What Is The Distance Between Two Planes?
The distance between two planes is the shortest, perpendicular gap between them - but it is only a meaningful, non-zero number when the two planes are parallel (they never intersect). If the planes are not parallel, they cross somewhere in space, so the shortest distance between them is zero. The distance is measured along the direction of the normal vector, the direction pointing straight out of each plane.
So the first question is always: are the two planes parallel? Only then does a distance formula apply. This idea extends the distance between two lines from 2D into three dimensions.
When Are Two Planes Parallel?
Write the two planes as:
$$P_1: a_1 x + b_1 y + c_1 z + d_1 = 0$$ $$P_2: a_2 x + b_2 y + c_2 z + d_2 = 0$$
The vector $(a, b, c)$ is the plane's normal - the direction perpendicular to the surface. Two planes are parallel exactly when their normals point the same way, which means the coefficients are proportional:
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
If these ratios are all equal, the planes are parallel and a distance can be found. If any ratio differs, the planes tilt relative to each other, they intersect, and the distance is zero. Always check this ratio first.
The Distance Formula for Parallel Planes
Once you know the planes are parallel, rewrite them so their normal coefficients are identical (scale one equation if needed):
$$P_1: ax + by + cz + d_1 = 0$$ $$P_2: ax + by + cz + d_2 = 0$$
Then the perpendicular distance between them is:
$$d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$$
Reading the formula:
$|d_1 - d_2|$ is the absolute difference of the two constant terms - the shift between the planes.
$\sqrt{a^2 + b^2 + c^2}$ is the length (magnitude) of the shared normal vector $(a, b, c)$.
Dividing by that length converts the raw shift into a true perpendicular distance.
The formula is really just the distance between point and plane applied to any point on one plane, measured to the other. The key precondition - matching $a$, $b$, $c$ - is what makes the numerator a clean $|d_1 - d_2|$. For the standard statement, see Wolfram MathWorld's entry on point-plane and plane distances.
Where The Formula Comes From, "One Point Bridges The Gap"
The distance between two parallel planes is worked out by a neat trick: pick any single point on the first plane, then measure its distance to the second plane. Because the planes are parallel, that distance is the same wherever you pick the point - the gap is constant.
Take a point $(x_0, y_0, z_0)$ on $P_1$, so $ax_0 + by_0 + cz_0 = -d_1$.
Its distance to $P_2: ax + by + cz + d_2 = 0$ is $\dfrac{|ax_0 + by_0 + cz_0 + d_2|}{\sqrt{a^2 + b^2 + c^2}}$.
Substitute $ax_0 + by_0 + cz_0 = -d_1$ to get $\dfrac{|-d_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$.
That is the whole derivation, and it explains why the coefficients must match first - otherwise $ax_0 + by_0 + cz_0$ would not simplify to $-d_1$. This is the same reasoning that sits behind the distance of a point from a line, lifted into 3D. For the formal proof, see the Wikipedia article on distance from a point to a plane.
Examples Of Distance Between Two Planes
Each example builds from a matched-coefficient case to a scale-first case. Problem statements are in bold; the working is not.
Example 1
Find the distance between the parallel planes $2x - y + 2z + 3 = 0$ and $2x - y + 2z + 9 = 0$.
The coefficients already match, so $d_1 = 3$, $d_2 = 9$, and $(a, b, c) = (2, -1, 2)$:
$$d = \frac{|3 - 9|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{6}{\sqrt{4 + 1 + 4}} = \frac{6}{\sqrt{9}} = \frac{6}{3} = 2 \text{ units}$$
The planes are $2$ units apart.
Example 2
Find the distance between $x + 2y + 2z - 6 = 0$ and $x + 2y + 2z + 3 = 0$.
Coefficients match, $d_1 = -6$, $d_2 = 3$, normal $(1, 2, 2)$:
$$d = \frac{|-6 - 3|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{9}{\sqrt{1 + 4 + 4}} = \frac{9}{\sqrt{9}} = \frac{9}{3} = 3 \text{ units}$$
The distance is $3$ units.
Example 3: The tempting shortcut that misfires
A student is asked for the distance between $x + y + z = 1$ and $2x + 3y + z = 5$ and plugs straight into the formula: $\frac{|1 - 5|}{\sqrt{1 + 1 + 1}} = \frac{4}{\sqrt{3}}$.
The number looks clean, so it feels like the answer. But the formula was applied without the required first check. Compare the coefficient ratios:
$$\frac{1}{2}, \quad \frac{1}{3}, \quad \frac{1}{1}$$
These are not equal, so the planes are not parallel - they intersect. Two intersecting planes cross along a line, and along that line the gap is zero.
The correct answer is $d = 0$. The rescue is to always test $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ before touching the distance formula. Skip that test and the formula returns a meaningless number.
Example 4
Find the distance between $3x - 6y + 2z + 4 = 0$ and $3x - 6y + 2z - 10 = 0$.
Matched coefficients, $d_1 = 4$, $d_2 = -10$, normal $(3, -6, 2)$:
$$d = \frac{|4 - (-10)|}{\sqrt{3^2 + (-6)^2 + 2^2}} = \frac{14}{\sqrt{9 + 36 + 4}} = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 \text{ units}$$
The planes are $2$ units apart.
Example 5
Find the distance between $x - 2y + 2z = 1$ and $2x - 4y + 4z = 10$.
The coefficients are not identical, but they are proportional: $\frac{1}{2} = \frac{-2}{-4} = \frac{2}{4}$, so the planes are parallel. Scale the second equation down by $2$ so both share the normal $(1, -2, 2)$:
$$P_1: x - 2y + 2z - 1 = 0 \quad (d_1 = -1)$$ $$P_2: x - 2y + 2z - 5 = 0 \quad (d_2 = -5)$$
Now apply the formula:
$$d = \frac{|-1 - (-5)|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{4}{\sqrt{9}} = \frac{4}{3} \text{ units}$$
The distance is $\frac{4}{3}$ units. Scaling first was essential.
Example 6
Two parallel shelves in a 3D model lie on the planes $z = 2$ and $z = 7$. Find the gap between them.
Rewrite as $0x + 0y + z - 2 = 0$ and $0x + 0y + z - 7 = 0$. Normal $(0, 0, 1)$:
$$d = \frac{|-2 - (-7)|}{\sqrt{0 + 0 + 1^2}} = \frac{5}{1} = 5 \text{ units}$$
The shelves are $5$ units apart - which matches simple intuition, since the planes are just two heights.
Where The Distance Between Planes Earns Its Keep
Parallel-plane distance is a working tool in any field that deals with flat layers in three dimensions.
Engineering and manufacturing. The thickness of a slab, a sheet of glass, or a layer of material is the distance between two parallel plane surfaces. Tolerance checks compute exactly this gap.
Crystallography. Atoms in a crystal sit on parallel planes, and the spacing between those planes - the distance formula applied to atomic planes - determines how the crystal diffracts X-rays. That spacing is how crystal structures are measured.
3D graphics and games. Collision systems and level geometry constantly test the gap between parallel surfaces - a floor and a ceiling, two walls - to decide whether an object fits.
The deeper "why" is that a plane is fully described by its normal direction plus one offset, so once two planes share a normal, only the offset difference separates them. That single insight, that parallel planes differ by a constant, is what collapses the whole problem to $|d_1 - d_2|$ over the normal's length.
The Mistakes Students Make Most Often
Mistake 1: Skipping the parallel check
Where it slips in: The rusher sees two plane equations and plugs into the formula without testing whether the normals are proportional, the trap in Example 3.
Don't do this: Compute $\frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$ for planes that actually intersect.
The correct way: First test $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. If the ratios are unequal, the planes intersect and the distance is $0$ — the formula does not apply.
Mistake 2: Forgetting to match the coefficients first
Where it slips in: The planes are parallel but written with different-sized normals, and the student uses $d_1$ and $d_2$ straight from the unmatched equations, the situation in Example 5.
Don't do this: Use $d_1 = -1$ and $d_2 = 10$ from $x - 2y + 2z = 1$ and $2x - 4y + 4z = 10$ without scaling.
The correct way: Scale one equation so both share the identical normal $(a, b, c)$, then read off $d_1$ and $d_2$. The formula assumes matched coefficients.
Mistake 3: Mishandling the constant's sign
Where it slips in: Moving the constant across the equals sign changes its sign, and the second-guesser loses track of whether $d$ is $+1$ or $-1$.
Don't do this: Read $d_1 = 6$ from $2x - y + 2z - 6 = 0$ (it is $-6$).
The correct way: Put both planes in the form $ax + by + cz + d = 0$ with the constant on the left. Then $d$ is exactly the number that appears there, sign included. The absolute value $|d_1 - d_2|$ forgives a final sign, but the individual $d$ values must be read correctly.
Conclusion
The distance between two parallel planes is $\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$, once both share matching normal coefficients.
Two planes have a non-zero distance only when they are parallel; intersecting planes are zero apart.
Check parallelism first via $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, then scale to match coefficients.
The formula follows from taking one point on a plane and measuring its distance to the other.
Read each constant $d$ with its correct sign from the form $ax + by + cz + d = 0$.
To take the distance between two planes further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor for 3D coordinate geometry, or browse math classes online. Want structured practice with a live trainer? Try a free class.
A Practical Next Step
Work through these problems to solidify your understanding, checking parallelism before each distance calculation.
Find the distance between $x + 2y + 2z + 1 = 0$ and $x + 2y + 2z - 8 = 0$. (Answer to Question 1: $\frac{|1 - (-8)|}{3} = \frac{9}{3} = 3$ units.)
Are the planes $x + y + 2z = 3$ and $2x + 5y + z = 4$ parallel? (Answer to Question 2: no — the ratios $\frac{1}{2}, \frac{1}{5}, \frac{2}{1}$ differ, so distance is 0.)
Find the distance between $2x + 2y + z = 4$ and $4x + 4y + 2z = 20$. (Answer to Question 3: scale to $2x+2y+z-10=0$; $\frac{|-4-(-10)|}{3} = \frac{6}{3} = 2$ units.)
Read More
3D Geometry Shapes - the wider world of planes and solids in three dimensions.
Coordinate Geometry - how points, lines, and planes are described by coordinates.
Coordinate Plane - the grid that extends into three dimensions for planes.
Points and Lines - the basic objects that planes are built from.
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