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 $\displaystyle \vec{c} = \vec{a} + \vec{b}$. The given is that

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

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

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

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

$\displaystyle \vec{a}(\vec{a} + \vec{b}) = \vec{c} \cdot \vec{a}$

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

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

Re: Proof: Converse of Pythagoras' theorem using vectors

Quote:

Originally Posted by

**MathCrusader** 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 $\displaystyle \vec{c} = \vec{a} + \vec{b}$. The given is that

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

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

From the given

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

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

But that mean $\displaystyle \vec{b} \cdot\vec{a}=0$

Re: Proof: Converse of Pythagoras' theorem using vectors

Thank you. It works just as well with $\displaystyle \vec{c} = \vec{a} + \vec{b}$ as with $\displaystyle \vec{c} = \vec{b} - \vec{a}$

Re: Proof: Converse of Pythagoras' theorem using vectors

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