Find the shortest distance between a point in 3D space and a plane, using the standard formula D = |ax0 + by0 + cz0 + d| / sqrt(a2 + b2 + c2). Enter your point coordinates and the plane equation coefficients below to see the perpendicular distance and full working.
In 3D geometry, a plane can be described by the equation ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane (a vector perpendicular to its surface) and d is a constant. Given a point P at (x0, y0, z0), the shortest distance from that point to the plane is always measured along the perpendicular, and is given by the formula:
D = |a·x0 + b·y0 + c·z0 + d| / √(a² + b² + c²)
The numerator substitutes the point's coordinates into the left-hand side of the plane equation and takes the absolute value, since distance cannot be negative. The denominator is the magnitude (length) of the plane's normal vector. Dividing the two gives the true perpendicular distance, regardless of which side of the plane the point sits on.
Take the point P(1, 2, 3) and the plane x + y + z - 6 = 0, so a = 1, b = 1, c = 1, d = -6.
| Step | Calculation | Result |
|---|---|---|
| Substitute point into plane equation | (1)(1) + (1)(2) + (1)(3) + (-6) | 0 |
| Take absolute value | |0| | 0 |
| Normal vector magnitude | √(1² + 1² + 1²) | √3 ≈ 1.7321 |
| Distance | 0 / 1.7321 | 0 |
The result is 0, which makes sense because (1, 2, 3) satisfies x + y + z - 6 = 0 exactly (1 + 2 + 3 - 6 = 0), meaning the point lies directly on the plane.
If you only know three points that lie on the plane rather than the equation itself, you can derive a, b, c and d first. Form two vectors by subtracting one point from the other two, then take their cross product to get the normal vector (a, b, c). Substitute any one of the three points into ax + by + cz + d = 0 and solve for d. Once you have all four coefficients, use this calculator to find the distance from any other point to that plane.
What is the formula for distance from a point to a plane? D = |ax0 + by0 + cz0 + d| / sqrt(a² + b² + c²), where (x0, y0, z0) is the point and ax + by + cz + d = 0 is the plane equation.
What if the point lies on the plane? The distance is 0, because substituting the point's coordinates into the plane equation gives exactly zero.
How do I find a, b, c and d from three points on the plane? Use the cross product of two vectors formed from the three points to get the normal vector (a, b, c), then substitute one point into the plane equation to solve for d.
Sources: Standard analytic geometry formula for the perpendicular distance from a point to a plane, as presented in university-level vector calculus and coordinate geometry references.
This calculator provides a mathematical result based on the coordinates and plane equation you enter. Double-check your plane equation is in the standard form ax + by + cz + d = 0 before entering coefficients.
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