Visual walkthrough — Lines and planes in 3D — vector equations

Before Step 1, three words we will use constantly. A position vector is just an arrow drawn from the fixed origin $O$ to a point — it names a point by "how to walk there from $O$". We write the origin's point as $O$ and use bold letters like $\mathbf{a}$, $\mathbf{r}$ for position vectors. A displacement is an arrow from one point to another (not necessarily starting at $O$); it says "go this far, this way".

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Step 1 — Name a point with an arrow

WHAT. We fix an origin $O$ and pick a special point $A$ on the object we want to describe. The arrow from $O$ to $A$ is the position vector $\mathbf{a}$.

WHY. Algebra needs numbers, and space has no numbers by itself. Choosing $O$ gives every point an address: the point $A$ is the arrow $\mathbf{a}$. Without this we could not turn a picture into an equation.

PICTURE. The amber arrow is $\mathbf{a}$: tail at $O$, head at $A$.

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Step 2 — A line is one direction you may re-scale

WHAT. Add a second arrow $\mathbf{d}$, the direction we are allowed to travel. We require $\mathbf{d}\ne 0$ (a zero arrow points nowhere, so it could not define a line).

WHY. "Line" means: from $A$ you may only walk forward or backward along one fixed heading. That single heading is exactly what $\mathbf{d}$ stores. Every point of the line is reached by stretching or flipping this one arrow.

PICTURE. Cyan arrow $\mathbf{d}$ sits at $A$. The faint white line is every stretch of $\mathbf{d}$ laid end-to-end through $A$.

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Step 3 — The parameter $t$: how far along we walk

WHAT. Introduce the scalar (an ordinary number) $t$. The scaled arrow $t\mathbf{d}$ is "walk $t$ copies of $\mathbf{d}$". Then the point we land on has position $\mathbf{r}=\underset{\text{start at }A}{\underbrace{\mathbf{a}}}+\underset{\text{walk }t\text{ steps of }\mathbf{d}}{\underbrace{t\mathbf{d}}}.$

WHY this tool — a parameter and not another equation? We want one dial that sweeps out the whole line. Turning the dial $t$ from $-\infty$ to $+\infty$ visits every point exactly once. This is parametrisation: trade "an equation the points satisfy" for "a formula that produces the points". It never divides by anything, so vertical or awkward lines cause no trouble.

PICTURE. Three positions: $t=0$ lands on $A$ (amber), $t=1$ one full $\mathbf{d}$ past it, $t=-1$ one full $\mathbf{d}$ behind it.

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Step 4 — WHY that formula is forced, not guessed

WHAT. Take any point $P$ on the line, position $\mathbf{r}$. The displacement from $A$ to $P$ is $\overrightarrow{AP}=\mathbf{r}-\mathbf{a}$ (head minus tail). Because $P$ lies on the line, this displacement points along $\mathbf{d}$ — it is parallel to $\mathbf{d}$.

WHY. Two parallel arrows are always scalar multiples of one another: one is a stretched copy of the other. So $\mathbf{r}-\mathbf{a}=t\mathbf{d}$ for some number $t$. Rearranging gives back $\mathbf{r}=\mathbf{a}+t\mathbf{d}$. We did not invent the formula — the meaning of "on the line" produced it.

PICTURE. The green displacement $\mathbf{r}-\mathbf{a}$ lies exactly on top of a scaled $\mathbf{d}$; the label shows it equals $t\mathbf{d}$.

Recall Why "head minus tail"?

$\overrightarrow{AP}$ means "start at $A$, end at $P$". To go from $A$ to $P$ you undo the walk to $A$ (subtract $\mathbf{a}$) and do the walk to $P$ (add $\mathbf{r}$): $\overrightarrow{AP}=\mathbf{r}-\mathbf{a}$.

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Step 5 — Line through two points (a special case, not a new rule)

WHAT. Given two points $A(\mathbf{a})$ and $B(\mathbf{b})$ we have no direction handed to us — so we build one: the displacement $\mathbf{d}=\mathbf{b}-\mathbf{a}$ points from $A$ straight to $B$.

WHY. Any arrow lying along the line works as $\mathbf{d}$; the arrow from $A$ to $B$ obviously does. Substituting: $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{a})=(1-t)\mathbf{a}+t\mathbf{b}.$ Here $t=0$ gives $\mathbf{a}$ (you are at $A$), $t=1$ gives $\mathbf{b}$ (at $B$), and $0\le t\le 1$ traces only the segment between them; outside that range you extend the full line.

PICTURE. $A$, $B$, the built direction $\mathbf{b}-\mathbf{a}$, and the shaded $0\le t\le 1$ segment.

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Step 6 — A plane needs two directions

WHAT. Now allow two non-parallel arrows $\mathbf{u}$ and $\mathbf{v}$ pinned at $A$. A point is $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v},s,t\in \mathbb{R}.$

WHY two dials. A flat sheet has two independent ways to slide. One dial $s$ slides along $\mathbf{u}$, the other dial $t$ slides along $\mathbf{v}$; combining them reaches every point of the sheet. They must be non-parallel — if $\mathbf{v}$ were a copy of $\mathbf{u}$ you'd only get the line of Step 3 back, never a full sheet.

PICTURE. The parallelogram grid spanned by $s\mathbf{u}$ (cyan) and $t\mathbf{v}$ (amber) tiling the plane through $A$.

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Step 7 — The cleaner idea: one arrow that points across the plane

WHAT. Two directions are clumsy. Instead attach a single arrow $\mathbf{n}$ that is perpendicular to the whole sheet — the normal. For any in-plane point $P$, the displacement $\mathbf{r}-\mathbf{a}$ lies flat in the plane, so it is perpendicular to $\mathbf{n}$.

WHY the dot product enters here. We need a numerical test for "perpendicular". The scalar (dot) product is exactly that tool: $\mathbf{x}\cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}|\cos \theta$, and this is zero precisely when the angle $\theta$ is $90^{\circ }$ (because $\cos 90^{\circ }=0$). No other elementary operation reads off perpendicularity so directly. So "$P$ is in the plane" becomes the single clean equation $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0.$

PICTURE. The white sheet, the amber normal $\mathbf{n}$ standing straight up, and a green in-plane displacement meeting it at a right angle.

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Step 8 — Bridging the two plane views with the cross product

WHAT. If we only have $\mathbf{u}$ and $\mathbf{v}$, we manufacture the normal with the cross product: $\mathbf{n}=\mathbf{u}\times \mathbf{v}.$

WHY this tool. The cross product is defined to output a vector perpendicular to both of its inputs at once — precisely the property a normal must have. So it converts "two in-plane directions" into "one across-plane arrow" in a single stroke. Writing $\mathbf{n}=(a,b,c)$, the dot-product equation $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ expands to the Cartesian plane $a(x-x_0)+b(y-y_0)+c(z-z_0)=0\Rightarrow ax+by+cz=d,$ where the coefficients $(a,b,c)$ are the normal — read straight off.

PICTURE. $\mathbf{u}$, $\mathbf{v}$ in the sheet; $\mathbf{n}=\mathbf{u}\times \mathbf{v}$ shooting perpendicular out of both.

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Step 9 — Degenerate cases you must never skip

WHAT & WHY. The formulas quietly assume some things are non-zero. When they aren't, here is what happens and what to do.

1. $\mathbf{d}=0$ (line). No direction $\Rightarrow$ no line, just the single point $A$. Forbidden — that is why the definition demands $\mathbf{d}\ne 0$.
2. A zero denominator in symmetric form. If $\mathbf{d}=(a,b,0)$ you cannot write $\frac{z-z_0}{0}$. State $z=z_0$ separately (the line is level in $z$) and use symmetric form on the other two coordinates only.
3. $\mathbf{u}\parallel \mathbf{v}$ (plane). Parallel spanning arrows give $\mathbf{u}\times \mathbf{v}=0$ — a zero "normal", meaning the two directions collapsed to one and you only made a line, not a plane. Choose genuinely non-parallel directions (or three non-collinear points).
4. Distance to a plane, $\mathbf{p}$ on the plane. Then $\mathbf{p}\cdot \mathbf{n}-d=0$, so the distance is $0$ — the formula still works, no special case needed.

PICTURE. Left: $\mathbf{u}\parallel \mathbf{v}$ collapse — the parallelogram flattens to a line, $\mathbf{n}$ vanishes. Right: the horizontal line where $z=z_0$ is stated separately.

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The one-picture summary

Everything on this page in a single frame: a point $A$ (from Step 1) grows a line by one arrow $\mathbf{d}$ (Steps 2–4), grows a plane by two arrows $\mathbf{u},\mathbf{v}$ (Step 6), and that plane is nailed flat by the normal $\mathbf{n}=\mathbf{u}\times \mathbf{v}$ tested with a dot product (Steps 7–8).

Recall Feynman retelling — say it to a friend

Start by planting a flag at a point $A$; the arrow from the origin to that flag is $\mathbf{a}$, its address. To make a line, hand yourself one arrow $\mathbf{d}$ and a dial $t$: "start at the flag, then walk $t$ arrows forward (or backward)." That single sentence is $\mathbf{r}=\mathbf{a}+t\mathbf{d}$, and it's forced, because any trip from $A$ to another point on the line is just a stretched copy of $\mathbf{d}$. To make a plane, you need two sliding arrows and two dials — the floor of a room needs two ways to slide. But there's a slicker trick: instead of describing the floor by how you slide across it, describe it by one arrow pointing straight up out of it, the normal $\mathbf{n}$. A point is on the floor exactly when walking to it from $A$ never rises or falls relative to that up-arrow — and "no rise or fall" is measured by the dot product being zero. If you only had the two sliding arrows, the cross product $\mathbf{u}\times \mathbf{v}$ manufactures that up-arrow for you. Watch the zeros: a line needs a non-zero $\mathbf{d}$; a plane needs non-parallel arrows, or the up-arrow collapses to nothing.

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Connections

• Lines and planes in 3D — vector equations — the parent this walkthrough derives.
• Vectors and scalar (dot) product — the perpendicular test in Step 7.
• Cross product — manufactures the normal in Step 8.
• Vector projection — turns Step 7's dot product into the distance formulas.
• Parametrisation and parameters — the dials $t,s$ of Steps 3 and 6.
• Intersection of lines and planes — where these equations meet next.
• Systems of linear equations — each Cartesian plane is one linear equation.