Find the perpendicular distance from a point to a plane in 3D space.
Last updated: June 2026 | By Patchworkr Team
Enter the plane coefficients and point coordinates as real numbers.
For a plane written as ax + by + cz + d = 0, the perpendicular distance to a point is:
The denominator is the length of the plane normal vector, which must not be zero.
1. Enter the plane coefficients a, b, c, and d.
2. Enter the point coordinates x0, y0, and z0.
3. Read the live perpendicular distance.
For plane 2x - 3y + 6z - 12 = 0 and point (1, 2, 3):
numerator = |2*1 + (-3)*2 + 6*3 - 12| = |2 - 6 + 18 - 12| = 2
denominator = sqrt(2^2 + (-3)^2 + 6^2) = 7
distance = 2/7
No. That would not define a valid plane.
No. The absolute value makes the result non-negative.
Yes, but only through the formula ax0 + by0 + cz0 + d.
Yes. Then the distance is 0.
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