Finding the intersection between a triangle and a parabola in 3-dimensional space The problem can be stated using the following vector equation
$$o_1+tw+t^2g=o_2+au+bv$$
where $o_1$ is a point on the curve (the starting point of the projectile), $w$ is a vector facing in the direction of the curve (the direction of launch), $g$ is a vector pointing down (gravity), $o_2$ is one of the triangles vertices and $u$ and $v$ are vectors from that vertex to the other two.
Solving the equation should give us $t$, $a$ and $b$.
Doing this on paper quickly got out of hand so I've plugged it into Mathematica to get the solution and it's huge so I'll only post the equation for one of the possible solutions for $t$.
$$\frac{(-u[3])v[2]w[1] + u[2]v[3]w[1] + u[3]v[1]w[2] -
      u[1]v[3]w[2] - u[2]v[1]w[3] + u[1]v[2]w[3] -
      \sqrt{-4((-g)u[3]v[1] + gu[1]v[3])(o[3]u[2]v[1] - 
          p[3]u[2]v[1] - o[2]u[3]v[1] + p[2]u[3]v[1] - o[3]u[1]v[2] +
          p[3]u[1]v[2] + o[1]u[3]v[2] - p[1]u[3]v[2] + o[2]u[1]v[3] -
          p[2]u[1]v[3] - o[1]u[2]v[3] + p[1]u[2]v[3]) +
        (u[3]v[2]w[1] - u[2]v[3]w[1] - u[3]v[1]w[2] + u[1]v[3]w[2] +
          u[2]v[1]w[3] - u[1]v[2]w[3])^2}}{2((-g)u[3]v[1] + gu[1]v[3])}$$
the problem here is that I have the added condition that the denominator must not be zero, but in case it is zero, I have no idea what to do. I don't know what the geometric interpretation is of that number being zero so I don't see how assuming it's zero simplifies my problem to something I can solve.
Now for the actual question, is there a different approach to solving this that produces a cleaner solution? Perhaps I can somehow get away with manipulating vectors without dropping down to their coordinates? Maybe there's an interesting way I can transform my data into something workable?
 A: Try choosing a convenient set of basis vectors. I'll rename things slightly, so that my basis consists of vectors $u$, $v$, $w:=u\times v$, my direction vector is $d$, gravity vector $g$, initial position vector $p$, and triangle vertex $q$.
Now, given this vector equation ...
$$p + t d + t^2 g = q + a u + b v \tag{1}$$
... we take the dot product of each side with $w$ to get ...
$$t^2 w\cdot g+ t w\cdot d + w\cdot (p-q) = 0 \tag{2}$$
... which we easily solve for $t$:
$$t = \frac{-w\cdot d \pm \sqrt{ (w\cdot d)^2 - 4(w\cdot (p-q))(w\cdot g)}}{2w\cdot g} \tag{3}$$
To simplify, let's take $p = 0$, and let's define ...
$$D = \angle wd \qquad Q := \angle wq \qquad G := \angle wg$$
... so that we can write ...
$$\begin{align}
t &= \frac{-|w||d|\cos D \pm \sqrt{ |w|^2|d|^2\cos^2D + 4|w|^2|q||g|\cos Q\cos G}}{2|w||g|\cos G} \\
&= \frac{-|d|\cos D \pm \sqrt{ |d|^2\cos^2D + 4|q||g|\cos Q\cos G}}{2|g|\cos G}
\end{align}\tag{4}$$
The geometric interpretation of a zero denominator here is clear: either the gravity vector has zero magnitude, or $G$ is a right angle (in which case, the pull of gravity is parallel to the plane of the triangle). In either case, the quadratic in $(2)$ is only linear in $t$ (provided $w\cdot d \neq 0$), which is easier to solve, anyway. 
