Visual walkthrough — Vector addition — triangle law, parallelogram law

1.1.8 · D2 · Physics › Measurement, Vectors & Kinematics › Vector addition — triangle law, parallelogram law

Hum do arrows add karte hain aur sirf ek hi sawaal poochte hain: jahan maine shuru kiya wahan se jahan main ruka, woh seedha arrow kitna lamba hai, aur kis direction mein point karta hai?

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Step 1 — Do arrows, ek shared corner

KYA. Hum do vectors $\vec{A}$ aur $\vec{B}$ ko ek hi point $O$ se draw karte hain. Yahan arrow ka matlab hai ek safar: uski length batati hai kitna door, uski direction batati hai kis taraf. Dono ke beech ka space ka wedge angle $\theta$ (theta, ek Greek letter jo "unke beech ka angle" ka naam hai) hai, jo $\vec{A}$ se $\vec{B}$ ki taraf counter-clockwise sweep karta hai.

Tail-to-tail kyun shuru karein? Kyunki jawab decide karne wala ekmaatra geometry piece dono arrows ke beech ki opening hai. Agar hum $A$ ($\vec{A}$ ki length), $B$ ($\vec{B}$ ki length), aur $\theta$ (opening) jaante hain, toh baaki sab forced hai.

PICTURE. $O$ se nikalte dono arrows aur unke beech ka shaded wedge $\theta$ dekho; chhota curved arrow woh counter-clockwise direction dikhata hai jismein hum measure karte hain.

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Step 2 — $\vec{A}$ ko flat rakho: measure karne ke liye ek ruler banao

KYA. Hum poori picture ko rotate karte hain taaki $\vec{A}$ ek horizontal line (the x-axis) ke along flat ho jaaye aur $O$ horizontal aur vertical lines ke crossing point (the origin) pe baithe. Poori picture rotate karna jawab ko kuch nahi badalta — arrows apni lengths aur apna opening $\theta$ rakhte hain.

KYUN. Horizontal-and-vertical grid se hum kisi bhi point ko do plain numbers se describe kar sakte hain: kitna right (use $x$ kaho) aur kitna up (use $y$ kaho). Numbers hum add kar sakte hain; slanted arrows nahi. Yahi woh trick hai jo geometry ko arithmetic mein badal deti hai — Resolution of a vector into components ke peeche yahi idea hai. Kyunki $\vec{A}$ ab $+x$ ki taraf point karta hai aur $\theta$ us se counter-clockwise measure hota hai, $\vec{B}$ upar aur side ki taraf jhukta hai — grid ke upper half mein.

PICTURE. $\vec{A}$ ab bottom ke along chalta hai; uski tip point $(A,0)$ pe hai — $A$ kitna right, $0$ kitna up.

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Step 3 — $\vec{B}$ ko "right" aur "up" mein toro

KYA. $\vec{B}$ ki tip se hum x-axis pe ek seedhi vertical line drop karte hain. Is se ek right triangle banta hai jiska slanted side (hypotenuse) $\vec{B}$ khud hai, aur angle $\theta$ $O$ pe baitha hai.

Yeh do pieces kyun? Hum chahte hain $\vec{B}$ (kitna right) aur (kitna up) mein likha jaaye taaki hum use $\vec{A}$ ke numbers ke saath add kar sakein. Woh tool jo angle-plus-hypotenuse ko right/up pair mein read karta hai woh trigonometry hai:

• $\cos \theta =\frac{\text{adjacent (horizontal side)}}{\text{hypotenuse}}$ — toh horizontal side $=B\cos \theta$.
• $\sin \theta =\frac{\text{opposite (vertical side)}}{\text{hypotenuse}}$ — toh vertical side $=B\sin \theta$.

"Right" part ke liye cosine kyun? Cosine measure karta hai ki arrow ka kitna part $\vec{A}$ ki direction ke along point karta hai; sine measure karta hai kitna across point karta hai. Yahi exactly woh split hai jo hum chahte hain. Kyunki $\theta \in [0^{\circ },180^{\circ }]$, up-part $B\sin \theta \ge 0$ (the tip of $\vec{B}$ base line ke neeche kabhi nahi jaati), jabki right-part $B\cos \theta$ positive ho sakta hai (agar $\theta <90^{\circ }$), zero (at $90^{\circ }$), ya negative (agar $\theta >90^{\circ }$, yaani $\vec{B}$ backward jhukta hai).

PICTURE. $\vec{B}$ ke neeche dotted right triangle: base $B\cos \theta$, height $B\sin \theta$.

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Step 4 — $\vec{B}$ ko $\vec{A}$ ki tip pe slide karo (head-to-tail)

KYA. $\vec{B}$ ki length aur direction same rakho, lekin use slide karo taaki uski tail $\vec{A}$ ke head pe baithe. Yahi triangle law hai: "pehli wali khatam hone ke baad safar jaari rakho." Ek arrow ko bina ghumaaye slide karna use nahi badalta.

KYUN. $\vec{A}$ chalne ke baad aap $(A,0)$ pe ho. Wahan se $\vec{B}$ chalna aapki right-position mein $B\cos \theta$ add karta hai aur up-position mein $B\sin \theta$ add karta hai. Toh woh jagah jahan aap finally pahunchte ho — resultant ka head — yeh hai:

PICTURE. $\vec{A}$ base ke along, $\vec{B}$ uski tip pe baitha, aur finish point $(R_x,R_y)$ mark kiya.

Yahan $R_x$ start se finish tak ka total rightward distance hai aur $R_y$ total upward distance. $O$ se $(R_x,R_y)$ tak ka seedha arrow resultant $\vec{R}=\vec{A}+\vec{B}$ hai, aur uski length hum $R=|\vec{R}|$ kehte hain.

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Step 5 — Pythagoras: seedha-ghar arrow ki length

KYA. Resultant $R_x$ right aur $R_y$ up tak pahunchta hai. Woh dono ek right triangle ke legs banaate hain; $\vec{R}$ hypotenuse hai, aur uski length $R=|\vec{R}|$ hai.

Pythagoras kyun aur kuch nahi? Kyunki $R_x$ (horizontal) aur $R_y$ (vertical) construction se right angle pe milte hain — grid square hai. Woh theorem jo do perpendicular legs ko straight-line distance mein badalta hai exactly $(\text{leg}{)}^2+(\text{leg}{)}^2=(\text{hypotenuse}{)}^2$ hai.

PICTURE. Right triangle $O$–(foot)–(head), legs $R_x$, $R_y$ aur hypotenuse $R$ ke saath.

$R=\sqrt{R_x^2+R_y^2}=\sqrt{\underset{R_x^2}{\underbrace{(A+B\cos \theta {)}^2}}+\underset{R_y^2}{\underbrace{(B\sin \theta {)}^2}}}$

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Step 6 — Multiply out karo, aur ek term vanish dekho

KYA. Hum square expand karte hain. $(A+B\cos \theta {)}^2$ expand karne se teen pieces milte hain; $(B\sin \theta {)}^2$ ek aur add karta hai:

$R^2=\underset{\text{from }{A}^2}{\underbrace{A^2}}+\underset{\text{cross term}}{\underbrace{2AB\cos \theta }}+\underbrace{B^2{\cos }^2\theta }+\underbrace{B^2{\sin }^2\theta }$

KYUN simplify hota hai. Last do mein $B^2$ common hai: $B^2({\cos }^2\theta +{\sin }^2\theta )$. Aur ${\cos }^2\theta +{\sin }^2\theta =1$ hamesha — yeh har angle ke andar chhote unit triangle pe Pythagoras lagana hai. Toh woh dono ek single $B^2$ mein collapse ho jaate hain.

PICTURE. Chaar terms ka colour-coded map: do "clean" squares $A^2$, $B^2$, aur orange cross term $2AB\cos \theta$ jo saari angle information carry karta hai.

Yeh Law of cosines ka disguise mein roop hai; andar ka $\cos \theta$ exactly woh quantity hai jo ek Scalar (dot) product deta.

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Step 7 — $\vec{R}$ kis taraf point karta hai?

KYA. Hum $\alpha$ (alpha) chahte hain, woh angle jo $\vec{R}$ base arrow $\vec{A}$ ke saath banata hai, $\vec{A}$ se counter-clockwise measure kiya bilkul $\theta$ ki tarah. Step 5 ke right triangle mein, $\alpha$ ke opposite wali side $R_y$ hai aur $\alpha$ ke adjacent wali side $R_x$ hai.

Tangent kyun? Hum dono legs jaante hain lekin koi hypotenuse-angle nahi. Woh ek trig ratio jo opposite over adjacent se bana hai — bina hypotenuse ya kuch aur ki zaroorat ke — tangent hai. Yeh precisely jawaab deta hai "yeh arrow base ke upar kitna steep hai?"

PICTURE. Angle $\alpha$ $O$ pe $\vec{A}$ aur $\vec{R}$ ke beech tucked, $R_y$ (opposite) aur $R_x$ (adjacent) label kiye hue.

$\alpha$ wapas padhna — aur ${\tan }^{-1}$ akela kyun kaafi nahi. Ordinary inverse-tangent ${\tan }^{-1}$ (jise $\arctan$ bhi likhte hain) jawaab deta hai "is tangent wala angle kaun sa hai?", lekin woh sirf apni principal range $(-90^{\circ },+90^{\circ })$ mein hi jawaab de sakta hai — woh angles jo plane ke right half mein point karte hain. Hamaara resultant, magar, up-and-left bhi point kar sakta hai (jab $R_x<0$), aur ek bare ${\tan }^{-1}$ use galat half mein fold kar dega. Saaf fix ek aisi function hai jo dono $R_y$ aur $R_x$ ko alag-alag dekhti hai (unke individual signs, sirf unka ratio nahi) aur ek hi shot mein correct wedge return karti hai. Woh function $atan2$ kehlati hai:

Last do lines woh degenerate "flat resultant" outcomes hain; neeche ka middle picture dikhata hai yeh kyun hota hai.

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Step 8 — Har edge case, drawn

KYA. Hum $\theta$ ko $0^{\circ }$ se $180^{\circ }$ tak sweep karte hain aur dekhte hain $R$ kaise badalta hai. Sirf cross term $2AB\cos \theta$ hilta hai, kyunki $\cos \theta$ $+1$ se $-1$ tak slide karta hai.

Yeh check kyun karein? Ek formula jo aap stress-test nahi kar sakte woh formula nahi hai jis par aap trust karte hain. Har limit common sense se match karni chahiye.

PICTURE. Teen stacked diagrams — aligned, perpendicular, opposed — resultant har baar redrawn ke saath.

$\theta$	$\cos \theta$	$R=\sqrt{A^2+B^2+2AB\cos \theta }$	$\alpha =atan2(R_y,R_x)$	matlab
$0^{\circ }$	$+1$	$\sqrt{(A+B{)}^2}=A+B$	$0^{\circ }$ ($R_y=0,R_x>0$)	arrows aligned → sabse lamba
$90^{\circ }$	$0$	$\sqrt{A^2+B^2}$	quadrant I ($R_x=A>0$)	perpendicular → plain Pythagoras
$180^{\circ }$	$-1$	$\sqrt{(A-B)^2}=	A-B	$

$\theta =0^{\circ }$ aur $\theta =180^{\circ }$ pe $R_y=B\sin \theta =0$ hai — resultant base line ke along flat rehta hai, toh $atan2$ $0^{\circ }$ return karta hai (head $+\vec{A}$ ke along, jab $R_x>0$) ya $180^{\circ }$ (head $-\vec{A}$ ke along, jab $R_x<0$), woh do "$R_y=0$" outcomes jo Step 7 mein listed hain.

Toh resultant $|A-B|\le R\le A+B$ mein trapped hai. Agar additionally $A=B$ ho, toh $\theta =180^{\circ }$ pe dono arrows cancel ho jaate hain aur $R=0$ — woh degenerate zero vector, ek aisa safar jo wahan khatam hota hai jahan shuru hua tha (aur $\alpha$ undefined hai, kyunki zero-length arrow kahin point nahi karta).

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Ek-picture summary

Upar ki saari cheez ek diagram mein rehti hai: $\vec{A}$ ko flat rakho, $\vec{B}$ ke components drop karo, right triangle close karo, $R$ Pythagoras se padho aur $\alpha$ $atan2$ se.

Recall Feynman: ek 12-saal ke bacche ko batao

Do arrows ko ek hi dot pe tip-tails mein rakho. Page ghumaao taaki pehla arrow flat ek ruler ki tarah lie kare, aur hamesha angles us ruler se anti-clockwise jaake measure karo. Ab doosra arrow us se kisi slant pe jhukta hai — us slant arrow ko "kitna sideways" (yeh cosine part hai) aur "kitna upar" (yeh sine part hai, jo positive rehta hai kyunki arrow upar jhukta hai) mein tod do. Doosra arrow pehle ke end pe stack karo: ab aap sideways $A+B\cos \theta$ aur upar $B\sin \theta$ chale ho. Seedhi line ghar tak khiincho. Uski length $R$ flat-triangle rule $a^2+b^2=c^2$ se milti hai — woh deta hai $R=\sqrt{A^2+B^2+2AB\cos \theta }$ — aur uska tilt hai "up-part over sideways-part." Agar home-arrow seedha upar point kare, uska tilt exactly $90^{\circ }$ hai; agar woh left ki taraf peeche jhuke, toh $90^{\circ }$ aur $180^{\circ }$ ke beech hai — $atan2$ tool sort out karta hai kaun sa, sirf yeh dekhkar ki "sideways" positive hai ya negative. Doosre arrow ko lined-up se opposite tak swing karo aur home-arrow $A+B$ se $|A-B|$ tak shrink ho jaata hai. Yahi poori kahaani hai.

Recall Quick self-test

Cross term saari angle info kyun carry karta hai? ::: $A^2$ aur $B^2$ kabhi nahi badlate; sirf $2AB\cos \theta$ $\theta$ par depend karta hai. ${\cos }^2\theta +{\sin }^2\theta =1$ kahaan use hota hai? ::: Step 6 mein, $B^2{\cos }^2\theta +B^2{\sin }^2\theta$ ko $B^2$ mein collapse karne ke liye. Direction ke liye tangent kyun, sine kyun nahi? ::: Hum dono legs $R_x,R_y$ jaante hain lekin hypotenuse nahi; tangent = opposite/adjacent exactly wahi use karta hai. Sirf ek cross-term sign kyun aata hai? ::: Kyunki $\theta \in [0^{\circ },180^{\circ }]$ $\sin \theta \ge 0$ rakhta hai, toh $R_y$ upar rehta hai; saare sign changes $\cos \theta$ ke andar chhup jaate hain. $R$ (bina arrow ke) versus $\vec{R}$ ka kya matlab hai? ::: $\vec{R}$ resultant vector hai; $R=|\vec{R}|$ sirf uski length hai. Bare ${\tan }^{-1}$ se kaam kyun nahi chalta, aur kya replace karta hai use? ::: ${\tan }^{-1}$ sirf $(-90^{\circ },90^{\circ })$ return karta hai toh woh leftward ($R_x<0$) arrows galat jagah rakhta hai aur $R_x=0$ pe zero se divide karta hai; $atan2(R_y,R_x)$ dono signs use karke correct wedge deta hai.

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Connections

• Resolution of a vector into components — Steps 2–3 resolution hi hain.
• Vectors — components and unit vectors — $\hat{i},\hat{j}$ form mein $(R_x,R_y)$ bookkeeping.
• Law of cosines — Step 6 ka result rewritten.
• Scalar (dot) product — cross term ke andar $\cos \theta$ supply karta hai.
• Subtraction of vectors and the difference vector — $\theta =180^{\circ }$ edge case.
• Relative velocity — boats aur rivers ke liye same triangle.
• Vector addition — triangle law, parallelogram law — parent topic.