Distance from Point to Plane Calculator

The shortest distance from a point to a plane is the perpendicular (or orthogonal) distance — the length of the line segment dropped from the point at a right angle to the plane. No other path between the point and the plane is shorter than this perpendicular.

Given a plane Ax + By + Cz + D = 0 and a point M(a, b, c), the perpendicular distance L is: L = |A·a + B·b + C·c + D| / √(A² + B² + C²). The numerator measures how far the point deviates from the plane, and the denominator normalises by the magnitude of the normal vector.

Enter the four plane equation coefficients A, B, C, and D (from the standard form Ax + By + Cz + D = 0), then enter the three point coordinates a, b, and c. The calculator instantly computes the perpendicular distance and displays all intermediate components.

The xy-plane has the equation z = 0, which in standard form is 0x + 0y + 1z + 0 = 0 (A=0, B=0, C=1, D=0). The distance from any point (a, b, c) to the xy-plane is simply |c| — the absolute value of its z-coordinate.

Write x + y = 0 as 1x + 1y + 0z + 0 = 0 (A=1, B=1, C=0, D=0). The numerator is |1·1 + 1·1 + 0·1 + 0| = 2, and the denominator is √(1² + 1² + 0²) = √2. So the distance is 2/√2 = √2 ≈ 1.4142.

To find the distance from the origin (0, 0, 0) to the plane Ax + By + Cz + D = 0, substitute a=0, b=0, c=0 into the formula. This simplifies to |D| / √(A² + B² + C²), so the distance equals the absolute value of D divided by the magnitude of the normal vector.

A distance of zero means the point lies exactly on the plane. In other words, the point's coordinates satisfy the plane equation Ax + By + Cz + D = 0, making the numerator |Aa + Bb + Cc + D| equal to zero.

No. If A, B, and C are all zero, the denominator √(A² + B² + C²) becomes zero, making the distance undefined. A valid plane must have at least one non-zero coefficient among A, B, and C; otherwise, it does not represent a real plane in 3D space.