Cross product — formula, direction (right-hand rule), torque - area calculation: Visual walkthrough

1.1.12 · D2 · Physics › Measurement, Vectors & Kinematics › Cross product — formula, direction (right-hand rule), torque

Pehli line se pehle, yeh pura vocabulary hai jo hum build karenge, ek ek karke:

Ek arrow (ek vector) — ek length aur ek direction.
Do arrows ke beech ka angle $θ$ .
Ek tilted patch ke liye base $\times$ height ka idea.
$sin θ$ word — yeh kahan se aata hai aur yeh ( $cos θ$ ki jagah) kyun aata hai.
Answer kahan point karta hai, right hand se.
Weird cases mein kya hota hai (parallel arrows, ek zero arrow, angle $9 0^{\circ}$ ke baad).

Jo prerequisites hum use karenge: Vectors — addition, components, unit vectors (arrow aur uski length ka matlab) aur, end mein contrast ke liye, Dot product — formula, projection, work calculation.

Step 1 — Do arrows, tail se tail

KYA: Do arrows, $A$ aur $B$ , ek hi point se draw karo. Woh shared starting point tail kehlata hai.

KYUN: Cross product measure karta hai ki do arrows ek common corner se kitna spread apart hain. Agar woh alag jagahon se start karte toh spread measure karne ke liye koi single corner nahi hota — isliye hum unhe hamesha slide karte hain jab tak unki tails touch na karain.

PICTURE: Figure mein, dono arrows ek hi dot se nikalte hain. Magenta wala $A$ hai, violet wala $B$ hai. Unke beech ka opening woh angle hai jise hum aage name karenge.

Figure — Cross product — formula, direction (right-hand rule), torque - area calculation

Step 2 — Angle $θ$ ko naam dena

KYA: Do arrows ke beech ke wedge ko symbol $θ$ se label karo.

KYUN: Cross product ke baare mein har fact is par depend karta hai ki do arrows kitne open hain, nahi ki woh page par kaise rotate hain. $θ$ woh ek number hai jo "kitna open" capture karta hai.

PICTURE: Orange arc wedge ke andar baitha hai. Jab arc patla hota hai arrows almost agree karte hain; jab arc ek quarter-turn hota hai woh perpendicular hain; jab yeh half-turn hota hai woh opposite taraf point karte hain.

Step 3 — Parallelogram complete karna

KYA: $B$ ki ek copy slide karo taaki uski tail $A$ ke head par ho, aur $A$ ki ek copy $B$ ke head par. Charon arrows ek filled patch enclose karte hain.

KYUN: Area woh natural cheez hai jo "do spread-apart arrows" produce karte hain. Ek akela number (jaise dot product deta hai) describe nahi kar sakta ki woh kitni surface sweep karte hain — isliye hum woh patch dekhte hain jo woh bound karte hain.

PICTURE: Peach-shaded region parallelogram hai. Uska bottom edge $A$ hai (hum ise base kahenge). Hume abhi bhi uski height chahiye — woh agle step mein aayegi, aur wahin $sin θ$ paida hota hai.

Step 4 — Height drop karo, aur $sin θ$ appear hota hai

KYA: $B$ ke head se, $A$ ko carry karne wali line par seedha perpendicular (right angle par) line neeche drop karo. Us dropped line ki length height $h$ hai. Yeh ek right-angled triangle banata hai jiska slanted side (the hypotenuse) $B$ khud hai, length $B$ ke saath.

$sin$ kyun aur kuch nahi: Ek right triangle mein, $sin θ$ define hota hai $sin θ = \frac{opposite}{hypotenuse} .$ Yahan angle $θ$ ke opposite wali side bilkul hamari height $h$ hai, aur hypotenuse $B$ ki length $B$ hai. To $opp / hyp sin θ = \frac{h}{B} ⟹ the height h = length of B B how much of B is "upward" sin θ .$ Hum $sin$ isliye use karte hain kyunki $sin$ woh tool hai jo "ek slanted length aur uska angle" ko "us length ka upright part" mein badalta hai — aur upright part hi height hoti hai.

PICTURE: Navy dashed segment $h$ hai. Dekho kaise yeh chhote right triangle ki vertical leg hai, $B$ slanted top ke roop mein aur $θ$ bottom-left corner par tucked.

Step 5 — Base ko height se multiply karo

KYA: Base ( $A$ ) aur height ( $B sin θ$ ) ko area rule ke saath rakh do.

KYUN: Hume ek single positive number chahiye tha jo kahe "yeh arrows kitna spread karte hain". Tilted box ka area woh number hai, aur Step 4 ne abhi hume uske do ingredients diye.

PICTURE: Figure usi parallelogram ko do baar shade karta hai — ek baar tilted box ke roop mein, ek baar same base $A$ aur same height $B sin θ$ ke upright rectangle mein "shear" karke. Shearing (top edge ko sideways slide karna) kabhi bhi area nahi badalta, jo visual proof hai ki base × height tab bhi kaam karta hai jab box tilted ho.

Base

A

kyun hai na ki

B

Humne

A

ko base choose kiya;

B

choose karne par same area milta hai kyunki

B \times (A sin θ)

same product hai.

Height

B sin θ

kyun hai?

Right triangle mein, height = opposite side = hypotenuse

B

times

sin θ

Step 6 — Edge case: parallel arrows ( $θ = 0^{\circ}$ )

KYA: $θ \to 0^{\circ}$ hone do. Tab $sin 0^{\circ} = 0$ , isliye $A B sin θ = 0$ .

KYUN yeh zero hona chahiye: Arrows ek doosre ke upar lie karte hain to height $h = B sin θ$ zero hai — "box" koi area nahi wali flat sliver hai. Ek flat sliver kuch bhi enclose nahi karta.

PICTURE: Magenta aur violet arrows almost ek hi line par hain; shaded region ek hairline tak thin ho gayi hai. Neeche, exactly $θ = 0$ par wahi cheez: ek single line, area $0$ .

Isliye parallel vectors give $A \times B = 0$ . Yeh $θ = 18 0^{\circ}$ par bhi sach hai (arrows opposite), kyunki $sin 18 0^{\circ} = 0$ bhi hai — box waahan bhi flat hai.

Step 7 — Edge case: $9 0^{\circ}$ ke baad, aur ek zero arrow

KYA — obtuse angle: $θ$ ko $9 0^{\circ}$ aur $18 0^{\circ}$ ke beech lo. Ab $B$ se perpendicular ka foot $A$ ke extension par, tail ke peeche girata hai. Height abhi bhi $B sin θ$ hai, aur kyunki $sin θ$ $0 < θ < 18 0^{\circ}$ ke liye positive rehta hai, area ek sensible positive number rehta hai. Koi sign trouble nahi — dot product ke unlike, jiska $cos θ$ $9 0^{\circ}$ ke baad negative ho jaata hai.

KYA — zero arrow: Agar kisi bhi arrow ki length $0$ hai (maan lo $B = 0$ , ek mere point), to $A B sin θ = A \cdot 0 \cdot sin θ = 0$ . Ek point ka koi direction nahi hai spread karne ke liye, to koi box nahi aur koi area nahi.

KYUN dono dikhana: Reader ko koi aisa angle ya input kabhi nahi milna chahiye jo humne cover nahi kiya. Dono milake, Steps 6–7 har case cover karte hain: $θ = 0$ , $0 < θ < 9 0^{\circ}$ , $θ = 9 0^{\circ}$ (maximum, $sin = 1$ ), $9 0^{\circ} < θ < 18 0^{\circ}$ , $θ = 18 0^{\circ}$ , aur ek degenerate zero-length arrow.

PICTURE: Left panel: ek obtuse wedge, perpendicular $A$ ke dashed backward extension par drop hota hua, height abhi bhi upward aur positive drawn. Right panel: $B$ ek dot mein shrink, box gayab.

Step 8 — Answer kahan point karta hai?

KYA: Perpendicular arrow ko parallelogram se attach karo, thumb ki side par.

KYUN yeh matter karta hai: Isliye cross product torque (Torque and rotational equilibrium), angular momentum (Angular momentum — L = r × p) aur magnetic force (Magnetic force — F = qv × B) ke liye useful hai: har ek ko space mein ek axis chahiye, sirf ek size nahi. Arrows swap karna ( $B \times A$ ) ungliyan doosri taraf curl karta hai aur thumb flip karta hai — anti-commutativity ki visual wajah, $A \times B = - B \times A$ .

PICTURE: Peach parallelogram flat baitha hai; navy arrow usse bahar nikal raha hai, ek choti curl-arrow ke saath jo ungliyan $A \to B$ jaate dikhati hai.

Ek-picture summary

Upar ki sab cheez, ek hi sheet par: ek tail se do arrows, angle $θ$ , completed parallelogram, dashed height $B sin θ$ , area label $A B sin θ$ , aur navy thumb-arrow $A \times B$ ke along bahar nikal raha hai.

Recall Feynman retelling — poora walkthrough simple shabdon mein

Do arrows rakh lo taaki unke peechle sire touch karen. Unke beech ka gap angle $θ$ hai. Ab woh tilted box banao jo in do arrows ne fence kiya hai — woh ek parallelogram hai. Yeh pata karne ke liye ki woh kitna bada hai tumhe uska base aur seedha-upright height chahiye. Base bas pehle arrow ki length hai, $A$ . Height doosre arrow ki puri length nahi hai, kyunki woh arrow lean karta hai; woh sirf uska upward part hai. Woh tool jo ek leaning length se upward part nikalta hai $sin θ$ hai, isliye height $B sin θ$ hai. Multiply karo: area $= A B sin θ$ . Jab arrows ek hi taraf point karte hain koi gap nahi, koi height nahi, koi box nahi — area zero, isliye parallel arrows cross karke kuch nahi bante. Jab woh square-on hote hain ( $9 0^{\circ}$ ) box sabse mota hota hai aur area sabse bada. $9 0^{\circ}$ ke baad yeh phir se patla hota hai, abhi bhi positive, koi minus signs ki chinta nahi. Aakhir mein, ek area ko pata hona chahiye ki woh kaun sa face dikhata hai: apne right hand ki ungliyan pehle arrow se doosre par curl karo, aur tumhara thumb us taraf point karta hai. Size box se, direction tumhare thumb se — milke yeh $A \times B$ hai.

Recall Quick self-check

Parallelogram ki height $B, θ$ ke terms mein? ::: $B sin θ$ $∣ A \times B ∣$ ki value jab $θ = 9 0^{\circ}$ ? ::: $A B$ (maximum, kyunki $sin 9 0^{\circ} = 1$ ) Value jab $θ = 0^{\circ}$ ya $18 0^{\circ}$ ? ::: $0$ (box flat hai) $sin$ kyun naki $cos$ ? ::: Hume $B$ ka upright (perpendicular) part chahiye; $sin$ = opposite/hyp woh height deta hai, jabki $cos$ flat part deta hai. Direction kahan se aati hai? ::: Right-hand rule — ungliyan $A \to B$ , thumb answer hai.

Connections

Cross product — formula, direction (right-hand rule), torque - area calculation — parent topic
Vectors — addition, components, unit vectors — arrows aur lengths ka matlab
Dot product — formula, projection, work calculation — $cos θ$ wala cousin, contrast ke liye
Torque and rotational equilibrium — jahan thumb-direction ek spin axis ban jaata hai
Angular momentum — L = r × p aur Magnetic force — F = qv × B — same $sin θ$ -and-thumb machinery