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

1.1.12 · D3 · Physics › Measurement, Vectors & Kinematics › Cross product — formula, direction (right-hand rule), torque

Yeh page har tarah ke cross-product problems ka drill hai jo aap exam mein dekh sakte ho. Numbers chhune se pehle, hum saare cases ka ek map banate hain, phir aise examples karte hain jo us map ke har square ko cover karein. Agar aapne abhi tak magnitude rule $∣ A \times B ∣ = A B sin θ$ ya determinant formula nahi padha, toh pehle Cross product — formula, direction (right-hand rule), torque - area calculation mein jaake build karo.

Tools jo aap har example mein use karoge

Drills se pehle, chaliye un do notations ko pin down karte hain jo har box mein appear hoti hain, taaki koi cheez define hone se pehle use na ho.

Formula

$3 \times 3$ cofactor expansion (yeh pattern memorise karo) $A = (A_{x}, A_{y}, A_{z})$ aur $B = (B_{x}, B_{y}, B_{z})$ ko cross karne ke liye hum determinant ko uski top row ke along expand karte hain: $A \times B = \hat{i} A_{x} B_{x} \hat{j} A_{y} B_{y} \hat{k} A_{z} B_{z} = + \hat{i} A_{y} B_{y} A_{z} B_{z} - \hat{j} A_{x} B_{x} A_{z} B_{z} + \hat{k} A_{x} B_{x} A_{y} B_{y}$ Kaise padhein: $\hat{i}, \hat{j}, \hat{k}$ mein se har ek ke liye aap uski apni column cover karte ho, phir baaki bache $2 \times 2$ box ko cross-multiply karte ho as (top-left × bottom-right) − (top-right × bottom-left). Poora likha jaye toh: $A \times B = (A_{y} B_{z} - A_{z} B_{y}) \hat{i} - (A_{x} B_{z} - A_{z} B_{x}) \hat{j} + (A_{x} B_{y} - A_{y} B_{x}) \hat{k}$ Signs hain $+, -, +$ — beech wala ( $\hat{j}$ ) term hamesha minus carry karta hai. Yeh alternating sign "checkerboard" cofactor pattern hai Determinants — 3×3 expansion se; yahan yahi hai jo cross product ko anti-commutative banata hai.

Scenario matrix

Har cross-product problem in cells mein se ek hai. Columns batate hain kya badlta hai; rows batate hain task kis type ka hai.

Cell	Case class	Usmein kya special hai	Example
A	Perpendicular, in-plane ( $θ = 9 0^{\circ}$ )	$sin θ = 1$ , magnitude maximum $= A B$ hai	Ex 1
B	Parallel / antiparallel ( $θ = 0^{\circ}$ ya $18 0^{\circ}$ )	$sin θ = 0$ , result hai zero vector	Ex 2
C	General 3-D components	teeno output components nonzero	Ex 3
D	Sign / order flip (anti-commutativity)	$A \times B = - B \times A$	Ex 4
E	Geometric: triangle / parallelogram area	use karo $\tfrac12	\vec A\times\vec B
F	Angle-and-magnitude form (components nahi diye)	directly use karo $A B sin θ$	Ex 6
G	Real-world word problem (torque)	"push at a distance" ko $r \times F$ mein translate karo	Ex 7
H	Exam twist: solve for an unknown	cross product ko ulta chalao	Ex 8
I	Limiting / degenerate: zero input	zero vector product ko khatam kar deta hai	Ex 9

Example 1 — Cell A: perpendicular vectors (maximum case)

Worked example Do vectors right angle par

$A = 3 \hat{i}$ aur $B = 4 \hat{j}$ . $A \times B$ nikalo, aur confirm karo ki iska magnitude maximum possible hai.

Forecast: direction guess karo (aapka thumb kis taraf point karega?) aur number bhi — aage padhne se pehle.

Angle identify karo. $A$ $x$ ke along hai, $B$ $y$ ke along, toh $θ = 9 0^{\circ}$ . Yeh step kyun? $sin θ$ size control karta hai; $9 0^{\circ}$ par yeh $1$ hota hai, jo maximum possible hai.
Magnitude. $∣ A \times B ∣ = A B sin 9 0^{\circ} = (3) (4) (1) = 12$ . Yeh step kyun? Yeh definition hai; $sin 9 0^{\circ} = 1$ ise peak value $A B$ banata hai.
Direction. Fingers $\hat{i}$ ke along, $\hat{j}$ ki taraf curl karo: thumb $+ \hat{k}$ (page se bahar) ke along point karta hai. Toh $A \times B = 12 \hat{k}$ . Yeh step kyun? Cycle $\hat{i} \times \hat{j} = \hat{k}$ exactly yahi thumb result encode karta hai.

Verify: determinant deta hai $(3) (4) \hat{k} = 12 \hat{k}$ . Magnitude $12 = A B$ . Kyunki $sin θ \leq 1$ hamesha, $12$ sach mein maximum hai. ✓

Neeche ki figure yeh case draw karti hai. Blue arrow $A = 3 \hat{i}$ hai $x$ ke along, orange arrow $B = 4 \hat{j}$ hai $y$ ke along; chhota square unka $9 0^{\circ}$ corner mark karta hai. Green rectangle woh parallelogram hai jo ve span karte hain — kyunki ye perpendicular hain yeh ek seedha rectangle hai area $3 \times 4 = 12$ ka, jo exactly $∣ A \times B ∣$ hai. Red dotted circle standard symbol hai ek arrow ke liye jo page se bahar aapki taraf aa raha ho: woh hai $A \times B = 12 \hat{k}$ . Notice karo kaise maximum area aur out-of-page thumb dono ek picture mein appear hote hain.

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

Example 2 — Cell B: parallel aur antiparallel (zero case)

Worked example Jab arrows ek hi direction mein point karte hain

(a) $A = 2 \hat{i} + \hat{j}$ , $B = 4 \hat{i} + 2 \hat{j}$ (parallel). (b) $C = 2 \hat{i} + \hat{j}$ , $D = - 6 \hat{i} - 3 \hat{j}$ (antiparallel). Dono cross products nikalo.

Forecast: koi width nahi wala parallelogram kaisa dikhta hai? Uska area kya hai?

(a) Notice karo $B = 2 A$ . Ye ek hi line par hain, $θ = 0^{\circ}$ . Yeh step kyun? Scalar multiple spot karna batata hai ki ye parallel hain bina angle compute kiye.
Determinant compute karo. $A \times B = \hat{i} 24 \hat{j} 12 \hat{k} 00 = (1 \cdot 0 - 0 \cdot 2) \hat{i} - (0 - 0) \hat{j} + (2 \cdot 2 - 1 \cdot 4) \hat{k} = (4 - 4) \hat{k} = 0$ . Yeh step kyun? $sin 0^{\circ} = 0$ zero result predict karta hai; components confirm karte hain.
(b) $D = - 3 C$ , toh $θ = 18 0^{\circ}$ . $sin 18 0^{\circ} = 0$ bhi, toh $C \times D = 0$ . Yeh step kyun? Antiparallel simply $θ = 18 0^{\circ}$ hai; $sin$ $0^{\circ}$ aur $18 0^{\circ}$ dono par zero hota hai.

Verify: parallel vectors se spanned parallelogram ek flat line hai — zero area — $0$ se match karta hai. (b) ka Determinant: $\hat{i} 2 - 6 \hat{j} 1 - 3 \hat{k} 00 = (2 \cdot - 3 - 1 \cdot - 6) \hat{k} = (- 6 + 6) \hat{k} = 0$ . ✓

Example 3 — Cell C: general 3-D vectors

Worked example Teeno components alive

$A = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $B = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$ . $A \times B$ nikalo.

Forecast: koi bhi component zero nahi hai, toh fully 3-D answer vector expect karo.

$\hat{i}$ component: $A_{y} B_{z} - A_{z} B_{y} = (2) (6) - (3) (5) = 12 - 15 = - 3$ . Yeh step kyun? $\hat{i}$ column cover karo aur baaki $2 \times 2$ ko cross-multiply karo — exactly upar ke cofactor formula ka pehla term.
$\hat{j}$ component: $- (A_{x} B_{z} - A_{z} B_{x}) = - [(1) (6) - (3) (4)] = - (6 - 12) = 6$ . Yeh step kyun? Middle term $+, -, +$ pattern ka built-in minus carry karta hai.
$\hat{k}$ component: $A_{x} B_{y} - A_{y} B_{x} = (1) (5) - (2) (4) = 5 - 8 = - 3$ . Yeh step kyun? Last cofactor phir positive hai ( $+, -, +$ ).

$A \times B = - 3 \hat{i} + 6 \hat{j} - 3 \hat{k}$

Verify (perpendicularity test): result dono inputs ke perpendicular hona chahiye, toh har dot product zero hona chahiye. $A \cdot (A \times B) = (1) (- 3) + (2) (6) + (3) (- 3) = - 3 + 12 - 9 = 0$ ✓ aur $B \cdot (A \times B) = (4) (- 3) + (5) (6) + (6) (- 3) = - 12 + 30 - 18 = 0$ ✓. (Same idea Dot product — formula, projection, work calculation se hai.)

Example 4 — Cell D: order matter karta hai (anti-commutativity)

Worked example Order swap karo, thumb flip ho jaata hai

Example 3 ke $A, B$ use karke, $B \times A$ compute karo aur compare karo.

Forecast: compute karne se pehle answer predict karo — rule kehta hai yeh pehle wale ka minus hona chahiye.

Rule recall karo. $B \times A = - (A \times B)$ . Yeh step kyun? Right-hand fingers ko $B$ se $A$ ki taraf curl karna (instead of $A$ se $B$ ki taraf) thumb ko opposite side par rakh deta hai.
Apply karo. $- (- 3 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 3 \hat{i} - 6 \hat{j} + 3 \hat{k}$ . Yeh step kyun? Same magnitude, reversed sense — har component negate karo.
Determinant se independent check. Example 3 ki matrix ki do rows swap karo; determinant ki do rows swap karne se uska sign flip hota hai (dekho Determinants — 3×3 expansion). Yeh step kyun? Confirm karta hai ki negation koi fluke nahi balki ek determinant property hai.

Verify: $B \times A = 3 \hat{i} - 6 \hat{j} + 3 \hat{k} = - (A \times B)$ . Magnitudes equal, directions opposite. ✓

Example 5 — Cell E: 3-D mein triangle ka area

Worked example Ek tilted triangle

Triangle with vertices $P (1, 0, 0)$ , $Q (0, 1, 0)$ , $R (0, 0, 1)$ . Iska area nikalo.

Forecast: yeh triangle space mein tilted hai — iska area simply ek plane mein base×height nahi hai. Cross product tilt ko automatically handle karta hai.

Same vertex $P$ se do edge vectors banao. $P Q = Q - P = (- 1, 1, 0)$ , $P R = R - P = (- 1, 0, 1)$ . Yeh step kyun? Cross product ke liye dono vectors ka shared tail chahiye; $P$ shared corner hai.
Cross karo. $P Q \times P R = \hat{i} - 1 - 1 \hat{j} 10 \hat{k} 01 = (1 \cdot 1 - 0 \cdot 0) \hat{i} - ((- 1) (1) - 0 \cdot (- 1)) \hat{j} + ((- 1) (0) - 1 \cdot (- 1)) \hat{k} = \hat{i} + \hat{j} + \hat{k}$ . Yeh step kyun? Result vector tilted triangle ke perpendicular hai — iska length parallelogram area hai.
Magnitude ko halve karo. $∣ P Q \times P R ∣ = 1^{2} + 1^{2} + 1^{2} = 3$ . Area $= \frac{1}{2} 3 \approx 0.866$ . Yeh step kyun? Triangle apne do edges se spanned parallelogram ka aadha hota hai.

Verify: yeh ek equilateral triangle hai side $2$ ke saath (jaise $∣ P Q ∣ = (- 1)^{2} + 1^{2} = 2$ ). Equilateral area $= \frac{3}{4} s^{2} = \frac{3}{4} (2) = \frac{3}{2} = \frac{1}{2} 3$ . ✓

Neeche ki figure 3-D mein yeh tilted triangle dikhati hai. Teen labelled corners $P, Q, R$ har axis ke along ek unit dur baithe hain; green shaded face triangle khud hai, clearly kisi bhi coordinate plane mein flat nahi pada. Blue aur orange arrows do edge vectors $P Q$ aur $P R$ hain shared corner $P$ se draw kiye gaye. Red arrow centre se nikalta hua $P Q \times P R$ hai: notice karo ki yeh green face ke perpendicular khada hai — yahi perpendicularity reason hai ki iska length tilted area measure karta hai bina hume kuch un-tilt kiye.

Example 6 — Cell F: angle-and-magnitude form (components nahi hain)

Worked example Sirf lengths aur ek angle diya hua

Do vectors ki magnitudes $A = 5$ aur $B = 8$ hain aur unke beech angle $θ = 3 0^{\circ}$ hai. $∣ A \times B ∣$ aur parallelogram area nikalo.

Forecast: components nahi hain, toh sirf $A B sin θ$ form apply hoti hai. Guess karo ki answer $0$ ke kareeb hai ya $A B = 40$ ke.

Sahi formula choose karo. Humare paas magnitudes aur ek angle hai, $x, y, z$ parts nahi, toh use karo $∣ A \times B ∣ = A B sin θ$ . Yeh step kyun? Determinant ko components chahiye jo humare paas nahi hain; $A B sin θ$ ko exactly wahi chahiye jo humare paas hai.
Substitute karo. $= 5 \cdot 8 \cdot sin 3 0^{\circ} = 40 \cdot \frac{1}{2} = 20$ . Yeh step kyun? $sin 3 0^{\circ} = \frac{1}{2}$ ek standard value hai.
Geometric meaning batao. Parallelogram area exactly yahi magnitude hai, $20$ square units. Yeh step kyun? $∣ A \times B ∣$ definition se hi parallelogram area hai.

Verify: $sin 3 0^{\circ} = 0.5$ , toh $40 \times 0.5 = 20$ . Kyunki $θ < 9 0^{\circ}$ , answer max $A B = 40$ se neeche hai, jaise expected tha. ✓

Example 7 — Cell G: real-world word problem (torque)

Worked example Wrench push karna

Ek mechanic wrench par force $F = (0, - 20, 0) N$ (seedha neeche) lagate hain handle ke end par jo bolt se $r = (0.3, 0, 0) m$ dur hai. Torque $τ = r \times F$ nikalo.

Forecast: bolt kis taraf turn karega — page mein jaayega ya bahar aayega? Compute karne se pehle guess karo.

Words ko vectors mein translate karo. "0.3 m handle ke along" $= r = (0.3, 0, 0)$ ; "20 N downward" $= F = (0, - 20, 0)$ . Yeh step kyun? Torque ke liye position aur force dono ko pivot se vectors ke roop mein chahiye.
Cross karo. $τ = \hat{i} 0.3 0 \hat{j} 0 - 20 \hat{k} 00 = (0 - 0) \hat{i} - (0 - 0) \hat{j} + ((0.3) (- 20) - 0) \hat{k} = - 6 \hat{k} N\cdotm$ . Yeh step kyun? $r$ $+ x$ ke along, $F$ $- y$ ke along, aur $\hat{i} \times (- \hat{j}) = - \hat{k}$ .
Sign interpret karo. $- \hat{k}$ page mein point karta hai → clockwise turning. Yeh step kyun? $τ$ ki direction woh axis hai jiske baare mein bolt rotate karta hai (dekho Torque and rotational equilibrium).

Verify: magnitude $∣ τ ∣ = r F sin 9 0^{\circ} = (0.3) (20) (1) = 6 N\cdotm$ , $∣ - 6 \hat{k} ∣ = 6$ se match karta hai. Units: m × N = N·m ✓.

Example 8 — Cell H: exam twist — ulta solve karo

Worked example Missing component nikalo

$A = (2, 0, 1)$ aur $B = (0, 3, b)$ . Given hai ki $A \times B$ ka $\hat{i}$ -component $- 3$ ke equal hai, $b$ nikalo.

Forecast: aapko cross-product machinery reverse chalani padegi — formula ko given number ke equal set karo aur solve karo.

$\hat{i}$ -component formula likho. $(A \times B)_{x} = A_{y} B_{z} - A_{z} B_{y} = (0) (b) - (1) (3) = - 3$ . Yeh step kyun? Sirf $\hat{i}$ -component constrained hai, toh sirf wahi formula chahiye.
Notice karo yeh already fixed hai. $(0) (b) - 3 = - 3$ har $b$ ke liye hold karta hai — constraint automatically satisfied hai. Yeh step kyun? $A_{y} = 0$ $b$ -term ko vanish kar deta hai, toh $b$ $\hat{i}$ -equation se unconstrained hai. "Twist" yeh hai ki answer hai "koi bhi $b$ ".
Ek alag component se sanity check. Agar instead unhone $\hat{j}$ constrain kiya hota: $(A \times B)_{y} = - (A_{x} B_{z} - A_{z} B_{x}) = - (2 b - 0) = - 2 b$ , jo $b$ ko pin down karta. Yeh dikhata hai ki kaunsi equation information carry karti hai. Yeh step kyun? Exam twists often yeh fact hide karte hain ki chosen component degenerate hai — hamesha doosra test karo.

Verify: $b = 5$ plug karo: $\hat{i}$ -comp $= (0) (5) - (1) (3) = - 3$ ✓; $b = - 1$ plug karo: $(0) (- 1) - 3 = - 3$ ✓. Dono kaam karte hain — confirm karta hai ki koi bhi $b$ ise satisfy karta hai.

Example 9 — Cell I: degenerate zero input

Recall Quick self-test (click to expand)

Yeh ek fold-out flashcard block hai: har line ek question state karti hai, aur ::: ke baad sab kuch hidden answer hai — isko cover karo, memory se answer karo, phir reveal karo.

Kaun se cell mein $sin θ = 1$ hota hai? ::: Cell A (perpendicular, $θ = 9 0^{\circ}$ ) Kaun se do cells zero vector dete hain? ::: Cell B (parallel/antiparallel) aur Cell I (zero input) Triangle area pane ke liye aap magnitude ko kis se multiply karte ho? ::: $\frac{1}{2}$ Computed $A \times B$ correct hai yeh kaise check karte ho? ::: Use $A$ aur $B$ se dot karo; dono $0$ hone chahiye (perpendicularity) Agar vectors ka order swap karo, toh result... ::: sign flip ho jaata hai (anti-commutative)

Connections

Cross product — formula, direction (right-hand rule), torque - area calculation (parent — woh theory jisko yeh examples drill karte hain)
Dot product — formula, projection, work calculation (perpendicularity check ke roop mein use hua)
Determinants — 3×3 expansion (cofactor pattern aur row-swap sign flip)
Vectors — addition, components, unit vectors ( $\hat{i}, \hat{j}, \hat{k}$ define karna aur points se edge vectors banana)
Torque and rotational equilibrium (Example 7)
Angular momentum — L = r × p (same $r \times (\dots)$ structure)
Magnetic force — F = qv × B (ek aur physical cross product)