# Angle between two n-dimensional vectors that are not already defined in some basis.

There's something that's always bothered me in my encounters with linear algebra, it seems like every vector they ever give is already defined according to some basis: e.g. A=<2,-1,0,5,6>.

Imagine that I have a very large circular sheet of paper so that there is no obvious x and y (or i and j). I can put a dot somewhere on the paper and draw an arrow and call it V. Then, from the same dot, I can draw another arrow and call it e1 (my first basis vector). Now, without going into details, if I just have a piece of string and a straight-edge (no rulers, protractors or other pre-made measuring devices) I can measure the length of V in terms of the length of e1 (which I take to be 1-unit long), and I can use the string to make circles and I can use the string to measure arc-lengths in terms of e1 and I can figure out the angle (in radians) between V and e1. And, with just the string and straight-edge, I can even construct e2 perpendicular to e1.

I can figure out how to do all of this in 2-dimensions without relying on a pre-existing set of basis vectors, but it requires me to actually physically draw and measure things.

My question is how do you do this in n-dimensions? Suppose I have a vector in 4-dimensional space, but that space does not yet have any defined basis. So, I define a second vector as e1. To the best of my knowledge 4D paper and 4D string do not exist in the physical world, so I have to do it abstractly. How do I determine the relative length of my two vectors and the angle between them?

Thanks.

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I'm not sure what your exact question is, but this may help: en.wikipedia.org/wiki/Dot_product#Geometric_interpretation –  wj32 Oct 19 '12 at 0:45

Well, you can take the vector space made up of continuous real-valued functions on the closed interval $[0,1],$ then say that the length of any single vector $f(x)$ is defined to be $$|f| \; = \; \sqrt{\int_0^1 \; f^2(x) \, dx \;}$$ and the inner product between two vectors $f,g$ is $$\langle f,g \rangle \; = \; \int_0^1 \; f(x) g(x) \, dx \;$$ and it all works, absolutely everything works, with no evident basis.
One could also point out that the dot product, working as it does in $\mathbb R^n,$ is a theorem by induction on the (finite) dimension, with the basic fact being the Pythagorean Theorem.
@user45177, you are incorrect in assuming that the $x^n$ make an "orthonormal basis." This is not the case. $$\langle x^m, x^n \rangle \; = \; \frac{1}{1 + m + n}.$$ The Taylor series for $e^x$ does not tell you anything useful about the inner product of $e^x$ and some fixed $x^m.$ –  Will Jagy Oct 19 '12 at 1:47