Source: Homework Set 2, §12.5
Tags: planes, normal vectors, distance between planes, line of intersection, perpendicular planes, 3D geometry, multivariable calculus
Difficulty: Intermediate Prerequisites: Vectors, dot product, cross product, equations of lines in 3D (§12.1–12.4).
Planes are the fundamental flat surfaces in three-dimensional space, and working with them is one of the core skills in Calc III. This material builds directly on your knowledge of vectors, dot products and cross products. You will need these tools constantly in later sections on partial derivatives, surface integrals and flux. If you are comfortable finding normal vectors and writing the equation of a plane from a point and a normal, you are ready for this.
Planes in 3D are defined by a normal vector and a point. The distance between parallel planes comes from projecting any connecting vector onto the shared normal. Finding a plane through the intersection of two other planes uses a parameter to combine their equations, then applies additional constraints like perpendicularity.
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 sticking straight out of the surface.
Equation of a plane
The standard form is ax + by + cz = d, where ⟨a, b, c⟩ is the normal vector. In simple terms, this equation collects every point in 3D space that sits on one particular flat surface.
Parallel planes
Two planes are parallel when their normal vectors are scalar multiples of each other. They never intersect, just like parallel lines in 2D.
Distance between parallel planes
The shortest (perpendicular) distance between two parallel planes. Computed by picking any point on one plane and projecting the vector to the other plane onto the unit normal.
Line of intersection of two planes
When two non-parallel planes meet, they share a line. This line’s direction is given by the cross product of the two normal vectors.
Perpendicularity condition (planes)
Two planes are perpendicular when their normal vectors are perpendicular, i.e. when the dot product of their normals equals zero.
Two planes are parallel when they share the same normal vector (their equations have the same coefficients on x, y, z).
To find the distance between parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
Pick any point P₀ on one plane (set two variables to zero and solve for the third).
Use the point-to-plane distance formula with P₀ and the other plane.
Example: for 3x – 2y + 4z = 12 and 3x – 2y + 4z = 4, the normal is ⟨3, –2, 4⟩. A point on the first plane is (4, 0, 0). The distance is |3(4) – 2(0) + 4(0) – 4| / √(9 + 4 + 16) = |12 – 4| / √29 = 8/√29.
Shortcut: when both planes have the form ax + by + cz = d, the distance is simply |d₁ – d₂| / √(a² + b² + c²).
The intersection of two planes P₁: f(x,y,z) = 0 and P₂: g(x,y,z) = 0 is a line.
Any plane through that line can be written as f(x,y,z) + λ g(x,y,z) = 0 for some parameter λ.
To satisfy an additional constraint (e.g. perpendicularity to a third plane), find the normal of the combined plane in terms of λ, then apply the constraint to solve for λ.
Example: planes x – z = 1 and y + 2z = 3. The family is (x – z – 1) + λ(y + 2z – 3) = 0, which gives normal ⟨1, λ, –1 + 2λ⟩. For perpendicularity to x + y – 2z = 1 (normal ⟨1, 1, –2⟩), set the dot product to zero: 1 + λ + (–1 + 2λ)(–2) = 0, giving 1 + λ + 2 – 4λ = 0, so λ = 1. The plane is x + y + z = 4.
Two planes are perpendicular when their normals satisfy n₁ · n₂ = 0.
This condition appears constantly in exam problems that ask you to find a plane subject to a geometric constraint.
When asked to find a plane perpendicular to a given plane, you know one thing about the unknown normal: its dot product with the given normal is zero. Combine this with other information (e.g. the plane passes through a known line) to pin down the answer.
\text{Point-to-plane distance: } D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\text{Distance between parallel planes } ax+by+cz=d_1 \text{ and } ax+by+cz=d_2\text{: } D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\text{Perpendicularity condition: } \mathbf{n}_1 \cdot \mathbf{n}_2 = 0\text{Family of planes through intersection of } P_1 \text{ and } P_2\text{: } P_1 + \lambda P_2 = 0The distance between parallel planes turns up in engineering tolerances: the gap between two parallel surfaces of a machined part is exactly this calculation. Computer graphics uses plane equations constantly to determine which side of a surface a camera or light source sits on, which is how rendering engines decide what to draw and what to shade.
Students often forget to check that both plane equations have the same coefficients before using the shortcut |d₁ – d₂| / |n|. If one equation is a scaled version of the other (e.g. 6x – 4y + 8z = 8 vs 3x – 2y + 4z = 4), you must divide through first to match coefficients.
Students sometimes pick a point on the wrong plane. It does not matter which plane you pick the point from, as long as you plug it into the other plane’s equation.
When using the family-of-planes method f + λg = 0, students sometimes forget to move the constants to the right side before reading off the normal vector.
Confusing “perpendicular planes” with “perpendicular normals” in the wrong direction. Two planes are perpendicular when their normals are perpendicular (dot product = 0), not when the normals are parallel.
⚠️ The point-to-plane distance formula is a near-certain exam question. Know it cold.
⚠️ The family-of-planes technique (f + λg = 0) appears in problems that combine intersection with a constraint. Practice setting up the λ equation from a perpendicularity or containment condition.
⚠️ Make sure you can extract a normal vector from a plane equation by inspection: the coefficients on x, y, z give you the normal directly.
True or False: The normal vector to the plane 2x – 3y + z = 7 is ⟨2, –3, 1⟩. (True)
Fill in the blank: Two planes are parallel if and only if their normal vectors are ______ of each other. (scalar multiples)
True or False: To find the distance between parallel planes, you can pick any point on either plane. (True)
Fill in the blank: The dot product of two perpendicular vectors is ______. (zero)
True or False: The family of planes through the intersection of P₁ and P₂ is written as P₁ + λP₂ = 0 for all values of λ. (True)
Q: Find the distance between the planes 2x + y – 2z = 5 and 2x + y – 2z = –1.
A: The normals match, so use |d₁ – d₂| / |n| = |5 – (–1)| / √(4 + 1 + 4) = 6/3 = 2.
Q: The planes x + 2y + z = 3 and 2x – y + 3z = 1 intersect in a line. What is the direction vector of that line?
A: Take the cross product of the normals: ⟨1, 2, 1⟩ × ⟨2, –1, 3⟩ = ⟨(2)(3) – (1)(–1), (1)(2) – (1)(3), (1)(–1) – (2)(2)⟩ = ⟨7, –1, –5⟩.
Q: Find the equation of a plane through the intersection of x + y = 2 and y + z = 1 that passes through the origin.
A: The family is (x + y – 2) + λ(y + z – 1) = 0. Plugging in (0, 0, 0): (–2) + λ(–1) = 0, so λ = –2. The plane is x + y – 2 – 2(y + z – 1) = 0, i.e. x – y – 2z = 0.
Q: Are the planes 3x + y – z = 4 and x – y + 2z = 1 perpendicular?
A: Dot product of normals: (3)(1) + (1)(–1) + (–1)(2) = 3 – 1 – 2 = 0. Yes, they are perpendicular.
This connects directly to space curves (§13.1–13.2): you will need plane equations to determine when a tangent line to a curve is parallel to a given plane. It also feeds into later work on tangent planes to surfaces (§14.4) and surface integrals (§16.7), where the normal vector plays the same role at every point of a curved surface.
Related Terms / Search Tags
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