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Math Help - Orthogonal Vectors (Calc. 3)

  1. #1
    Newbie mishdo's Avatar
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    Exclamation Orthogonal Vectors (Calc. 3)

    The question is: Suppose the vectors (a+b) and (a-b) are orthogonal. What, if anything, can you conclude about a and b?

    So far all I know is that since they are orthogonal, the dot product of (a+b).(a-b)=0. I'm not so sure what to do after that. I've been trying to substitute a=<a1,a2,a3> and b=<b1,b2,b3> to do the dot product and somehow get a.b=0 but no such luck. I don't even think that approach is right. Is there another way to approach this problem?

    Thanks!!
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  2. #2
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    Re: Orthogonal Vectors (Calc. 3)

    You can also write (a+b).(a-b) as:

    (a+b)(a-b) = aa - ab + ba - bb .

    Do you know if the dot product is commutative? That will help you to simplify the above expression.
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    Re: Orthogonal Vectors (Calc. 3)

    Quote Originally Posted by mishdo View Post
    The question is: Suppose the vectors (a+b) and (a-b) are orthogonal. What, if anything, can you conclude about a and b?
    Can you show that
    (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|-\|\vec{b}\|~?
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  4. #4
    Newbie mishdo's Avatar
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    Re: Orthogonal Vectors (Calc. 3)

    Quote Originally Posted by Plato View Post
    Can you show that
    (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|-\|\vec{b}\|~?
    I had a feeling it had to do with magnitude. I tried that out, but I'm a little stuck. So far I have:

    a1^2+a2^2+a3^2-b1^2-b2^2-b3^2=\|\vec{a}\|-\|\vec{b}\|

    I can see the magnitude formula in there, but can't figure out how to pull it out
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  5. #5
    Newbie mishdo's Avatar
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    Re: Orthogonal Vectors (Calc. 3)

    Quote Originally Posted by SammyS View Post
    You can also write (a+b).(a-b) as:

    (a+b)(a-b) = aa - ab + ba - bb .

    Do you know if the dot product is commutative? That will help you to simplify the above expression.
    Thanks, I tried it but I sadly didn't get anywhere
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  6. #6
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    Re: Orthogonal Vectors (Calc. 3)

    Quote Originally Posted by mishdo View Post
    Thanks, I tried it but I sadly didn't get anywhere
    Looking an SammyS's post # 2, the dot product there is commutative. That is,

    \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a  }.

    That leads to a simplification. Can you see it? That should lead you to be able to obtain Plato's result.
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