So far we have talked about geometric vectors, which are like arrows with magnitude and direction. This is an intuitive geometric interpretation, and it’s useful for some things, but there’s a lot more to vectors than just that. In the broader context of vectors, we talk about vector spaces. There are many different vector spaces, but we will only worry about the Euclidean ones. A Euclidean vector is a collection of real numbers. In `RR^2` or two-space, there are two components; in `RR^3` or three-space, there are three.
The notation we use for a vector in two-space is `vec v = [Delta x,Delta y]`, where the deltas go from the vector’s tail to its tip. Say we wanted to designate a vector from the origin to point P(3,−2). This is simply `vec(OP) = [3,-2]`. Since we often interpret our Euclidean vectors in the Cartesian coordinate system, we also call them Cartesian vectors.
What if you know points A and B, and want a vector that takes you from A to B? We can simplify this problem using the identity that we learned in the section on addition. Consider this:
`vec(AB) = vec(AO) + vec(OB) = -vec(OA) + vec(OB) = vec(OB) - vec(OA)`.
In words, you can get a vector from A to B by doing `vec(PB)-vec(PA)`, where P is some common reference point (usually the origin). You just need to remember that it’s tip minus tail and not the other way around.
How do we add or subtract Euclidean vectors? It’s actually very easy—just do the operation to each component separately:
`[a,b] + [c,d] = [a+c,b+d]`.
It works the same way for subtraction. Scalar multiplication distributes:
`k[x,y] = [kx,ky]`.
There are a few advantages of this over the geometric vectors we were using before. For one thing, it’s a lot less work. Also, it’s much easier to be precise: you could add a hundred vectors this way without breaking a sweat. Try that with geometric vectors—your answer will be buried so deep in trig functions and square roots that you will be forced to round off.
To get the magnitude of `vec v = [x,y]`, you can use the Pythagorean theorem:
`|vec v| = sqrt(x^2 + y^2)`.
You can find the direction of `vec v` by drawing a triangle and using the inverse tangent function. Then, you can state the vector with the usual magnitude-direction representation—for example, `[4\ "m",3\ "m"]` becomes 5 m [E 30º N]. Going the other way (from magnitude-direction to x and y components) is called resolving the vector, and you can do it by sketching a right triangle and using sine and cosine.
All of this is straightforward in `RR^3` as well. You just use one more component. Addition, subtraction, and scalar multiplication work the same. To get the magnitude, just include `z^2` in the sum.
Normalizing a geometric vector is easy: just change the magnitude to 1. It’s a little bit more work with these Euclidean vectors. Say we want the unit vector parallel to `vec v = [3,5,6]`:
`hat v = vec v/|vec v| = ([3,5,6])/(sqrt(3^2+5^2+6^2)) = 1/sqrt(70)[3,5,6]`.
That’s it. You can distribute the coefficient if you want, but there is really no need—why write `sqrt(70)` three times instead of once?