Distance from a Point to a Plane (3D) Notes

Published on: September 29, 2026

This Distance from a Point to a Plane (3D) Calculator measures the shortest gap between a point and a plane, and shows the working line by line. Substituting the point into the plane equation measures how far off it is, and dividing by the length of the normal turns that into a real distance. The absolute value is taken because a side of the plane is not a direction. Following the same pattern every time keeps the numerator and the normal apart.

Step-by-step method

  1. Set up the problem.
  2. Substitute into the distance formula.
  3. Simplify to get the distance.
See Example 1 Hide Example 1

Problem

\(P = \left(1,\; 2,\; 3\right),\quad 2x - 3y + 4z = 10\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Substitute into the distance formula.
  3. Simplify to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: The point is \(P = \left(1,\; 2,\; 3\right)\) and the plane is \(2x - 3y + 4z = 10\).

\(P = \left(1,\; 2,\; 3\right),\quad 2x - 3y + 4z = 10\)

Step 2

Step 2 - Write the distance formula.

In this problem: This measures the perpendicular distance from the point to the plane.

\(D = \frac{\left|Ax_{0} + By_{0} + Cz_{0} - d\right|}{\sqrt{A^{2} + B^{2} + C^{2}}}\)

Step 3

Step 3 - Substitute into the distance formula.

In this problem: Substitute the point and coefficients into the distance formula.

\(D = \frac{\left|2 \cdot 1 + \left(-3\right) \cdot 2 + 4 \cdot 3 - 10\right|}{\sqrt{2^{2} + \left(-3\right)^{2} + 4^{2}}}\)

Step 4

Step 4 - Simplify to get the distance.

In this problem: The distance is \(\frac{2 \sqrt{29}}{29}\).

\(D = \frac{2}{\sqrt{29}} = \frac{2 \sqrt{29}}{29} \approx 0.37\)

Final Answer

\(D = \frac{2 \sqrt{29}}{29} \approx 0.37\)
See Example 2 Hide Example 2

Problem

\(P = \left(0,\; 0,\; 0\right),\quad x + y + z = 6\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Substitute into the distance formula.
  3. Simplify to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: The point is \(P = \left(0,\; 0,\; 0\right)\) and the plane is \(x + y + z = 6\).

\(P = \left(0,\; 0,\; 0\right),\quad x + y + z = 6\)

Step 2

Step 2 - Write the distance formula.

In this problem: This measures the perpendicular distance from the point to the plane.

\(D = \frac{\left|Ax_{0} + By_{0} + Cz_{0} - d\right|}{\sqrt{A^{2} + B^{2} + C^{2}}}\)

Step 3

Step 3 - Substitute into the distance formula.

In this problem: Substitute the point and coefficients into the distance formula.

\(D = \frac{\left|1 \cdot 0 + 1 \cdot 0 + 1 \cdot 0 - 6\right|}{\sqrt{1^{2} + 1^{2} + 1^{2}}}\)

Step 4

Step 4 - Simplify to get the distance.

In this problem: The distance is \(2 \sqrt{3}\).

\(D = \frac{6}{\sqrt{3}} = 2 \sqrt{3} \approx 3.46\)

Final Answer

\(D = 2 \sqrt{3} \approx 3.46\)
See Example 3 Hide Example 3

Problem

\(P = \left(2,\; 1,\; -1\right),\quad x + 2y + 2z = 3\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Substitute into the distance formula.
  3. Simplify to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: The point is \(P = \left(2,\; 1,\; -1\right)\) and the plane is \(x + 2y + 2z = 3\).

\(P = \left(2,\; 1,\; -1\right),\quad x + 2y + 2z = 3\)

Step 2

Step 2 - Write the distance formula.

In this problem: This measures the perpendicular distance from the point to the plane.

\(D = \frac{\left|Ax_{0} + By_{0} + Cz_{0} - d\right|}{\sqrt{A^{2} + B^{2} + C^{2}}}\)

Step 3

Step 3 - Substitute into the distance formula.

In this problem: Substitute the point and coefficients into the distance formula.

\(D = \frac{\left|1 \cdot 2 + 2 \cdot 1 + 2 \cdot \left(-1\right) - 3\right|}{\sqrt{1^{2} + 2^{2} + 2^{2}}}\)

Step 4

Step 4 - Simplify to get the distance.

In this problem: The distance is \(\frac{1}{3}\).

\(D = \frac{1}{3} = \frac{1}{3} \approx 0.33\)

Final Answer

\(D = \frac{1}{3} \approx 0.33\)
See Example 4 Hide Example 4

Problem

\(P = \left(3,\; 0,\; 4\right),\quad 2x - y + 2z = 5\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Substitute into the distance formula.
  3. Simplify to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: The point is \(P = \left(3,\; 0,\; 4\right)\) and the plane is \(2x - y + 2z = 5\).

\(P = \left(3,\; 0,\; 4\right),\quad 2x - y + 2z = 5\)

Step 2

Step 2 - Write the distance formula.

In this problem: This measures the perpendicular distance from the point to the plane.

\(D = \frac{\left|Ax_{0} + By_{0} + Cz_{0} - d\right|}{\sqrt{A^{2} + B^{2} + C^{2}}}\)

Step 3

Step 3 - Substitute into the distance formula.

In this problem: Substitute the point and coefficients into the distance formula.

\(D = \frac{\left|2 \cdot 3 + \left(-1\right) \cdot 0 + 2 \cdot 4 - 5\right|}{\sqrt{2^{2} + \left(-1\right)^{2} + 2^{2}}}\)

Step 4

Step 4 - Simplify to get the distance.

In this problem: The distance is \(3\).

\(D = \frac{9}{3} = 3\)

Final Answer

\(D = 3\)