Given two skew lines, is there always a point of view from which they look perpendicular? Is this hypothesis about skew lines true?

For any two skew lines, there exists a plane such that the projections of these lines on that plane are two lines that are orthogonal.

(Meaning, there must be a point of view where if you look at the lines you will see them perpendicular.)

My guess: If it's true, it's probably an alternative form of an already known theorem. But I just came up with it on my own and don't really know its name.
 A: The question can be reduced to this one:
Being given two non collinear vectors $u$ and $v$ (assumed unit norm WLOG), is it possible to find a plane such that $u':=P(u) \perp v':=P(v)$ where $P$ is the orthogonal projection onto this plane ?
We can answer affirmatively, and even say that there are an infinite number of such planes.
Here is how such planes can be obtained (see figure below).
Consider the following ("Gram-Schmidt" type) operation;
$$v \ \text{changed into} \ v':=v-\lambda u$$
where $\lambda$ is such that $v' \perp u$. Otherwise said such that
$$(v-\lambda u)\cdot u=0 \ \iff \ \lambda=u \cdot v$$
Then plainly take any vector plane "passing through" (containing) vector $v'$ (and not orthogonal to $u$) : the orthogonal projection of the rectangle defined by $(u,v')$ is a rectangle (not a general parallelogram) defined by $(u',v')$: this means that from the point of view of an observer situated at infinity, say on the left (in the case of the picture), vectors $(u, v)$ are seen as orthogonal vectors $(u',v')$.
Remark: inverting the roles of $u$ and $v$ would give another family of planes.

A: Yes there always does exist such a plane, and in fact if you don't care about using only orthogonal projections, a plane parallel to both lines is sufficient with the correct angle and direction between the projection and the plane. The angle of the projection to the plane should be the same as the acute angle between the orthogonal projections of the lines onto a plane parallel to each. The angle should also be coming from the direction the same angle from each of the orthogonal projections I mentioned earlier.
Alternatively if you want to restrict your discussion only to orthogonal projections, consider a plane parallel to both lines, the orthogonal projections of those lines onto the plane, the non-obtuse angle between those projections (we'll call this angle theta), and a plane intersecting this first plane at an angle of 90-theta degrees, such that this new plane intersects both lines with the same angle. The orthogonal projections of these skew lines onto this new plane will also be orthogonal.
