hungryhippo
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Given a parametric equation of a curve, how would you show that it is contained in a plane?
In 3-D space, a plane is defined by a point, and a vector that is orthogonal to the plane. For this question type, you're only given the parametric equation.
How I would appraoch this problem, would use t=0 to find a point on the curve that exists. Then, would I just find another point, in which it doesn't exist on the curve, since two nonparallel vectors are needed. Then, I would find the normal (n) by taking the cross product of those points...Would this be the way to go? If so, would I then just conclude that since the curve contains the point with parameter t=0, it is contained in a plane?

In desparate need of help on this
Thanks in advance
In 3-D space, a plane is defined by a point, and a vector that is orthogonal to the plane. For this question type, you're only given the parametric equation.
How I would appraoch this problem, would use t=0 to find a point on the curve that exists. Then, would I just find another point, in which it doesn't exist on the curve, since two nonparallel vectors are needed. Then, I would find the normal (n) by taking the cross product of those points...Would this be the way to go? If so, would I then just conclude that since the curve contains the point with parameter t=0, it is contained in a plane?

In desparate need of help on this
Thanks in advance