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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:


I have tried deriving this from the formula for the distance of a point to a plane, but with no success. Any suggestions?

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Your three planes are parallel, so the only question is one of relative distances. Hint: Consider the point on the first plane with x = y = 0. What's its distance to the second plane? Now consider the point with x = y = 0 on your proposed third plane. What's its distance from the second plane? (Note: Checking any points on these planes would do, but taking x = y = 0 is convenient.) – Blue Apr 13 '11 at 6:56
I know it works with examples, but the point of this is to derive it for any a, b, and c without specific points. – Bartlomiej Lewandowski Apr 13 '11 at 7:01
The argument is a valid one for your purposes. (The distance between any two parallel planes is the distance between any point on one of them to the other.) If you didn't already know the $(d_1+d_2)/2$ part of the solution, you could take your third plane to be $a x + b y + c z + d_3=0$, find the distance from the $x=y=0$ point to the second plane, and force that to be equal to half the distance between the first two planes. Don't like using $x=y=0$? Call them $x_0$ and $y_0$ and solve for $z$; the strategy of the argument still works, though the equations are messier. – Blue Apr 13 '11 at 7:10
(and @theo): For completeness, one should note that $c$ could be zero, in which case the $x=y=0$ point doesn't help; however, given that we're talking about actual planes, we are assured that $a$, $b$, and $c$ cannot all be zero, so we can always make the argument work. (That said, @Theo's approach is better.) – Blue Apr 13 '11 at 7:34
up vote 3 down vote accepted

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 $$ax + by + cz + \frac{d_1 + d_2}{2} = 0.$$

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Checking distances from the origin is even more convenient than (as in my hint-comment) checking from the points with $x=y=0$. :) (Depending on @coolbartek's level, use of vector terminology may be somewhat sophisticated. Nevertheless, the argument works just as well for someone who simply has access to the point-to-plane distance formula (which, of course, follows from the vector stuff).) – Blue Apr 13 '11 at 7:28
@Day Late Don: This may be true, but everything else I could think of just blurs the apparent simplicity. – t.b. Apr 13 '11 at 7:34
I was proposed to use the distance to plane formula ax1 + by2 + cz3 + d / sqrt(a^2 + b^2 + c^2) – Bartlomiej Lewandowski Apr 13 '11 at 20:22

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 $\pi_m$.

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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.

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It isn't clear enough for me, sorry. Any way to derive that (d1+d2)/2=d3? a, b and c are same for all 3 planes. – Bartlomiej Lewandowski Apr 13 '11 at 6:58
What is $d_3$? Anyway, in my answer the first sentence is simply rearranging the 3 equations into more workable forms. Another way to get that the third plane lies in the middle of the other two would be to show $z_3 = \frac{z_1 + z_2}{2}$, which is a simple matter of calculation. – Alex Becker Apr 13 '11 at 7:23
You are assuming that both $a$ and $c$ are non-zero. – t.b. Apr 13 '11 at 7:25
@Theo: Yes, but if this is the case then the problem reduces to an even simpler version in lower dimension. – Alex Becker Apr 13 '11 at 7:43

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