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I'm having some trouble with a question involving an ellipse tilted in 3D space. The vector equation of the ellipse is

$r(t) = (\cos (t), \sin (t), c \sin(t))$

where c is any real number. The question asks me to prove that the curve corresponding to this vector equation is an ellipse. I have only ever dealt with conic sections in the xy-plane, so I have no idea how to go about doing this.

My only idea is to use a rotation matrix to bring the ellipse back into a plane that is parallel to the xy-plane, and then show that the parametric equations of the resulting curve satisfy $((x-m)/a)^2 + ((y-n)/b)^2 = 1$. But we haven't covered rotations in $\mathbb{R}^3$, so this leads me to believe that there is a much simpler solution...

Any help is appreciated. Thanks!

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Hint: Notice that $r(t) = \cos(t) \mathbf{u} + \sin(t) \mathbf{v}$, where $\mathbf{u} = (1,0,0)$, and $\mathbf{v} = (0,1,c)$, so $r(t)$ always lies in the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. Any point in that plane can be represented as $x \mathbf{u} + y \mathbf{v}$ and you can find an equation for the $x$'s and $y$'s corresponding to points on your curve. Why does that equation represent an elipse? (Hint for that part: think about the angle between $\mathbf{u}$ and $\mathbf{v}$ and about the lengths of $\mathbf{u}$ and $\mathbf{v}$.)

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Omar's answer is a good one. I'll just give some geometric intuition about the hint in his answer because apparently after 9 hours when I'm writing this answer the question is still unsolved. If you already know that two vectors $\mathbf{a}=(a_1,\cdots,a_n)$ and $\mathbf{b}=(b_1,\cdots,b_n)$ in $\mathbb{R}^n$ are orthogonal if their dot product defined by $a_1b_1 + \cdots + a_nb_n$ is zero, then you can see that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal in $\mathbb{R}^3$ and $|\mathbf{u}|=1$ and $|\mathbf{v}| = \sqrt{1+c^2}$. The $x$'s and $y$'s in Omar's answer are $x=\cos(t)$ and $y=\sin(t)$. Therefore, if we look at the curve as we live inside the $\mathbf{u}-\mathbf{v}$ plane, it must be a circle. Inside the $\mathbf{u}-\mathbf{v}$ plane our coordinates $\mathbf{u}$ and $\mathbf{v}$ are orthogonal but $\mathbf{v}$ has been scaled by $\sqrt{1+c^2}$. So, if we scale back $\mathbf{v}$ to $\hat{\mathbf{v}}$ which is a unit vector obtained from $\mathbf{v}$ we see that our new coordinates will be $x' = \cos(t)$ and $y' = \frac{1}{\sqrt{1+c^2}}\sin(t)$ and this is an ellipse in the plane spanned by unit vectors $\mathbf{u}$ and $\hat{\mathbf{v}}$. You can write the equation of this plane where the ellipse lies by finding the normal vector of the plane using cross product as $\mathbf{n} = \mathbf{u} \times \hat{\mathbf{v}}=\frac{1}{\sqrt{1+c^2}}(0,-c,1)$ and using the point $r(0) = (1,0,0)$ which gives the plane $z-cy=0$ after simplification. (The equation of the plane is calculated by using the fact $\mathbf{n}.(\mathbf{r}-\mathbf{r_0})=0$)

Now the geometric picture becomes completely clear. Your ellipse always lies in the plane $z=cy$ depending on what $c$ is. If you define the coordinates of the plane $z=cy$ to be $x'$ and $y'$ then $x' = \cos(t)$ and $y' = \frac{1}{\sqrt{1+c^2}} \sin(t)$

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