Distance from Point to Plane Calculator
Calculate the perpendicular distance from a 3D point to a plane given by standard equation, normal vector, or three points with step-by-step steps.
Understanding Distance from a Point to a Plane
In three-dimensional analytic geometry, the distance from a point to a plane is the length of the shortest line segment connecting the point to the plane. Because the shortest path is always perpendicular (orthogonal) to the plane surface, this is also called the perpendicular distance.
Finding this distance is essential in linear algebra, 3D computer graphics, physics simulations, robotics path planning, and architectural modeling.
The Standard Distance Formula
Consider a plane defined by the general linear Cartesian equation:
$$Ax + By + Cz + D = 0$$where $\vec{n} = (A, B, C)$ is the normal vector perpendicular to the plane and $D$ is the constant offset. For any point $P(x_0, y_0, z_0)$ in 3D space, the perpendicular distance $d$ is given by:
$$d = \frac{|A x_0 + B y_0 + C z_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$Vector Formulation
If the plane passes through a known point $P_p(x_p, y_p, z_p)$ with a normal vector $\vec{n} = (A, B, C)$, the vector from $P_p$ to the target point $P$ is $\vec{v} = P - P_p = (x_0 - x_p, y_0 - y_p, z_0 - z_p)$.
The distance equals the absolute scalar projection of $\vec{v}$ onto the unit normal vector $\hat{n}$:
$$d = \frac{|\vec{v} \cdot \vec{n}|}{\|\vec{n}\|} = \frac{|(x_0 - x_p)A + (y_0 - y_p)B + (z_0 - z_p)C|}{\sqrt{A^2 + B^2 + C^2}}$$Plane Defined by Three Points
When a plane is defined by three non-collinear points $P_1, P_2, P_3$:
- Form two direction vectors lying in the plane: $\vec{v_1} = P_2 - P_1$ and $\vec{v_2} = P_3 - P_1$.
- Compute the normal vector via cross product: $\vec{n} = \vec{v_1} \times \vec{v_2} = (A, B, C)$.
- Determine $D = -(A x_1 + B y_1 + C z_1)$.
- Apply the standard perpendicular distance formula.
Finding the Orthogonal Projection Point ($P'$)
The closest point on the plane to $P(x_0, y_0, z_0)$ is the orthogonal projection $P'$. It can be calculated by moving along the normal vector direction:
$$P' = P - \left(\frac{A x_0 + B y_0 + C z_0 + D}{A^2 + B^2 + C^2}\right)\vec{n}$$Related 3D Geometry and Distance Tools
Explore other spatial geometry calculators in our collection: 3D Distance Calculator, 2D Distance Calculator, Midpoint Calculator, and Dot Product Calculator.
Frequently Asked Questions
What does a distance of zero mean?
If $d = 0$, the numerator $|Ax_0 + By_0 + Cz_0 + D| = 0$, which proves that the point $P(x_0, y_0, z_0)$ lies directly on the plane surface.
What is the difference between signed distance and absolute distance?
Absolute distance ($d \ge 0$) measures physical separation regardless of orientation. Signed distance keeps the sign of the numerator: positive values mean the point lies on the side the normal vector points toward, while negative values indicate the opposite side.
Why can the coefficients A, B, and C not all be zero?
If $A = B = C = 0$, the normal vector has zero magnitude ($\|\vec{n}\| = 0$), which does not define a geometric plane in 3D space, and division by zero is undefined.
How do you calculate distance from the origin (0, 0, 0) to a plane?
Substituting $(0, 0, 0)$ simplifies the numerator to $|D|$, resulting in the streamlined origin distance formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.