The Cartesian Equation of a Plane

A normal to the plane

A normal to a plane is a vector at right angles to the plane. It is perpendicular to every direction that lies in the plane.

If the plane is $r=a+\lambda b+\mu c$r = a + λb + μc, a normal n is perpendicular to both directions, so n · b = 0 and n · c = 0. Then it is perpendicular to every direction $\lambda b+\mu c$λb + μc in the plane as well, since $n\cdot \left(\lambda b+\mu c\right)=\lambda \left(n\cdot b\right)+\mu \left(n\cdot c\right)=0+0=0$n · (λb + μc) = λ(n · b) + μ(n · c) = 0 + 0 = 0.

One normal fixes how the plane faces. Any multiple of it, such as 2n or −n, is a normal too.

The plane seen edge-on, so that it looks like a line. The vector r − a, from one point of the plane to another, lies along it, and the normal n meets it at a right angle.

Every point gives the same number

Take any point r of the plane. The vector r − a joins two points of the plane, so it lies in the plane, and the normal is perpendicular to it: (r − a) · n = 0. Multiplying out, r · n − a · n = 0, so r · n = a · n.

The same thing happens if you dot the vector equation with n: $r\cdot n=\left(a+\lambda b+\mu c\right)\cdot n=a\cdot n+\lambda \left(b\cdot n\right)+\mu \left(c\cdot n\right)$r · n = (a + λb + μc) · n = a · n + λ(b · n) + μ(c · n). Both b · n and c · n are 0, so the $\lambda$λ and $\mu$μ terms vanish, and r · n = a · n whatever $\lambda$λ and $\mu$μ are.

The number d

Since a · n is one fixed number for the whole plane, give it a name, d. The plane is then r · n = d: the set of all points whose position vector has dot product d with the normal.

For the plane through (1, 2, 3) with normal (2, 1, −1), d = (1, 2, 3) · (2, 1, −1) = 2 + 2 − 3 = 1, so the plane is r · (2, 1, −1) = 1.

Edge-on again: the plane is the line at height 2, and n = (0, 1) is a normal of length 1. The points of the plane at (−2, 2), (1, 2) and (3, 2) all have r · n = 2, so the plane is r · n = 2.

Opening out the dot product

Write r as (x, y, z). Then (x, y, z) · (2, 3, 1) = 7 multiplies place by place and adds: 2x + 3y + z = 7. This is the Cartesian equation of the plane: one equation in x, y and z.

In general, with $n=\left(n_1,n_2,n_3\right)$n = (n₁, n₂, n₃), the plane r · n = d is $n_{1}x+n_{2}y+n_{3}z=d$n₁x + n₂y + n₃z = d. For the plane above, r · (2, 1, −1) = 1 becomes 2x + y − z = 1. Check that the point (1, 2, 3) is on it: 2 × 1 + 2 − 3 = 1.

Reading the normal back off

The coefficients of x, y and z are the components of a normal, signs included. The plane 4x − y + 2z = 9 has normal (4, −1, 2); writing (4, 1, 2) drops the minus and gives a different direction.

A missing variable has coefficient 0. The plane z = 3 is 0x + 0y + 1z = 3, with normal (0, 0, 1): it is horizontal, 3 above the floor. The plane x + y = 4 has normal (1, 1, 0), which is level, so the plane itself is vertical, like a wall.

Multiplying the whole equation by a number gives the same plane. 8x − 2y + 4z = 18 is 4x − y + 2z = 9 doubled, and its normal (8, −2, 4) is 2(4, −1, 2), along the same line.

The plane r · n = d, with n a normal of length 1. Tilt n and the plane turns with it, staying at right angles, and the arrow r − a lying in the plane stays square to n. Slide d and the plane moves along n without turning: a different d is a parallel plane.

Is a point on the plane?

Substitute the point and see whether it gives d. For 2x + 3y + z = 7, the point (1, 1, 2) gives 2 + 3 + 2 = 7, so it lies on the plane. The point (2, 1, 1) gives 4 + 3 + 1 = 8, which is not 7, so it does not.

Same normal, different d

Planes with the same normal face the same way, so they are parallel; d says which one of them you mean. 2x + 3y + z = 7 and 2x + 3y + z = 10 are parallel planes.

With d = 0 the plane passes through the origin, since (0, 0, 0) gives 0. So 2x + 3y + z = 0 is the plane parallel to 2x + 3y + z = 7 that goes through the origin. It is a different plane; leaving out the constant changes where the plane is.

From the vector equation

To turn $r=\left(1,0,2\right)+\lambda \left(1,1,0\right)+\mu \left(0,1,1\right)$r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 1) into a Cartesian equation, find a normal as the vector product of the two directions: (1, 1, 0) × (0, 1, 1) = (1 × 1 − 0 × 1, 0 × 0 − 1 × 1, 1 × 1 − 1 × 0) = (1, −1, 1).

Then d = a · n = (1, 0, 2) · (1, −1, 1) = 1 + 0 + 2 = 3, and the plane is x − y + z = 3. Check with another point of the plane: $\lambda =1$λ = 1 and $\mu =2$μ = 2 give (2, 3, 4), and 2 − 3 + 4 = 3.

The usual mistakes

Dropping a minus sign. The normal of 5x − 2y + 3z = 4 is (5, −2, 3), not (5, 2, 3).

Taking the constant as a component. In 5x − 2y + 3z = 4, the 4 says where the plane is, not which way it faces; the normal has only the three coefficients.

Throwing the constant away. (x, y, z) · (5, −2, 3) = 4 is 5x − 2y + 3z = 4; writing = 0 gives the parallel plane through the origin.

Dropping the coefficients. x + y + z = 4 has normal (1, 1, 1), a different plane altogether.

Is a glass panel flat?

Four points need not lie in one plane. In the application below, three corners of a roof panel give two edges, their vector product gives the normal, one corner gives d, and the fourth corner is substituted to decide whether the panel is flat.