Visual walkthrough — Resolution of vectors — into components (any axes)

1.1.9 · D2 · Physics › Measurement, Vectors & Kinematics › Resolution of vectors — into components (any axes)

Hum assume karte hain ki tumne pehle kabhi angle-wala arrow nahi dekha. Hum ek single dot se shuru karte hain.

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Step 1 — Vector ek picture ke roop mein kya hota hai?

KYA. Hum ek arrow draw karte hain. Yeh ek point se shuru hota hai (ise origin, $O$ kehte hain) aur ek tip par khatam hota hai. Arrow mein do cheezein baked in hain: yeh kitna lamba hai, aur yeh kis taraf point karta hai.

KYU. Ek arrow ko pieces mein kaatne se pehle, humein agree karna hai ki poora arrow hai kya. Length + direction — bas itna hi hai, kuch chhupa nahi.

PICTURE. Lal arrow dekho. Hum ise $\vec{A}$ kehte hain (upar ka chhota arrow sirf yeh batata hai ki "yeh ek vector hai, ordinary number nahi"). Iske length ko hum $A=|\vec{A}|$ likhte hain — seedhe vertical bars ka matlab hai "kitna lamba," aur length kabhi negative nahi hoti.

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Step 2 — Arrow ko ek grid par pin karo: angle $\theta$

KYA. Hum do number-lines bichhaate hain jo $O$ par perfect right angle par cross karti hain: horizontal x-axis aur vertical y-axis. Phir hum positive x-axis se angle measure karte hain, anticlockwise ghoomte hue, arrow tak. Us angle ko hum $\theta$ kehte hain (Greek letter "theta").

KYU. Angle tab tak meaningless hai jab tak tum yeh naa batao ki "kahan se measure kar rahe ho?" Hum x-axis ko apna zero-line choose karte hain aur anticlockwise ko positive — ek fixed rule taaki sabko same number mile. $\theta$ ek single number hai jo kehta hai "kis taraf."

PICTURE. Lal arrow kaale x-axis ke upar angle $\theta$ par jhuka hua hai. Agar $\theta =0$ ho toh arrow x ke saath flat lie karta hai; jaise-jaise $\theta$ badhta hai arrow upar ki taraf tip hota hai.

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Step 3 — Ek perpendicular drop karo: ek right triangle janam leta hai

KYA. $\vec{A}$ ki tip se hum ek seedhi line seedha neeche x-axis par drop karte hain. Yeh x-axis se right angle par milti hai. Ab hamare paas ek triangle hai.

KYU. Right triangle woh ek shape hai jisme side-lengths aur angles fixed ratios se locked hote hain (yahi toh trigonometry hai). Right triangle manufacture karke hum "length + angle" ko "do seedhe sides" mein convert kar lete hain — jo exactly woh do-number picture hai jo hum chahte hain.

PICTURE. Teen sides appear hoti hain:

• tedhi side (hypotenuse) arrow hi hai, length $A$;
• flat bottom side x ke saath lie karti hai — uski length $A_x$ rakho (horizontal reach);
• vertical side y ke parallel lie karti hai — uski length $A_y$ rakho (vertical reach).

Right-angle bottom-right corner par hai (chhota square). $\theta$ $O$ par hai.

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Step 4 — Cosine horizontal side ko kyun pakadta hai

KYA. Hum claim karte hain $A_x=A\cos \theta$. Dekhte hain cosine kahaan se aata hai.

YEH TOOL KYU — cosine hi kyun, kuch aur kyun nahi? Hamare paas ek right triangle hai aur hum ek angle ($\theta$) aur hypotenuse ($A$) jaante hain. Hum woh side chahte hain jo angle ko touch karti hai. Cosine exactly us ratio ke liye define kiya gaya hai jo in teeno cheezein connect karta hai: $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}.$ Yeh woh tool hai jo bana hai yeh jawab dene ke liye ki "angle ko hug karne wali side kitni lambi hai, hypotenuse ke fraction mein?" Koi doosra function exactly woh sawaal nahi answer karta.

PICTURE. Triangle mein, adjacent side $A_x$ hai aur hypotenuse $A$ hai, toh $\underset{\text{shrink factor}}{\underbrace{\cos \theta }}=\frac{\stackrel{\text{horizontal side}}{\overbrace{A_x}}}{\underset{\text{arrow length}}{\underbrace{A}}}⟹A_x=A\cos \theta .$

$\cos \theta$ ko $0$ aur $1$ ke beech ek shrink factor ki tarah padho: yeh arrow ki length ka woh fraction hai jo horizontal reach ke roop mein bachta hai. Jab arrow flat ho ($\theta =0$), $\cos 0=1$ aur $A$ poora horizontal ho jaata hai. Jab arrow seedha upar point kare ($\theta =90^{\circ }$), $\cos 90^{\circ }=0$ aur kuch bhi horizontal nahi.

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Step 5 — Sine vertical side ko kyun pakadta hai

KYA. Hum claim karte hain $A_y=A\sin \theta$.

YEH TOOL KYU. Same triangle, lekin ab hum woh side chahte hain jo $\theta$ ke saamne hai (vertical wali). "Opposite over hypotenuse" ke liye bana function sine hai: $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{A_y}{A}⟹A_y=A\sin \theta .$

PICTURE. Vertical side upar jaati hai jaise arrow tip hota hai. $\sin \theta$ ko climb factor ki tarah padho: $\theta =0$ par arrow flat hai, $\sin 0=0$, zero climb; $\theta =90^{\circ }$ par, $\sin 90^{\circ }=1$, poori length vertical hai.

$\underset{\text{vertical side}}{\underbrace{A_y}}=\underset{\text{arrow length}}{\underbrace{A}}\cdot \underset{\text{climb factor}}{\underbrace{\sin \theta }}$

Notice karo yeh sundar trade-off: jaise-jaise $\theta$ badhta hai, $\cos \theta$ ghatta hai (kam horizontal) bilkul usi tarah jaise $\sin \theta$ badhta hai (zyada vertical). Arrow ka "stuff" x-side se y-side par shift hota hai lekin kabhi gaayab nahi hota.

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Step 6 — Arrow ko rebuild karo, aur Pythagoras se check karo

KYA. Do sides $A_x$ (x ke saath) aur $A_y$ (y ke saath), tip-to-tail rakh do, tumhe exactly $O$ se $\vec{A}$ ki tip tak le jaati hain. Toh $\vec{A}$ hai woh do pieces add hokar: $\vec{A}=A_x\hat{x}+A_y\hat{y}.$ Yahan $\hat{x}$ aur $\hat{y}$ unit vectors hain — exactly $1$ length ke chhote arrows jo x aur y ke saath point karte hain (dekho Unit vectors and Cartesian coordinates). $A_x\hat{x}$ likhne ka matlab hai "x-direction mein $A_x$ steps chalo."

KYU. Yahi toh payoff hai: tedha arrow aur do seedhe arrows ek hi journey hain. Resolve karne se vector nahi badla — bas usse re-describe kiya.

PICTURE. Horizontal lal piece follow karo, phir vertical lal piece; tum tip par land karte ho. Confirm karne ke liye ki do numbers original length rebuild karte hain, Pythagoras use karo (right-triangle side lengths ke liye tool): $A=\sqrt{A_x^2+A_y^2}.$ Aur angle recover karne ke liye, hum poochte hain "kis angle mein vertical-over-horizontal $A_y/A_x$ ke barabar hai?" — woh sawaal ${\tan }^{-1}$ (arctangent) se answer hota hai (dekho Dot product & scalar projection projection viewpoint ke liye): $\tan \theta =\frac{A_y}{A_x}.$

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Step 7 — Har quadrant: kya hota hai jab arrow doosri taraf jhuke?

KYA. Ab tak arrow upar-aur-daayein point kar raha tha (angles $0$–$90^{\circ }$, jise Quadrant I kehte hain). Kya ho agar yeh kahin aur point kare? Formulas $A_x=A\cos \theta ,A_y=A\sin \theta$ tab bhi hold karte hain — kyunki cos aur sin pehle se hi har angle ke liye correct sign carry karte hain. Chalte hain chaar cases dekhte hain.

KYU. Ek student jo sirf Quadrant I draw karta hai woh jaise hi koi force left ya down point kare signs galat kar dega. Components ke signs batate hain kis taraf har piece point karta hai, toh humein chaaon cover karna zaroori hai.

PICTURE (chaar arrows, ek per quadrant):

Quadrant	$\theta$ range	$\cos \theta$	$\sin \theta$	$A_x$	$A_y$	arrow points
I	$0$–$90^{\circ }$	$+$	$+$	right	up	up-right
II	$90$–$180^{\circ }$	$-$	$+$	left	up	up-left
III	$180$–$270^{\circ }$	$-$	$-$	left	down	down-left
IV	$270$–$360^{\circ }$	$+$	$-$	right	down	down-right

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Step 8 — Degenerate cases: zero, aur axis-aligned arrows

KYA. Story ke corners:

• +x ke saath ($\theta =0$): $A_x=A\cos 0=A$, $A_y=A\sin 0=0$. Poora horizontal, koi vertical nahi. ✓
• Seedha upar ($\theta =90^{\circ }$): $A_x=A\cos 90^{\circ }=0$, $A_y=A\sin 90^{\circ }=A$. Poora vertical. ✓ Yahan $\tan \theta =A_y/A_x=A/0$ undefined hai — zero se division — jo maths ka polite tarika hai yeh kehne ka "arrow seedha upar hai, uska angle exactly $90^{\circ }$ hai."
• Zero vector ($A=0$): $A_x=A_y=0$, aur $\theta$ meaningless hai — ek dot ka koi direction nahi hota.

KYU. Yahi woh jagahein hain jahan naive formulas blast kar jaate hain ($A/0$ wala). Inhe pehle se jaanoge toh calculator error se kabhi surprised nahi hoge.

PICTURE. Teen special arrows: flat, vertical, aur akela dot.

Recall Corners khud check karo

Purely vertical arrow ka x ke saath component? ::: Zero ($A\cos 90^{\circ }=0$) Vertical arrow ke liye $\theta$, ${\tan }^{-1}$ se, undefined kyun hai? ::: Kyunki $A_x=0$ hai aur $A_y/A_x$ zero se divide karta hai; angle exactly $90^{\circ }$ hai Zero vector ke components? ::: Dono zero, aur uski koi direction nahi hoti

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Ek-picture summary

Upar sab kuch ek single frame mein: ek lal arrow, uska right triangle, cos-side aur sin-side labelled, rebuild, aur corners mein sign-table reminder.

Recall Feynman retelling — poora walkthrough plain words mein

Ek corner se shuru hota ek tedha lal arrow imagine karo. Main uski tip se seedha neeche ek plumb-line drop karta hoon — ab maine arrow ko ek right triangle ke andar box kar diya. Us box ka flat bottom hai "arrow kitna daayein tak pahuncha," aur tall side hai "kitna upar." Cosine bas $0$ aur $1$ ke beech ek shrink-factor hai jo batata hai ki arrow ki length ka kitna fraction sideways point karta hai, aur sine mujhe woh fraction batata hai jo upar point karta hai. Arrow ki length ko un do factors se multiply karo aur mere paas mere do numbers aa gaye. Pehle number se daayein chalo, phir doosre se upar, aur main exactly arrow ki tip par pahunch jaata hoon — proof ki do seedhe pieces hain tedha arrow. Pythagoras length rebuild karta hai, aur "kis angle mein yeh up-over-right ratio hai?" (arctan) direction rebuild karta hai — jab tak main signs dekhun ki main kis quadrant mein hoon, kyunki arctan akele up-left ko down-right se nahi bataa sakta. Flat arrow: poora sideways. Vertical arrow: poora upar, aur uski slope division-by-zero hai, jo maths ka yeh polite tarika hai ki "yeh seedha upar hai." Ek dot ka koi direction nahi hota. Yahi resolution hai, shuru se ant tak.

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Connections

• Resolution of vectors — into components (any axes) (parent)
• Unit vectors and Cartesian coordinates
• Dot product & scalar projection
• Vectors — addition (parallelogram & triangle law)
• Inclined plane dynamics
• Projectile motion
• Equilibrium of concurrent forces

Concept Map