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Homework Help: Principal normal line to helix is normal to z axis

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data
    DoCarmo Section 1.5 problem 1 part d. Show that the lines containing n(s) and passing through a(s) [a is the curve, and n(s) is the unit normal vector] meet the z axis under a constant angle of pi/2.
    Helix: a(s) = (a cos(s/c_, a sin(s/c), b*s/c), so I computed n(s) = (cos(s/c),sin(s/c),0).

    2. Relevant equations



    3. The attempt at a solution
    I think the principal normal containing n(s) and a(s) is : L(t) = (t-s)*n(s) + a(s). THis is because L'(t) = n(s), and L(s) = a(s), but when I try to show d/dt ( L(t) . z/Norm(L(t))) = Cos(theta) =0, I keep getting that it is not zero, due to the z-component of a(s), and the norm of L(t), which I compute to be ((Norm a)^2+(t-s)^2)^0.5, by orthogonality of n(s) and a(s).

    Is there something I am donig wrong?

    Thanks,
     
  2. jcsd
  3. Jan 25, 2012 #2

    LCKurtz

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    Since the z axis is in the direction of ##k=\langle 0,0,1\rangle##, isn't it obvious by the dot product with n(s) that they are perpendicular? That only leaves the question of whether the line L(t) actually hits the z axis. Can you find a t value for which ##L(t) = \langle 0,0,k\rangle## for some k, no matter what s is?
     
  4. Jan 26, 2012 #3
    It's obvious to me that n(s) is perpendicular to z-axis, but I guess I don't really understand the following: The line L(t) must be in the direction of the vector n(s), but when we take the dot of L(t) with e_3 (the third basis vector in the canonical basis for R^3), we don't get 0 since the line is translated by the vector a(s0).

    So: how is the line L(t) actually in the direction of n(s) if L(t) . e_3 =/= 0, but n(s). e_3 =0? I guess I dont understand the term "direction" well enough, since the angle between L(t) and z axis is precisely ArcCos(L(t).e_3/Norm(L(t)))?

    Thanks
     
  5. Jan 26, 2012 #4

    HallsofIvy

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    That makes no sense. a vector "translated" is still the same vector.
     
  6. Jan 26, 2012 #5

    LCKurtz

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    L(t) is the position vector from the origin to the point on the line and it isn't perpendicular to the z axis. L(t) is the sum of two vectors: from the orgin to a(s) then from a(s) in a direction perpendicular to the z axis, which is given by n(s). The whole point of the problem is that if you go from a point on the curve in the direction of n(s) it will hit the z axis. It might help you to draw a picture of a helix going around the z axis. Pick a point on the helix an draw a position vector to that point then a line from that point perpendicularly towards the z axis. L(t) is a vector from the origin to a point on that line which varies along that line as t changes.
     
  7. Jan 26, 2012 #6
    This clarifies it! Thank You !
     
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