I've got a question concerning how to proof the Pythagoras theorem using the following assumption:
$x$ is perpendicular to $y$ (if and only if) $||x+y||^2 = ||x||^2 + ||y||^2$,
where $x$ and $y$ are vectors.
I have a basic understanding of linear algebra, however I'm a beginner with this. The question provides hints how to prove the above mentioned equation.
Namely that I should use the properties of the dot product and the definition of the norm of a vector.
Those being symmetry, scaling and distributivity as the dot product properties and the norm of a vector being the squared root of the dot product between the same vector.
I was thinking about using the fact that if a vector is perpendicular to another vector the dot product between those vectors should be 0. But that is not provided as a hint so I'm not sure. I know the under lying thought behind it is the cosine rule for vectors, that being:
$$x\cdot y = ||x||\,||y|| \cos(\theta)$$
If the angle between the two vectors is perpendicular you should use $\cos(\pi/2)$ which is $0$ and $||x||\cdot 0 = 0$ and $||y||\cdot 0 = 0$ with the vectors not necessarily being $0$. Thus $x\cdot y = 0$.
How would I apply this to the equation I first mentioned to prove the Pythagoras theorem?
I have a few more thoughts on how I could prove this but I'm not sure if they're correct.
I hope someone could point me in the right direction.
Best Answer
$x,y$ are perpendicular if and only if $x\cdot y=0$. Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$. The middle two terms are zero if and only if $x,y$ are perpendiculat. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular.