|May2-07, 02:50 AM||#1|
prove that a curve lies in a plane
1. The problem statement, all variables and given/known data
let r(t) (a<t<b) be a smooth vector valued function. suppose that the nonzero vector n is perpendicular to r'(t) for all values of t. Prove that the curve with parametrization r(t) lies in a plane
2. Relevant equations
3. The attempt at a solution
i know this has somethign to do with cross product. infact i know that the cross product of n x r'(t) will give me a third vector which will define the plane which r(t) lies in, but I dont know how to show a proof of this.
Thanks for your help or advice
|May2-07, 05:46 AM||#2|
r(t) is the curve, r'(t) is how the curve moves, so, if r'(t) is penpendicular to n where n is some normal vector to a plane, then that means no component of the r'(t) vector is pointing "away" from the plane.
so, if it doesn't move off the plane means that tangent vectors at any three distinct time t's to the curve r(t) are coplanar.
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