Worked examples — Lines and planes in 3D — vector equations

Before anything else, three symbols you must own, because every example leans on them:

---

The scenario matrix

Every problem this topic can throw is one of these cells. The example that covers each cell is named in the last column.

#	Case class	The trap / twist it tests	Covered by
C1	Line through 2 points, all directions nonzero	plain symmetric form	Ex 1
C2	Line with a zero component in $\mathbf{d}$	you CANNOT divide by 0	Ex 2
C3	Plane from point + normal, negative constant	sign of $d$, distance sign	Ex 3
C4	Plane through 3 points	build normal via cross product	Ex 4
C5	Degenerate 3 points (collinear)	cross product $=0$ ⇒ no plane	Ex 5
C6	Point-to-plane distance, point on the "other side"	absolute value handles both sides	Ex 6
C7	Point-to-line distance in full 3D	why the cross product, not projection alone	Ex 7
C8	Line parallel to a plane vs hitting it	dot of $\mathbf{d}$ with $\mathbf{n}$: zero or not	Ex 8
C9	Word problem (drone flight path)	translate English → vectors	Ex 9
C10	Exam twist: two lines — parallel, intersecting, or skew?	limiting/degenerate classification	Ex 10

---

Example 1 — C1: Line through two points, clean case

Step 1 — direction. $\mathbf{d}=B-A=(5-2,7-1,10-4)=(3,6,6)$. Why this step? A line's direction is the displacement from one of its points to another — literally the arrow from $A$ to $B$.

Step 2 — vector form. Anchor at $A$, with $\mathbf{r}$ the position vector of a generic point on the line: $\mathbf{r}=(2,1,4)+t(3,6,6)$. Why this step? Every point is "start at $A$, walk $t$ copies of $\mathbf{d}$" — the parameter $t$ sweeps the whole line, and $\mathbf{r}$ is wherever you land.

Step 3 — parametric form (read off each coordinate of $\mathbf{r}=(x,y,z)$): $x=2+3t,y=1+6t,z=4+6t.$ Why this step? The vector equation is really three equations stacked — one per coordinate. Peeling them apart lets us plug in a single value of $t$ and read off a concrete point $(x,y,z)$.

Step 4 — symmetric form. Solve each for $t$ and equate (allowed because all of $3,6,6\ne 0$): $\frac{x-2}{3}=\frac{y-1}{6}=\frac{z-4}{6}.$ Why this step? Every point shares one $t$; setting the three "$t$ expressions" equal removes $t$.

Verify: Plug $t=1$ into parametric: $(2+3,1+6,4+6)=(5,7,10)=B$. ✓ And $t=0$ gives $A$. ✓

---

Example 2 — C2: Zero component, the divide-by-zero trap

Step 1 — spot the zero. $\mathbf{d}=(2,4,0)$: the third slot is $0$. Why this step? Symmetric form divides by each component. Division by $0$ is undefined — the formula silently breaks here.

Step 2 — what a zero direction MEANS. If $z=2+0\cdot t=2$ for all $t$, the line never changes height. It lives entirely on the flat sheet $z=2$. Why this step? Interpreting the zero geometrically tells us what to write instead.

Step 3 — correct symmetric form. Use symmetric form only on the coordinates with nonzero direction, and state the frozen one separately: $\frac{x-1}{2}=\frac{y-0}{4},z=2.$ Why this step? The two nonzero-direction coordinates still share the same $t$, so their ratios are equal; the frozen coordinate carries no $t$ at all, so it must be written as a standalone constraint rather than jammed into a fraction.

Verify: Take $t=1$: $x=3,y=4,z=2$. Check $\frac{3-1}{2}=1$ and $\frac{4-0}{4}=1$ — equal ✓ and $z=2$ ✓.

---

Example 3 — C3: Plane from point + normal, negative constant

Step 1 — normal form. Write $A$ as its position vector $\mathbf{a}=(2,-1,3)$. Then $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ gives $1(x-2)-2(y+1)+2(z-3)=0$. Why this step? Any in-plane arrow $\mathbf{r}-\mathbf{a}$ is perpendicular to $\mathbf{n}$, so its dot product with $\mathbf{n}$ is zero.

Step 2 — simplify. $x-2-2y-2+2z-6=0\Rightarrow x-2y+2z=10$. Why this step? The constant is $d=\mathbf{a}\cdot \mathbf{n}=2+2+6=10$. Here it landed positive despite the negative in $\mathbf{n}$.

Step 3 — distance from $P=(0,0,0)$, whose position vector is $\mathbf{p}=(0,0,0)$: $\text{dist}=\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}=\frac{|0-10|}{\sqrt{1+4+4}}=\frac{10}{3}.$ Why the absolute value? Distance is never negative; $|\cdot |$ handles whichever side of the plane $P$ sits on.

Verify: $\mathbf{n}=(1,-2,2)$, $|\mathbf{n}|=\sqrt{9}=3$; $10/3\approx 3.33$. Also $A$ satisfies the plane: $2-2(-1)+2(3)=2+2+6=10$ ✓.

---

Example 4 — C4: Plane through three points

Step 1 — two spanning directions. $\mathbf{u}=B-A=(1,-1,0)$, $\mathbf{v}=C-A=(0,1,2)$. Why this step? A plane is "one point plus two independent walking directions." These are the two directions.

Step 2 — normal via cross product. The Cross product gives the arrow perpendicular to both: $\mathbf{n}=\mathbf{u}\times \mathbf{v}=((-1)(2)-(0)(1),(0)(0)-(1)(2),(1)(1)-(-1)(0))=(-2,-2,1).$ Why this step? By definition $\mathbf{u}\times \mathbf{v}\perp \mathbf{u}$ and $\perp \mathbf{v}$ — exactly the normal we need.

Step 3 — Cartesian equation. Using point $A$: $-2(x-1)-2(y-1)+1(z-1)=0\Rightarrow -2x-2y+z=-3$, i.e. $2x+2y-z=3.$ Why this step? Once we have the normal, the plane is just "all points whose arrow from $A$ is perpendicular to $\mathbf{n}$" — i.e. $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Multiplying that dot product out and collecting terms turns the geometry into a single tidy Cartesian equation whose coefficients ARE the normal.

Verify: Each point must satisfy it. $A$: $2+2-1=3$ ✓. $B$: $4+0-1=3$ ✓. $C$: $2+4-3=3$ ✓.

---

Example 5 — C5: Degenerate — three points that DON'T make a plane

Step 1 — spanning directions. $\mathbf{u}=B-A=(1,1,1)$, $\mathbf{v}=C-A=(2,2,2)$. Why this step? Same recipe as Ex 4 — but watch the two directions.

Step 2 — cross product. $\mathbf{n}=\mathbf{u}\times \mathbf{v}=(1\cdot 2-1\cdot 2,1\cdot 2-1\cdot 2,1\cdot 2-1\cdot 2)=(0,0,0).$ Why this step? The cross product is $0$ exactly when the two directions are parallel. Here $\mathbf{v}=2\mathbf{u}$.

Step 3 — interpret. A zero normal means "no unique perpendicular exists." The three points are collinear — they all lie on the single line $\mathbf{r}=t(1,1,1)$. Three collinear points define infinitely many planes (a whole hinge of them), so there is no one plane. Why this step? The algebra ($\mathbf{n}=0$) is the warning light for a degenerate geometric setup.

Verify: Is $C$ on line through $A,B$? $C=(2,2,2)=2\cdot (1,1,1)$ — yes, $t=2$. Collinear confirmed ✓.

---

Example 6 — C6: Point-to-plane distance, both sides

Step 1 — set up. $\mathbf{n}=(2,2,-1)$, $d=3$, $|\mathbf{n}|=\sqrt{4+4+1}=3$. Why this step? The distance formula needs three ingredients: the normal $\mathbf{n}$ (read straight off the coefficients), the constant $d$ (the right-hand side), and the normal's length $|\mathbf{n}|$ (to convert a raw dot-product difference into an actual distance). Gathering them first keeps both point-computations clean.

Step 2 — signed value for $P$. With $\mathbf{p}=(3,3,3)$: $\mathbf{p}\cdot \mathbf{n}-d=(6+6-3)-3=6$. So ${\text{dist}}_P=\frac{|6|}{3}=2$. Why this step? The numerator $\mathbf{p}\cdot \mathbf{n}-d$ measures how far $P$'s "height along $\mathbf{n}$" exceeds the plane's height. Its sign tells you which side.

Step 3 — signed value for $Q$. With $\mathbf{q}=(1,1,1)$: $\mathbf{q}\cdot \mathbf{n}-d=(2+2-1)-3=-3$. So ${\text{dist}}_Q=\frac{|-3|}{3}=1$. Why this step? $Q$ gives a negative signed value while $P$ gave positive ⇒ they sit on opposite sides of the plane.

Verify: Signs $+6$ vs $-3$ differ ⇒ opposite sides ✓. Distances $2$ and $1$ are both positive as distances must be ✓.

---

Example 7 — C7: Point-to-line distance in 3D

Step 1 — the vector from a line point to $P$. Take $A=(0,0,0)$ (position vector $\mathbf{a}$) and $\mathbf{p}=(1,2,3)$: $\mathbf{p}-\mathbf{a}=(1,2,3)$. Why this step? Distance to a line is the perpendicular drop from $P$; we start from any known line point.

Step 2 — why the cross product. We want $|\mathbf{p}-\mathbf{a}|\sin \theta$ (the perpendicular height), where $\theta$ is the angle to the line. The Cross product magnitude gives exactly $|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$, so dividing by $|\mathbf{d}|$ isolates the height: $\text{dist}=\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}.$ Why not just project? Projection gives the along-line component; we need what's left over perpendicular — the cross product captures that directly.

Step 3 — compute. $(1,2,3)\times (1,0,0)=(2\cdot 0-3\cdot 0,3\cdot 1-1\cdot 0,1\cdot 0-2\cdot 1)=(0,3,-2)$. Magnitude $\sqrt{0+9+4}=\sqrt{13}$. And $|\mathbf{d}|=1$. So $\text{dist}=\sqrt{13}$. Why this step? This just executes the formula from Step 2 with our numbers: form the cross product, take its length, and divide by $|\mathbf{d}|$. Because $|\mathbf{d}|=1$ here, the cross-product length IS the distance — no rescaling needed.

Verify: Distance from $(1,2,3)$ to the $x$-axis should be $\sqrt{y^2+z^2}=\sqrt{4+9}=\sqrt{13}$ ✓ — the $x$-coordinate is irrelevant, as expected.

---

Example 8 — C8: Line parallel to a plane vs hitting it

Step 1 — the deciding test. A line is parallel to a plane when its direction is perpendicular to the normal, i.e. $\mathbf{d}\cdot \mathbf{n}=0$ (the direction has no component along $\mathbf{n}$, so it never gains "height"). Why this step? Moving along $\mathbf{n}$ is the only way to leave/approach the plane; $\mathbf{d}\cdot \mathbf{n}$ measures that motion.

Step 2 — test $L_1$. ${\mathbf{d}}_1\cdot \mathbf{n}=(2)(1)+(1)(2)+(-2)(2)=2+2-4=0$. So $L_1$ is parallel to the plane. Why this step? We apply the Step-1 test to $L_1$'s specific direction $(2,1,-2)$. Getting exactly $0$ confirms the direction has no along-normal component, so the line stays at constant height relative to $\mathbf{n}$ — it can never approach or leave the plane. Since $L_1$ passes through the origin and $0+0+0=0\ne 6$, that origin is off the plane, so $L_1$ never touches it.

Step 3 — test $L_2$. ${\mathbf{d}}_2\cdot \mathbf{n}=(1)(1)+0+0=1\ne 0$. Nonzero ⇒ $L_2$ crosses the plane. Find where: $x+0+0=6\Rightarrow t=6$, intersection at $(6,0,0)$. Why this step? Nonzero dot means the line steadily changes height along $\mathbf{n}$, so it must hit any given height exactly once — see Intersection of lines and planes.

Verify: $L_1$: ${\mathbf{d}}_1\cdot \mathbf{n}=0$ ✓ (parallel). $L_2$ intersection $(6,0,0)$: $6+0+0=6$ ✓.

---

Example 9 — C9: Word problem (drone flight)

Step 1 — model the path. Direction $\mathbf{d}=B-A=(30,40,60)$. Path: $\mathbf{r}=(0,0,0)+t(30,40,60)$, so $x=30t,y=40t,z=60t$, where $\mathbf{r}$ is the drone's position at "clock reading" $t$. Why this step? "Straight flight through two known points" is exactly a line — point + direction.

Step 2 — model the fence and find $t$. Fence is $z=50$. Set $60t=50\Rightarrow t=\frac{5}{6}$. Why this step? Crossing means the $z$-coordinate equals the fence height; solve for the single unknown $t$.

Step 3 — crossing point. $x=30\cdot \frac{5}{6}=25$, $y=40\cdot \frac{5}{6}=\frac{100}{3}\approx 33.33$, $z=50$. Point $=(25,33.33,50)$ m. Why this step? The value $t=\frac{5}{6}$ is the "clock reading" at which the drone reaches the fence; substituting it back into the path's $x$ and $y$ equations converts that time into the actual location in space where the crossing happens.

Step 4 — distance travelled to fence. $|\mathbf{d}|=\sqrt{30^2+40^2+60^2}=\sqrt{900+1600+3600}=\sqrt{6100}=10\sqrt{61}$. Distance $=t|\mathbf{d}|=\frac{5}{6}\cdot 10\sqrt{61}=\frac{25\sqrt{61}}{3}\approx 65.1$ m. Why this step? Distance flown is (fraction of the way $=t$) $\times$ (length of one direction step).

Verify: At $t=\frac{5}{6}$, $z=60\cdot \frac{5}{6}=50$ ✓ (on fence). $\frac{25\sqrt{61}}{3}\approx 65.1$ m, and $\sqrt{25^2+33.33^2+50^2}\approx \sqrt{625+1111+2500}=\sqrt{4236}\approx 65.1$ ✓.

---

Example 10 — C10: Exam twist — parallel, intersecting, or skew?

Step 1 — parallel check (both pairs). Directions parallel ⇔ one is a scalar multiple of the other. Pair A: $(2,2,0)=2(1,1,0)$ — parallel directions. Pair B: $(1,0,0)$ vs $(0,1,0)$ — not parallel. Why this step? Parallel directions collapse three possibilities to two (same line, or parallel-and-distinct).

Step 2 — resolve Pair A. Directions parallel: are they the same line? Check if $L_2$'s point $(0,5,0)$ lies on $L_1$: need $(0,5,0)=(1+t,t,0)$. From row 1, $t=-1$; then row 2 gives $t=5$. Contradiction ⇒ point not on $L_1$. So parallel and distinct (never meet). Why this step? Same direction + a shared point ⇒ identical line; same direction + no shared point ⇒ parallel distinct.

Step 3 — resolve Pair B (not parallel). Solve for a meeting point: $L_3$ gives $(t,0,0)$; $L_4$ gives $(0,s,1)$. Match: $t=0$, $0=s$, and $z$: $0=1$. The last is impossible ⇒ no intersection. Not parallel + no intersection = skew. Why this step? When directions aren't parallel, either the equations have a solution (they meet) or they're inconsistent (skew).

Verify: Pair A directions: $2\cdot (1,1,0)=(2,2,0)$ ✓ parallel; point test $t=-1$ vs $t=5$ inconsistent ✓ distinct. Pair B: the $z$-equation $0=1$ is false ✓ ⇒ skew.

---

Recall Active recall — cover the answers

• Direction component is $0$: how do you write symmetric form? ::: split that coordinate off as $z=\text{const}$; use ratios only for nonzero components.
• Cross product of two spanning directions is $0$ — what does it mean? ::: the points are collinear; no unique plane exists.
• One dot product decides line-vs-plane parallelism — which? ::: $\mathbf{d}\cdot \mathbf{n}=0$ means parallel; nonzero means it crosses.
• Two 3D lines, not parallel, never meet — what are they called? ::: skew.
• Which side of a plane is a point on? ::: sign of $\mathbf{p}\cdot \mathbf{n}-d$: positive vs negative = opposite sides.

---

Connections

• Lines and planes in 3D — vector equations — the parent toolkit these examples exercise.
• Cross product — builds normals (Ex 4, 5) and point-to-line distance (Ex 7).
• Vectors and scalar (dot) product — the parallel/side tests (Ex 3, 6, 8).
• Vector projection — geometry behind the distance formulas.
• Intersection of lines and planes — Ex 8, 9, 10.
• Systems of linear equations — solving for meeting parameters (Ex 8–10).
• Parametrisation and parameters — the roles of $t,s$.