Equations of planes (HL)

A plane, $\Pi$, is a flat surface, like an infinitely large square. It is the 2-D equivalent of a line. Two dimensional also means it requires two direction vectors to specify a plane.

Last edited: 2026-06-09 changed to proper vector notations

Contents

• Using two directional vectors
• Normal vector
• Vector equation of a plane
• Scalar (Cartesian) equation of a plane
• Wait, these are planes?

Using two directional vectors

where $\rvec{x_1,y_1,z_1}$ is a point on the plane and $\bm{d_1}$ and $\bm{d_2}$ are two direction (displacement) vectors on the plane that are in different directions.

Every point has a unique combination of $\lambda$ and $\mu$.

For example, here is an example of a plane

Consider $\lambda = 1$ and $\mu = -1$, we see that ${(-5, 0, 3)}$ is a point on the plane.

Normal vector

A normal vector $\bm n$ is one that is perpendicular to all displacement vectors on the plane. In particular, it can be found using the cross product of two direction vectors of the plane, meaning

where $k$ means the normal vector is any non-zero scalar multiple of the cross product of the two direction vectors.

Each displacement vector with respect to some point on the plane $(x_0, y_0, z_0)$ can be represented by $\rvec{x - x_0, y - y_0, z - z_0}$

Perpendicular vectors dot to $0$, so

For the aforementioned plane, the cross product of the two direction vectors is

so the plane could also be written as

Vector equation of a plane

We start with

If we have $\bm n = \rvec{a,b,c}$, then rearranging the dot product , we get

So our plane above becomes

In the formula booklet, this is written as

where $\bm r$ is the general point, $\bm n$ is the normal vector, $\bm a$ is a particular point on the plane.

Scalar (Cartesian) equation of a plane

Simplifying the dot products, we get

For our plane thus far, this is

or equivalently

To calculate distance between a plane and a point, it is useful to keep in mind the constant $d$ is a dot product.

Note that unlike other forms, the Cartesian equation of a plane does not provide a point.

To find a point $P(x_0, y_0, z_0)$ on the plane, guess the component whose corresponding $a, b, c$ is not zero yet closest to zero. This ensures smallest denominator of fractions we have to work with. Then solve a system of two equations.

Because of the likelihood of working with fractions, you are very likely to show that a point is on the plane, or otherwise be given a point.

Wait, these are planes?

The $xy$-plane is represented by $z = 0$. Similarly $xz$-plane is $y = 0$, and $yz$-plane is $x = 0$.

So in 2-D, $x = 0$ is a line, but in 3-D it is a plane.

Normally, the plane $ax + by + cz = d$ passes through the points ${\left(\frac da, 0, 0\right)}$, ${\left(0, \frac db, 0, \right)}$, and ${\left(0, 0, \frac dc\right)}$ on the respective axes. However for the plane $y = 7$, both $a$ and $c$ are $0$, meaning $y = 7$ intersects neither the $x$-axis nor the $z$-axis. Hence $y = 7$ is parallel to the $xz$-plane.