Difficulty: Intermediate | Prerequisites: Vectors, dot product, equations of lines in 3D (Chapter 12.1–12.3)
Planes are the fundamental flat surfaces of three-dimensional space, and learning to write their equations is the gateway to nearly everything else in Calculus III. You need planes to describe tangent surfaces, constraint boundaries, and the geometry of optimisation problems. If you are comfortable with the dot product and the idea that a normal vector pins down a plane's orientation, you are ready. If "normal vector" still feels vague, revisit Section 12.3 first.
A plane in 3D is fully determined by a point it passes through and a normal vector perpendicular to it. Parallel planes share the same normal vector and differ only by a constant. The point-to-plane distance formula lets you measure how far apart two parallel planes sit, and combining plane inequalities carves out solid regions (tetrahedra, octahedra) you can sketch by finding intercepts.
Normal vector
A vector perpendicular to every vector that lies in a given plane. It defines the plane's orientation. Think of it as an arrow sticking straight out of a tabletop.
Scalar equation of a plane
The standard form ax + by + cz = d, where (a, b, c) is the normal vector and d is a constant determined by a specific point on the plane. In simple terms, this is the "address" of a flat surface in 3D.
Parallel planes
Two planes whose normal vectors are scalar multiples of each other (they point in the same or exactly opposite directions). Parallel planes never intersect. Think of two floors in a building: same tilt, different height.
Distance between parallel planes
The perpendicular (shortest) distance separating two parallel planes. Computed by picking any point on one plane and measuring its distance to the other using the point-to-plane formula.
Point-to-plane distance
The shortest distance from a point P = (x0, y0, z0) to a plane ax + by + cz = d, given by |ax0 + by0 + cz0 - d| / sqrt(a^2 + b^2 + c^2). In simple terms, drop a perpendicular from the point to the plane and measure its length.
Half-space
The region on one side of a plane. An inequality such as x + 2y + 3z <= 4 selects exactly one half-space. Think of it as "everything on this side of the wall."
First octant
The region of 3D space where x >= 0, y >= 0, and z >= 0. It is the 3D equivalent of the first quadrant in 2D.
A plane through point P0 = (x0, y0, z0) with normal vector n = (a, b, c) has equation a(x - x0) + b(y - y0) + c(z - z0) = 0.
Expanding gives the scalar form ax + by + cz = d, where d = ax0 + by0 + cz0.
Special cases: the plane y = 2 has normal vector (0, 1, 0) and passes through any point with y-coordinate 2. The plane 2x - y + z = 0 has normal vector (2, -1, 1) and passes through the origin.
Two planes are parallel when they share the same normal vector (a, b, c). Their equations differ only in the constant on the right side: ax + by + cz = d1 and ax + by + cz = d2.
To find all planes parallel to ax + by + cz = d and at distance k from it, use the distance formula |d1 - d2| / sqrt(a^2 + b^2 + c^2) = k. Solve for d2. There will always be two solutions (one on each side).
Example: planes parallel to y = 2 and 4 units away. The normal is (0, 1, 0), so |d2 - 2| / 1 = 4 gives d2 = 6 or d2 = -2. The planes are y = 6 and y = -2.
Example: planes parallel to 2x - y + z = 0 and 3 units away. |d2 - 0| / sqrt(4 + 1 + 1) = 3 gives |d2| = 3sqrt(6). The planes are 2x - y + z = 3sqrt(6) and 2x - y + z = -3sqrt(6).
The distance from point (x0, y0, z0) to the plane ax + by + cz = d is |ax0 + by0 + cz0 - d| / sqrt(a^2 + b^2 + c^2).
This formula comes from projecting the vector from any point on the plane to (x0, y0, z0) onto the unit normal.
The numerator is the absolute value of plugging the point into the left side of the equation minus the right side. The denominator normalises the normal vector.
A single plane equation divides space into two half-spaces. An inequality like ax + by + cz <= d selects one.
Combining with x >= 0, y >= 0, z >= 0 restricts to the first octant.
Example: the region x + 2y + 3z <= 4 in the first octant is a tetrahedron. Find the intercepts: (4, 0, 0), (0, 2, 0), (0, 0, 4/3). Connect them to sketch.
The region |x| + 2|y| + 3|z| <= 4 is symmetric across all coordinate planes. In the first octant it reduces to x + 2y + 3z <= 4 (a tetrahedron). Reflecting through the coordinate planes produces an octahedron-like solid with 8 triangular faces.
Find the intercepts on each axis by setting the other two variables to zero.
Plot those intercept points on the axes.
Connect them with straight edges to outline the triangular face.
For absolute-value inequalities, sketch the first-octant piece and reflect it through each coordinate plane.
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0Point-normal form of a plane through (x0, y0, z0) with normal (a, b, c).
ax + by + cz = dScalar (general) equation of a plane.
D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}Distance from point (x0, y0, z0) to the plane ax + by + cz = d.
\text{Distance between parallel planes} = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}Distance between ax + by + cz = d1 and ax + by + cz = d2.
Plane equations describe any flat boundary in engineering and physics: the surface of a wall, a floor, a wing cross-section reference plane. When an architect says "the ceiling is 4 metres above the floor," that is two parallel planes with a known distance between them. The distance formula also appears in computer graphics, where the engine needs to know how far a point (say, a light source) is from a surface to compute shading.
Students often think that two planes with the same normal vector are the same plane. They are not; they are parallel, and they coincide only if their constants also match.
Students sometimes forget the absolute value in the distance formula and get a negative distance. Distance is always non-negative.
When finding planes at a given distance, students frequently find only one plane instead of two. There is always a plane on each side.
Students confuse the normal vector with a direction vector that lies in the plane. The normal vector is perpendicular to the plane, not parallel to it.
⚠️ The point-to-plane distance formula is a near-certain exam question. Memorise it and practise plugging in quickly.
⚠️ "Find all planes parallel to ... and at distance ..." is a classic problem type. Always expect two answers.
⚠️ Sketching planar regions in the first octant (finding intercepts, drawing the triangle) is explicitly flagged as exam-relevant in this homework set.
⚠️ The absolute-value version (|x| + 2|y| + 3|z| <= 4) tests whether you can handle symmetry across coordinate planes. Expect to describe the full solid, not just the first octant piece.
True or false: the planes 3x + y - z = 5 and 6x + 2y - 2z = 7 are parallel. (True: the second normal is 2 times the first.)
The plane y = 2 has normal vector ______. (Answer: (0, 1, 0))
True or false: the distance from the origin to the plane x + y + z = 3 is 3. (False: it is 3/sqrt(3) = sqrt(3).)
How many planes are parallel to a given plane and exactly 5 units away? (Answer: exactly 2.)
The intercepts of x + 2y + 3z = 6 on the coordinate axes are ______, ______, and ______. (Answer: (6,0,0), (0,3,0), (0,0,2).)
Q: Find the equations of all planes parallel to 2x - y + z = 0 that are 3 units away from it.
A: The normal vector is (2, -1, 1) with magnitude sqrt(6). Any parallel plane has the form 2x - y + z = d. The distance formula gives |d|/sqrt(6) = 3, so d = 3sqrt(6) or d = -3sqrt(6). The planes are 2x - y + z = 3sqrt(6) and 2x - y + z = -3sqrt(6).
Q: Describe the region in the first octant where x + 2y + 3z <= 4. What shape is it?
A: It is a tetrahedron (a solid with four triangular faces). The vertices are the origin (0,0,0) and the three axis intercepts: (4,0,0), (0,2,0), and (0,0,4/3). The flat face on the plane x + 2y + 3z = 4 is a triangle; the other three faces lie on the coordinate planes.
Q: A point P = (1, 2, 3) and the plane 3x + 4y - 12z = 7 are given. Find the distance from P to the plane.
A: Plug in: |3(1) + 4(2) - 12(3) - 7| / sqrt(9 + 16 + 144) = |3 + 8 - 36 - 7| / sqrt(169) = |-32| / 13 = 32/13.
Q: Describe the full solid defined by |x| + 2|y| + 3|z| <= 4.
A: In the first octant it is the tetrahedron x + 2y + 3z <= 4 with x, y, z >= 0. Reflecting through each coordinate plane produces a solid with 8 triangular faces, symmetric about all three coordinate planes. It is a stretched octahedron (each axis intercept is at a different distance from the origin: 4, 2, and 4/3).
Plane equations feed directly into Section 12.5 (lines and planes in space) and into the cross product, which gives you a quick way to find a normal vector from two direction vectors. Later, in Chapter 14, tangent planes to surfaces generalise the idea of a tangent line, and the normal vector reappears as the gradient. The distance formula also underpins the definition of a parabolic surface as an equidistant locus (see the Quadric Surfaces notes).
Tags: plane equation, normal vector, scalar equation of a plane, parallel planes, distance between planes, point-to-plane distance, half-space, first octant, tetrahedron, octahedron, absolute value inequality, 3D regions, MATH 215, Calculus III, Stewart Multivariable Calculus, Section 12.4, Section 12.5