Why has the scalar product anything to do with angles?

Question

As a maths student I use scalar products incredibly often and have gotten used to the fact that $<a,b>=0 $ is equivalent to $a\perp b$. The generalizations on function spaces, etc. are also quite useful (Orthonormal basis have really nice properties). But I realized that I couldn't explain why the scalar product has anything to do with orthogonality.

And looking the scalar product up, it seems it has even more information about angles. Since apparently $<x,y>=||x||\cdot||y|| \text{cos}(\alpha)$ with $\alpha$ being the angle between x and y.

Why does multiplying the individual scalars end up telling us so much about angles? I am mostly interested in an intuition. As I think we did a proof in linear algebra at some point but I only remember that I was disappointed by it in the sense that I still had no idea why this relation existed afterwards. But if you know about a very instructive proof that would be interesting as well

And lastly, is it even possible to get an intuition for scalar products on function spaces, or is that a generalization you can only get used to, but not really understand?

Answer 1

The notion of orthogonality is defined only on inner product vector spaces and the definition is:

two vectors $x,y$ are orthogonal if $\langle x,y\rangle=0$

From this we can define what an ''angle'' is in any vector space with an inner product and the definition generalize the usual definition of angle between two oriented segments in a plane.

Answer 2

Physicists use the scalar product in the way you stated, $<x,y>=||x|| ||x||\cos(\alpha)$, so the scalar product was born from the notion of angle.

In mathematics, things are the other way around: if you have studied linear algebra, and metrics in vector spaces, you will now that, given a metric $T_2$ in a vector space $E$, you $\textit{define}$ the scalar product as $<e,e'>=T_2(e,e')$, then the norm as $||e||=+\sqrt{T_2(e,e)}$, and then the angle as $\cos\left({\widehat{e,e'}}\right)=\frac{<e,e'>}{||e|| ||e'||}$, so the definiton of angle comes from the scalar product.