What Is The Distance Between A Point And A Plane?
The distance between a point and a plane is the length of the shortest line segment joining the point to the plane. That shortest segment is always perpendicular to the plane - it runs along the plane's normal vector, the direction pointing straight out of the surface. Any slanted path from the point to the plane is longer, so the perpendicular drop is the one that counts.
If the point happens to sit on the plane, the distance is simply zero. This is the 3D partner of the distance of a point from a line in the plane.
The Formula And What Each Term Means
For a point $P(x_0, y_0, z_0)$ and a plane $ax + by + cz + d = 0$, the distance is:
$$d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$$
Reading the formula piece by piece:
The numerator $|ax_0 + by_0 + cz_0 + d|$ is what you get by substituting the point's coordinates into the left side of the plane equation, then taking the absolute value. It measures how far off the plane the point is.
The denominator $\sqrt{a^2 + b^2 + c^2}$ is the length of the normal vector $(a, b, c)$ - the direction perpendicular to the plane.
The absolute value keeps the distance positive; the sign inside only tells you which side of the plane the point is on.
If the point lies on the plane, then $ax_0 + by_0 + cz_0 + d = 0$, so the numerator is zero and the distance is zero - exactly as it should be.
Where The Formula Comes From, "Follow The Normal"
The formula is not arbitrary; it comes from projecting the gap onto the plane's normal direction. The reasoning runs like this:
The vector $(a, b, c)$ points perpendicular to the plane, so the shortest route from $P$ to the plane runs along it.
Substituting $P$ into $ax + by + cz + d$ gives a value proportional to how far $P$ sits from the plane, measured in "normal units."
Dividing by the normal's length $\sqrt{a^2 + b^2 + c^2}$ converts those units into an ordinary length.
That division is the crucial step - without it, a plane written with large coefficients would report a falsely large distance. The absolute value at the end drops the side-of-plane sign, since distance is a positive quantity. This mirrors the derivation of the distance between two planes, which is just this formula reused. For the formal proof, see the Wikipedia article on distance from a point to a plane.
Examples of Distance Between Point and Plane
Each example builds from a direct substitution to a fuller task. Problem statements are in bold; the working is not.
Example 1
Find the distance from the point $P(1, 2, 5)$ to the plane $3x + 4y + z + 7 = 0$.
Read off $(a, b, c, d) = (3, 4, 1, 7)$ and $(x_0, y_0, z_0) = (1, 2, 5)$:
$$d = \frac{|3(1) + 4(2) + 1(5) + 7|}{\sqrt{3^2 + 4^2 + 1^2}}$$ $$= \frac{|3 + 8 + 5 + 7|}{\sqrt{9 + 16 + 1}} = \frac{23}{\sqrt{26}} \text{ units}$$
The distance is $\dfrac{23}{\sqrt{26}} \approx 4.51$ units.
Example 2
Find the distance from $P(4, -4, 3)$ to the plane $2x - 2y + 5z + 8 = 0$.
Substitute $(a, b, c, d) = (2, -2, 5, 8)$ and the point:
$$d = \frac{|2(4) + (-2)(-4) + 5(3) + 8|}{\sqrt{2^2 + (-2)^2 + 5^2}}$$ $$= \frac{|8 + 8 + 15 + 8|}{\sqrt{4 + 4 + 25}} = \frac{39}{\sqrt{33}} \text{ units}$$
The distance is $\dfrac{39}{\sqrt{33}} \approx 6.79$ units.
Example 3: The tempting shortcut that misfires
A student finds the distance from $P(1, 1, 1)$ to the plane $2x + 3y + 6z - 5 = 0$ and forgets the denominator, reporting the numerator alone.
They compute:
$$|2(1) + 3(1) + 6(1) - 5| = |2 + 3 + 6 - 5| = 6$$
and write "distance $= 6$ units." The number looks like an answer, so it is easy to stop there. But that is only the numerator — the raw value has not yet been converted into a true length. The plane's normal $(2, 3, 6)$ has length $\sqrt{4 + 9 + 36} = 7$, and skipping it inflates the distance sevenfold.
The correct calculation divides by the normal's length:
$$d = \frac{6}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{6}{\sqrt{49}} = \frac{6}{7} \text{ units}$$
The distance is $\frac{6}{7}$ units, not $6$. The rescue is to never treat the numerator as the answer — always divide by $\sqrt{a^2 + b^2 + c^2}$.
Example 4
Find the distance from the origin $O(0, 0, 0)$ to the plane $x + 2y + 2z - 9 = 0$.
With the origin, the substitution is short:
$$d = \frac{|1(0) + 2(0) + 2(0) - 9|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-9|}{\sqrt{9}} = \frac{9}{3} = 3 \text{ units}$$
The plane is $3$ units from the origin.
Example 5
Show that the point $P(2, -1, 3)$ lies on the plane $x + y + z - 4 = 0$.
Substitute the point into the numerator:
$$|2 + (-1) + 3 - 4| = |0| = 0$$
The numerator is $0$, so the distance is $0$. A distance of zero means the point lies on the plane.
Example 6
A drone at position $(3, 0, 4)$ must stay clear of a flat glass wall on the plane $z = 0$ (the ground plane $0x + 0y + z = 0$). How far is it from the wall?
Rewrite the wall as $0x + 0y + z + 0 = 0$, normal $(0, 0, 1)$:
$$d = \frac{|0(3) + 0(0) + 1(4) + 0|}{\sqrt{0^2 + 0^2 + 1^2}} = \frac{4}{1} = 4 \text{ units}$$
The drone is $4$ units above the ground - which matches intuition, since the point's height is exactly $4$.
Where The Distance Earns Its Keep
The point-to-plane distance is a workhorse formula across three-dimensional fields.
3D graphics and collision detection. Game engines constantly ask "how far is this object from that surface?" to decide whether a ball has hit a wall or a character has clipped through a floor. Each test is a point-to-plane distance.
Robotics and navigation. A robot arm or a self-driving car keeps a safe standoff from flat obstacles by computing the perpendicular distance from its position to each surface plane.
Computer-aided design. Checking whether a drilled hole or a mounted part clears a face means measuring the perpendicular gap from a point to a plane, exactly this formula.
The deeper "why" is that a plane divides space, and the normal direction is the one axis along which "how far off the plane" has a single, unambiguous meaning. Every slanted path double-counts sideways motion that does not bring you closer; the perpendicular strips that away. That is why the shortest distance always follows the normal. For the broader treatment, see Wolfram MathWorld's entry on point-plane distance.
The Mistakes Students Make Most Often
Mistake 1: Forgetting to divide by the normal's length
Where it slips in: The rusher computes the numerator $|ax_0 + by_0 + cz_0 + d|$ and reports it as the distance, the trap in Example 3.
Don't do this: Write "distance $= 6$" when $6$ is only the numerator.
The correct way: Always divide by $\sqrt{a^2 + b^2 + c^2}$. The numerator alone is proportional to the distance but is measured in the wrong units until scaled by the normal's length.
Mistake 2: Mishandling the plane's constant sign
Where it slips in: The plane is given as $ax + by + cz = k$ rather than $= 0$, and the second-guesser uses $d = k$ instead of $d = -k$.
Don't do this: Read $d = 9$ from $x + 2y + 2z = 9$.
The correct way: Move everything to one side first: $x + 2y + 2z - 9 = 0$, so $d = -9$. The absolute value forgives a final sign, but the substitution must use the correct $d$.
Mistake 3: Dropping the absolute value
Where it slips in: The substitution gives a negative value and the memoriser reports a negative distance.
Don't do this: Report a distance of $-\frac{9}{3} = -3$.
The correct way: Apply the absolute value bars. The sign only records which side of the plane the point is on; the distance itself is always positive.
Conclusion
The distance between a point and a plane is $\dfrac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
It is the shortest, perpendicular distance, measured along the plane's normal.
Divide by the normal's length $\sqrt{a^2 + b^2 + c^2}$ - the numerator alone is not the distance.
A distance of zero means the point lies on the plane.
Read the constant $d$ with its correct sign, and keep the absolute value so the distance stays positive.
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A Practical Next Step
Work through these problems to solidify your understanding, dividing by the normal's length each time.
Find the distance from $(1, 0, 2)$ to the plane $2x + y + 2z - 3 = 0$. (Answer to Question 1: $\frac{|2 + 0 + 4 - 3|}{3} = \frac{3}{3} = 1$ unit.)
Find the distance from the origin to $x + 2y + 2z + 6 = 0$. (Answer to Question 2: $\frac{6}{3} = 2$ units.)
Does the point $(1, 1, 1)$ lie on the plane $x + y + z - 3 = 0$? (Answer to Question 3: numerator $= 0$, so yes - distance is 0.)
Read More
Distance Between Two Lines - perpendicular distance for parallel lines in space.
3D Geometry Shapes - the wider world of points, planes, and solids in three dimensions.
Coordinate Geometry - how points and planes are described by coordinates.
Coordinate Plane - the grid that extends into three dimensions.
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