Equations for normal, osculating, and rectifying planes

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To find the equations for the normal, osculating, and rectifying planes for the curve defined by r(t) = (cos t)i + (sin t)j - k at t = pi/4, the tangent vector T(pi/4) has been calculated as (-√2/2)i + (√2/2)j + 0k, while the normal vector N(pi/4) is (-√2/2)i + (-√2/2)j + 0k, and the binormal vector B is 0i + 0j + k. The discussion emphasizes that with these vectors, deriving the equations for the three planes should be straightforward, as the normal vector provides the necessary orientation and the point r(pi/4) gives a location on the planes. A suggested resource for further understanding is the Wikipedia page on Frenet-Serret formulas. The key takeaway is that the calculations for the planes can be easily derived using the vectors and the specified point.
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We are being asked to find the eq's for normal, osculating, and rectifying planes for the following equation:
r(t) (cos t)i + (sin t)j - k @ t=pi/4
I have already found the following:
T(pi/4) = (-√2/2)i + (√2/2)j = 0k

N(pi/4) = (-√2/2)i + (-√2/2)j + 0k

B = 0i + 0j + k

But, I don't know where to start for those three equations.
 
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cue928 said:
We are being asked to find the eq's for normal, osculating, and rectifying planes for the following equation:
r(t) (cos t)i + (sin t)j - k @ t=pi/4
Should be r(t) = ...
cue928 said:
I have already found the following:
T(pi/4) = (-√2/2)i + (√2/2)j = 0k
Should be T(pi/4) = (-√2/2)i + (√2/2)j + 0k

Note that I didn't check your math, just obvious typos.

cue928 said:
N(pi/4) = (-√2/2)i + (-√2/2)j + 0k

B = 0i + 0j + k

But, I don't know where to start for those three equations.
Try this link to a wikipedia page that discusses these planes- http://en.wikipedia.org/wiki/Frenet–Serret_formulas. You have all three vectors, so getting an equation of any of the three planes should be easy, since you have a normal to the plane and can easily get the point that's on all three planes (the point at r(pi/4)).
 
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