Proof: Converse of Pythagoras' theorem using vectors

**MathCrusader** · November 6th 2012, 04:33 AM

I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:

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.

**Plato** · November 6th 2012, 05:47 AM

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:

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 $\vec{b} \cdot\vec{a}=0$

**MathCrusader** · November 6th 2012, 12:41 PM

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

**Chalama123** · November 8th 2012, 07:11 PM

How To Prove Trapezoid Median Theorem Using Vectors?

exede satellite internet

Thread: Proof: Converse of Pythagoras' theorem using vectors

LinkBack

Thread Tools

Display

Proof: Converse of Pythagoras' theorem using vectors

Re: Proof: Converse of Pythagoras' theorem using vectors

Re: Proof: Converse of Pythagoras' theorem using vectors

Re: Proof: Converse of Pythagoras' theorem using vectors

Similar Math Help Forum Discussions

is there converse for side splitter theorem?

Pythagorean Theorem and Converse

converse to thales theorem

Converse of Fermat's Little Theorem (particular case)

proving the converse theorem

Search Tags