Vector inner product as measure of closeness between vectors.

Question

Is this true because we can measure the alignment between 2 vectors through computing the angle between them via $a^T \cdot b = \|a\| \cdot \|b\| \cdot \cos(\theta)$ ? Or is there another reason?

Answer 1

Here is a very informal way of looking at this using the polarisation identity $\langle x , y \rangle = {1 \over 4} (\|x+y\|^2 - \|x-y\|^2)$ (in a real vector space).

Suppose $x,y$ are unit vectors, then we have $\|x+y\|^2 \le 4$ and similarly $\|x-y\|^2 \le 4$.

Further suppose $\langle x , y \rangle \approx 1$, then the polarisation identity shows that $\|x+y\|^2 - \|x-y\|^2 \approx 4$ or $\|x-y\|^2 \approx 0$ or equivalently $x \approx y$.

Similarly, if $\langle x , y \rangle \approx -1$, we get $x \approx -y$.

For arbitrary non zero vectors $a,b$, if the quantity $\langle {a \over \|a\|}, {\frac{b}{||b||}} \rangle $ is close to $\pm 1$ then we see that, roughly speaking, $a$ and $b$ are aligned in some way.

Answer 2

This can even be taken as the definition of the angle $\widehat{(u,v)}$ between two vectors:

$$\widehat{(u,v)}:=\arccos\Big(\frac{\langle u,v\rangle}{||u||\cdot||v||}\Big)\in[0,\pi].$$