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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
tryst with freedom
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Cyrus
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1 Answers1

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$"d_1=d_2$ 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:

  1. Draw the two parallel planes! (let's call the two planes $\pi_1$ and $\pi_2$)
  2. Pick one point on each plane, construct a vector from these two points [ assume coordinates of the point]
  3. Dot this vector with the normal of the plane [ why? look at the picture!]

enter image description here

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

tryst with freedom
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