Distance From Point to Plane Calculator
Free Distance point plane Calculator for coordinate geometry. Enter values to get step-by-step solutions with formulas and graphs.
Formula
D = |ax0 + by0 + cz0 + d| / sqrt(a^2 + b^2 + c^2)
Where (x0, y0, z0) is the point, ax + by + cz + d = 0 is the plane equation, and (a, b, c) is the normal vector. The absolute value gives unsigned distance; without it, you get the signed distance indicating which side of the plane the point is on.
Worked Examples
Example 1: Distance from Point to Plane
Problem: Find the distance from point P(1, 2, 3) to the plane 2x + 3y + 6z - 14 = 0.
Solution: D = |2(1) + 3(2) + 6(3) + (-14)| / sqrt(4 + 9 + 36)\n= |2 + 6 + 18 - 14| / sqrt(49)\n= |12| / 7\n= 12/7 = 1.7143\nSigned distance is positive, so the point is on the normal side.\nFoot of perpendicular: t = -12/49\nFoot = (1 - 2*12/49, 2 - 3*12/49, 3 - 6*12/49) = (0.5102, 1.2653, 1.5306)
Result: Distance: 1.7143 | Foot: (0.5102, 1.2653, 1.5306) | Positive side
Example 2: Point on the Plane
Problem: Verify that point (7, 0, 0) lies on the plane 2x + 3y + 6z - 14 = 0.
Solution: D = |2(7) + 3(0) + 6(0) - 14| / sqrt(4 + 9 + 36)\n= |14 - 14| / 7\n= 0/7 = 0\nThe distance is 0, confirming the point lies on the plane.\nThe x-intercept of this plane is -(-14)/2 = 7, which matches.
Result: Distance: 0 | Point lies exactly on the plane
Frequently Asked Questions
What is the distance from a point to a plane?
The distance from a point to a plane is the shortest distance between a given point in 3D space and a flat plane surface. This shortest distance is always measured along the line perpendicular (normal) to the plane passing through the given point. Unlike the distance between two points which is simply a straight line, the point-to-plane distance requires understanding the orientation of the plane. The formula uses the equation of the plane in general form (ax + by + cz + d = 0) and produces the length of the perpendicular from the point to the plane. This concept is fundamental in computational geometry, computer graphics, and many engineering applications.
What formula calculates the distance from a point to a plane?
The distance from a point P(x0, y0, z0) to the plane ax + by + cz + d = 0 is given by the formula: D = |ax0 + by0 + cz0 + d| / sqrt(a^2 + b^2 + c^2). The numerator substitutes the point coordinates into the plane equation and takes the absolute value. The denominator is the magnitude of the normal vector (a, b, c) of the plane, which normalizes the result. Without the absolute value, you get the signed distance, which tells you on which side of the plane the point lies. This formula is derived by projecting the vector from any point on the plane to the given point onto the unit normal vector of the plane.
What is the signed distance and what does it tell us?
The signed distance is the distance from a point to a plane without taking the absolute value: D_signed = (ax0 + by0 + cz0 + d) / sqrt(a^2 + b^2 + c^2). When the signed distance is positive, the point lies on the same side of the plane as the normal vector (a, b, c). When negative, the point lies on the opposite side. When zero, the point lies exactly on the plane. The signed distance is particularly useful in computer graphics for determining which side of a surface an object is on, in collision detection for determining penetration depth, and in classification algorithms like support vector machines where the decision boundary is a hyperplane.
How do you find the foot of the perpendicular from a point to a plane?
The foot of the perpendicular is the point on the plane that is closest to the given point. To find it, move from the given point P along the direction of the negative normal vector by the signed distance. The formulas are: foot_x = x0 - a*t, foot_y = y0 - b*t, foot_z = z0 - c*t, where t = (ax0 + by0 + cz0 + d) / (a^2 + b^2 + c^2). This t parameter represents how far along the normal direction you need to travel to reach the plane. The foot point can be verified by checking that it satisfies the plane equation and that the vector from the foot to the original point is parallel to the normal vector. This calculation is essential in projection operations and closest-point queries.
What is the normal vector of a plane and why is it important?
The normal vector of a plane ax + by + cz + d = 0 is the vector n = (a, b, c), which is perpendicular to every vector lying in the plane. It defines the orientation of the plane in 3D space and is directly read from the coefficients of x, y, and z in the plane equation. The normal vector is important because it determines the direction of the perpendicular distance, allows you to classify which side of the plane a point is on, and is used extensively in computer graphics for lighting calculations (how light reflects off surfaces). Two planes are parallel if and only if their normal vectors are parallel (scalar multiples of each other), and perpendicular if their normal vectors have a dot product of zero.
What are axis intercepts of a plane and how are they calculated?
The axis intercepts of a plane are the points where the plane crosses the x, y, and z coordinate axes. The x-intercept is found by setting y = 0 and z = 0 in the plane equation, giving x = -d/a (if a is not zero). Similarly, the y-intercept is y = -d/b and the z-intercept is z = -d/c. These intercepts are useful for visualizing and sketching the plane in 3D space, as they provide three concrete points through which the plane passes. If a coefficient is zero, the plane is parallel to that axis and has no intercept on it. The intercept form of the plane equation is x/p + y/q + z/r = 1, where p, q, r are the x, y, and z intercepts respectively.