Further Maths/Lessons/Vectors/Equations of Lines & Planes in 3D
Core Pure · Vectors · Year 1

Equations of Lines & Planes in 3D

A line needs a point and a direction; a plane needs a point and a normal — vector equations describe both with almost the same idea.

14 min read 3 worked examples

The vector equation of a line

Definition 1.1 — vector equation of a line

A line through the point with position vector $\mathbf{a}$, in the direction of vector $\mathbf{d}$, has equation $$\mathbf{r}=\mathbf{a}+t\mathbf{d}, \qquad t\in\mathbb{R}$$ where $\mathbf{r}=(x,y,z)$ is the position vector of a general point on the line.

A 3D coordinate system showing a straight line passing through a labelled point A, with a direction vector d drawn as an arrow along the line and a general point r on the line
Fig. 1 — the line r = a + td: starting at point A, moving along direction d by a variable amount t traces out every point on the line.
Example 1 · Finding a line through two points
Find the vector equation of the line through $A(1,2,-1)$ and $B(3,-1,4)$.

Direction vector: $\vec{AB}=B-A=(2,-3,5)$.

$$\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+t\begin{pmatrix}2\\-3\\5\end{pmatrix}$$

The vector and Cartesian equations of a plane

Definition 1.2 — Cartesian equation of a plane

A plane through point $\mathbf{a}=(x_0,y_0,z_0)$ with normal vector $\mathbf{n}=(n_1,n_2,n_3)$ satisfies $\mathbf{n}\cdot(\mathbf{r}-\mathbf{a})=0$, giving the Cartesian form $$n_1x+n_2y+n_3z=d, \qquad d=\mathbf{n}\cdot\mathbf{a}$$

A 3D coordinate system showing a flat plane, with a normal vector n drawn perpendicular to the plane from a point A on the plane, and a dashed line showing the perpendicular relationship
Fig. 2 — the normal vector n is perpendicular to every direction that lies within the plane, which is what defines the plane's equation.
Example 2 · Finding the equation of a plane
Find the Cartesian equation of the plane through $(2,1,-3)$ with normal vector $(1,-2,4)$.

$d=\mathbf{n}\cdot\mathbf{a}=1(2)+(-2)(1)+4(-3)=2-2-12=-12$.

$$x-2y+4z=-12$$
Example 3 · Finding a plane through three points
Find the Cartesian equation of the plane through $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$.

Two direction vectors in the plane: $\vec{AB}=(-1,1,0)$, $\vec{AC}=(-1,0,1)$.

Normal $\mathbf{n}=\vec{AB}\times\vec{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&1&0\\-1&0&1\end{vmatrix}=(1(1)-0(0))\mathbf{i}-((-1)(1)-0(-1))\mathbf{j}+((-1)(0)-1(-1))\mathbf{k}=(1,1,1)$.

$d=\mathbf{n}\cdot A=1(1)+1(0)+1(0)=1$.

$$x+y+z=1$$
Common mistake

The direction vector of a line must be found by subtracting position vectors ($B-A$), not by using either point's coordinates directly as the direction.

Exam-style questions

Question 1 5 marks

Find the vector equation of the line through $P(4,-1,2)$ parallel to $\mathbf{d}=(1,2,-2)$, and determine whether the point $(6,3,-2)$ lies on this line.

$\mathbf{r}=\begin{pmatrix}4\\-1\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\-2\end{pmatrix}$ (M1, A1)

$x$: $4+t=6\Rightarrow t=2$. Check $y$: $-1+2(2)=3$ ✓. Check $z$: $2-2(2)=-2$ ✓. (M1, A1)

All three coordinates are consistent with $t=2$, so the point does lie on the line. (A1)

Question 2 5 marks

Find the Cartesian equation of the plane through $(0,2,-1)$ with normal vector $(3,-1,2)$, and state whether the point $(1,1,1)$ lies on this plane.

$d=3(0)+(-1)(2)+2(-1)=-4$, so the plane is $3x-y+2z=-4$ (M1, A1)

Substituting $(1,1,1)$: $3(1)-1+2(1)=4\neq-4$ (M1, A1)

The point does not lie on the plane. (A1)

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