Scalar product and geometric interpretation of formula

Question

I just studied the inner product space and wanted to find out any geometric explanation of the equality:

$\langle x,y\rangle=\frac{||x||^2+||y||^2-||x-y||^2}{2}. $

I know that the scalar product gives the length of the orthogonal projection of one vector $x$ over the other vector $y$. Furthermore I understand the geometric picture of the distributive law, which has to be applied twice:

$||x-y||^2=\langle x-y,x-y\rangle =\langle x,x-y\rangle-\langle y ,x-y\rangle= ||x||^2+||y||^2-2\langle x,y\rangle.$

Nevertheless, this does not convince myself to get an global geometric picture/explanation of my first equality.

Is there anyone who has it clear? Thank you very much.

Answer 1

Please bear in mind that while the scalar product has a geometric motivation, it is generally hard to understand why $\langle x, y \rangle = k $ except for when $x \perp y $ and when $x $ is a scalar multiple of $y $.

Having that said, remember that if you want to project $x $ over some vector $u $ then the projection $p $ is:

$$\frac {\langle x, u \rangle}{||u||^2}u = p $$

And therefore the only way your identity could make sense is, in my opinion, by substituting in the formula above and check that it can reduce to the same thing. Perhaps you could start by checking heuristically that the formula holds for orthogonal $x, y $ and for parallel $x, y $.

Answer 2

That formula is nothing but the cosine law, because vector $x-y$ is the third side of a triangle of sides $x$ and $y$: $$ |x-y|^2=|x|^2+|y|^2-2|x||y|\cos\theta= |x|^2+|y|^2-2\langle x,y\rangle. $$ And cosine law has a geometric explanation, see here for instance.