Difficulty: Intermediate | Prerequisites: MATH 215 §12.1–12.3 (vectors, dot product, equations of planes)
This material sits at the junction of plane geometry and distance formulas in three dimensions. You need a working knowledge of normal vectors, the general equation of a plane, and the point-to-plane distance formula before any of this will land. The payoff is large: these techniques reappear in optimisation, computer graphics, and every applied setting where you need to know how far something is from a flat surface. If you missed the earlier lectures on dot products and plane equations, revisit those first.
Parallel planes share a normal vector and differ only in their constant term. The distance between two parallel planes is found by normalising that constant difference by the magnitude of the normal. An equidistant surface from a point and a plane turns out to be a paraboloid, derived by setting the point-to-point distance equal to the point-to-plane distance and simplifying.
Normal vector
A vector perpendicular to a plane. For the plane ax + by + cz = d, the normal vector is n = ⟨a, b, c⟩. Think of it as the arrow that sticks straight out of the surface.
Parallel planes
Two planes whose normal vectors are scalar multiples of one another. In practice, this means their equations have proportional x, y, z coefficients. In simple terms, they face the same direction and never intersect.
Distance between parallel planes
The perpendicular gap between two planes that share a normal vector. Computed by taking the absolute difference of their constant terms and dividing by the magnitude of the normal. Think of it as the shortest ruler you could fit between the two flat surfaces.
Equidistant surface (locus)
The set of all points in space that are the same distance from two given geometric objects. When one object is a point and the other is a plane, the resulting surface is a paraboloid.
Paraboloid
A bowl-shaped quadric surface. In this context, it arises as the set of points equidistant from a fixed point (the focus) and a fixed plane (the directrix). In simple terms, it is the 3D version of a parabola, swept around its axis of symmetry.
A plane of the form y = k is parallel to y = 2 for every value of k. The normal vector is n = ⟨0, 1, 0⟩ and the distance between y = 2 and y = k is |k − 2|.
Setting |k − 2| = 4 gives k = 6 or k = −2.
The two planes parallel to y = 2 and 4 units away are y = 6 and y = −2.
For a general plane ax + by + cz = d, any parallel plane has the form ax + by + cz = d′ for some different constant d′.
The distance between the two parallel planes ax + by + cz = d₁ and ax + by + cz = d₂ is |d₁ − d₂| / √(a² + b² + c²).
Example: planes parallel to 2x − y + z = 0 have the form 2x − y + z = d. Setting |d − 0| / √(4 + 1 + 1) = 3 gives |d| = 3√6, so d = 3√6 or d = −3√6.
The set of all points P = (x, y, z) equidistant from a fixed point and a fixed plane defines a paraboloid.
To derive it, set the distance from P to the point equal to the distance from P to the plane, then square both sides and simplify.
Worked approach for the point (5, 0, 0) and the plane y = 2:
Distance from (x, y, z) to (5, 0, 0): √((x − 5)² + y² + z²)
Distance from (x, y, z) to the plane y = 2: |y − 2|
Setting these equal and squaring: (x − 5)² + y² + z² = (y − 2)²
Expanding: x² − 10x + 25 + y² + z² = y² − 4y + 4
Simplifying: x² − 10x + z² + 4y + 21 = 0
This is a paraboloid (the y² terms cancel, leaving a linear y term alongside quadratic x and z terms).
Recognising the surface type: when you square both sides and one squared variable cancels, the result is a paraboloid opening in the direction of that variable. The axis of the paraboloid is perpendicular to the original plane.
\text{Distance between parallel planes } ax + by + cz = d_1 \text{ and } ax + by + cz = d_2\colon \quad D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\text{Distance from point } (x_0, y_0, z_0) \text{ to plane } ax + by + cz = d\colon \quad D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\text{Equidistant surface (focus-directrix definition):} \quad \sqrt{(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2} = |\text{signed distance to plane}|Squaring and simplifying always removes one squared variable, producing a paraboloid.
The distance-between-planes formula is used in manufacturing tolerances, where two parallel surfaces must be held within a specified gap. Paraboloids appear in satellite dish design and radio telescope construction: the reflective surface is shaped so that all incoming parallel signals reflect to a single focal point, which is precisely the focus-directrix relationship you derive in this problem set.
Students often forget there are two planes at a given distance from a reference plane (one on each side). Always account for the ± in the absolute value.
When finding the distance between parallel planes, the planes must have identical coefficients on x, y, z before you apply the formula. If one is 2x − y + z = 0 and you write a parallel plane as 4x − 2y + 2z = d, you must divide through first, or adjust the formula accordingly.
The equidistant surface from a point and a plane is a paraboloid, not a plane. Students sometimes expect a plane because the equidistant surface between two points is a plane. The geometry changes when one object is a plane.
When squaring both sides of the distance equation, students sometimes forget to expand fully. Missed cross-terms lead to incorrect surface identification.
⚠️ The distance formula between parallel planes is a staple exam question. Know it cold and be ready to apply it in both directions: given the distance, find the plane; given two planes, find the distance.
⚠️ Identifying a surface from its equation (paraboloid, ellipsoid, hyperboloid, etc.) is explicitly flagged as a testable skill in this homework. Practice completing the square and recognising which squared terms survive.
⚠️ The focus-directrix definition of a paraboloid is the 3D extension of the 2D parabola definition. If the exam asks you to derive an equidistant surface from a point and a plane, this is the method.
True or false: every plane parallel to 3x + 2y − z = 7 has a normal vector of ⟨3, 2, −1⟩.
Fill in the blank: the distance between the planes x + y + z = 1 and x + y + z = 4 is ____.
True or false: the set of points equidistant from a point and a plane always forms a plane.
Fill in the blank: for the plane y = 2, the two planes 4 units away are y = ____ and y = ____.
True or false: when deriving the equidistant surface from (5, 0, 0) and y = 2, the y² terms cancel.
Answers: 1. True. 2. 3/√3 = √3. 3. False (it is a paraboloid). 4. 6, −2. 5. True.
Q: Find all planes parallel to 2x − y + z = 0 that are 3 units away from it.
A: Parallel planes have the form 2x − y + z = d. The distance is |d|/√(4 + 1 + 1) = |d|/√6 = 3, so d = ±3√6. The planes are 2x − y + z = 3√6 and 2x − y + z = −3√6.
Q: Find the equation of the surface consisting of all points equidistant from the point (5, 0, 0) and the plane y = 2. What type of surface is it?
A: Set √((x − 5)² + y² + z²) = |y − 2|. Square both sides: (x − 5)² + y² + z² = (y − 2)². Expand and simplify: x² − 10x + 25 + z² + 4y − 4 = 0, i.e. x² − 10x + z² + 4y + 21 = 0. This is a paraboloid (opening in the −y direction).
Q: The planes y = k and y = 2 are 7 units apart. Find k.
A: |k − 2| = 7, so k = 9 or k = −5.
Q: Why does squaring both sides of the equidistant condition remove the y² term in problem 2?
A: The left side contributes y² from the point-to-point distance. The right side contributes y² from (y − 2)². When you move everything to one side, the y² terms cancel exactly, leaving a linear y term. This is the signature of a paraboloid rather than an ellipsoid or hyperboloid.
This connects to §12.5 (quadric surfaces) because the equidistant surface turns out to be a paraboloid, one of the standard quadric surfaces you classify by completing the square. It also connects to §12.1 (three-dimensional coordinate systems), since the entire framework rests on distance formulas in R³. In later chapters on optimisation and Lagrange multipliers, the distance-to-a-plane calculation reappears as a constraint.
Tags: parallel planes, distance between planes, normal vector, equidistant surface, paraboloid, focus-directrix, MATH 215, Calculus III, §12.4, §12.5, quadric surfaces, point-to-plane distance, completing the square, University of Michigan