not using the magnitudes anymore. I'm using the vectors. And this isn't just multiplying two numbers. It's a vector multiplication that follows a very specific format. In this case, it's that thing where I take the x components and multiply them plus the y components and multiply them. So this dot product here for my work is exactly equivalent numerically to taking the magnitudes and multiplying them. But then I also have to multiply the angle in there. So if I'm given the angle, it's more convenient to use this form where I'm taking the cosine of the angle. But if I'm given everything in component form, it's much more convenient to use the dot product. In the end, though, regardless of which way you do this, you don't have two vectors anymore. You simply have a scalar. So sometimes this dot product is also called the scalar product. So let's do an example of that. So here's an example where I've got my force written out as a vector and my displacement written out as a vector. Well, first I want to take the x components, the 2 in the i-hat for my force and my 4 in the i-hat for my displacement. And that 2 newtons times 4 meters. And then my j, 3 newtons and minus 1 meters gives me my other part. When I multiply these, I've got then 8 newton meters for the contribution of the work in the x direction and minus 3 newton meters for my j direction. And putting those together, I then have 5 newton meters. So I don't have a vector anymore. I've just got a number. Now remember, we could also express that 5 newton meters as 5 joules because joule is equivalent to a newton meter. It's the other unit we can use for work. So there's an example of dot products and how we use them to find the work if we've got a constant force.