Intersection of two moving objects There are two objects. The first object moves with speed $U_1$ from known point $A$ to known point $B$. The second object has speed $U_2$ and starts from known point $C$. 
What is the direction the second object must have in order to "collide" with the first object. The direction must be defined by the collision point.
http://i.imgur.com/3oY7hJi.png
 A: We can describe the trajectory of the first object as
$$
r_1(t) = A (1 - \lambda) + B \lambda \quad U_1 T_1 = \lvert AB \rvert \quad \lambda = t / T_1 = \frac{U_1}{\lvert AB \rvert} t
$$
Similar we get
$$
r_2(t) = C (1 - \mu) + D \mu \quad U_2 T_2 = \lvert CD \rvert \quad \mu = t / T_2 = \frac{U_2}{\lvert CD \rvert} t
$$
for the second object moving from $C$ to some point $D$, needing time $T_2$.
Interesting it gets if $D$ is the collision point:
$$
D = r_1(T_2) = r_2(T_2) 
$$
for $T_2 \le T_1$.
This gives
$$
D = A (1-\lambda) + B \lambda = A + \lambda (B-A)
\quad \lambda 
= \frac{U_1}{\lvert AB \rvert} T_2
$$
so with the unit vectors (the first known, the second sought):
$$
e_A = \frac{B-A}{\lvert AB \rvert} \quad
e_D = \frac{D-C}{\lvert CD \rvert} = (\cos(\phi), \sin(\phi)) \\
D = C + e_D \lvert CD \rvert = C + e_D U_2 T_2 = A + e_A U_1 T_2
$$

Per component we get:
\begin{align}
C_x - A_x &= (e_{A,x} U_1 - \cos(\phi) U_2) T_2 \quad (*) \\
C_y - A_y &= (e_{A,y} U_1 - \sin(\phi) U_2) T_2
\end{align}
where we equate for $T_2$:
$$
\frac{C_x - A_x}{e_{A,x} U_1 - \cos(\phi) U_2} =
\frac{C_y - A_y}{e_{A,y} U_1 - \sin(\phi) U_2} \quad (**)
$$
Assuming $A \ne C$ this is one equation in one unkown $\phi$, so it could be
solvable, if the given $A$, $C$, $U_1$, $U_2$ permit it.
Alternatively one could use
$$
\cos(\phi) = e_{D,x} \quad \sin(\phi) = e_{D,y} \quad e_{D,x}^2 + e_{D,y}^2 = 1 
$$
in equation $(**)$ and try to solve for $e_{D,x}$ or $e_{D,y}$.
Changing the coordinate system can simplify the above. E.g.
one might translate and rotate such that $A_x = C_x$. 
This would simplify $(*)$ to 
$$
\cos(\phi) = \frac{U_1}{U_2} e_{A,x}
$$
A: This diagram shows a geometric solution to finding the desired direction of $\vec U_2$. The idea here is that we have changed the entire frame of reference by adding vector $-\vec U_1$ to all velocities. In this frame, point $A$ is stationary and point $C$ has been given a velocity of $-\vec U_1$. We want to add an additional velocity vector $\vec U_2$ to point $C$ to aim it at stationary point $A$.

As the diagram shows, there may be two different directions of $\vec U_2$ that would suffice. This will happen if the angle $\theta$ between vectors $-\vec U_1$ and $\overrightarrow{CA}$ is acute and $|\vec U_1|\sin\theta<|\vec U_2|<|\vec U_1|$. It should be clear what conditions must hold for there to be no solutions. (This discussion is very similar to explaining the cases in solving the triangle in SSA for trigonometry class.)
A: Problems like this are easily visualized in a coordinate frame in which
one of the objects is at fixed coordinates.
In this case, we can choose a transformed coordinate frame that moves with $A$,
that is, the coordinates of $A$ in this coordinate frame never change.
If $C$ had zero velocity in the original coordinate frame
(the one in which $A$ has velocity $U_1$ in the direction of $B$),
then in the transformed coordinate frame $C$ would have velocity $-U_1$,
that is, its velocity would have the same magnitude as $U_1$ but
would be in the exact opposite direction.
Any velocity $U_2$ that we give to $C$ in the original frame,
in any direction, is added to the velocity $-U_1$ 
in the transformed coordinate frame.
The resultant velocity in the transformed coordinate frame,
$U_2 - U_1$, can be a vector from $C$ to any point on the circle
shown in the figure below.
(Note that $U_2 - U_1$ is the vector difference of two velocity vectors,
not merely the scalar difference of two speeds.)

Choose the direction of the vector $U_2$ so that $U_2 - U_1$ is in
the direction from $C$ to $A$.
You can find such a direction (if there is one) by finding
the intersection of the circle in the figure with the ray
from $C$ through $A$.
If $|U_2| > |U_1|$ then $C$ will lie inside the circle, as shown in
the figure above, and there will be exactly one solution.
If $|U_2| < |U_1|$ then $C$ will lie outside the circle and there will be
either two solutions or none at all.
If you require the object starting at $C$ to intercept the object starting
at $A$ before that object finishes traveling from $A$ to $B$, then you
may compute how long it takes the object to reach $B$ in the original
coordinate system
(using the distance from $A$ to $B$ and the magnitude of $U_1$)
and how long it takes the object at $C$ to reach $A$ in the transformed
coordinate system (using the distance from $C$ to $A$ and the magnitude of
the vector $U_2 - U_1$)
to determine which event happens first.
