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

4.5.5 · D2 · Maths › Linear Algebra (Full) › Lines and planes in 3D — vector equations

Step 1 se pehle, teen words jo hum baar baar use karenge. Ek position vector bas ek aisa arrow hai jo fixed origin $O$ se kisi point tak khicha gaya ho — yeh ek point ko "$O$ se wahan kaise pahuncho" ke zariye naam deta hai. Origin ke point ko hum $O$ likhte hain aur position vectors ke liye bold letters jaise $\mathbf{a}$, $\mathbf{r}$ use karte hain. Ek displacement ek point se doosre point tak ka arrow hota hai (zaruri nahi ki $O$ se shuru ho); yeh kehta hai "itna aur is direction mein jao".

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Step 1 — Ek point ko ek arrow se naam do

KYA. Hum ek origin $O$ fix karte hain aur us object par ek khaas point $A$ chunte hain jise hum describe karna chahte hain. $O$ se $A$ tak ka arrow position vector $\mathbf{a}$ hai.

KYUN. Algebra ko numbers chahiye, aur space mein apne aap koi numbers nahi hote. $O$ chunne se har point ko ek address milta hai: point $A$ hi arrow $\mathbf{a}$ hai. Iske bina hum ek picture ko equation mein nahi badal sakte.

PICTURE. Amber arrow $\mathbf{a}$ hai: tail $O$ par, head $A$ par.

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Step 2 — Ek line ek aisi direction hai jise aap re-scale kar sakte ho

KYA. Ek doosra arrow $\mathbf{d}$ add karo, yeh direction hai jisme hum travel karne ki permission rakhte hain. Hum maante hain $\mathbf{d}\ne 0$ (ek zero arrow kahi nahi point karta, isliye woh koi line define nahi kar sakta).

KYUN. "Line" ka matlab hai: $A$ se aap sirf ek fixed heading ke aage ya peeche chal sakte ho. Woh ek heading bilkul wahi hai jo $\mathbf{d}$ store karta hai. Line ka har point is ek arrow ko stretch ya flip karke pahuncha jaata hai.

PICTURE. Cyan arrow $\mathbf{d}$ $A$ par placed hai. Faint white line har stretch of $\mathbf{d}$ hai jo $A$ se hoke end-to-end rakhi gayi hai.

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Step 3 — Parameter $t$: hum kitna aage chalte hain

KYA. Scalar (ek ordinary number) $t$ introduce karo. Scaled arrow $t\mathbf{d}$ ka matlab hai "$\mathbf{d}$ ke $t$ copies chalo". Tab jis point par hum land karte hain uski position hai $\mathbf{r}=\underset{\text{start at }A}{\underbrace{\mathbf{a}}}+\underset{\text{walk }t\text{ steps of }\mathbf{d}}{\underbrace{t\mathbf{d}}}.$

KYUN yeh tool — ek parameter, koi aur equation nahi? Hum chahte hain ek dial jo poori line sweep kare. Dial $t$ ko $-\infty$ se $+\infty$ tak ghumaane se har point exactly ek baar milta hai. Yahi parametrisation hai: "ek equation jo points satisfy karte hain" ki jagah "ek formula jo points produce karta hai" use karo. Yeh kabhi kisi cheez se divide nahi karta, isliye vertical ya awkward lines se koi pareshani nahi.

PICTURE. Teen positions: $t=0$ pe $A$ par land karo (amber), $t=1$ pe ek poora $\mathbf{d}$ aage, $t=-1$ pe ek poora $\mathbf{d}$ peeche.

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Step 4 — KYUN woh formula forced hai, guess nahi

KYA. Line par koi bhi point $P$ lo, position $\mathbf{r}$. $A$ se $P$ tak ka displacement $\overrightarrow{AP}=\mathbf{r}-\mathbf{a}$ hai (head minus tail). Kyunki $P$ line par lie karta hai, yeh displacement $\mathbf{d}$ ke saath point karta hai — yeh $\mathbf{d}$ ke parallel hai.

KYUN. Do parallel arrows hamesha ek doosre ke scalar multiples hote hain: ek doosre ka stretched copy hota hai. Isliye kisi number $t$ ke liye $\mathbf{r}-\mathbf{a}=t\mathbf{d}$. Rearrange karne par $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ wapas milta hai. Humne formula invent nahi kiya — "line par hona" ke meaning ne use produce kiya.

PICTURE. Green displacement $\mathbf{r}-\mathbf{a}$ exactly scaled $\mathbf{d}$ ke upar lie karta hai; label dikhata hai ki woh $t\mathbf{d}$ ke barabar hai.

Recall "Head minus tail" kyun?

$\overrightarrow{AP}$ ka matlab hai "$A$ se shuru karo, $P$ par khatam karo". $A$ se $P$ tak jaane ke liye tum $A$ tak ki walk undo karte ho ($\mathbf{a}$ subtract karo) aur $P$ tak ki walk karte ho ($\mathbf{r}$ add karo): $\overrightarrow{AP}=\mathbf{r}-\mathbf{a}$.

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Step 5 — Do points se hoke line (ek special case, koi nayi rule nahi)

KYA. Do points $A(\mathbf{a})$ aur $B(\mathbf{b})$ diye hain, hamare paas koi direction nahi hai — toh hum ek banate hain: displacement $\mathbf{d}=\mathbf{b}-\mathbf{a}$ $A$ se seedha $B$ ki taraf point karta hai.

KYUN. Line ke saath lie karne wala koi bhi arrow $\mathbf{d}$ ke roop mein kaam karta hai; $A$ se $B$ tak ka arrow obviously karta hai. Substitute karne par: $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{a})=(1-t)\mathbf{a}+t\mathbf{b}.$ Yahan $t=0$ se $\mathbf{a}$ milta hai (tum $A$ par ho), $t=1$ se $\mathbf{b}$ milta hai ($B$ par), aur $0\le t\le 1$ sirf unke beech ka segment trace karta hai; us range ke bahar tum poori line extend karte ho.

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

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Step 6 — Ek plane ko do directions chahiye

KYA. Ab $A$ par do non-parallel arrows $\mathbf{u}$ aur $\mathbf{v}$ allow karo. Ek point hai $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v},s,t\in \mathbb{R}.$

KYUN do dials. Ek flat sheet mein slide karne ke do independent tarike hote hain. Ek dial $s$ $\mathbf{u}$ ke saath slide karta hai, doosra dial $t$ $\mathbf{v}$ ke saath; unhe combine karne se sheet ka har point milta hai. Woh non-parallel hone chahiye — agar $\mathbf{v}$ $\mathbf{u}$ ki copy hoti toh sirf Step 3 wali line milti, kabhi poori sheet nahi.

PICTURE. Parallelogram grid jo $s\mathbf{u}$ (cyan) aur $t\mathbf{v}$ (amber) se spanned hai, $A$ se hoke plane tile kar raha hai.

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Step 7 — Ek cleaner idea: ek arrow jo plane ke across point kare

KYA. Do directions clumsy hain. Iske bajaye ek single arrow $\mathbf{n}$ attach karo jo poori sheet ke perpendicular ho — normal. Plane mein kisi bhi point $P$ ke liye, displacement $\mathbf{r}-\mathbf{a}$ plane mein flat lie karta hai, isliye woh $\mathbf{n}$ ke perpendicular hai.

KYUN yahan dot product aata hai. Hume "perpendicular" ke liye ek numerical test chahiye. Scalar (dot) product bilkul wahi tool hai: $\mathbf{x}\cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}|\cos \theta$, aur yeh zero theek tab hota hai jab angle $\theta$ $90^{\circ }$ ho (kyunki $\cos 90^{\circ }=0$). Koi doosra elementary operation perpendicularity ko itni seedha nahi padhta. Toh "$P$ plane mein hai" ek single clean equation ban jaata hai $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0.$

PICTURE. White sheet, amber normal $\mathbf{n}$ seedha khada hua, aur ek green in-plane displacement right angle par milta hua.

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Step 8 — Do plane views ko cross product se bridge karna

KYA. Agar hamare paas sirf $\mathbf{u}$ aur $\mathbf{v}$ hain, toh hum cross product se normal banate hain: $\mathbf{n}=\mathbf{u}\times \mathbf{v}.$

KYUN yeh tool. Cross product defined hi is liye hai ki woh apne dono inputs ke perpendicular ek vector output kare — exactly woh property jo ek normal mein honi chahiye. Toh yeh "do in-plane directions" ko ek hi stroke mein "ek across-plane arrow" mein convert kar deta hai. $\mathbf{n}=(a,b,c)$ likhne par, dot-product equation $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ Cartesian plane tak expand hoti hai $a(x-x_0)+b(y-y_0)+c(z-z_0)=0\Rightarrow ax+by+cz=d,$ jahan coefficients $(a,b,c)$ hi normal hain — seedha padhkar.

PICTURE. $\mathbf{u}$, $\mathbf{v}$ sheet mein; $\mathbf{n}=\mathbf{u}\times \mathbf{v}$ dono ke perpendicular bahar shoot karta hua.

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Step 9 — Degenerate cases jo aap kabhi skip nahi kar sakte

KYA & KYUN. Formulas quietly kuch cheezein non-zero assume karte hain. Jab woh nahi hote, yeh hota hai aur kya karna hai.

1. $\mathbf{d}=0$ (line). Koi direction nahi $\Rightarrow$ koi line nahi, sirf single point $A$. Forbidden — isliye definition mein $\mathbf{d}\ne 0$ ki demand hai.
2. Symmetric form mein zero denominator. Agar $\mathbf{d}=(a,b,0)$ hai toh tum $\frac{z-z_0}{0}$ nahi likh sakte. $z=z_0$ alag se batao (line $z$ mein level hai) aur sirf doosre do coordinates par symmetric form use karo.
3. $\mathbf{u}\parallel \mathbf{v}$ (plane). Parallel spanning arrows se $\mathbf{u}\times \mathbf{v}=0$ milta hai — ek zero "normal", matlab do directions collapse hokar ek ho gayi aur tumne plane nahi balki sirf ek line banayi. Genuinely non-parallel directions chuno (ya teen non-collinear points).
4. Plane se distance, $\mathbf{p}$ plane par. Tab $\mathbf{p}\cdot \mathbf{n}-d=0$, toh distance $0$ hai — formula phir bhi kaam karta hai, koi special case nahi.

PICTURE. Left: $\mathbf{u}\parallel \mathbf{v}$ collapse — parallelogram ek line mein flatten ho jaata hai, $\mathbf{n}$ vanish ho jaata hai. Right: horizontal line jahan $z=z_0$ alag se bataya jaata hai.

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Ek-picture summary

Is page ki har cheez ek single frame mein: ek point $A$ (Step 1 se) ek arrow $\mathbf{d}$ se line banata hai (Steps 2–4), do arrows $\mathbf{u},\mathbf{v}$ se plane banata hai (Step 6), aur woh plane normal $\mathbf{n}=\mathbf{u}\times \mathbf{v}$ se flat nail ho jaata hai, dot product se test hota hai (Steps 7–8).

Recall Feynman retelling — kisi dost ko batao

Shuru karo ek point $A$ par flag lagane se; origin se us flag tak ka arrow $\mathbf{a}$ hai, uska address. Line banane ke liye, apne aap ko ek arrow $\mathbf{d}$ aur ek dial $t$ do: "flag par shuru karo, phir $t$ arrows aage (ya peeche) chalo." Woh ek sentence hi $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ hai, aur yeh forced hai, kyunki $A$ se line ke kisi doosre point tak ka koi bhi safar sirf $\mathbf{d}$ ka stretched copy hai. Plane banane ke liye tumhe do sliding arrows aur do dials chahiye — ek room ke floor ko slide karne ke do tarike chahiye. Lekin ek slicker trick hai: floor ko is tarah describe karne ki jagah ki tum uske across kaise slide karte ho, use ek arrow se describe karo jo seedha uske oopar point kare, normal $\mathbf{n}$. Ek point floor par hai exactly tab jab $A$ se uski taraf chalna kabhi bhi us up-arrow ke relative upar ya neeche na jaaye — aur "koi rise ya fall nahi" dot product ke zero hone se measure hota hai. Agar tumhare paas sirf do sliding arrows the, toh cross product $\mathbf{u}\times \mathbf{v}$ woh up-arrow tumhare liye manufacture karta hai. Zeros ka dhyan rakho: ek line ko non-zero $\mathbf{d}$ chahiye; ek plane ko non-parallel arrows chahiye, ya up-arrow nothing mein collapse ho jaata hai.

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Connections

• Lines and planes in 3D — vector equations — parent jise yeh walkthrough derive karta hai.
• Vectors and scalar (dot) product — Step 7 mein perpendicular test.
• Cross product — Step 8 mein normal manufacture karta hai.
• Vector projection — Step 7 ke dot product ko distance formulas mein convert karta hai.
• Parametrisation and parameters — Steps 3 aur 6 ke dials $t,s$.
• Intersection of lines and planes — jahan yeh equations aage milti hain.
• Systems of linear equations — har Cartesian plane ek linear equation hai.