I have two points on the X axis, A and B, which are connected to the two points, C and D on the sketch plane parallel to XY plane. I have a point E which lies at distance h from D point in Y direction (Fig 1).
I need to define a line segment UV which is perpendicular to AC, U point lies on the extension of the AC line and when rotated about the AC line point V hits point E. The UV segment is initially parallel to the YZ Plane (Fig 2).
I would like to know position of the U point on the AC line, the length of the UV segment and the angle of rotation for V to hit E.
I think a solution could be obtained by solving for intersection of the plane defined by the C, D and E points and the cone with an apex in point A and with a base defined by the UV. But I don’t know how to set up the equation for the cone.
Let $d=\overline{AC}$ and $t=\overline{CU}$. Then $$ U=C+{t\over d}(C-A). $$ Vector $E-U$ must be perpendicular to vector $C-A$, so their dot product vanishes: $(E-U)\cdot(C-A)=0$, that is: $\bigl(E-C-{t\over d}(C-A)\bigr)\cdot(C-A)=0$, or: $$ (E-C)\cdot(C-A)-{t\over d}d^2=0, $$ whence: $$ t={1\over d}(E-C)\cdot(C-A). $$ Once you have computed $t$ the rest easily follows, as $UV=UE$.
From your picture one gets $C-A=(0,26,50)$ and $E-C=(10,h,0)$, so that $d=\sqrt{3176}$ and $t=26h/\sqrt{3176}=13h/\sqrt{794}$.