Law of cosines for $n$ dimensional vectors

Question

The law of cosine is $$\|b-a\|^{2}=\|b\|^{2}+\|a\|^{2}-2\|b\|\|a\| \cos \theta $$

Is there a way to prove this when vectors $a, b$ have more than 3 components?

For 3 or less components one can prove it using geometry but what about higher dimensions?

Answer 1

This is obviously trivially independent of the dimension.. If you square the difference b-a you get the square of b minus the square of a, minus 2 times the the inner product of a and b.

If you want a geometric interpretation, imagine you want to find a circle circumference. You need to approximate its lenght by regular polygons whose sides you relate to the radius using the inner product and if you use cosine in its definition and take as angle to be in radians you get the correct approximation and in the limit you get 2 pi which you can again approximate using archimedes method.

Answer 2

In any dimension, the norm is related to the scalar product by : $$\| a\|^2 = a\cdot a$$

and the angle $\theta$ between two vectors $a,b$ is defined by : $$a\cdot b = \|a\|\|b\|\cos \theta$$

Therefore, : \begin{align} \|a- b\|^2 &= (a-b)\cdot (a-b) \\ &= a\cdot a - a \cdot b - b\cdot a + b\cdot b \\ &= \|a\|^2 + \|b\|^2 - 2 a\cdot b\\ &= \|a\|^2 + \|b\|^2 - 2 \| a\|\|b\|\cos \theta \end{align}