How Do You Write the Cartesian Equation of a Plane?

To write the Cartesian equation of a plane, identify a normal vector $\mathbf{n}=(a,b,c)$ and a point $P(x_0,y_0,z_0)$ on the plane. Then use

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0,$$

which can be rearranged into the Cartesian form $ax+by+cz=d$.

Here, $d=ax_0+by_0+cz_0$, because substituting the known point must satisfy the plane equation. Any non-zero scalar multiple represents the same plane.

This is IB Mathematics: Analysis and Approaches HL content from AHL 3.17: Vector equations of a plane. It may be assessed on Paper 1, Paper 2, or Paper 3.

The normal vector is perpendicular to every direction vector within the plane. Therefore, if $\mathbf{r}=(x,y,z)$ is the position vector of any point on the plane and $\mathbf{r}_0=(x_0,y_0,z_0)$ is a known position vector, then

$$(\mathbf{r}-\mathbf{r}_0)\cdot\mathbf{n}=0.$$

Expanding this dot product gives the Cartesian equation. Geometrically, the dot product is zero because the displacement $\mathbf{r}-\mathbf{r}_0$ lies in the plane and is perpendicular to $\mathbf{n}$. This is the mechanism behind both forms.

For example, suppose the plane contains $P(1,2,-1)$ and has normal vector $\mathbf{n}=(2,-1,3)$. Then

$$2(x-1)-(y-2)+3(z+1)=0.$$

Expanding and simplifying gives

$$2x-y+3z=-3.$$

Check: substituting $P(1,2,-1)$ gives $2-2-3=-3$, so the point lies on the plane. The coefficient vector $(2,-1,3)$ also matches the stated normal vector.

A common misconception is to use a direction vector in place of the normal vector. Direction vectors lie within the plane; the coefficients of $x$, $y$, and $z$ must come from a vector perpendicular to the plane.

In an IB response, show how the normal vector was obtained, substitute a known point, and simplify fully. If the normal is found by a cross product, retain that working because it can earn method credit even if a later arithmetic error occurs.