I have 3 parallel planes and the normal vector 'W'(A,B,C) and this is normal to all the 3 planes.
Top plane equation is Ax+By+Cz+D=k
Middle plane equation is Ax+By+Cz+D=0
Below plane equation is Ax+By+Cz+D=-k
How to prove that that Top and Below planes are equal distance from middle plane?
On
HINT
A simple way is to proceed as follow
$$(x,y,z)=t(A,B,C)$$
A shorter alternative is to consider the dot product between any vector $OP=(x,y,z)$ on the plane with the normal unitary vector to the plane $\hat n=(a,b,c)$ and observe that
$$|(x,y,z)\cdot (a,b,c)|=|ax+by+cz|$$
represents the distance between the plane and the origin.
This kind of approach leads to the formula for the distance between $P_0=(x_0,y_0,z_0)$ and a plane $Ax+By+Cz+D=0$
$$d_{\pi P_0}=\frac{|Ax_0+By_0+C_0z+D|}{\sqrt{A^2+B^2+C^2}}$$
On
Since the distance between parallel planes is measured perpendicularly to them, you could take any line that’s parallel to the common normal of these planes, compute its intersection with them, and compare distances. However, this problem can be solved by inspection.
Rewrite the equations in the form $W\cdot(x,y,z) = k$. The left-hand side is equal to $\|W\|$ times the (signed) length of the projection of $(x,y,z)$ onto $W$, and the distance of these planes from the origin is measured parallel to $W$, so the constant term in these equations is proportional to the distances of the planes from the origin. The differences between the constant terms of the equations of the two outer planes and the middle one are $\pm k$, so these planes are the same distance from the central one.
Pick any point on the middle plane.
I would pick $P(0,0,-D/C)$ assuming that $C\ne 0$
Use the distance from P to the other planes.
$$ d_1 = \frac {|0+0+C(-D/C)+D-k|}{\sqrt {A^2+B^2+C^2}} =\frac {|-k|}{\sqrt {A^2+B^2+C^2}}$$
$$d_2= \frac {|0+0+C(-D/C)+D+k|}{\sqrt {A^2+B^2+C^2}}= \frac {|k|}{\sqrt {A^2+B^2+C^2}}$$
Thus $d_1=d_2$