Calculating the vector to intercept a falling object Given two objects with known positions in a three dimensional space, of which:


*

*One is a ballistic projectile (that is, it is free-falling with constant gravitational acceleration), and 

*The other (the "catcher") is not subject to gravity and may travel in any direction with some constant speed (and instantaneous acceleration),


I want to calculate the vector that the catcher must travel along in order to intercept the falling projectile.
I've got as far as formulating the problem into equations as follows:
$$
\mathbf{P_i} = \mathbf{P_p}  + \mathbf{U_p} t + \frac{1}{2} \mathbf{G} t^2
$$
$$
\mathbf{P_i} = \mathbf{P_c} + \mathbf{V_c} t
$$
where:


*

*$\mathbf{P_i}$ is a 3-dimensional vector describing the position of intercept

*$t$ is a scalar representing the time of intercept relative to the start time.

*$\mathbf{P_c}$ and $\mathbf{P_p}$ are (known) 3-dimensional vectors describing the initial positions of the catcher and the projectile, respectively.

*$\mathbf{U_p}$ is a 3-dimensional vector describing the initial velocity of the projectile.

*$\mathbf{G}$ is a 3-dimensional vector describing the acceleration of the projectile due to gravity.

*$\mathbf{V_c}$ is a 3-dimensional vector describing the velocity of the catcher required to reach $\mathbf{P_i}$ at time $t$. Its magnitude is known, but not its direction.


Obviously, finding any one of $t$, $\mathbf{P_i}$ or $\mathbf{V_c}$ is sufficient to solve the problem.
Combining the two equations above to eliminate $\mathbf{P_i}$ gives me a quadratic equation in $t$, but with vector co-efficients, which I haven't a clue how to even begin solving. Can anyone help me out? Is there something missing from or incorrect about my approach here?
 A: You do not need to solve for $t$. You want to solve for $V_c$. This gives the required velocity vector for the catcher to catch it at an unspecified time $t$ later.
If you are instead given $V_c$ then you could solve for $t$. However the vast majority of values of $V_c$ would result in no solution.
A: To tackle the difficulty of dealing with a vector system, we can exploit the fact that we know all of the vectors except ${\bf V}_c$, and that we know the magnitude $V_c := \left\vert\left\vert{\bf V}_c\right\vert\right\vert$ to produce a scalar equation in $t$.
As indicated in the question, at the time $t$ we can equate the positions of the projectile and the catcher:
$${\bf P}_p + {\bf U}_p t + \frac{1}{2} {\bf G} t^2 = {\bf P}_c + {\bf V}_c t.$$
Rearranging gives $${\bf V}_c t = {\bf Q}(t)$$
where ${\bf Q}$ is some quadratic polynomial in $t$ whose coefficients are known (in fact, ${\bf Q}$ is just the position of the projective relative to the initial position of the catcher). Thus, if we take the dot product of each side with itself we get
$$V_c^2 t^2 = ({\bf V}_c t) \cdot ({\bf V}_c t) = {\bf Q}(t) \cdot {\bf Q}(t).$$
But now, as desired, we have produced a scalar equation in $t$, namely (after rearranging), $${\bf Q}(t) \cdot {\bf Q}(t) - V_c^2 t^2 = 0,$$ whose coefficients are all known. This is quartic in $t$, so we may have up to four solutions (and we can always find exact expressions for them, though in general these are not very nice). Since the leading coefficient of the l.h.s. is $\frac{1}{4} {\bf G} \cdot {\bf G} > 0$, we can test whether there are any solutions by finding its critical points (a quartic always has at least one) and checking whether it is $\leq 0$ for at least one of them.
