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A silly question about unit tangent and unit normal

  1. Sep 29, 2011 #1
    I know how to obtain the unit tangent, it is very easy. But for the case of unit normal, I am confused.

    Usually we need to know [itex]\vec{T'}(t)[/itex] and |[itex]\vec{T'}(t)[/itex]|
    in order the find [itex]\vec{N}(t)[/itex].

    However are there any quicker method to do it? Since I saw the textbook do it without step, it seems there is a quicker method.

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  3. Sep 29, 2011 #2


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    What's the dot product of T and N?
  4. Sep 29, 2011 #3

    But it is not enough.
    Since [itex]\vec{N}[/itex]=a[itex]\hat{i}[/itex]+b[itex]\hat{j}[/itex]
    There are 2 unknows, a and b.

    Dot product yields only one equation, this is not enough to get [itex]\vec{N}[/itex]
  5. Sep 29, 2011 #4


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    You said N was a unit normal. The length of N gives you another equation for a and b.

    The "easy" way to get the answer is to draw a picture of the tangent vector, and rotate it through 90 degrees to get the normal vector. It should then be clear than if the tangent vector is (a, b), the normal vector must be (-b, a).
  6. Sep 29, 2011 #5
    Oh! Thanks a lot!
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