# How to prove the Pythagoras theorem using vectors

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.

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I can't help but feel that something here is circular... –  Ｊ. Ｍ. Nov 23 '10 at 14:04

$x,y$ 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.

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Ok now I get it. Using distributivity on the left side I'd get (x.(x+y))+(y.(x+y)) right? Then using distributivity again I'd get (x.x)+(x.y)+(y.x)+(y.y) and because the x.y=0 as I mentioned that results in (x.x)+(y.y) which is the same as the right side of the equation. Can I mark two answers as the answer? Because both helped me realize how to get the answer. Thanks! –  Rope Nov 23 '10 at 14:44

The definition of $||x||$ for vectors is: $$||x|| = \sqrt{x\cdot x}.$$

So, you have that \begin{align*} ||x+y||^2 &= (x+y)\cdot(x+y) &\text{(by definition)}\\ &= x\cdot x + x\cdot y + y\cdot x + y\cdot y &\text{(by distributivity)}\\ &= x\cdot x + y\cdot y + 2(x\cdot y) &\text{(by symmetry)}\\ &= ||x||^2 + ||y||^2 + 2(x\cdot y) &\text{(by definition)}\\ &= ||x||^2 + ||y||^2 + 2||x||\,||y||\cos(\theta), \end{align*} where $\theta$ is the angle between $x$ and $y$.

This holds in any case.

So, $||x+y||^2 = ||x||^2 + ||y||^2$ if and only if $2||x||\,||y||\cos(\theta)=0$. One possibility is $||x||=0$; another is $||y||=0$; and the final one is $\cos(\theta)=0$. So the equality holds if and only if one of the following happens:

1. $||x||=0$;
2. $||y||=0$; or
3. $\cos(\theta)=0$ where $\theta$ is the angle between $x$ and $y$, $x\neq 0$, $y\neq 0$.
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"angle", Arturo, "angle". :) –  Ｊ. Ｍ. Nov 23 '10 at 15:37
@J.M. If the devil is in the details, can't the angels be in between the vectors? Thanks. –  Arturo Magidin Nov 23 '10 at 15:57

If I understood the question correctly, you have to prove that $$\|x+y\|^2 = \|x\|^2 + \|y\|^2$$ if and only if $x$ and $y$ are perpendicular. That is, as you noted, $x \cdot y = 0$.

Hint: You can use the following formulas: $$\|x\|^2 = x \cdot x$$ $$(x+y) \cdot z = x \cdot z + y \cdot z$$

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Yes I have to prove that equation because, it is stated that it is the Pythagoras theorem expressed in vectors on the symbolic level. I started out with exactly what you hinted about. But how do I use the fact that x and y have to perpendicular? Or do I not use it at all and just go solving the equation by pattern matching the equation? –  Rope Nov 23 '10 at 14:17
There is not much to do after expanding $\|x+y\|^2$ using the dot product and those two rules I gave you. Then you just notice that the result is equal to $\|x\|^2 + \|y\|^2$ if and only if $x \cdot y = 0$. –  J. J. Nov 23 '10 at 14:24
If I were to use the distributivity rule you gave as a hint wouldn't that result in (x.(x+y))+(y.(x+y))? If I use it on the left side of the equation? –  Rope Nov 23 '10 at 14:35
Yes, and then you can use it again (well, strictly speaking you need to know that distributivity works for the second argument as well) to get $x \cdot x + x \cdot y + y \cdot x + y \cdot y = \|x\|^2 + x \cdot y + y \cdot x + \|y\|^2$. –  J. J. Nov 23 '10 at 15:57