Acceleration of a ball on a plane I have a plane ($ax+by+cz+d=0$) in a 3D world, and a gravity vector $\vec{g}$ (say it's $[0, 0, -9.81]$.) How would I find the acceleration vector of an object on this plane, ignoring friction?

What you want to do is project $\vec{g}$ on the plane. 
First we will consider a new plane: $ax+by+cz=0$. This plane is parallel to our previous one, so our result won't change, but it goes through the origin. 
Then we need a normal vector of this plane. In this case, we're lucky because we have this particular form of the equation of the plane. A normal vector is in this case $\vec{n}=[a,b,c]$.
Now we will create a line going through $\vec{g}$, and parallel to $\vec{n}$. Then we get the following parameter-equation (I don't know the proper term):
$$\vec{x}=\lambda\vec{n}+\vec{g}$$
Where $\lambda$ is the parameter. The intersection of this line and our plane will be $\vec{g}$ projected on our plane.
Let's call this intersection-point $\vec{p}$. For $\vec{p}$ it must be true that:
$$\vec{p}\text{ is on our line: }ap_x+bp_y+cp_z=\vec{p}\cdot\vec{n}=0$$
$$\vec{p}\text{ is on our plane: }\lambda\vec{n}+\vec{g}=\vec{p}$$
Plugging the second equation into the first we get:
$$(\lambda\vec{n}+\vec{g})\cdot\vec{n}=\lambda\vec{n}\cdot\vec{n}+\vec{g}\cdot\vec{n}=0$$
$$\Rightarrow\lambda=-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}$$
Plugging this into the first equation gives us:
$$\vec{p}=\vec{g}-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}\vec{n}$$
In your example ($\vec{g}=[0, 0, -9.81]$) it looks like:
$$\vec{p}=\begin{bmatrix}
\frac{9.81c}{a^2+b^2+c^2}a\\
\frac{9.81c}{a^2+b^2+c^2}b\\
\frac{9.81c}{a^2+b^2+c^2}c-9.81\\
\end{bmatrix}$$
