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Full disclosure, this is for a Calculus III graded homework set--though we are allowed to use any resources available to complete it.

I feel I have a good understanding of space curves, though my understanding of the equation of planes leaves a lot to be desired, as well as transformations on space curves.

The full wording of the question is this: Find the vector equation of the 3-dimensional space curve r(t) that represents an ellipse whose center is at (7,3,-5) that lines in the plane 2x-y+z=6. Also verify that the ellipse lies in the given plane(I have no problem with the last part)

I know that the equation of the ellipse will have to be of the form: $f(a\cos(t),b\sin(t),z)$. Technically, the a and b are unnecessary because a circle is just a special case of an ellipse, but I'd prefer to find it in general and then just plug in some constants after the fact--doing a circle is cheap, in my opinion.

I know that the plane is defined by its normal vector, <2,-1,1>, and that D=6 defines its shift from the origin. My issue I'm not sure to incorporate this shift into the equation of the ellipse, and I'm not quite sure how to incorporate the normal vector. I'm planning on finding the vector equation for an ellipse with the given center and then "forcing" it to always be orthogonal to the normal vector--is this the correct approach? I can do most of this on my own, but I'm not sure of where to start.

Thank you in advance for your help.

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There is a typo somewhere. The indicated center does not lie in the given plane, even when $-7$ is replaced by $-7y$. – Christian Blatter Mar 13 '14 at 20:43

Some hints:

The ellipse is not determined by the given data. You only know its center ${\bf c}=(7,3,-5)$ and the normal vector ${\bf n}=(2,-7,1)$ (?) of the plane it lies in. The most general ellipse satisfying these conditions is obtained as follows: Choose any two linearly independent vectors ${\bf a}$ and ${\bf b}$ which are orthogonal to ${\bf n}$ and consider the parametric representation $$t\mapsto {\bf c}+\cos t\ {\bf a}+\sin t\ {\bf b}\qquad(0\leq t\leq 2\pi)\ .$$

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First, I would find a basis for the plane. Let $\hat u, \hat v$ be orthonormal vectors parallel to this plane. The normal vector, $\hat n$, should then obey $\hat u \times \hat v = \pm \hat n$.

Now, forget about the normal vector or the translation for a moment. You have a $uv$-plane, and you can write the equation of an ellipse in this plane. It would be in the form of $a \hat u \cos t + b \hat v \sin t$ because, instead of using simple vectors like $\hat x$ or $\hat y$, we have to use the more complicated $\hat u, \hat v$ vectors to stay in this plane.

This gets us a plane that goes through the origin. Now we need to translate it. Let $r_c$ be the center of the ellipse then. You should be able to write that

$$r(t) =r_c + a \hat u \cos t + b \hat v \sin t$$

The challenge to you, then, will be finding the $\hat u, \hat v$ unit vectors. There's considerable freedom here; any pair of unit vectors that span the plane and are orthogonal will do.

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I'm still wrapping my head around this, but I think I understand it. I'm extremely glad I took linear algebra prior to taking Calc III. The unit vectors shouldn't be an issue, I'll just find two arbitrary points that lie on the plane, normalize the resultant vector(for u), and then do a cross product with the plane's normal vector(for v). Thank you very much for your help and detailed, helpful explanation. Edit:I just realized I'll also need to either normalize the normal vector defining my plane or the resultant from my cross product--no big deal, just correcting myself. Thanks again. – user2896564 Mar 13 '14 at 20:48
@user2896564: Right, you picked up on something my wording was unclear on. The vectors $\hat u, \hat v$ should be parallel to the plane (I made it sound like they should merely be position vectors that lie on the offset plane, which gives a very different, and wrong, answer). Your approach should give the right result. – Muphrid Mar 13 '14 at 20:52

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