# Proof: Converse of Pythagoras' theorem using vectors

• Nov 6th 2012, 04:33 AM
Proof: Converse of Pythagoras' theorem using vectors
I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:
http://draw.to/static/d/19raZB.png?v=531d0
Take notice that $\vec{c} = \vec{a} + \vec{b}$. The given is that

$|\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} + \vec{b}|^2$

The goal is to show that $v = 90^\circ$. I dotted $\vec{a}$ with $\vec{b}$, thus yielding

$\vec{a}\cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos v$

I then attemped to find an expression for $\vec{a}\cdot \vec{b}$ from the equation $\vec{a} + \vec{b} = \vec{c}$ by multiplying both sides by $\vec{a}$ and dot every vector with it.

$\vec{a}(\vec{a} + \vec{b}) = \vec{c} \cdot \vec{a}$
$|\vec{a}|^2 + |\vec{b}| \cdot |\vec{a}| \cdot \cos v = |\vec{c}|\cdot |\vec{a}| \cdot \cos \theta$

$\theta$ is the angle formed between $\vec{b}$ and $\vec{c}$. I then I figured that I could compare the equation with $|\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} + \vec{b}|^2$ and then identify the terms. This however did not work out as it lead to showing that $\vec{a}$ is the hypotenuse whereas $\vec{c}$ is not, thus contradicting the very premise of the proof.

Anyone got an idea?
PS: I've already seen the proof using congruence but that particular proof is rather irrelevant, so don't bother referencing it.
• Nov 6th 2012, 05:47 AM
Plato
Re: Proof: Converse of Pythagoras' theorem using vectors
Quote:

I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:
http://draw.to/static/d/19raZB.png?v=531d0
Take notice that $\vec{c} = \vec{a} + \vec{b}$. The given is that
$|\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} + \vec{b}|^2$
The goal is to show that $v = 90^\circ$. I dotted $\vec{a}$ with $\vec{b}$, thus yielding.

From the given
$\|\vec{c}\|^2=\|\vec{a}\|^2 +\|\vec{b}\|^2$ and $\vec{c}=\vec{b}-\vec{a}$.

\begin{align*}\|\vec{c}\|^2&= \vec{c}\cdot\vec{c}\\ &= (\vec{b}-\vec{a})\cdot (\vec{b}-\vec{a})\\ &= \|\vec{b}\|^2 -2\vec{b} \cdot\vec{a}+\|\vec{a}\|^2 \end{align*}.

But that mean $\vec{b} \cdot\vec{a}=0$
• Nov 6th 2012, 12:41 PM
Thank you. It works just as well with $\vec{c} = \vec{a} + \vec{b}$ as with $\vec{c} = \vec{b} - \vec{a}$