How to Find the Distance Between Two Planes

Written by

Malcolm McKinsey

January 11, 2023

Fact-checked by

Paul Mazzola

Distance between two parallel planes

The distance between two parallel planes is understood to be the shortest distance between their surfaces. Think about that; if the planes are not parallel, they must intersect, eventually. If they intersect, then at that line of intersection, they have no distance - 0 distance - between them. Part of your detective work is finding out if two planes are parallel.

The walls of your classroom, school or home are probably vertical to the floors and ceilings. That means two walls in a room are probably parallel to each other. If you could imagine those two walls extending infinitely in their two directions (width and length; not height), you would have two planes of a given distance apart. You could find the distance between those two planes mathematically.

Steps to find the distance between two planes

The five steps are as follows:

Write equations in standard format for both planes
Learn if the two planes are parallel
Identify the coefficients a, b, c, and d from one plane equation
Find a point ( ${x}_{1},{y}_{1},{z}_{1}$ ) in the other plane
Substitute for a, b, c, d, ${x}_{1}$ , ${y}_{1}$ and ${z}_{1}$ into the distance formula

Having previewed the steps, let's go through them.

Step 1: Write the equations for the planes in standard format, $ax+by+cz+d=0$ , with subscripts indicating the two planes, 1 and 2, in our generic equations:

{a}_{1}x+{b}_{1}y+{c}_{1}z+{d}_{1}=0

{a}_{2}x+{b}_{2}y+{c}_{2}z+{d}_{2}=0

Step 2: Are they intersecting or parallel? Remember, if they are intersecting, your work is done! The distance is 0. Compare the ratios of each plane's three identified points $(\frac{{a}_{1}}{{a}_{2}},\frac{{b}_{1}}{{b}_{2}},\frac{{c}_{1}}{{c}_{2}})$ . The two planes are parallel if their ratios are equal:

\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}

Step 3: Identify coefficients a, b, c, and d in one of the plane equations. Simply look at one equation and pull those coefficients, realizing ${a}_{1}$ is a (if you use the first plane), ${b}_{1}$ is b, and so on.

Step 4: Pick a point ( ${x}_{1},{y}_{1},{z}_{1}$ ) on the other plane. In our generic problem, we are using the second plane. Using a point with two 0 values makes your work easier: $x=0=y$ , which gives you:

{a}_{2}(0)+{b}_{2}(0)+{c}_{2}({z}_{1})+{d}_{2}=0

Solve for ${z}_{1}$ (the unknown third point).

Subtract ${d}_{2}$ from both sides:

{c}_{2}({z}_{1})+{d}_{2}-{d}_{2}=0-{d}_{2}

{c}_{2}({z}_{1})=0-{d}_{2}

Divide both sides by ${c}_{2}$ :

\frac{{c}_{2}({z}_{1})}{{c}_{2}}=\frac{-{d}_{2}}{{c}_{2}}

{z}_{1}=\frac{-{d}_{2}}{{c}_{2}}

Our three points are now clear:

{x}_{1}=0

{y}_{1}=0

{z}_{1}=\frac{-{d}_{2}}{{c}_{2}}

Do not get hung up on the variables in our generic example. You will have actual numbers in real equations, greatly simplifying the look of these equations.

Step 5: Substitute and plug the discovered values into the distance formula. You found a, b, c, and d in Step 3, above. They are the coefficients of one plane's equation. You found ${x}_{1}$ , ${y}_{1}$ , and ${z}_{1}$ in Step 4, above.

They are the coordinates of a point on the other plane. Plug those found values into the Point-Plane distance formula.

Point-plane distance formula

D=\frac{\left|a{x}_{1}+b{y}_{1}+c{z}_{1}+d\right|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}

Get free estimates from math tutors near you.

Finding the distance between two planes

Here are two equations for planes:

3x+4y+5z+9=0

9x+12y+15z-27=0

Go through your five steps:

Write equations in standard format for both planes - we already did that for you!
Learn if the two planes are parallel:
$\frac{3}{9}=\frac{4}{12}=\frac{5}{15}=\frac{1}{3}$
The two planes are parallel.
Identify the coefficients a, b, c, and d from one plane equation. Rather than repeat the choices from our generic example, we will use the second equation, giving us these values:
$a=9$
$b=12$
$c=15$
$d=-27$
Find a point ( ${x}_{1},{y}_{1},{z}_{1}$ ) in the other plane, which in this case is the first plane. Try to use $x=y=0$ for two of the point's coordinates, and solve for ${z}_{1}$ :
$3x+4y+5z+9=0$
$3(0)+4(0)+5({z}_{1})+9=0$
$5({z}_{1})+9=0$
Subtract ${d}_{1}$ from both sides:
$5({z}_{1})+9-9=0-9$
$5({z}_{1})=-9$
Divide both sides by ${c}_{1}$ :
$\frac{5({z}_{1})}{5}=\frac{-9}{5}$
${z}_{1}=\frac{-9}{5}=-1.8$
Our three points on the first plane are:
$x=0$
$y=0$
$z=\frac{-9}{5}=-1.8$
Substitute a, b, c, d, ${x}_{1}$ , ${y}_{1}$ , and ${z}_{1}$ in the distance formula.

The distance between our two planes is roughly 2.54.