Equation of a plane — normal form, intercept form, general form

3.6.8 · Maths › 3D Geometry

WHY a plane needs a normal

Derivation 1 — General (point–normal) form

KYA chahiye: ek equation jo point $r = (x, y, z)$ satisfy kare iff woh us plane pe ho jo ek known point $r_{0} = (x_{0}, y_{0}, z_{0})$ se guzarti hai aur jiska normal $n = (a, b, c)$ hai.

KAISE (first principles): $r_{0}$ se $r$ tak ka vector plane mein pada hai, isliye woh $n$ ke perpendicular hona chahiye: $n \cdot (r - r_{0}) = 0.$ Yeh step kyun? Perpendicular ⇒ dot product zero — yahi normal ki definition hai.

$n = (a, b, c)$ se expand karke: $a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0.$ Constants $d = - (a x_{0} + b y_{0} + c z_{0})$ collect karo: $a x + b y + cz + d = 0 (general form)$

Derivation 2 — Normal form (foot of perpendicular)

KYA: plane ko unit normal $\overset{n}{^}$ aur origin se perpendicular distance $p \geq 0$ use karke likho.

KAISE: Origin $O$ se plane pe ek perpendicular giraao; woh plane ko foot $N$ pe hit karta hai jahan $O N = p \overset{n}{^}$ (length $p$ , direction $\overset{n}{^}$ ). Plane pe kisi bhi point $r$ ke liye, $r - p \overset{n}{^}$ plane mein pada hai, isliye $\overset{n}{^}$ ke ⟂ hai: $\overset{n}{^} \cdot (r - p \overset{n}{^}) = 0 \Rightarrow \overset{n}{^} \cdot r - p = 1 (\overset{n}{^} \cdot \overset{n}{^}) = 0.$ $\overset{n}{^} \cdot \overset{n}{^} = 1$ kyun? Kyunki $\overset{n}{^}$ ek unit vector hai.

General → Normal convert karna: $a x + b y + cz = - d$ ko $\pm ∣ n ∣ = \pm a^{2} + b^{2} + c^{2}$ se divide karo, sign aise choose karo ki right side $p$ non-negative ho: $\frac{a x + b y + cz}{a ^{2} + b ^{2} + c ^{2}} = \frac{- d}{a ^{2} + b ^{2} + c ^{2}} = p .$

Derivation 3 — Intercept form

KYA: agar ek plane axes ko $(a, 0, 0)$ , $(0, b, 0)$ , $(0, 0, c)$ pe kaate — yeh intercepts $a, b, c$ hain.

KAISE: General $A x + B y + C z = k$ se shuru karo ( $k \neq = 0$ use karo, warna woh origin se guzarti hai aur uske finite intercepts nahi hote). Har axis point plug karo:

$(a, 0, 0)$ : $A a = k \Rightarrow A = k / a$ .
$(0, b, 0)$ : $B b = k \Rightarrow B = k / b$ .
$(0, 0, c)$ : $C c = k \Rightarrow C = k / c$ .

Substitute karo aur $k$ se divide karo: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$ $k$ se divide kyun? Right side ko 1 pe normalise karne ke liye, jisse intercepts directly read ho sakein.

Figure — Equation of a plane — normal form, intercept form, general form

Worked examples

Common mistake Classic traps ko steel-man karna

Trap A: "Normal form ka matlab sirf $a x + b y + cz = d$ hai." Yeh sahi kyun lagta hai: yeh clean dikhta hai. Fix: normal form ke liye unit normal zaroori hai — $l^{2} + m^{2} + n^{2} = 1$ — aur $p \geq 0$ . Pehle hamesha $\pm a^{2} + b^{2} + c^{2}$ se divide karo.

Trap B: galat sign choose karna taaki $p < 0$ ho jaaye. Yeh sahi kyun lagta hai: tum bas divide karte ho aur aage badh jaate ho. Fix: $p$ ek distance hai, $\geq 0$ honi chahiye. Agar $- d$ negative hai, toh $- a^{2} + b^{2} + c^{2}$ se divide karo taaki $p$ aur direction dono consistently flip ho jaayein.

Trap C: origin se guzarne wali plane pe intercept form force karna. Yeh sahi kyun lagta hai: har plane ke "intercepts hone chahiye." Fix: agar $d = 0$ toh plane $O$ se guzarti hai; $0$ ka intercept $\frac{x}{0}$ undefined banata hai. Intercept form ke liye $d \neq = 0$ zaroori hai.

Trap D: $a x + b y + cz + d = 0$ mein $(a, b, c)$ ko intercepts samajhna. Fix: wahan yeh normal components hain; intercepts tabhi aate hain jab tum $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1$ ki tarah rewrite karo.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek flat cardboard ka tukda hawa mein tair raha hai. Uss cardboard ke bahar se seedha ek pencil nikalo — woh pencil "normal" hai. Cardboard exactly un sabhi points ka set hai jinhe pencil across point karta hai, kabhi uske saath nahi. Yeh batane ke liye ki cardboard kahan hai, tumhe sirf itna chahiye: pencil kis taraf point kar rahi hai, aur cardboard kamre ke kone se kitni door hai. Equation $a x + b y + cz = d$ bas yahi likhti hai — " $(a, b, c)$ pencil hai, aur $d$ batata hai kitni door hai." "Intercept" waala version iske bajaye batata hai ki cardboard teeno walls se kahan paar nikalta hai. Same cardboard, use describe karne ke alag alag tarike.

Connections

Flashcards

a x + b y + cz + d = 0

equation tumhe plane ke baare mein immediately kya batati hai?

Vector

(a, b, c)

plane ke normal (perpendicular) hai.

General form ko normal form mein convert kaise karte hain?

a x + b y + cz = - d

\pm a^{2} + b^{2} + c^{2}

se divide karo, sign aise choose karo ki RHS

p \geq 0

ho; phir LHS coefficients direction cosines ban jaate hain.

Normal form

l x + m y + n z = p

mein

l, m, n

kaun si condition satisfy karte hain?

l^{2} + m^{2} + n^{2} = 1

(yeh unit normal ke direction cosines hain).

Normal form mein

p

kya represent karta hai?

Origin se plane ki perpendicular distance (

p \geq 0

Intercept form likho aur batao

a, b, c

ka kya matlab hai.

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

;

a, b, c

the

x, y, z

-axis intercepts hain.

Kab ek plane ka intercept form NAHI hota?

Jab woh origin se guzarti hai (

d = 0

Teen points

A, B, C

se normal kaise nikaalte hain?

n = A B \times A C

r_{0}

se guzarne wali plane ka point–normal form kya hai jiska normal

n

hai?

n \cdot (r - r_{0}) = 0

2 x - y + 2 z = 6

ke intercepts nikalo.

3, - 6, 3

(6 se divide karo).

General form mein

x, y, z

ke coefficients intercepts kyun nahi hote?

Yeh normal components hain; intercepts tabhi aate hain jab RHS ko 1 pe normalise karo.