Lines and Planes in ℝ³

The big idea. A line is just "a known point and a heading." Fix one point you are sure lies on the line, then say which way the line runs. Everything else on the line is reached by sliding away from that point in the chosen direction.

Let $\mathbf{p} = \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}$ be a known point on the line and let $\mathbf{d} = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} \neq \mathbf{0}$ be a direction vector — the arrow that points along the line. The scalar $t \in \mathbb{R}$ is the parameter: it tells you how many copies of $\mathbf{d}$ to add to $\mathbf{p}$.

Vector equation. Every point $\mathbf{x}$ on the line can be written $\mathbf{x} = \mathbf{p} + t\,\mathbf{d}, \qquad t \in \mathbb{R}.$ As $t$ runs over all real numbers, $\mathbf{x}$ traces the whole line. At $t=0$ you are at $\mathbf{p}$; positive $t$ moves one way, negative $t$ the other.

Parametric equations. Reading the vector equation one coordinate at a time gives three scalar equations: $x = p_1 + t\,d_1, \qquad y = p_2 + t\,d_2, \qquad z = p_3 + t\,d_3.$

Two things to remember.

• The direction vector is not unique: any nonzero scalar multiple $c\,\mathbf{d}$ (with $c\neq 0$) describes the same line. So $(2,4,4)$, $(1,2,2)$, and $(-1,-2,-2)$ are all valid directions for the same line.
• The point $\mathbf{p}$ is not unique either — any point on the line works as the anchor.

Finding a direction from two points. If $\mathbf{a}$ and $\mathbf{b}$ are two distinct points on a line, the arrow from one to the other points along the line: $\mathbf{d} = \mathbf{b} - \mathbf{a}.$ Then the line is $\mathbf{x} = \mathbf{a} + t(\mathbf{b}-\mathbf{a})$. (Notice $t=0$ gives $\mathbf{a}$ and $t=1$ gives $\mathbf{b}$.)

Is a given point on the line? Suppose the line is $\mathbf{x}=\mathbf{p}+t\,\mathbf{d}$ and you are handed a point $\mathbf{q}=(q_1,q_2,q_3)$. The point lies on the line exactly when there is a single value of $t$ that satisfies all three coordinate equations: $q_1 = p_1 + t\,d_1, \qquad q_2 = p_2 + t\,d_2, \qquad q_3 = p_3 + t\,d_3.$

Procedure. Solve for $t$ using one coordinate (any coordinate whose $d_i \neq 0$), then check that the same $t$ works in the other two.

• If all three agree $\Rightarrow$ the point is on the line.
• If even one disagrees $\Rightarrow$ it is off the line.

This is the most common mistake to avoid: a point can match in two coordinates and still fail in the third, so you must check all of them.

A line needed one direction; a plane is two-dimensional, so it needs two independent directions — or, cleverly, a single arrow that sticks straight out of it.

Parametric (point + two directions). Fix a point $\mathbf{p}$ on the plane and two direction vectors $\mathbf{u},\mathbf{v}$ that lie in the plane and are not parallel (neither is a scalar multiple of the other). Then every point of the plane is $\mathbf{x} = \mathbf{p} + s\,\mathbf{u} + t\,\mathbf{v}, \qquad s,t \in \mathbb{R}.$ Here $s$ and $t$ are two independent parameters — you need two knobs because a plane is a 2D sheet.

Normal form (point + normal). A cleaner description uses a normal vector $\mathbf{n}\neq\mathbf{0}$, an arrow perpendicular to the plane. A point $\mathbf{x}$ is on the plane exactly when the displacement $\mathbf{x}-\mathbf{p}$ is perpendicular to $\mathbf{n}$, i.e. their dot product is zero: $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p}) = 0.$

Scalar (Cartesian) form. Expanding the dot product with $\mathbf{n}=(n_1,n_2,n_3)$ and $\mathbf{x}=(x,y,z)$ gives the familiar single equation $n_1 x + n_2 y + n_3 z = d, \qquad \text{where } d = \mathbf{n}\cdot\mathbf{p} = n_1 p_1 + n_2 p_2 + n_3 p_3.$ The coefficients of $x,y,z$ are precisely the components of a normal vector — read a plane's normal straight off its equation.

Parametric $\to$ normal. Given two in-plane directions $\mathbf{u},\mathbf{v}$, a normal is their cross product, which is perpendicular to both: $\mathbf{n} = \mathbf{u}\times\mathbf{v} = \begin{bmatrix} u_2 v_3 - u_3 v_2 \\ u_3 v_1 - u_1 v_3 \\ u_1 v_2 - u_2 v_1 \end{bmatrix}.$ Then use $d = \mathbf{n}\cdot\mathbf{p}$ to get the scalar equation. (Swapping the order $\mathbf{v}\times\mathbf{u}$ flips the sign of $\mathbf{n}$ — still a valid normal, since $\mathbf{n}$ and $-\mathbf{n}$ describe the same plane.)

Three points $\to$ plane. Given points $\mathbf{a},\mathbf{b},\mathbf{c}$ (not all on one line), build two directions $\mathbf{u}=\mathbf{b}-\mathbf{a}$ and $\mathbf{v}=\mathbf{c}-\mathbf{a}$, then cross them.

Is a point on a plane? Plug its coordinates into the scalar equation $n_1 x + n_2 y + n_3 z = d$. If the left side equals $d$, it is on the plane; otherwise not. (No parameter-solving needed — one arithmetic check.)

Where does a line meet a plane? Substitute the line's parametric coordinates into the plane's scalar equation, which leaves a single equation in $t$. Solve for $t$, then plug that $t$ back into the line to get the intersection point.

• One solution for $t$ $\Rightarrow$ a single crossing point.
• No solution ($0 =$ nonzero) $\Rightarrow$ the line is parallel to the plane and misses it.
• Every $t$ works ($0=0$) $\Rightarrow$ the line lies entirely in the plane.