# Vector spaces

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 ${\mathbb{R}}^{2}$ or two-space, there are two components; in ${\mathbb{R}}^{3}$ or three-space, there are three.

The notation we use for a vector in two-space is $\overrightarrow{v}=[\mathrm{\Delta}x,\mathrm{\Delta}y]\text{,}$ 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 $\overrightarrow{OP}=[3,-2]\text{.}$ 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:

$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}\text{.}$

In words, you can get a vector from A to B by doing $\overrightarrow{PB}-\overrightarrow{PA}\text{,}$ 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:

$[a,b]+[c,d]=[a+c,b+d]\text{.}$

It works the same way for subtraction. Scalar multiplication distributes:

$k[x,y]=[kx,ky]\text{.}$

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 $\overrightarrow{v}=[x,y]\text{,}$ you can use the Pythagorean theorem:

$\left|\overrightarrow{v}\right|=\sqrt{{x}^{2}+{y}^{2}}\text{.}$

You can find the direction of $\overrightarrow{v}$ by drawing a triangle and using the inverse tangent function. Then, you can state the vector with the usual magnitude-direction representation—for example, [4m, 3m] 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 ${\mathbb{R}}^{3}$ 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 $\overrightarrow{v}=[\mathrm{3,5,6}]$:

$\hat{v}=\frac{\overrightarrow{v}}{\left|\overrightarrow{v}\right|}=\frac{[\mathrm{3,5,6}]}{\sqrt{{3}^{2}+{5}^{2}+{6}^{2}}}=\frac{1}{\sqrt{70}}[\mathrm{3,5,6}]\text{.}$

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?