Visual walkthrough — Unit vectors — î, ĵ, k̂; constructing unit vector

1.1.10 · D2 · Physics › Measurement, Vectors & Kinematics › Unit vectors — î, ĵ, k̂; constructing unit vector

Yeh page parent note ka picture-version hai, Unit vectors. Agar yahan koi word unfamiliar lage, toh woh usi step mein build hota hai jahan woh pehli baar aata hai.

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Step 1 — Ek arrow draw karo. Dekho yeh ek saath do kaam karta hai.

KYA. Hum gridded paper pe ek arrow draw karte hain aur use $\vec{A}$ kehte hain (letter ke upar chhota arrow-hat, $\stackrel{⃗}{}$, bas itna batata hai ki "yeh letter ek arrow hai, koi ordinary number nahi").

KYUN. Jab tak hum ek vector ko "kitna" aur "kis taraf" mein split nahi kar sakte, hamen pehle yeh dekhna hoga ki woh dono ko secretly bundle karta hai. Ek akeli arrow pehle se do sawaalon ka jawaab deti hai: woh kitni lambi hai? aur woh kahan point karti hai?

PICTURE. Amber arrow ko dekho. Uski length (tip kitni door hai tail se) "kitna" ka jawaab deti hai. Uska tilt "kis taraf" ka jawaab deta hai. Yeh do independent facts ek object mein chipke hue hain.

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Step 2 — Arrow ko axes pe rakho aur uske components padhho.

KYA. Hum $\vec{A}$ ki tail ko origin (woh corner jahan axes cross karti hain) pe slide karte hain aur uski tip se do dashed lines giraaate hain: ek seedha horizontal axis tak, ek seedha vertical axis tak. Isse do numbers milti hain, $A_x$ aur $A_y$.

KYUN. Grid paper sirf apni khud ki lines ke along cheezein measure kar sakta hai — rightward aur upward. Toh hum poochhte hain: arrow kitna door right jaata hai, aur kitna door upar? Woh do jawaab, $A_x$ aur $A_y$, components hain. Woh arrow ko completely rebuild karte hain, kyunki "right $A_x$ jao, phir upar $A_y$ jao" exactly tip pe land karta hai.

PICTURE. Cyan horizontal leg ki length $A_x$ hai; cyan vertical leg ki length $A_y$ hai; amber arrow woh slanted line hai jo corner band karti hai.

$\vec{A}=\underset{\text{kitna right}}{\underbrace{A_x}}\underset{\text{right-pointer}}{\underbrace{\hat{i}}}+\underset{\text{kitna upar}}{\underbrace{A_y}}\underset{\text{up-pointer}}{\underbrace{\hat{j}}}$

Yahan $\hat{i}$ ek length-1 arrow hai jo right point karta hai aur $\hat{j}$ ek length-1 arrow hai jo upar point karta hai — grid ke chhote fixed rulers. (Poori detail Vectors — components and resolution mein hai.)

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Step 3 — Arrow ke andar chhupa right triangle dekho.

KYA. Hum notice karte hain ki do legs $A_x$, $A_y$ aur khud arrow milke ek right triangle banate hain: do legs ek perfect $90^{\circ }$ corner pe milti hain, aur arrow slanted side (hypotenuse) hai.

KYUN. Right triangle woh ek shape hai jahan length ek clean rule follow karta hai (Pythagoras theorem). Kyunki hum arrow ki length chahte hain, hamen woh shape dhoondha tha jiska length compute karna hamen aata hai — aur woh yahan free mein milti hai.

PICTURE. Corner mein chhota square $90^{\circ }$ angle mark karta hai. Horizontal leg $A_x$ aur vertical leg $A_y$ perpendicular hain; amber hypotenuse $\vec{A}$ hai.

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Step 4 — Triangle ko magnitude formula mein badlo.

KYA. Hum Step 3 ke triangle pe Pythagoras apply karte hain.

WHY. Pythagoras woh tool hai jo "do perpendicular legs" ko "slanted side ki length" mein convert karta hai. Hum iske liye reach karte hain — na ki, maan lo, plain addition ke liye — precisely isliye kyunki legs perpendicular hain. ($A_x+A_y$ add karna length hogi sirf tab jab woh ek hi line pe hote, jo yeh nahi hain.)

PICTURE. Har side pe bane squares theorem ko visualise karte hain: hypotenuse pe area do leg-areas ke barabar hai.

$\underset{{\text{(arrow length)}}^2}{\underbrace{|\vec{A}{|}^2}}=\underset{{\text{right-leg}}^2}{\underbrace{A_x^2}}+\underset{{\text{up-leg}}^2}{\underbrace{A_y^2}}$

Dono sides ka square root lo. Length negative nahi ho sakti, toh hum positive root rakhte hain:

3D ke liye bas do baar karo: pehle flat diagonal $r=\sqrt{A_x^2+A_y^2}$, phir height $A_z$ ke saath combine karke $|\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2}$ milta hai. Same right-triangle move, stacked. Magnitude and direction of a vector dekho.

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Step 5 — Key idea: arrow ko uski apni length se shrink karo.

KYA. Hum poori arrow ko us single positive number $|\vec{A}|$ se divide karte hain jo humne abhi compute ki: $\hat{A}=\frac{\vec{A}}{|\vec{A}|}.$

KYUN. Hum chahte hain exactly $1$ length ka ek pointer jo abhi bhi same direction mein aim kare. Ek vector ko ek positive number se multiply karna kabhi use rotate nahi karta — woh sirf use stretch ya squash karta hai (ek negative number use flip kar deta; zero use destroy kar deta). Magic multiplier hai $\frac{1}{|\vec{A}|}$: woh positive hai, toh direction survive karta hai, aur woh length cancel karne ke liye sized hai.

PICTURE. Lamba amber arrow $\vec{A}$ aur chhota cyan arrow $\hat{A}$ exactly same line ke along hain; sirf length badi, $|\vec{A}|$ se ghatke $1$ ho gayi.

Nayi length check karo $\frac{1}{|\vec{A}|}$ constant ko length bars se bahar kheench ke: $|\hat{A}|=\left|\frac{\vec{A}}{|\vec{A}|}\right|=\frac{|\vec{A}|}{|\vec{A}|}=1.✓$

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Step 6 — Ek numeric case poora shuru se ant tak chalao.

KYA. $\vec{A}=3\hat{i}+4\hat{j}$ lo aur recipe grind karo.

KYUN. Numbers yeh pin down karte hain ki abstract steps actually length $1$ pe land karti hain.

PICTURE. $3$-$4$-$5$ triangle: right-leg $3$, up-leg $4$, hypotenuse $5$; unit vector wahi slant hai jo length $1$ tak shrink ho gayi.

$|\vec{A}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$ $\hat{A}=\frac{3\hat{i}+4\hat{j}}{5}=0.6\hat{i}+0.8\hat{j}$ $|\hat{A}|=\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1✓$

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Step 7 — Edge aur degenerate cases (koi scenario kabhi mat chhodna).

KYA / KYUN / PICTURE, teen cases ek saath:

• Negative components — $\vec{C}=-3\hat{i}-4\hat{j}$ down-left point karta hai. Uski length phir bhi $\sqrt{(-3{)}^2+(-4{)}^2}=5$ hai (squares signs maar dete hain), toh $\hat{C}=-0.6\hat{i}-0.8\hat{j}$. Minus signs rehte hain — woh "kis taraf" hain, aur unhe hatana silently pointer ko $180^{\circ }$ rotate kar deta.
• Ek axis vector — $\vec{E}=5\hat{j}$ mein $A_x=0$ hai. Tab $|\vec{E}|=\sqrt{0^2+5^2}=5$ aur $\hat{E}=\hat{j}$. $\hat{j}$ jaisa ek basis vector already ek unit vector hai; recipe bas ise confirm karti hai.
• Zero vector — $\vec{0}=0\hat{i}+0\hat{j}$ ki length $0$ hai. Recipe $0$ se division maangti hai, jo undefined hai. Geometrically arrow point karne ke liye koi arrow nahi hai, toh zero vector ki koi direction nahi aur koi unit vector nahi. Yeh woh ek input hai jise construction refuse karta hai.

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Ek-picture summary

Left se right padho: ek arrow length × direction carry karta hai; uski shadows giraaao components paane ke liye; woh components ek right triangle ki legs hain jiska hypotenuse-length Pythagoras se $\sqrt{A_x^2+A_y^2}$ hai; poori arrow ko us length se divide karo aur tum tilt rakhte ho lekin size reset hokar $1$ ho jaata hai — ek pure pointer $\hat{A}$.

Recall Feynman: poora walkthrough simple words mein

Maine grid paper pe ek slanted arrow draw ki. Woh do kaam kar rahi thi — mujhe kitna door aur kis taraf dono bata rahi thi. Unhe alag karne ke liye, maine measure kiya woh kitni door rightward gayi aur kitni door upward; woh do numbers uske do rulers pe uski shadows hain. Shadows ek square corner pe milti hain, toh arrow, right-shadow, aur up-shadow ek right triangle banate hain. Right triangles ke liye mujhe ek magic fact pata hai — do legs ko square karo, add karo, square-root karo — aur arrow ki asli length milti hai. Ab finishing move: main poori arrow ko exactly us length se shrink karta hoon. Ek positive number se shrink karna kabhi arrow ko spin nahi karta, sirf uska size badalta hai, toh meri chhoti arrow abhi bhi same taraf point karti hai lekin ab exactly 1 lambi hai. Woh length-1 arrow unit vector hai: ek stick jo sirf kis taraf jaanta hai, kabhi kitna door nahi. Woh ek arrow jis se yeh build nahi ho sakta woh length zero ki arrow hai — point karne ke liye kuch nahi hai, aur tum kuch se divide kar rahe hoge.

Recall Quick self-test

$-3\hat{i}-4\hat{j}$ ka unit vector kya hai? ::: $-0.6\hat{i}-0.8\hat{j}$ (length abhi bhi $5$ hai; signs rakhho). Kyun $|\vec{A}|$ se divide karte hain na ki kisi aur number se? ::: Sirf arrow ki apni length se divide karna result ko exactly $1$ length force karta hai jabki ek positive divisor direction rakhta hai. Kis vector ka koi unit vector nahi hai? ::: Zero vector $\vec{0}$ — $0$ se divide karna undefined hai aur woh kahi point nahi karta.

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Connections

• Unit vectors — î, ĵ, k̂; constructing unit vector (parent)
• Vectors — components and resolution
• Magnitude and direction of a vector
• Pythagoras theorem
• Vector addition — triangle and parallelogram laws
• Dot product
• Cross product
• Position vector and displacement