Finding a vector $\vec{a}$ that is perpendicular to a certain plane

Question

Consider a cylinder with elliptical perpendicular cross-sections to $z-$axis; $2x^2+y^2=1$. Now I 'm asked to find a vector $\vec{a}$ being perpendicular to which the cylinder has circular cross-sections.

First, I thought of finding a plane with $\vec{a}$ as its normal vector instead. So, the equation of it would be $ax+by+cz=d$ for some constants $a,b,c,d$. Then, I was to consider such that plane to be $ax+by+cz=0$. Because, it wouldn't change the generality of the question. So, I should find just proper constants $a,b$ and $c$.

As I imagine the cylinder, I thought that this last plane would include the $x-$ axis. (I'm not sure about it).

All these steps has remained unused to solve the problem. Any hint will be appreciated!

Answer 1

The ellipse has semi-axes $\frac{1}{\sqrt{2}}$ along the $x$ axis and $1$ along the $y$ axis.

The ellipse lies in the $x-y$ plane, so you need to rotate the plane of the ellipse until the semi-axes are equal.

Therefore rotate the plane $45^o$ about the $y$ axis so that the plane intersects with the cylinder in a circle of radius $1$.

Therefore the required normal vector is in the direction $$\left(\begin{matrix}-1\\0\\1\end{matrix}\right)$$

Answer 2

Hint: An ellipse can be transformed into a circle via scaling along its axes. What scale factors are needed to transform the given ellipse into a circle? Can you then interpret those scale factors as the angle that the plane you’re being asked to find needs to make with the $xy$-plane?

Equivalently, think about what happens to the cross-section when you tilt the cylinder. How much and in what direction would you need to tilt it to “flatten” the ellipse into a circle.