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Math Help - Vector Reflection

  1. #1
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    Vector Reflection

    First of all, sorry if i posted this in the wrong forum. This is for my Discrete preparation course for univerisity and there was no Discrete forum under the highschool heading

    Anyways here is the question:

    Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

    I tried showing it through the use of the projection of a vector formula but I keep getting stuck. Could any of you please offer some guidance or a suggestion on where to start?
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  2. #2
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    Quote Originally Posted by dimatt
    ...
    Anyways here is the question:
    Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

    Could any of you please offer some ... suggestion on where to start?
    Hello,

    I've attached an image to demonstrate, what I've done:

    1.: |\vec{a}|=|\vec{a'}|

    2.: \vec{a}+\vec{a'}=c \cdot \vec{b}
    (c is a constant factor)

    Therefore: \vec{a'}=c \cdot \vec{b}-\vec{a}

    My "solution" looks a little bit too easy.

    The other possibility I can think of is, that you use matrices.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Vector Reflection-spieg_vekt.gif  
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  3. #3
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    Quote Originally Posted by dimatt
    First of all, sorry if i posted this in the wrong forum. This is for my Discrete preparation course for univerisity and there was no Discrete forum under the highschool heading

    Anyways here is the question:

    Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

    I tried showing it through the use of the projection of a vector formula but I keep getting stuck. Could any of you please offer some guidance or a suggestion on where to start?
    Resolve \bf a into componets parallel and orthogonal to \bf b.

    Parrallel component:

    <br />
(\bf a . \bf b) \hat{\bf b}<br />

    Othogonal component:

    <br />
\bf a - (\bf a . \bf b) \hat{\bf b}<br />

    Hence:

    <br />
\bf c=(\bf a . \bf b) \hat{\bf b}-(\bf a - (\bf a . \bf b) \hat{\bf b})=2(\bf a . \bf b) \hat{\bf b}-\bf a<br />

    Which may be rewritten:

    <br />
\bf c=2\ \frac{\bf a . \bf b}{|\bf b|}\ \bf b-\bf a<br />

    which is the form required.

    RonL
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  4. #4
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    dimatt

    Wow, thanks guys. I totally forgot about the orthogonal vector

    edit- Why do you subtract the parallel from the orothogonal? Wouldnt that just give you the length of 'a'?
    Last edited by dimatt; April 14th 2006 at 07:34 PM.
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    ??
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  6. #6
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    Quote Originally Posted by dimatt
    Wow, thanks guys. I totally forgot about the orthogonal vector

    edit- Why do you subtract the parallel from the orothogonal? Wouldnt that just give you the length of 'a'?
    It gives a vector with the orthogonal component about b flipped
    from one side of b to the other, that is it is reflecting the orthogonal
    component in b.

    RonL
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  7. #7
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    Oh, i see where i went wrong. I kept looking at c as the projection of a on b; for some reason Thanks for all of the help.
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