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Math Help - Proof: Converse of Pythagoras' theorem using vectors

  1. #1
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    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:

    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.
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  2. #2
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    Re: Proof: Converse of Pythagoras' theorem using vectors

    Quote Originally Posted by MathCrusader View Post
    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
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  3. #3
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    Re: Proof: Converse of Pythagoras' theorem using vectors

    Thank you. It works just as well with \vec{c} = \vec{a} + \vec{b} as with \vec{c} = \vec{b} - \vec{a}
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    Re: Proof: Converse of Pythagoras' theorem using vectors

    How To Prove Trapezoid Median Theorem Using Vectors?





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