Find A and B for which ACB=90 degrees and AB minimum You are given an angle xOy=90 degrees and a point C which is not on the plane of xOy. Find a point A on Ox and a point B on Oy for which the angle ACB=90 degrees and AB will be minimum.
I ve tried some stuff, i noticed that C and O must be points of a sphere and therefore, we just need to create the smallest sphere which would imply the minimum AB. Cant get anywhere though
 A: Let $P$ be the plane of $xOy$. As you have observed, for any $A$ and $B$, the set of points $\{X \in \mathbb{R}^3 : |\angle AXB| = 90^\circ\}$ is a sphere $S_{AB}$ that has $AB$ as a diameter. Obviously it contains both $O$ and $C$, however, this implies that the center of the sphere $M$ (and the middle point of $AB$) belongs to a bisector plane $Q$ of $OC$. Now, as $AB \subset P$, we have that $M \in P \cap Q$.
Now, to get the smallest sphere that has a center at $P \cap Q$ and contains $O$, we need to minimize its radius, namely $|MO|$, however this happens for
$M$ being an orthogonal projection of $O$ at $P \cap Q$. Now, if we were to
draw an intersection $S_{AB} \cap P$ (blue) then it would be a circle that would
determine $\{A\} = S_{AB} \cap P \cap Ox$ and $\{B\} = S_{AB} \cap P \cap Oy$. 
Finally, there is the special case of $P \parallel Q$, where there is no solution.

I hope this helps $\ddot\smile$
Edit: It was wrong, fixed now.
A: You could also use an approach by coordinate geometry which shall be convenient because we are mostly dealing with right angled triangles.
Let the point $O$ be the Origin and the lines $OX$ and $OY$ be the $x$ and $y$ axis respectively.
Point $A = (x,0)$, Point $B = (0,y)$ and Point $C=(h,k,l)$
Given that $|∠ACB| = 90$∘,
$$AC^2+BC^2=AB^2$$
$$[(h-x)^2+k^2+l^2] + [h^2+(k-y)^2+l^2] = x^2+y^2$$
$$h^2+k^2+l^2=h.x+k.y$$
$$h.x+k.y=d^2$$where $d=OC$
Now, we need to minimise $|AB|$ i.e. minimise $(x^2+y^2)$
Substituting $y=\frac{d^2-h.x}{k}$,
we need to minimse the quadratic function $$f(x)=x^2 + (\frac{d^2-h.x}{k})^2$$
which after a simple calculation yields $$x=h.(\frac{d^2}{h^2+k^2})$$
$$x=h.\sec^2\theta$$
$$y=k.\sec^2\theta$$
where $\theta$ is the angle that $OC$ makes with the $xy$ plane.
Also, the minimum distance $$|AB| = \sqrt{x^2+y^2}$$
$$=|d\sec\theta|$$
when $$x_∘=h.\sec^2\theta$$
$$y_∘=k.\sec^2\theta$$
