Prove that with 2 parallel planes, the one in between is given by I would like to prove that having two planes
$$ax+by+cz+d_1 = 0 \quad\text{and}\quad
ax+by+cz+d_2 = 0$$
you can automatically have a plane with equal distance from each plane that looks like this:
$$ax+by+cz+\frac{d_1+d_2}{2}=0.$$
I have tried deriving this from the formula for the distance of a point to a plane, but with no success. Any suggestions?
 A: Think about the geometric meaning of the scalar product $(a,b,c)\cdot(x,y,z) = ax + by  +cz$.
The vector $(a,b,c)$ is orthogonal to the two planes and has length $\sqrt{a^2 + b^2 + c^2} \neq 0$ (if this were zero, the equations wouldn't give planes since then $a=b=c=0$).
Therefore the (signed) distance of the two planes to the origin is given by $-\frac{d_1}{\sqrt{a^2 + b^2 + c^2}}$ and $-\frac{d_2}{\sqrt{a^2 + b^2 + c^2}}$, respectively. So the signed distance of the plane lying in the middle must be $-\frac{(d_1+d_2)/2}{\sqrt{a^2 + b^2 + c^2}}$, in other words, the plane in the middle must be given by the equation
$$ax + by + cz + \frac{d_1 + d_2}{2} = 0.$$
A: The "plane in between" $\pi_m$ is the locus of all midpoints of segments having one endpoint on $\pi_1$ and the other on $\pi_2$. Now the linear function $f(x,y,z):=a x+ by + c z$ assumes the constant value $-d_1$ on $\pi_1$ and the constant value $-d_2$ on $\pi_2$, so it has to assume the value $-(d_1+d_2)/2$ on $\pi_m$.
A: These equations can be rewritten as $z_1 = \frac{-a(x + \frac{d_1}{a}) - by}{c}$ and $z_2 = \frac{-a(x + \frac{d_2}{a}) - by}{c}$, while the third one becomes $z_3 = \frac{-a(x + \frac{d_1 + d_2}{2a}) - by}{c}$. Recall that for a constant $c$, $f(x,y) \rightarrow f(x + c,y)$ translates the graph of the function left by $c$ units. Combining these facts should give you your answer.
