Plane — Basic Equations

A cricket pitch is flat. The ground crew has levelled it so that every point on the surface lies in the same plane. If you placed a long, perfectly straight ruler anywhere on the pitch, it would touch the surface at every point along its length — no gaps, no bumps.

Now imagine the pitch tilted. One end is raised, the other lowered, like a ramp. It is still flat — it is still a plane — but its orientation in space has changed. The pitch and the level ground beneath it are two different planes, tilted at an angle to each other.

How do you describe such a surface mathematically? A line in 3D needs a point and a direction. A plane needs more — it is two-dimensional, so it has infinitely many directions within it. Specifying which directions lie in the plane is cumbersome. A cleaner approach is to specify the one direction that is perpendicular to the plane. That single perpendicular direction — called the normal — pins down the orientation of the plane completely.

The normal to a plane

A normal to a plane is any vector that is perpendicular to every vector lying in the plane.

If you know one normal vector \vec{n}, every scalar multiple k\vec{n} is also a normal. The direction matters; the magnitude does not. The unit normal \hat{n} = \frac{\vec{n}}{|\vec{n}|} is the normal with magnitude 1.

A tilted cricket pitch and the level ground have different normals. The level ground has a vertical normal (pointing straight up). The tilted pitch has a normal that leans sideways. The angle between the two normals is the angle between the two planes — but that is a story for the article on Plane — Distances and Angles.

Vector equation of a plane (normal form)

Here is the key idea: a point R with position vector \vec{r} lies on the plane if and only if the vector from a known point P_0 on the plane to R is perpendicular to the normal \vec{n}.

Let \vec{a} be the position vector of a known point P_0 on the plane, and let \vec{n} be the normal. The vector \vec{P_0 R} = \vec{r} - \vec{a} lies in the plane. For R to be on the plane, this vector must be perpendicular to \vec{n}:

Expanding:

The right side \vec{a} \cdot \vec{n} is a fixed number — call it d. So:

Reading the equation. The equation \vec{r} \cdot \vec{n} = d says: take any point \vec{r} in space, dot it with the fixed normal \vec{n}, and check whether you get d. If yes, the point is on the plane. If not, it is off the plane. The number d is a "threshold" that selects exactly one flat surface from the family of all planes with normal \vec{n}.

The geometric meaning of d. When \vec{n} is a unit vector, d is the signed distance from the origin to the plane, measured along the normal. If d > 0, the origin is on the opposite side of the plane from the direction of \vec{n}. If d = 0, the plane passes through the origin.

The unit normal form

When you write the equation with a unit normal \hat{n}:

the number p is the perpendicular distance from the origin to the plane. This is the cleanest form for computing distances and is sometimes called the normal form or Hesse normal form of the plane equation.

Cartesian equation

Write \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} and \vec{n} = a\hat{i} + b\hat{j} + c\hat{k}. The vector equation \vec{r} \cdot \vec{n} = d becomes:

This is the Cartesian equation of the plane. The coefficients a, b, c are the components of the normal vector.

Going the other way. Given the equation 3x - 2y + 5z = 7, you can read off the normal immediately: \vec{n} = 3\hat{i} - 2\hat{j} + 5\hat{k}.

How many planes are there with a given normal?

Infinitely many. The normal \vec{n} fixes the tilt of the plane, but not its position. Changing d slides the plane along the normal direction without rotating it. All planes with normal \vec{n} are parallel to each other, stacked at different heights.

The equation ax + by + cz = 0 always gives the plane through the origin with normal (a, b, c).

Plane through three points

Three non-collinear points P_1 = (x_1, y_1, z_1), P_2 = (x_2, y_2, z_2), P_3 = (x_3, y_3, z_3) determine a unique plane — just as three legs determine the position of a table on an uneven floor.

The method. Two vectors lying in the plane are \vec{P_1 P_2} and \vec{P_1 P_3}. The normal to the plane is their cross product:

Once you have \vec{n}, the equation of the plane is \vec{n} \cdot (\vec{r} - \vec{a}_1) = 0, where \vec{a}_1 is the position vector of P_1.

The derivation. Any point R in the plane can be reached from P_1 by a linear combination of \vec{P_1 P_2} and \vec{P_1 P_3}. The normal \vec{n} = \vec{P_1 P_2} \times \vec{P_1 P_3} is perpendicular to both of these vectors, and hence perpendicular to every vector in the plane. The condition for R to lie in the plane is (\vec{r} - \vec{a}_1) \cdot \vec{n} = 0 — the vector from P_1 to R must be perpendicular to the normal.

The determinant form

Expanding the condition (\vec{r} - \vec{a}_1) \cdot (\vec{P_1 P_2} \times \vec{P_1 P_3}) = 0 gives a scalar triple product. This can be written as a determinant:

This is a useful form for JEE problems because it directly gives the equation of the plane without requiring you to compute the cross product separately.

Why does it work? The rows of the determinant are three vectors: \vec{P_1 R}, \vec{P_1 P_2}, \vec{P_1 P_3}. The determinant is their scalar triple product, which is zero precisely when the three vectors are coplanar — meaning R lies in the plane of P_1, P_2, P_3.

Intercept form

When a plane cuts all three coordinate axes at non-zero points, the equation takes a particularly clean shape.

Let the plane cut the x-axis at (a, 0, 0), the y-axis at (0, b, 0), and the z-axis at (0, 0, c). These three points determine the plane.

Substitute into the general equation Ax + By + Cz = D:

• At (a, 0, 0): Aa = D, so A = D/a
• At (0, b, 0): Bb = D, so B = D/b
• At (0, 0, c): Cc = D, so C = D/c

Substituting back and dividing by D:

Reading the equation. Each fraction is the coordinate divided by the intercept on that axis. Setting y = z = 0 gives x = a; setting x = z = 0 gives y = b; setting x = y = 0 gives z = c. The plane "uses up" exactly one unit of each fraction to reach the value 1.

This form is the 3D analogue of the intercept form \frac{x}{a} + \frac{y}{b} = 1 for a line in 2D, which you have seen in Straight Line Forms.

When does the intercept form fail?

The intercept form requires all three intercepts to be non-zero. If the plane passes through the origin, or if it is parallel to one of the axes (meaning it never cuts that axis), the intercept form does not apply. In those cases, use the general Cartesian form ax + by + cz = d instead.

For example, the plane z = 5 is parallel to the xy-plane. It has a z-intercept of 5, but it never touches the x-axis or the y-axis, so the intercept form is not available.

Converting between forms

Here is a quick summary of the four forms and how to move between them.

From intercept to Cartesian. Multiply \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 through by abc to get bcx + acy + abz = abc. The normal is (bc, ac, ab).

From Cartesian to vector. Write \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} and the equation becomes \vec{r} \cdot \vec{n} = d.

From three points to Cartesian. Compute \vec{n} = \vec{P_1 P_2} \times \vec{P_1 P_3}, read off (a, b, c), and find d = ax_1 + by_1 + cz_1.

From Cartesian to intercept. Divide both sides by d (assuming d \neq 0): \frac{x}{d/a} + \frac{y}{d/b} + \frac{z}{d/c} = 1. The intercepts are d/a, d/b, d/c.

Two worked examples

Common confusions

• "A plane in 3D is defined by two equations." No — a plane is defined by a single equation ax + by + cz = d. A single equation in three unknowns removes one degree of freedom, leaving a two-dimensional surface. Two equations would give the intersection of two planes, which is a line (one-dimensional).

• "The normal to a plane is unique." The direction of the normal is unique (up to sign), but the vector is not — \vec{n} and 2\vec{n} are both normals. Any non-zero scalar multiple of the normal is also a normal. What matters is the direction, not the magnitude.

• "Three points always determine a plane." Only if the three points are non-collinear. Three collinear points lie on infinitely many planes (any plane containing the line through them will do). The cross-product test \vec{P_1 P_2} \times \vec{P_1 P_3} \neq \vec{0} is the check for non-collinearity.

• "The intercept form works for every plane." It does not. Planes through the origin (d = 0) and planes parallel to a coordinate axis (with an infinite intercept) cannot be written in intercept form. The general Cartesian form is always available; the intercept form is a convenient special case.

• "d in ax + by + cz = d is the distance from the origin." Only when (a, b, c) is a unit vector. In general, the distance from the origin is \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}. You have to divide by |\vec{n}| first.

Going deeper

Parametric equation of a plane

The normal form \vec{r} \cdot \vec{n} = d describes the plane as the set of points satisfying a single constraint. There is a dual description: the parametric form.

If P_0 is a point on the plane and \vec{v}_1, \vec{v}_2 are two non-parallel vectors lying in the plane, then every point on the plane is:

This is the plane analogue of the parametric equation of a line \vec{r} = \vec{a} + \lambda\vec{b}. A line has one parameter (\lambda); a plane has two (s and t).

You can always convert from parametric to normal form by computing \vec{n} = \vec{v}_1 \times \vec{v}_2 and forming \vec{r} \cdot \vec{n} = \vec{a}_0 \cdot \vec{n}.

Why "three points determine a plane"

In the language of linear algebra, a plane through the origin is a 2-dimensional subspace of \mathbb{R}^3. Its normal is the 1-dimensional orthogonal complement. To specify a 2-dimensional subspace, you need two linearly independent vectors — and two vectors are determined by three points (two edges of the triangle they form).

A plane not through the origin is an affine subspace — a translate of a 2-dimensional subspace. The third point pins down the translation.

The formula with the determinant — the scalar triple product equalling zero — is saying: the three vectors from P_1 to R, from P_1 to P_2, and from P_1 to P_3 must be linearly dependent (coplanar). The determinant is the test for linear dependence.

The general equation and its degrees of freedom

The equation ax + by + cz = d has four constants, but not four degrees of freedom. Multiplying all of a, b, c, d by the same non-zero constant gives the same plane (the same set of points). So the plane has 4 - 1 = 3 essential parameters.

This matches the geometry: to fix a plane, you need three independent pieces of information. Three non-collinear points (nine coordinates, minus six constraints from the three collinearity-avoiding conditions) give exactly three degrees of freedom. Three intercepts give exactly three numbers. A unit normal (two parameters, since it has unit length) plus a perpendicular distance (one parameter) give three.

Normal form and the Hesse normal form

The German mathematician Ludwig Hesse introduced the convention of writing the plane equation with a unit normal: \vec{r} \cdot \hat{n} = p, where p \geq 0 is the perpendicular distance from the origin. This form is especially useful for computing the distance from any point to the plane: the signed distance from a point Q with position vector \vec{q} to the plane is \vec{q} \cdot \hat{n} - p. You will use this extensively in the article on Plane — Distances and Angles.

Connection to lines in 3D

A line in 3D is the intersection of two planes. If you have two plane equations a_1 x + b_1 y + c_1 z = d_1 and a_2 x + b_2 y + c_2 z = d_2, the set of points satisfying both simultaneously is a line — provided the two normals are not parallel. The direction of this line is \vec{n}_1 \times \vec{n}_2, perpendicular to both normals. This gives yet another way to write the equation of a line in 3D: as the intersection of two planes.

Where this leads next

You now have the four standard ways to write the equation of a plane. The next questions are about measurement — distances from points to planes, angles between planes, and the angle between a line and a plane.

• Plane — Distances and Angles — perpendicular distance from a point to a plane, angle between two planes, angle between a line and a plane.
• Straight Line in 3D — Equations — the three forms of a line in 3D, including the two-plane intersection form.
• Dot Product — the operation that defines the normal condition \vec{r} \cdot \vec{n} = d.
• Cross Product — the tool that computes the normal from two in-plane vectors.
• 3D Coordinates — the foundation: coordinate axes, octants, and the distance formula in three dimensions.