Showing that the intersection of a cylinder and a plane is an ellipse

Question

One of the questions in my homework was:

"Show that the curve $\vec{r}(t)=\cos t \vec{i}+\sin t \vec{j}+(1-\cos t)\vec{k}$ is an ellipse by showing that it is the intersection of a cylinder and a plane. Find equations for the cylinder and the plane."

It is easy to see that the curve is the intersection of the cylinder $x^2 + y^2 =1$ and the plane $x+z=1$.

Maybe I'm misunderstanding, but the question makes it sound like the hard part is showing that the curve is the intersection of a cylinder and a plane and that once they are found it is obvious that the curve is an ellipse. I have no idea how I can prove that the curve is an ellipse. Since it is not parallel to the $xy$ plane (or any other "conventional" planes), I am having a hard time showing that it satisfies the equation of an ellipse.

Edit: I should add that I have not taken a linear algebra course yet, so please bear that in mind when posting a solution/hint. (I’m only adding this because many of the multivariable calculus hints online assume I am familiar with linear algebra.)

Answer 1

Sometimes a figure really helps craft the algebra:

Answer 2

You need to rotate coordiate system as $z_1 = (x+z)/\sqrt2$, $x_1=(x-z)/\sqrt2$, $y_1=y$, in this coordinate system the equations are: $$ x+z=z_1\sqrt2=1,\\ x^2+y^2 = 2(x_1+z_1)^2 +y_1^2=1 $$

In the plane $z_1=1/\sqrt2$, the equation of intersection reads: $$ 2\left(x_1+\frac1{\sqrt2}\right)^2+y_1^2=1, $$ which is a general equation for ellipse.