Lines and Planes in Vector and Parametric Form

The vector form of a line

Definition (line in vector form). Let $\mathbf{p} \in \mathbf{R}^n$ and let $\mathbf{d} \in \mathbf{R}^n$ with $\mathbf{d} \neq \mathbf{0}$. The line through $\mathbf{p}$ with direction $\mathbf{d}$ is the set

The equation

is the vector form of $L$. The scalar $t$ is the parameter, $\mathbf{p}$ is the position vector of one point of $L$ — the point reached at $t = 0$ — and $\mathbf{d}$ is a direction vector for $L$.

Read the definition as an instruction. Stand at $\mathbf{p}$; choose a real number $t$; travel the displacement $t\mathbf{d}$. As $t$ runs through $\mathbf{R}$ the resulting point $\mathbf{r}$ traces out the whole of $L$: positive $t$ moves one way along the line, negative $t$ the other, and $|t|$ measures the distance travelled in units of $\|\mathbf{d}\|$.

The structure: a fixed offset plus a span

Lesson 2 defined $\operatorname{span}\{\mathbf{d}\} = \{\, t\mathbf{d} : t \in \mathbf{R} \,\}$ and showed that for $\mathbf{d} \neq \mathbf{0}$ this is a line through the origin. Comparing that with the definition above,

A line in vector form is therefore a span that has been translated by a fixed offset. Two cases arise.

• If $\mathbf{p} \in \operatorname{span}\{\mathbf{d}\}$ — in particular if $\mathbf{p} = \mathbf{0}$ — then $L = \operatorname{span}\{\mathbf{d}\}$ exactly, and $L$ passes through the origin.
• If $\mathbf{p} \notin \operatorname{span}\{\mathbf{d}\}$ then $L$ misses the origin, and $L$ is not a span of anything.

The second statement needs its reason, and the reason is short. Every span contains $\mathbf{0}$ (Lesson 2: take every coefficient to be zero). Now $\mathbf{0} \in L$ would mean $\mathbf{p} + t\mathbf{d} = \mathbf{0}$ for some $t$, that is $\mathbf{p} = (-t)\mathbf{d} \in \operatorname{span}\{\mathbf{d}\}$. So $\mathbf{p} \notin \operatorname{span}\{\mathbf{d}\}$ forces $\mathbf{0} \notin L$, and a set that omits $\mathbf{0}$ cannot be a span.

Non-example (why $\mathbf{d} \neq \mathbf{0}$ is required). If $\mathbf{d} = \mathbf{0}$ then $\mathbf{p} + t\mathbf{0} = \mathbf{p}$ for every $t$, so $\{\,\mathbf{p} + t\mathbf{0} : t \in \mathbf{R}\,\} = \{\mathbf{p}\}$, a single point. Writing a parameter into an equation does not by itself produce a line; the parameter must actually move the point.

Two points, and why one line has many equations

The line through two points. Let $\mathbf{a} \neq \mathbf{b}$ in $\mathbf{R}^n$ and put $\mathbf{d} = \mathbf{b} - \mathbf{a}$, which is non-zero precisely because $\mathbf{a} \neq \mathbf{b}$. Then

is a line containing $\mathbf{a}$ (at $t = 0$) and $\mathbf{b}$ (at $t = 1$). The displacement $\mathbf{b}-\mathbf{a}$ is the journey from $\mathbf{a}$ to $\mathbf{b}$ in the sense of Lesson 1. Values $t \in [0,1]$ sweep the segment from $\mathbf{a}$ to $\mathbf{b}$; values outside $[0,1]$ continue past the ends.

Worked example. For $\mathbf{a} = (1,-2,4)^{\mathsf{T}}$ and $\mathbf{b} = (4,1,-2)^{\mathsf{T}}$, $\mathbf{b}-\mathbf{a} = (3,3,-6)^{\mathsf{T}}$, so the line through them is

The equation is not unique

One line has infinitely many vector equations, because two choices are free: which point of the line serves as the offset, and which non-zero multiple of the direction serves as $\mathbf{d}$.

Take the line above, use $\mathbf{b}$ as the offset, and rescale the direction by $-\tfrac{1}{3}$ to get $(-1,-1,2)^{\mathsf{T}}$:

The two equations share no number, yet describe the same set. In components the first gives $(1+3t,\; -2+3t,\; 4-6t)$ and the second gives $(4-u,\; 1-u,\; -2+2u)$. Equating first components gives $u = 3-3t$. Substituting into the second component of the second form: $1-(3-3t) = -2+3t$, which matches. Into the third: $-2+2(3-3t) = 4-6t$, which also matches. Since $t \mapsto 3-3t$ is a bijection of $\mathbf{R}$ onto $\mathbf{R}$, every point produced by one equation is produced by the other, and the two sets are equal.

Criterion. The equations $\mathbf{r} = \mathbf{p}+t\mathbf{d}$ and $\mathbf{r} = \mathbf{p}'+u\mathbf{d}'$ describe the same line if and only if $\mathbf{d}' = \lambda\mathbf{d}$ for some $\lambda \neq 0$ and $\mathbf{p}' - \mathbf{p} \in \operatorname{span}\{\mathbf{d}\}$. In words: the directions are parallel, and the second offset is a point of the first line. Both conditions are needed — parallel directions alone give parallel lines, which may be distinct.

Misconception. Two very different-looking equations may describe the same line, and two nearly identical ones may not. Compare $(1,3)^{\mathsf{T}} + t(2,-4)^{\mathsf{T}}$ with $(1,3)^{\mathsf{T}} + t(4,-4)^{\mathsf{T}}$: they share an offset and differ in one entry, but $(4,-4)^{\mathsf{T}}$ is not a scalar multiple of $(2,-4)^{\mathsf{T}}$, so these are two different lines crossing at $(1,3)$. Never compare vector equations by appearance; apply the criterion.

Component, Cartesian and symmetric forms

Writing $\mathbf{r} = (x,y,z)^{\mathsf{T}}$ and equating components turns one vector equation into $n$ scalar parametric equations. In $\mathbf{R}^3$, with $\mathbf{p} = (p_1,p_2,p_3)^{\mathsf{T}}$ and $\mathbf{d} = (d_1,d_2,d_3)^{\mathsf{T}}$,

In $\mathbf{R}^2$. Eliminate $t$. If $d_1 \neq 0$ then $t = (x-p_1)/d_1$, so

a line of gradient $m = d_2/d_1$. For $\mathbf{p} = (1,2)^{\mathsf{T}}$ and $\mathbf{d} = (4,-8)^{\mathsf{T}}$ this gives $m = -8/4 = -2$ and $y = 2 - 2(x-1) = -2x + 4$. If $d_1 = 0$ the elimination is impossible: the line is vertical, $x = p_1$, and has no equation of the form $y = mx + c$. Vector form covers vertical lines with no special case, which is one practical reason to prefer it.

In $\mathbf{R}^3$. If $d_1, d_2, d_3$ are all non-zero, solving each parametric equation for $t$ and equating gives the symmetric form

For $\mathbf{p} = (2,-1,3)^{\mathsf{T}}$ and $\mathbf{d} = (1,4,-2)^{\mathsf{T}}$ this reads $\dfrac{x-2}{1} = \dfrac{y+1}{4} = \dfrac{z-3}{-2}$. If some $d_i = 0$ that coordinate is constant and is recorded as a separate equation: for $\mathbf{p} = (1,5,-2)^{\mathsf{T}}$ and $\mathbf{d} = (3,0,1)^{\mathsf{T}}$ the line is $\dfrac{x-1}{3} = \dfrac{z+2}{1}$ together with $y = 5$.

Collinearity

Definition. Points of $\mathbf{R}^n$ are collinear if some single line contains all of them. Any two points are collinear; the content of the notion begins at three.

Theorem (collinearity test). Let $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbf{R}^n$ with $\mathbf{a} \neq \mathbf{b}$. Then $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are collinear if and only if there is a scalar $k \in \mathbf{R}$ with

Plain reading. Start at $\mathbf{a}$ and record two journeys: the one to $\mathbf{b}$ and the one to $\mathbf{c}$. The three points lie on one line exactly when the second journey is a rescaling of the first — same direction up to sign and length.

Proof. Suppose $\mathbf{c}-\mathbf{a} = k(\mathbf{b}-\mathbf{a})$. Since $\mathbf{a} \neq \mathbf{b}$, the vector $\mathbf{d} = \mathbf{b}-\mathbf{a}$ is non-zero, so $L: \mathbf{r} = \mathbf{a} + t\mathbf{d}$ is a line. It contains $\mathbf{a}$ at $t = 0$, $\mathbf{b}$ at $t = 1$, and $\mathbf{c} = \mathbf{a} + k\mathbf{d}$ at $t = k$. Hence all three are collinear.

Conversely, suppose all three lie on a line $\mathbf{r} = \mathbf{p} + t\mathbf{d}$ with $\mathbf{d} \neq \mathbf{0}$, say $\mathbf{a} = \mathbf{p}+t_a\mathbf{d}$, $\mathbf{b} = \mathbf{p}+t_b\mathbf{d}$, $\mathbf{c} = \mathbf{p}+t_c\mathbf{d}$. Subtracting, $\mathbf{b}-\mathbf{a} = (t_b-t_a)\mathbf{d}$ and $\mathbf{c}-\mathbf{a} = (t_c-t_a)\mathbf{d}$. Because $\mathbf{a} \neq \mathbf{b}$ we have $t_b \neq t_a$, so we may set $k = (t_c-t_a)/(t_b-t_a)$ and obtain $\mathbf{c}-\mathbf{a} = k(\mathbf{b}-\mathbf{a})$. $\blacksquare$

The hypothesis $\mathbf{a} \neq \mathbf{b}$ cannot be dropped. Take $\mathbf{a} = \mathbf{b} = (0,0)^{\mathsf{T}}$ and $\mathbf{c} = (1,1)^{\mathsf{T}}$. These three points are collinear — the line $\mathbf{r} = t(1,1)^{\mathsf{T}}$ contains all of them — yet $\mathbf{b}-\mathbf{a} = \mathbf{0}$, so $k(\mathbf{b}-\mathbf{a}) = \mathbf{0} \neq \mathbf{c}-\mathbf{a}$ for every $k$ and the criterion returns the wrong verdict. With coincident points the test says nothing.

Procedure. Compute $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Read off the candidate $k$ from the first component in which $\mathbf{b}-\mathbf{a}$ is non-zero, then check that the same $k$ works in every remaining component. One mismatch is decisive.

Worked example 1 (collinear). $\mathbf{a} = (2,-1,0)^{\mathsf{T}}$, $\mathbf{b} = (5,1,4)^{\mathsf{T}}$, $\mathbf{c} = (-4,-5,-8)^{\mathsf{T}}$. Then $\mathbf{b}-\mathbf{a} = (3,2,4)^{\mathsf{T}}$ and $\mathbf{c}-\mathbf{a} = (-6,-4,-8)^{\mathsf{T}}$. The first component gives $k = -6/3 = -2$; then $-2 \cdot 2 = -4$ and $-2 \cdot 4 = -8$, both correct. The points are collinear.

Worked example 2 (not collinear). $\mathbf{a} = (1,1,2)^{\mathsf{T}}$, $\mathbf{b} = (4,3,5)^{\mathsf{T}}$, $\mathbf{c} = (7,5,9)^{\mathsf{T}}$. Then $\mathbf{b}-\mathbf{a} = (3,2,3)^{\mathsf{T}}$ and $\mathbf{c}-\mathbf{a} = (6,4,7)^{\mathsf{T}}$. The first component gives $k = 2$; the second agrees, since $2 \cdot 2 = 4$; the third fails, since $2 \cdot 3 = 6 \neq 7$. Not collinear.

Misconception. Example 2 is the standard trap: two of the three components agreed. A scalar multiple is an all-or-nothing relation between whole vectors, so partial agreement is not evidence. Check every component.

Planes in vector form

Definition (parallel vectors). Two vectors $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$ are parallel if one is a scalar multiple of the other. They are non-parallel otherwise; in particular non-parallel vectors are both non-zero, since $\mathbf{0} = 0\,\mathbf{u}$ is a multiple of everything.

Definition (plane in vector form). Let $\mathbf{p} \in \mathbf{R}^n$ and let $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$ be non-parallel. The plane through $\mathbf{p}$ with directions $\mathbf{u}, \mathbf{v}$ is

Lesson 2 showed $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a plane through the origin exactly when $\mathbf{u},\mathbf{v}$ are non-parallel, and collapses to a line when they are parallel. The offset $\mathbf{p}$ translates that plane, and as before $\Pi$ contains $\mathbf{0}$ — hence is itself a span — if and only if $\mathbf{p} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$.

The plane through three points

Construction. Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be non-collinear. Put $\mathbf{u} = \mathbf{b}-\mathbf{a}$ and $\mathbf{v} = \mathbf{c}-\mathbf{a}$. Then

is a plane containing $\mathbf{a}$ (at $s=t=0$), $\mathbf{b}$ (at $s=1, t=0$) and $\mathbf{c}$ (at $s=0, t=1$).

Why non-collinearity is exactly the right hypothesis. Non-collinear points are in particular distinct, so $\mathbf{u} \neq \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$. Given $\mathbf{a} \neq \mathbf{b}$, the collinearity theorem says the three points are collinear if and only if $\mathbf{v} = \mathbf{c}-\mathbf{a}$ is a scalar multiple of $\mathbf{u} = \mathbf{b}-\mathbf{a}$. Non-collinearity is therefore precisely the statement that $\mathbf{v}$ is not a multiple of $\mathbf{u}$; and since $\mathbf{u} \neq \mathbf{0}$, $\mathbf{u}$ is then not a multiple of $\mathbf{v}$ either. So the three points are non-collinear if and only if $\mathbf{u},\mathbf{v}$ are non-parallel — the exact condition the definition of a plane demands.

Non-example (collinear points give no plane). If $\mathbf{a},\mathbf{b},\mathbf{c}$ are collinear with $\mathbf{a} \neq \mathbf{b}$, then $\mathbf{v} = k\mathbf{u}$ and every $s\mathbf{u}+t\mathbf{v} = (s+tk)\mathbf{u}$ is a multiple of $\mathbf{u}$. The set $\mathbf{a} + \operatorname{span}\{\mathbf{u},\mathbf{v}\} = \mathbf{a} + \operatorname{span}\{\mathbf{u}\}$ is the line through the three points. Two parameters were written down, but only one dimension was obtained. Three points determine a plane only when they are non-collinear.

Worked example. $\mathbf{a} = (1,0,2)^{\mathsf{T}}$, $\mathbf{b} = (3,1,2)^{\mathsf{T}}$, $\mathbf{c} = (0,2,5)^{\mathsf{T}}$. Then $\mathbf{u} = (2,1,0)^{\mathsf{T}}$ and $\mathbf{v} = (-1,2,3)^{\mathsf{T}}$. These are non-parallel: matching first components would need $\lambda = -2$ in $\mathbf{u} = \lambda\mathbf{v}$, but then the second component would give $1 = -4$. So the points are non-collinear and

Testing membership. A point $\mathbf{q}$ lies in $\Pi$ if and only if $\mathbf{q}-\mathbf{a} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$, which is a linear system in the two unknowns $s,t$ — the same reduction Lesson 2 used for span membership, applied to $\mathbf{q}-\mathbf{a}$ rather than $\mathbf{q}$.

Take $\mathbf{q} = (6,0,-1)^{\mathsf{T}}$. The third components give $2 + 0s + 3t = -1$, so $t = -1$. The second give $0 + s + 2t = 0$, so $s = 2$. Check the first: $1 + 2(2) + (-1)(-1) = 6$. Consistent, so $\mathbf{q} \in \Pi$ at $(s,t) = (2,-1)$.

Take $\mathbf{q}' = (6,0,0)^{\mathsf{T}}$. The third give $2+3t = 0$, so $t = -\tfrac{2}{3}$; the second give $s = \tfrac{4}{3}$; the first then give $1 + \tfrac{8}{3} + \tfrac{2}{3} = \tfrac{13}{3} \neq 6$. Inconsistent, so $\mathbf{q}' \notin \Pi$.

Perpendicularity: exactly as much dot product as is needed

Definition (dot product). For $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$,

The output is a scalar, not a vector.

The one fact used here. $\mathbf{u}\cdot\mathbf{v} = 0$ means $\mathbf{u}$ and $\mathbf{v}$ are perpendicular (equivalently, orthogonal). Take this as the working definition of perpendicular for now. Its agreement with the schoolroom notion of a right angle, together with norms, angles, projections, orthogonal complements and the rest of the theory, is developed in Section 9; nothing beyond the displayed fact is used in this section.

Normal form of a plane in $\mathbf{R}^3$

Let $\Pi$ be a plane in $\mathbf{R}^3$ through the point $\mathbf{p}$, and let $\mathbf{n} \neq \mathbf{0}$ be perpendicular to every displacement that lies within $\Pi$; such an $\mathbf{n}$ is a normal to $\Pi$. A point $\mathbf{r}$ lies in $\Pi$ exactly when the displacement $\mathbf{r}-\mathbf{p}$ lies within $\Pi$, so

Writing $\mathbf{n} = (a,b,c)^{\mathsf{T}}$ and $\mathbf{r} = (x,y,z)^{\mathsf{T}}$ and expanding gives the Cartesian form

Finding a normal from two directions. $\mathbf{n}$ must satisfy $\mathbf{n}\cdot\mathbf{u} = 0$ and $\mathbf{n}\cdot\mathbf{v} = 0$: two linear equations in three unknowns, which always have a non-zero solution.

For $\mathbf{u} = (2,1,0)^{\mathsf{T}}$ and $\mathbf{v} = (-1,2,3)^{\mathsf{T}}$ the conditions are $2n_1 + n_2 = 0$ and $-n_1 + 2n_2 + 3n_3 = 0$. The first gives $n_2 = -2n_1$; substituting, $-n_1 - 4n_1 + 3n_3 = 0$, so $3n_3 = 5n_1$. Choosing $n_1 = 3$ gives $\mathbf{n} = (3,-6,5)^{\mathsf{T}}$. Check: $\mathbf{n}\cdot\mathbf{u} = 6-6+0 = 0$ and $\mathbf{n}\cdot\mathbf{v} = -3-12+15 = 0$.

With $\mathbf{p} = (1,0,2)^{\mathsf{T}}$, $d = \mathbf{n}\cdot\mathbf{p} = 3 + 0 + 10 = 13$, so the plane of the previous block is

The three original points confirm it: $\mathbf{a} = (1,0,2)$ gives $3-0+10 = 13$; $\mathbf{b} = (3,1,2)$ gives $9-6+10 = 13$; $\mathbf{c} = (0,2,5)$ gives $0-12+25 = 13$. So does the member $(6,0,-1)$: $18-0-5 = 13$. The non-member $(6,0,0)$ gives $18 \neq 13$, agreeing with the parametric test.

A caution about dimension. In $\mathbf{R}^3$ a single linear equation describes a plane; in $\mathbf{R}^2$ the equation $ax+by = e$ describes a line. One equation removes one dimension, so normal form as stated is a fact about $\mathbf{R}^3$ specifically: a plane sitting inside $\mathbf{R}^4$ is not cut out by one equation. Section 2 turns this counting into a theorem.