Distance Formulas for Planes

In short

The perpendicular distance from a point (x_0, y_0, z_0) to a plane ax + by + cz = d is \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}. From this single formula, you can derive the foot of the perpendicular, the mirror image of the point in the plane, and the distance between two parallel planes.

A cricket ball is hit high into the air. Below it, the ground is flat \u2014 a plane. At any instant, how far is the ball from the ground? Not the distance along its curved trajectory, and not the horizontal distance to the nearest fielder. The shortest distance. The distance straight down.

That shortest distance is the perpendicular distance from the point (the ball) to the plane (the ground). It is the length of the line segment that drops from the point to the plane at a right angle. No other path from the point to the plane is shorter.

This single measurement \u2014 "how far is a point from a plane?" \u2014 turns out to be the foundation for several other questions. Where exactly does the perpendicular hit the plane? (That is the foot of the perpendicular.) Where is the point's mirror image on the other side of the plane? (That is the image or reflection.) How far apart are two parallel planes? (That is just the perpendicular distance from any point on one plane to the other.)

All four questions have clean formulas, and they all flow from one derivation.

The distance formula: derivation

Take a point P = (x_0, y_0, z_0) and a plane \pi : ax + by + cz = d.

The normal vector of the plane is \vec{n} = (a, b, c). The shortest path from P to the plane is along the normal direction \u2014 perpendicular to the plane. So the perpendicular distance is the length of the projection of a certain vector onto \vec{n}.

Pick any point Q on the plane. (For concreteness, you could pick the point where the plane crosses the x-axis, or any other convenient point \u2014 the final answer does not depend on which Q you choose.) The vector from Q to P is \vec{QP} = P - Q.

The perpendicular distance from P to the plane is the component of \vec{QP} along \vec{n}:

d_\perp = \frac{|\vec{QP} \cdot \vec{n}|}{|\vec{n}|}

Now compute \vec{QP} \cdot \vec{n}. Since Q = (q_1, q_2, q_3) lies on the plane, we have aq_1 + bq_2 + cq_3 = d. So:

\vec{QP} \cdot \vec{n} = a(x_0 - q_1) + b(y_0 - q_2) + c(z_0 - q_3)

= (ax_0 + by_0 + cz_0) - (aq_1 + bq_2 + cq_3)

= (ax_0 + by_0 + cz_0) - d

The coordinates of Q cancel out completely. That is why the formula does not depend on which point on the plane you chose \u2014 the plane equation absorbs Q into the constant d.

Distance of a point from a plane

The perpendicular distance from the point (x_0, y_0, z_0) to the plane ax + by + cz = d is

D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}

Reading the formula. The numerator is |ax_0 + by_0 + cz_0 - d| \u2014 plug the point's coordinates into the plane equation, subtract d, and take the absolute value. The denominator is \sqrt{a^2 + b^2 + c^2} \u2014 the magnitude of the normal vector. The absolute value in the numerator is there because distance is always non-negative; the sign of ax_0 + by_0 + cz_0 - d tells you which side of the plane the point is on, but for the distance you only care about how far, not which side.

If you have ever seen the 2D formula for the distance from a point (x_0, y_0) to a line ax + by = c \u2014 which is \dfrac{|ax_0 + by_0 - c|}{\sqrt{a^2 + b^2}} \u2014 this is the exact same structure, extended to three dimensions. The pattern generalises to any number of dimensions: plug the point into the equation, take the absolute value, divide by the magnitude of the coefficient vector.

The perpendicular distance $D$ from point $P$ to the plane. The foot $F$ is where the perpendicular meets the plane. The dashed line from $P$ to $F$ is parallel to the normal $\vec{n}$ and has length $D$.

Foot of the perpendicular

The formula above gives you the distance. But what if you want the actual point where the perpendicular from P hits the plane? That point is called the foot of the perpendicular, and you will often see it denoted F.

The foot lies on the line through P in the direction of \vec{n} = (a, b, c). Any point on this line can be written as:

F = P + t\,\vec{n} = (x_0 + at, \; y_0 + bt, \; z_0 + ct)

for some scalar t. The foot is the specific point on this line that also lies on the plane. So substitute into the plane equation:

a(x_0 + at) + b(y_0 + bt) + c(z_0 + ct) = d

ax_0 + by_0 + cz_0 + t(a^2 + b^2 + c^2) = d

t = \frac{d - (ax_0 + by_0 + cz_0)}{a^2 + b^2 + c^2}

Once you have t, plug it back into the parametric expression for F to get the coordinates. Notice that |t| \cdot |\vec{n}| = |t| \cdot \sqrt{a^2 + b^2 + c^2}, which equals the perpendicular distance D. The sign of t tells you which direction along the normal you have to travel: if t < 0, the point P is on the side of the plane where ax + by + cz > d; if t > 0, it is on the other side.

Foot of the perpendicular

The foot of the perpendicular from (x_0, y_0, z_0) to the plane ax + by + cz = d is

F = \left(x_0 + at, \; y_0 + bt, \; z_0 + ct\right)

where t = \dfrac{d - (ax_0 + by_0 + cz_0)}{a^2 + b^2 + c^2}.

The foot is useful beyond just answering "where does the perpendicular land." It is the closest point on the plane to P, which matters in optimisation problems. And it is the midpoint between P and the image of P in the plane \u2014 which leads to the next formula.

Image of a point in a plane

The image (or reflection) of a point P in a plane is the point P' on the other side of the plane, at the same perpendicular distance. If you think of the plane as a mirror, P' is where you see P's reflection.

The foot F is the midpoint of P and P'. So:

F = \frac{P + P'}{2} \quad \Longrightarrow \quad P' = 2F - P

In coordinates:

P' = (2(x_0 + at) - x_0, \; 2(y_0 + bt) - y_0, \; 2(z_0 + ct) - z_0)

P' = (x_0 + 2at, \; y_0 + 2bt, \; z_0 + 2ct)

where t is the same value as in the foot formula.

Image of a point in a plane

The image of (x_0, y_0, z_0) in the plane ax + by + cz = d is

P' = (x_0 + 2at, \; y_0 + 2bt, \; z_0 + 2ct)

where t = \dfrac{d - (ax_0 + by_0 + cz_0)}{a^2 + b^2 + c^2}.

The image is exactly the foot of the perpendicular, but "doubled." You travel from P to the foot (t units along the normal), and then travel the same distance again beyond the foot (t more units) to reach the image. The total displacement from P to P' is 2t \cdot \vec{n}.

The image $P'$ is the reflection of $P$ in the plane. The foot $F$ is the midpoint: $P$ and $P'$ are equidistant from the plane, on opposite sides. To find $P'$, travel from $P$ to $F$ and then the same distance beyond.

Distance between parallel planes

Two parallel planes have the same normal direction but different constants. Take \pi_1 : ax + by + cz = d_1 and \pi_2 : ax + by + cz = d_2 (same coefficients, different right sides).

The distance between them is simply the perpendicular distance from any point on one plane to the other. Pick a point on \pi_1 \u2014 any point will do, since all points on a parallel plane are the same distance from the other plane.

But you can get the answer even without picking a point. Any point on \pi_1 satisfies ax_0 + by_0 + cz_0 = d_1. Plugging into the distance formula for the plane \pi_2:

D = \frac{|ax_0 + by_0 + cz_0 - d_2|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}

Distance between parallel planes

The distance between the parallel planes ax + by + cz = d_1 and ax + by + cz = d_2 is

D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}

This only works if both plane equations have the same coefficients (a, b, c). If the second plane is written as 2ax + 2by + 2cz = d_2', you first need to divide through by 2 to make the coefficients match, giving d_2 = d_2'/2, and only then apply the formula.

The formula says something geometrically clear: the gap between parallel planes is proportional to the difference in their constants, scaled by the normal's magnitude. If the normal is a unit vector (|\vec{n}| = 1), the distance is simply |d_1 - d_2|.

Two parallel planes with the same normal direction. The perpendicular distance between them depends only on the difference in their constants, $|d_1 - d_2|$, divided by the magnitude of the shared normal vector.

Worked examples

Example 1: Distance and foot of perpendicular

Problem. Find the perpendicular distance from P = (2, 3, -1) to the plane 2x - y + 2z = 6, and find the foot of the perpendicular.

Step 1. Identify the plane's normal and the formula's ingredients.

a = 2, \; b = -1, \; c = 2, \; d = 6

Why: reading straight from the equation 2x - y + 2z = 6.

Step 2. Compute the numerator of the distance formula.

ax_0 + by_0 + cz_0 - d = 2(2) + (-1)(3) + 2(-1) - 6 = 4 - 3 - 2 - 6 = -7

Why: plug the point into the plane equation and subtract d. The sign tells you the point is on the side where ax + by + cz < d.

Step 3. Compute the denominator.

\sqrt{a^2 + b^2 + c^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

Why: a clean square root \u2014 the numbers were chosen for this. The normal (2, -1, 2) has magnitude 3.

Step 4. Distance.

D = \frac{|-7|}{3} = \frac{7}{3}

Why: the absolute value drops the sign. The distance is 7/3 units.

Step 5. Foot of perpendicular. Compute t. From Step 2, ax_0 + by_0 + cz_0 = -1.

t = \frac{d - (ax_0 + by_0 + cz_0)}{a^2 + b^2 + c^2} = \frac{6 - (-1)}{9} = \frac{7}{9}

F = \left(2 + 2 \cdot \frac{7}{9}, \; 3 + (-1) \cdot \frac{7}{9}, \; -1 + 2 \cdot \frac{7}{9}\right) = \left(\frac{32}{9}, \; \frac{20}{9}, \; \frac{5}{9}\right)

Result: Distance = \dfrac{7}{3}. Foot = \left(\dfrac{32}{9}, \dfrac{20}{9}, \dfrac{5}{9}\right).

A 2D cross-section (setting $z = 0$): the trace of the plane $2x - y + 2z = 6$ on the $xy$-plane is $2x - y = 6$. The point $P$ sits above-left of this line. The dashed red segment is the perpendicular from $P$ to the plane's trace. In the full 3D picture, $F$ is the point $\left(\frac{32}{9}, \frac{20}{9}, \frac{5}{9}\right)$ on the plane.

Verify: F should satisfy the plane equation. 2(32/9) - (20/9) + 2(5/9) = 64/9 - 20/9 + 10/9 = 54/9 = 6. Correct.

Example 2: Image of a point in a plane

Problem. Find the image (reflection) of P = (1, 2, 3) in the plane x + y + z = 3.

Step 1. Identify the ingredients.

a = 1, \; b = 1, \; c = 1, \; d = 3, \quad P = (1, 2, 3)

Why: reading directly from the equation.

Step 2. Compute ax_0 + by_0 + cz_0.

1(1) + 1(2) + 1(3) = 6

Why: this tells you how far the point is (in the signed sense) from the plane.

Step 3. Compute t.

t = \frac{d - (ax_0 + by_0 + cz_0)}{a^2 + b^2 + c^2} = \frac{3 - 6}{1 + 1 + 1} = \frac{-3}{3} = -1

Why: t is negative because the point is on the side where ax + by + cz > d, so you travel in the -\vec{n} direction to reach the plane.

Step 4. Compute the image P' = (x_0 + 2at, \; y_0 + 2bt, \; z_0 + 2ct).

P' = (1 + 2(1)(-1), \; 2 + 2(1)(-1), \; 3 + 2(1)(-1)) = (-1, 0, 1)

Why: the image is the foot doubled. You travel from P to the foot (t = -1 along \vec{n}), then continue the same distance beyond the foot.

Step 5. Verify. The foot F should be the midpoint of P and P':

F = \frac{(1, 2, 3) + (-1, 0, 1)}{2} = (0, 1, 2)

Check that F lies on the plane: 0 + 1 + 2 = 3. Correct. And P' should be at the same distance from the plane as P: |(-1) + 0 + 1 - 3|/\sqrt{3} = |-3|/\sqrt{3} = 3/\sqrt{3} = \sqrt{3}. For P: |1 + 2 + 3 - 3|/\sqrt{3} = 3/\sqrt{3} = \sqrt{3}. Same distance. Correct.

Result: Image = (-1, 0, 1).

A 2D cross-section (setting $z = 0$): the plane's trace is $x + y = 3$. The point $P = (1, 2)$ is reflected to $P' = (-1, 0)$, with the foot $F = (0, 1)$ as the midpoint. In the full 3D picture, $P = (1, 2, 3)$ reflects to $P' = (-1, 0, 1)$ through the foot $F = (0, 1, 2)$ on the plane.

The image is the "mirror twin" of the original point. Every coordinate shifted by 2at, 2bt, 2ct respectively \u2014 the same displacement in each direction, scaled by the corresponding component of the normal. The plane acts as a perfect mirror: the foot is the midpoint, and the perpendicular distance is preserved.

Common confusions

"The distance can be negative." It cannot. Distance is always a non-negative number. The quantity ax_0 + by_0 + cz_0 - d can be negative (telling you which side of the plane the point is on), but the distance formula takes the absolute value. If you need the signed distance (for example, to determine which side a point is on), drop the absolute value \u2014 but then you are computing a signed quantity, not a distance.
"To find the foot, I just drop the perpendicular by setting one coordinate to zero." That finds a point on the plane, not the foot. The foot is the specific point on the plane nearest to P, and it lies along the line from P in the direction of the normal. You must parametrise this line and solve for the parameter.
"The image is twice the foot." The image is 2F - P, not 2F. The foot is the midpoint of P and the image, so the image is P' = 2F - P. Writing it as P + 2t\vec{n} (where t is the parameter that gives the foot) is the cleanest way.
"The distance between parallel planes is |d_1 - d_2|." Only if the normal vector is a unit vector. In general, it is |d_1 - d_2| / |\vec{n}|. And both plane equations must have the same coefficient vector (a, b, c) \u2014 if one is scaled differently, you need to normalise first.
"I can use this formula for the distance between a point and a line." This specific formula is for point-to-plane distance. The formula for the distance from a point to a line in 3D is different (it involves a cross product). The structure is related but not identical.

Going deeper

If you came here for the four distance-related formulas and the worked examples, you have them \u2014 you can stop here. What follows explores the geometric meaning of the signed distance, a vector derivation that makes everything coordinate-free, and a neat application to finding the equation of a bisector plane.

The signed distance and half-spaces

The expression ax_0 + by_0 + cz_0 - d (without the absolute value) is called the signed distance (up to a scaling factor). Its sign partitions all of 3D space into two half-spaces:

Points where ax + by + cz > d are on the "positive" side of the plane (the side the normal points toward).
Points where ax + by + cz < d are on the "negative" side.
Points where ax + by + cz = d are on the plane itself.

This signed version is essential for the theory of bisector planes. If you have two planes \pi_1 and \pi_2, a point is on the angle bisector plane if and only if its signed distances to \pi_1 and \pi_2 are equal (or equal in magnitude but opposite in sign). That is why the bisector plane article starts from this signed distance.

The vector form

There is a coordinate-free way to state the distance formula that makes its geometric meaning transparent. If \vec{r_0} is the position vector of the point and \hat{n} is the unit normal to the plane, and p is the perpendicular distance of the plane from the origin (so the plane equation is \hat{n} \cdot \vec{r} = p), then:

D = |\hat{n} \cdot \vec{r_0} - p|

This says: project the point onto the normal direction, and compare with where the plane sits. The difference is the distance. No square roots in sight, because the normal is already a unit vector.

This form also makes it obvious why parallel planes have distance |p_1 - p_2| when written in unit-normal form \u2014 the two constants p_1 and p_2 are the perpendicular distances of the two planes from the origin, and the gap between the planes is the difference.

An application: the locus equidistant from two parallel planes

Given two parallel planes ax + by + cz = d_1 and ax + by + cz = d_2, what is the locus of points equidistant from both?

Set the distances equal:

\frac{|ax + by + cz - d_1|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|ax + by + cz - d_2|}{\sqrt{a^2 + b^2 + c^2}}

The denominators cancel, leaving |ax + by + cz - d_1| = |ax + by + cz - d_2|. Let u = ax + by + cz. Then |u - d_1| = |u - d_2|, which gives u = \dfrac{d_1 + d_2}{2}.

So the locus is the plane ax + by + cz = \dfrac{d_1 + d_2}{2} \u2014 the plane exactly halfway between the two given planes. Its constant is the average of the two original constants. This is the 3D analogue of the midpoint of a segment: the equidistant set from two parallel planes is the plane that splits the gap evenly.

The plane equidistant from two parallel planes sits exactly halfway between them. Its constant is the average of the two original constants: $d = (d_1 + d_2)/2$.

The distance formula in non-standard form

Some textbooks write the plane equation as ax + by + cz + d' = 0 (with d' on the left instead of d on the right). In that convention, the distance formula becomes

D = \frac{|ax_0 + by_0 + cz_0 + d'|}{\sqrt{a^2 + b^2 + c^2}}

The only difference is the sign of the constant. If your plane is 2x - 3y + z - 5 = 0, then d' = -5, and the formula uses ax_0 + by_0 + cz_0 + (-5) = ax_0 + by_0 + cz_0 - 5 in the numerator. Both conventions give the same distance \u2014 just be consistent about which form you are using.

Where this leads next

The distance formula is one of the most-used results in 3D geometry. The natural continuations:

Bisector Planes \u2014 using the signed distance to find planes that bisect the angle between two given planes.
Angle and Intersection \u2014 the angle between planes and lines, and the line where two planes meet.
Plane \u2014 Basic Equations \u2014 the general equation, normal form, three-point form, and intercept form.
Straight Line in 3D \u2014 Angles and Distances \u2014 the analogous distance formulas for lines in 3D (perpendicular distance from a point to a line, shortest distance between skew lines).
Distance Formulas \u2014 the 2D versions of point-to-line distance, for comparison and review.