Distance Between Two Planes: Formula and Examples

Distance Between Two Planes: Formula and Examples

TL;DR

The distance between two parallel planes ( ax + by + cz + d_1 = 0 ) and ( ax + by + cz + d_2 = 0 ) is ( \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} ), provided their normal coefficients match.

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. 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.

When Are Two Planes Parallel?

Write the two planes as:

P1: ( a_1 x + b_1 y + c_1 z + d_1 = 0 )
P2: ( 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:

P1: ( ax + by + cz + d_1 = 0 )
P2: ( 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:

The formula is really just the distance between point and plane applied to any point on one plane, measured to the other.

Examples Of Distance Between Two Planes

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}{3} = 2 \text{ units} )

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}} = 3 \text{ 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:

( d = \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. The correct answer is ( d = 0 ).

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 ):

( d = \frac{|4 - (-10)|}{\sqrt{3^2 + (-6)^2 + 2^2}} = 2 \text{ units} )

Example 5

Find the distance between ( x - 2y + 2z = 1 ) and ( 2x - 4y + 4z = 10 ).

The coefficients are not identical, but they are proportional, so the planes are parallel:

( d = \frac{|-1 - (-5)|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{4}{3} \text{ units} )

Example 6

Find the distance between planes ( z = 2 ) and ( z = 7 ).

( d = 5 \text{ units} )

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.

  1. Engineering and manufacturing.
  2. Crystallography.
  3. 3D graphics and games.

The Mistakes Students Make Most Often

Mistake 1: Skipping the parallel check

Mistake 2: Forgetting to match the coefficients first

Mistake 3: Mishandling the constant's sign

Conclusion