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Math Help - Easy vector question

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Easy vector question

    I was rifling through some USSR problem books (with no solutions!) and I found this one, its really easy, but I was dissatisfied with my proof. I think it can be done with less work.

    Prove that the principal normal unit vector, denoted \bold{N(t)} does not exist for a linear function \bold{r(t)}

    My proof is, dont look if you dont want to

















    \bold{r(t)}=\langle{at+b,ct+d,et+f\rangle}

    so \bold{r'(t)}=\langle{a,c,e}\rangle

    and

    ||\bold{r'(t)}||=\sqrt{a^2+c^2+e^2}

    So the tangent unit vector is

    \bold{T(t)}=\frac{\langle{a,c,e}\rangle}{\sqrt{a^2  +c^2+e^2}}

    So now

    \bold{T'(t)}=\langle{0,0,0\rangle}

    and ||\bold{T'(t)}||=0

    \therefore\bold{N(t)}=\frac{\bold{T'(t)}}{||\bold{  T'(t)}||} does not exist because there will be a division by zero \blacksquare
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  2. #2
    Super Member Aryth's Avatar
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    It looks right and it makes sense. The Tangent Vector Function's components are all constants, and therefore it has to have a magnitude of zero, and in order for the PUV to exist, ||\bold{T'(t)}|| can't be zero. That's the way I'd have proven it...
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