 So, remember, a vector tells us how to get from one point to another, but if you want to get from one point to another, you'll go some distance in some direction, and so we often say that a vector is a directed distance. And because it is a directed distance, we can ask questions like how far, that's our distance, and what direction, that's our directed. For example, suppose I travel along the vector 5 negative 12, how far do you travel? Now the important idea here is the vector itself tells us to travel five units to the right, and twelve units downward. But here's a useful idea to keep in mind. Going in circles is going nowhere. It doesn't matter how far you've gone, what really matters is where you are relative to where you started from. And in this case, the actual distance we travel is the hypotenuse of the right triangle. And so we can find that hypotenuse, the square of that distance, well that's five squared plus twelve squared, and so our distance will be, and this will be positive because we want it to be a distance. And this leads to the following idea. The Pythagorean theorem generalizes to higher dimensions, and this gives us the following. The magnitude of a vector is going to be the square root of the sum of the squares of the components. And this is just a generalization of the Pythagorean theorem to higher dimensions. So for example, let's find the magnitude of the vector 3 negative 1, 5. So definitions are the whole of mathematics, all else is commentary. Let's pull in our definition of the magnitude. So the magnitude will be the square root of the sum of the squares of the components. And so that will be, now remember a vector has both a magnitude and a direction, and to make it easier to compare two vectors, it's helpful if we standardize their length. And so we introduce the following concept. A unit vector is a vector with a magnitude of 1. So let's find a unit vector that points in the same direction as 1 for negative 3. Well, let's think about how we might do this. Remember that mv is the result of repeating vm times. But this won't change the direction, only the distance we go. And so thinking about this, we want to find m where the magnitude of mv is equal to 1. Now we also might throw in one other idea if m is less than 0, then m times the vector will have us going in the opposite direction, so we'll require that m be greater than 0. Will this work? I don't know, but remember it's better to try something than to try nothing. So let's see what happens. We want the magnitude of m times the vector to be equal to 1. Well, we have the scalar multiplication here, so let's go ahead and do what we can when we multiply m times the vector we get. But now we have the magnitude of a vector, and we know how to find the magnitude, it's the square root of the sum of the squares of the components. We can do a little algebra. We can do a little more than that. Now remember that we are assuming that m is greater than 0, so this m squared can be removed outside of the square root, and we can solve for m. And so that tells us that 1 over square root 26 times the vector will be a vector going in the same direction that has magnitude 1, so our vector will be