Distance Between Point and Plane: Formula & Examples

TL;DR

The distance between a point ((x_0,y_0,z_0)) and a plane (ax+by+cz+d=0) is the shortest, perpendicular distance, given by [d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}]. This article derives the formula, explains what every term does, works six examples, and shows that a distance of zero means the point lies on the plane.

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 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 crucial; 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.

Examples of Distance Between Point and Plane

Each example builds from a direct substitution to a fuller task. Problem statements are in bold;

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}} \approx 4.51 \text{ units}]

Example 2

Find the distance from (P(4,−4,3)) to the plane (2x−2y+5z+8=0).

[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}} \approx 6.79 \text{ 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]

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}]

Example 4

Find the distance from the origin (O(0,0,0)) to the plane (x+2y+2z−9=0).

[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}]

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]

Example 6

A drone at position ( (3,0,4) ) must stay clear of a flat glass wall on the plane (z=0). How far is it from the wall?

[d = \frac{|0(3)+0(0)+1(4)+0|}{\sqrt{0^2+0^2+1^2}} = \frac{4}{1} = 4 \text{ units}]

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.

• 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 Mistakes Students Make Most Often

Mistake 1: Forgetting to divide by the normal's length

Mistake 2: Mishandling the plane's constant sign

Mistake 3: Dropping the absolute value

Conclusion

• The distance between a point and a plane is given by (\frac{|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.