Equation of a line in 3D — vector, symmetric, parametric forms

---

1. Vector form — the master equation

WHY this works (derivation from scratch): Let $P$ be any point on the line and $A$ a fixed point on it. The vector $\vec{AP}$ lies along the line, so it must be parallel to the direction $\vec{b}$. Two parallel vectors are scalar multiples of each other: $\vec{AP} = \lambda\,\vec{b}.$ Now write $\vec{AP}$ using position vectors from origin $O$: $\vec{AP} = \vec{OP} - \vec{OA} = \vec{r} - \vec{a}.$ Substitute: $\vec{r} - \vec{a} = \lambda\,\vec{b}\;\;\Longrightarrow\;\; \boxed{\vec{r} = \vec{a} + \lambda\,\vec{b}}.$ Why this step? Because "being on the line" means the displacement from a fixed point is a multiple of the direction — there's nothing else a line is.

---

2. Parametric form — break the vector into coordinates

Let $\vec{a} = (x_1,y_1,z_1)$ and $\vec{b}=(a,b,c)$ (direction ratios). Write $\vec{r}=(x,y,z)$.

(x,y,z)=(x_1+\lambda a,\; y_1+\lambda b,\; z_1+\lambda c)$$ > [!formula] Parametric equations > $$x = x_1+\lambda a,\qquad y=y_1+\lambda b,\qquad z=z_1+\lambda c$$ > The numbers $a,b,c$ are the ==direction ratios== (DRs) of the line. **HOW to use it:** plug a value of $\lambda$ → get one point. $\lambda=0$ gives the anchor $A$; $\lambda=1$ gives $A+\vec{b}$, etc. --- ## 3. Symmetric (Cartesian) form — eliminate $\lambda$ > [!intuition] WHY eliminate $\lambda$ > $\lambda$ is "scaffolding." If we solve each parametric equation for $\lambda$, all three expressions equal the same $\lambda$, so they equal each other — and $\lambda$ vanishes. From the parametric equations (assuming $a,b,c\neq 0$): $$\lambda=\frac{x-x_1}{a},\qquad \lambda=\frac{y-y_1}{b},\qquad \lambda=\frac{z-z_1}{c}.$$ Set them equal: > [!formula] Symmetric form > $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\;(=\lambda)$$ > Read the **point** $(x_1,y_1,z_1)$ from the numerators, the **direction ratios** $(a,b,c)$ from the denominators. > [!mistake] When a denominator is zero > Wrong instinct: "write $\frac{z-z_1}{0}$ and move on." It *feels* fine because the pattern looks symmetric — but division by 0 is meaningless. > **Fix:** a zero DR means the line is **constant in that coordinate**. If $c=0$, write > $$\frac{x-x_1}{a}=\frac{y-y_1}{b},\qquad z=z_1.$$ > The third coordinate is frozen; it never changes along the line. --- ## 4. Line through two points > [!definition] Two-point form > A line through points $A(\vec{a})$ and $B(\vec{b})$ has direction $\vec{b}-\vec{a}$, so > $$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}).$$ > Cartesian: > $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}.$$ **WHY:** the direction "from $A$ to $B$" *is* $\vec{AB}=\vec{b}-\vec{a}$. The anchor is just $A$. Same master equation, new costume. --- ## 5. Worked examples > [!example] Example 1 — Build all three forms > Find vector, parametric & symmetric forms of the line through $A(2,-1,3)$ parallel to $\vec{b}=2\hat i+3\hat j-\hat k$. > > **Vector:** $\vec{r}=(2\hat i-\hat j+3\hat k)+\lambda(2\hat i+3\hat j-\hat k)$. > *Why?* Master equation: anchor $+$ $\lambda\cdot$direction. > > **Parametric:** $x=2+2\lambda,\; y=-1+3\lambda,\; z=3-\lambda$. > *Why?* Read components of $\vec{r}$ separately. > > **Symmetric:** $\dfrac{x-2}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{-1}$. > *Why?* Solve each parametric for $\lambda$ and equate. Note $y-(-1)=y+1$. > [!example] Example 2 — Two points > Line through $P(1,0,2)$ and $Q(4,5,2)$. > > Direction $=\vec{PQ}=(4-1,\,5-0,\,2-2)=(3,5,0)$. > *Why this step?* Direction is the displacement between the points. > > Since $c=0$ (z-DR is zero), the line lives in plane $z=2$: > $$\frac{x-1}{3}=\frac{y-0}{5},\qquad z=2.$$ > *Why?* The $z$-coordinate of both points is $2$, so it can't change. > [!example] Example 3 — Read off info from symmetric form > Given $\dfrac{x+1}{4}=\dfrac{y-2}{-1}=\dfrac{z}{3}$, state a point and direction. > > Match $\dfrac{x-x_1}{a}$: numerator $x+1=x-(-1)\Rightarrow x_1=-1$. Similarly $y_1=2$, and $z=z-0\Rightarrow z_1=0$. > **Point** $(-1,2,0)$, **direction** $(4,-1,3)$. > *Why?* Numerators encode the point, denominators the DRs — always check signs by rewriting as $x-x_1$. > [!example] Example 4 — Forecast-then-Verify > **Forecast:** Does point $(6,1,0)$ lie on $\vec r=(0,3,2)+\lambda(2,-1,-1)$? Guess first. > **Verify:** Solve $x$: $6=0+2\lambda\Rightarrow\lambda=3$. Check $y$: $3+3(-1)=0\neq 1$. ✗ > So the point is **not** on the line (one $\lambda$ must satisfy *all three* equations simultaneously). *Why?* The same parameter controls all coordinates. --- > [!mistake] Steel-manning three classic traps > 1. **"Direction ratios must be the actual point displacements."** Feels right because two-point form uses $(x_2-x_1)$ etc. But *any* scalar multiple of the direction works — $(2,3,-1)$ and $(4,6,-2)$ give the same line. **Fix:** DRs are unique only up to a nonzero scalar. > 2. **"Symmetric form's numerators are the direction."** Wrong — numerators give the **point**, denominators give the **direction**. **Fix:** always rewrite as $\frac{x-x_1}{a}$ and label both. > 3. **"A point is on the line if it fits one equation."** Feels sufficient. **Fix:** the *same* $\lambda$ must satisfy all three; check it everywhere. --- > [!recall]- Feynman: explain to a 12-year-old (click to reveal) > Imagine you're standing on a spot in a giant room and pointing your arm in some direction. The line is **every spot you'd hit if you walked straight along your arm**, forwards or backwards, any number of steps. The spot you start on is the "point," your arm's direction is the "direction." If you take $\lambda$ steps (negative = backwards), you land on the line. That's the whole secret — start point plus a direction. > [!mnemonic] Remember the structure > **"PADDLE"** → **P**oint **A**nchor, **D**irection **D**rives, **L**ambda **E**xplores. > And for symmetric form: **"Top = where you ARE (point), Bottom = where you GO (direction)."** --- ## #flashcards/maths What two pieces of data uniquely determine a line in 3D? ::: A point on the line and a direction vector parallel to it. Vector equation of a line through $\vec a$ parallel to $\vec b$? ::: $\vec r = \vec a + \lambda\vec b$, $\lambda\in\mathbb R$. Why is $\vec r-\vec a=\lambda\vec b$? ::: Because $\vec{AP}$ lies along the line so it is parallel to $\vec b$, hence a scalar multiple of it. In symmetric form, what do the numerators encode? ::: The fixed point $(x_1,y_1,z_1)$ on the line. In symmetric form, what do the denominators encode? ::: The direction ratios $(a,b,c)$ of the line. How do you handle a zero direction ratio, e.g. $c=0$? ::: Drop that fraction and write $z=z_1$ (the coordinate is constant) alongside the other equal ratios. Direction of the line through points $A$ and $B$? ::: $\vec b-\vec a$ (i.e. $\vec{AB}$). Are direction ratios unique? ::: No — only up to a nonzero scalar multiple; $(2,3,-1)$ and $(4,6,-2)$ describe the same direction. How to test if a point lies on a parametric line? ::: Find $\lambda$ from one coordinate and check the SAME $\lambda$ satisfies all three equations. What point do you get at $\lambda=0$? ::: The anchor point $\vec a$ itself. --- ## Connections - [[Vectors — scalar multiplication & parallel vectors]] (the engine behind $\vec{AP}=\lambda\vec b$) - [[Direction ratios and direction cosines]] - [[Angle between two lines in 3D]] - [[Shortest distance between two skew lines]] - [[Equation of a plane in 3D]] - [[Coplanarity of two lines]] - [[Distance of a point from a line in 3D]] ## 🖼️ Concept Map ```mermaid flowchart TD A[Line in 3D] -->|needs| PT[Anchor point A] A -->|needs| DIR[Direction vector b] PT -->|combine| VEC[Vector form r = a + lambda b] DIR -->|combine| VEC VEC -->|derived from| AP[AP parallel to b] VEC -->|read coordinates| PAR[Parametric form] PAR -->|x1 y1 z1| POINT[Point coords] PAR -->|a b c| DR[Direction ratios] PAR -->|eliminate lambda| SYM[Symmetric form] DR -->|denominators| SYM POINT -->|numerators| SYM VEC -->|lambda is| SCAL[Scalar parameter] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, 3D mein ek line ko fix karne ke liye sirf do cheezein chahiye: ek **point** (jahan se start karte ho) aur ek **direction** (kis taraf jaa rahe ho). Bas itna hi. Isi soch se saari forms nikalti hain. Master equation hai $\vec r=\vec a+\lambda\vec b$ — yahan $\vec a$ tumhara starting point hai, $\vec b$ direction hai, aur $\lambda$ ek "kitne steps" wala number hai jo positive ya negative kuch bhi ho sakta hai. > > Derivation simple hai: agar $P$ line par koi bhi point hai, to $\vec{AP}$ line ke saath-saath jaata hai, yaani $\vec b$ ke parallel. Parallel matlab scalar multiple, isliye $\vec{AP}=\lambda\vec b$. Ab $\vec{AP}=\vec r-\vec a$ likho, to seedha $\vec r=\vec a+\lambda\vec b$ aa jaata hai. Parametric form mein hum bas x, y, z alag-alag likh dete hain. Symmetric form mein har equation se $\lambda$ nikaal kar barabar kar dete hain, to $\lambda$ gayab ho jaata hai aur milta hai $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$. > > Do important baatein yaad rakhna. Pehla: symmetric form mein **upar (numerator) point** hota hai aur **neeche (denominator) direction**. Doosra: agar koi denominator zero hai (jaise $c=0$), to division mat karo — seedha likho $z=z_1$, kyunki us coordinate mein line move hi nahi karti. Aur agar check karna hai ki koi point line par hai ya nahi, to ek hi $\lambda$ se teeno equations satisfy honi chahiye — sirf ek se kaam nahi chalega. Isko samjhloge to poora 3D Geometry ka chapter aasaan lagega, kyunki angle, distance, coplanarity sab isi base par bante hain. ![[audio/3.6.06-Equation-of-a-line-in-3D-—-vector,-symmetric,-parametric-forms.mp3]]