Deriving the formula for distance between two parallel planes Show that the distance between parallel planes, $ax + by + cz + d_1 = 0$,  and  $ax + by + cz + d_2 = 0$ is,
$$
D = \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}
$$
Solution:
(a) Get the distance from point (x, y, z) to the plane, $ax+by+cz+d_1 = 0$.
$$
 D = \frac{|ax+by+cz +d_1|}{\sqrt{a^2+b^2+c^2}}
$$

(b) Get the value of ax + by + cz.

*

*$d_1 = d_2$ since $-d_1 = ax+by+cz$, and $-d_2 = ax+by+cz.$

(c) Substitute ax+by+cz into $-d_2$, since, $-d_2 = ax+by+cz$
$$
 D= \frac{|ax+by+cz +d_1|}{\sqrt{a^2+b^2+c^2}} 
$$
$$
 D= \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}} 
$$
QED
 A: 
$"d_1=d_2$ since$ −d_1=ax+by+cz$, and $−d_2=ax+by+cz."$

This is clearly wrong. Since the planes are parallel, by definition there is no $ (x,y,z)$ triplet which satisfies both conditions at once.
Proof sketch:

*

*Draw the two parallel planes!  (let's call the two planes $\pi_1$ and $\pi_2$)

*Pick one point on each plane, construct a vector from these two points [ assume coordinates of the point]

*Dot this vector with the normal of the plane [ why? look at the picture!]


Legend:

*

*orange vector is the one connecting the point on each of the parallel planes

*Blue is the normal (same for both planes)


The general solution:
Assume two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$  then the vector connecting them is:
$$ \vec{r} = (x_2 - x_1, y_2 - y_1 , z_2 - z_1)$$
Now notice the equation in G Cab's comment:
$$ a(x_2 - x_1)  + b(y_2 - y_1) + c(z_2 - z_1) = d_2 - d_1$$
We can write this as:
$$ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \cdot \begin{bmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{bmatrix} = d_2 - d_1$$
Now normalize the unit vector :
$$ \frac{1}{\sqrt{a^2 + b^2 +c^2} } \begin{bmatrix} a \\ b \\ c \end{bmatrix} \cdot \begin{bmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{bmatrix}= \frac{d_2 - d_1}{\sqrt{a^2 + b^2 +c^2} }$$
This quantity can be geometrically interpreted as the distance the perpendicular distance between two planes and hence:
$$ D= = \frac{d_2 - d_1}{\sqrt{a^2 + b^2 +c^2} }$$
You can take modulus on both sides but actually, the sign has meaning as is discussed in this stack post
