Cross product — formula, geometric meaning (area), right-hand rule: Worked examples

4.5.3 · D3 · Maths › Linear Algebra (Full) › Cross product — formula, geometric meaning (area), right-han

Intuition Yeh page kis kaam ki hai

Parent note ne bataya kya hai cross product aur kyun kaam karta hai. Yeh page ek shooting gallery hai: har tarah ka problem jo exam (ya real world) tumpe fire kar sakta hai, woh sab yahan zero se worked out hain. Agar koi case yahan mile jo tumne pehle kabhi nahi dekha, toh woh ek bug hai — mujhe batao. Hum cover karte hain har sign, har zero, har degenerate input, limits, ek word problem, aur ek exam twist.

Shuru karne se pehle ek rule, taaki koi symbol undefined na rahe:

$a \times b$ ka matlab hai " $a$ aur $b$ ka cross product" — yeh do arrows leta hai aur ek naya arrow nikalta hai jo dono ke perpendicular khada hota hai. Uski length dono arrows ke beech ki area hai us flap ki jo woh span karte hain.
$∣ v ∣$ matlab hai arrow $v$ ki length: agar $v = (x, y, z)$ hai toh $∣ v ∣ = x^{2} + y^{2} + z^{2}$ (3D mein Pythagoras — origin se tip tak seedhi-line distance).
$θ$ (theta) dono arrows ke beech ka angle hai, us corner par measure kiya jahan woh milte hain.

Neeche sab kuch parent par depend karta hai: the master formula $a \times b = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) .$

Scenario matrix

Cross product ka har problem inhi cells mein se kisi ek mein aata hai. Neeche har example us cell ke saath tagged hai jise woh hit karta hai.

Cell	Situation	Yahan kaata hai	Example
A	Plain in-plane vectors ( $z = 0$ )	$\hat{k}$ result ka sign order ke saath flip hota hai	Ex 1
B	Fully 3D general vectors	Middle-sign trap; perpendicular check	Ex 2
C	Degenerate: parallel / anti-parallel	Output hai $0$ (zero flap)	Ex 3
D	Degenerate: ek input $0$ hai	Output hai $0$ (koi flap possible nahi)	Ex 3
E	Limiting angle: $θ \to 0$ aur $θ \to 9 0^{\circ}$	Length $0$ se $	\vec a
F	3D mein triangle ki area	Parallelogram ki half; ek vertex se edges	Ex 5
G	Unit direction (ek plane ka normal)	Length-1 arrow pane ke liye length se divide karo	Ex 6
H	Word problem: torque	$τ = r \times F$ , right-hand sign	Ex 7
I	Order-swap / anticommutativity	$b \times a = - (a \times b)$	Ex 8
J	Exam twist: missing component dhundho	Solve karo taaki do vectors perpendicular/parallel ho jaayein	Ex 9

Setup figure — arrow aur flap kaisa dikhta hai

Figure — Cross product — formula, geometric meaning (area), right-hand rule

Picture dekho: do kaale arrows $a$ aur $b$ pizza slice ki tarah khulete hain, unke beech ka grey patch parallelogram hai (woh "flap"), aur laal arrow seedha uske upar khada hai. Lamba laal arrow = bada flap. Yahi poori kahani hai; har example sirf yeh padhta hai ki kitna lamba aur kis taraf.

Ex 1 — Cell A: do flat vectors

Ex 2 — Cell B: full 3D, middle sign ka dhyan rakho

Worked example Statement

$a = (2, - 1, 3)$ , $b = (0, 4, - 2)$ . $a \times b$ compute karo, phir prove karo ki yeh dono ke perpendicular hai.

Forecast: koi shortcut nahi — space mein ek tilted flap. Answer genuinely 3D vector hoga. Bas ensure karo ki middle component $a_{3} b_{1} - a_{1} b_{3}$ use kare (wahan ek minus chhupta hai).

Step 1 — Pehla component $a_{2} b_{3} - a_{3} b_{2}$ . Kyun? Yeh formula ka $\hat{i}$ slot hai. $(- 1) (- 2) - (3) (4) = 2 - 12 = - 10.$ Step 2 — Middle component $a_{3} b_{1} - a_{1} b_{3}$ . Yeh exact order kyun? Determinant ka middle cofactor ek minus carry karta hai; $a_{1} b_{3} - a_{3} b_{1}$ likhna exam ka #1 error hai. $(3) (0) - (2) (- 2) = 0 + 4 = 4.$ Step 3 — Teesra component $a_{1} b_{2} - a_{2} b_{1}$ . Kyun? $\hat{k}$ slot. $(2) (4) - (- 1) (0) = 8 - 0 = 8.$ Toh $a \times b = (- 10, 4, 8)$ .

Verify (perpendicularity hamaara safety net hai): $a \cdot (- 10, 4, 8) = 2 (- 10) + (- 1) (4) + 3 (8) = - 20 - 4 + 24 = 0 ✓$ $b \cdot (- 10, 4, 8) = 0 (- 10) + 4 (4) + (- 2) (8) = 0 + 16 - 16 = 0 ✓$ Dono zero → result sach mein dono inputs ke perpendicular hai. Confidence earned.

Ex 3 — Cells C & D: degenerate cases (zero output)

Worked example Statement

Forecast: (C) do arrows ek hi taraf point karte hain, toh koi flap nahi — zero area — expect $0$ . (D) ek arrow ki koi length hi nahi, toh koi flap exist nahi kar sakta — expect $0$ .

Step 1 — Case C. Kyun compute karein? Algebra ko dekhne ke liye ki har term cancel ho jaata hai. Note karo $b = 2.5 a$ (parallel). $a \times b = (2 \cdot 5 - 2 \cdot 5, 2 \cdot 5 - 2 \cdot 5, 2 \cdot 5 - 2 \cdot 5) = (0, 0, 0) .$ Yeh $sin θ = sin 0 = 0$ se match karta hai: parallel vectors zero area dete hain. Step 2 — Case D. Kyun? Har slot mein zero times kuch bhi. $u \times v = (0 \cdot 3 - 0 \cdot (- 1), 0 \cdot 7 - 0 \cdot 3, 0 \cdot (- 1) - 0 \cdot 7) = (0, 0, 0) .$

Verify: dono $0$ dete hain. $a \times b = 0$ hone ka ek hi tarika hai — vectors parallel hों ya ek zero ho — exactly yeh do degenerate cases. ✓

Ex 4 — Cell E: limiting angles

Worked example Statement

$∣ a ∣ = 5$ , $∣ b ∣ = 2$ fix karo. Jaise angle $θ$ unke beech $0^{\circ}$ se $9 0^{\circ}$ tak slide karta hai, cross-product length $∣ a ∣∣ b ∣ sin θ$ kaise behave karti hai? $θ = 0^{\circ}, 3 0^{\circ}, 9 0^{\circ}$ par values do.

Forecast: $0^{\circ}$ par woh overlap karte hain → zero flap → length $0$ . $9 0^{\circ}$ par woh perpendicular hain → sabse bada possible rectangle → length $5 \times 2 = 10$ . Beech mein, kuch $0$ se $10$ tak badhta hua.

Step 1 — $θ = 0^{\circ}$ . Kyun? "No flap" edge test karta hai. $5 \cdot 2 \cdot sin 0^{\circ} = 10 \cdot 0 = 0$ . Step 2 — $θ = 3 0^{\circ}$ . Kyun? Ek midpoint sanity check. $sin 3 0^{\circ} = \frac{1}{2}$ , toh $10 \cdot \frac{1}{2} = 5$ . Step 3 — $θ = 9 0^{\circ}$ . Kyun? Maximum test karta hai. $sin 9 0^{\circ} = 1$ , toh $10 \cdot 1 = 10$ .

Figure mein laal curve exactly $10 sin θ$ hai: shuru mein flat-zero, beech mein sabse steep, $10$ par peak. Isko dot product ke $10 cos θ$ (dotted) se contrast karo, jo parallel hone par maximum hota hai — "coS saath mein, Sin sideways".

Verify: $0^{\circ}, 3 0^{\circ}, 9 0^{\circ}$ par $0, 5, 10$ . Kabhi $∣ a ∣∣ b ∣ = 10$ se zyada nahi. ✓

Ex 5 — Cell F: 3D mein triangle ki area

Worked example Statement

Triangle with vertices $A (1, 1, 0)$ , $B (4, 1, 0)$ , $C (1, 5, 2)$ . Uski area nikalo.

Forecast: base $A B$ ki length $3$ hai aur flat hai; $C$ floor se utha hua hai, toh true area flat guess $\frac{1}{2} \cdot 3 \cdot 4 = 6$ se thodi zyada hogi. $6$ se slightly above koi number expect karo.

Step 1 — Ek vertex se do edges. Ek vertex se kyun? Triangle exactly us parallelogram ka half hai jo woh do edges span karte hain, toh humein dono edges same corner se chahiye. $A B = B - A = (3, 0, 0), A C = C - A = (0, 4, 2) .$ Step 2 — Cross product. Kyun? Uski length parallelogram area hai. $A B \times A C = (0 \cdot 2 - 0 \cdot 4, 0 \cdot 0 - 3 \cdot 2, 3 \cdot 4 - 0 \cdot 0) = (0, - 6, 12) .$ Step 3 — Length, phir half karo. Half kyun? Triangle $= \frac{1}{2}$ parallelogram. $∣ (0, - 6, 12) ∣ = 0 + 36 + 144 = 180 = 65 .$ $Area = \frac{1}{2} \cdot 65 = 35 \approx 6.708.$

Verify: $35 \approx 6.71$ , flat-triangle guess $6$ se thoda upar kyunki $C$ lifted hai — forecast se match karta hai. ✓

Ex 6 — Cell G: ek plane ka unit normal

Worked example Statement

Ek length-1 vector nikalo jo $a = (1, 0, 1)$ aur $b = (0, 1, 1)$ dono ke perpendicular ho.

Forecast: cross product ek perpendicular deta hai; phir main use length se divide karke length $1$ tak shrink kar dunga. Direction dono se door tilt karna chahiye.

Step 1 — Direction ke liye cross product. Kyun? Yahi woh machine hai jo perpendicular produce karti hai. $a \times b = (0 \cdot 1 - 1 \cdot 1, 1 \cdot 0 - 1 \cdot 1, 1 \cdot 1 - 0 \cdot 0) = (- 1, - 1, 1) .$ Step 2 — Length. Kyun? Jaanne ke liye ki kis se divide karna hai. $∣ (- 1, - 1, 1) ∣ = 1 + 1 + 1 = 3 .$ Step 3 — Divide (normalise) karo. Length se divide kyun? Vector ko $1/∣ v ∣$ se scale karna uski direction rakhta hai lekin uski length exactly $1$ set kar deta hai — ek "unit vector". $\overset{n}{^} = \frac{1}{3} (- 1, - 1, 1) = (- \frac{1}{3}, - \frac{1}{3}, \frac{1}{3}) .$

Verify: $∣ \overset{n}{^} ∣^{2} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$ ✓, aur $\overset{n}{^} \cdot a = - \frac{1}{3} + 0 + \frac{1}{3} = 0$ ✓ (perpendicularity scaling ke baad bhi rehti hai). (Negative $\overset{n}{^}$ bhi equally valid hai — plane ke do faces hote hain.)

Ex 7 — Cell H: torque (word problem)

Worked example Statement

Ek wrench arm bolt se tumhari haath tak $r = (0.3, 0, 0)$ metres (east) direction mein point karta hai. Tum $F = (0, 50, 0)$ newtons (north) se push karte ho. Torque $τ = r \times F$ nikalo aur batao bolt kis taraf turn karta hai.

Forecast: tum eastward arm par north push karte ho — bolt counter-clockwise upar se dekhe jaane par ghoomta hai, toh torque upar ( $+ \hat{k}$ ) point karta hai. Size $\approx 0.3 \times 50 = 15$ N·m.

Step 1 — Cross product. Cross kyun aur dot kyun nahi? Torque poochta hai "kitna twist?", jo push ke perpendicular part par depend karta hai — exactly yahi $sin θ$ (cross) measure karta hai. $τ = r \times F = (0 \cdot 0 - 0 \cdot 50, 0 \cdot 0 - 0.3 \cdot 0, 0.3 \cdot 50 - 0 \cdot 0) = (0, 0, 15) .$ Step 2 — Direction interpret karo. Kyun? "Kis taraf ghoomta hai" ka jawab dene ke liye. $τ$ $+ \hat{k}$ (upar) point karta hai; right-hand rule se upward torque matlab counter-clockwise from above.

Verify: $15$ N·m, upar point karta hua — exactly forecast. (Dekho Torque and Angular Momentum.) Units: metre × newton = newton·metre ✓.

Ex 8 — Cell I: order matters (anticommutativity)

Ex 9 — Cell J: exam twist (missing component dhundho)

Worked example Statement

$t$ ki woh value nikalo jo $a = (2, t, 1)$ ko $c = a \times b$ ke perpendicular banaye jahan $b = (1, 0, - 1)$ … trick — $a$ hamesha $a \times b$ ke perpendicular hota hai! Asli sawaal: $t$ nikalo taaki $a$ , $b$ ke parallel ho jaaye.

Forecast: parallel matlab "same direction, scaled". $a = k b$ ke liye $2 = k (1)$ , $1 = k (- 1)$ chahiye — lekin $k = 2$ aur $k = - 1$ clash karte hain, toh koi single $t$ unhe parallel nahi bana sakta. Cross product yeh confirm karega ki woh kabhi $0$ nahi banega.

Step 1 — Cross product se parallelism test karo. Kyun? Parallel $⟺ a \times b = 0$ . $t$ ke terms mein compute karo. $a \times b = (t (- 1) - 1 \cdot 0, 1 \cdot 1 - 2 (- 1), 2 \cdot 0 - t \cdot 1) = (- t, 3, - t) .$ Step 2 — Har component ko zero set karo. Kyun? Zero vector ke liye teeno vanish hone chahiye. Middle component $3 \neq = 0$ hai chahe $t$ kuch bhi ho. Toh $a \times b$ kabhi $0$ nahi ho sakta.

Verify: middle slot mein constant $3$ prove karta hai ki $a ∥ b$ kisi bhi $t$ ke liye impossible hai. "Twist" yeh hai ki answer hai no solution — aur cross product yeh instantly visible bana deta hai. ✓

Recall Woh ek quantity jo bataati hai ki cross product degenerate (zero) hai?

Magnitude $∣ a ∣∣ b ∣ sin θ$ exactly tab zero hoti hai jab $sin θ = 0$ (parallel/anti-parallel) ya koi length $0$ ho. Degenerate condition ::: $a \times b = 0 ⟺$ vectors parallel HÓN YA ek zero vector ho.

Recall Kisi bhi computed cross product ko verify karne ka fast check?

Use dono inputs ke saath dot karo. The check ::: $(a \times b) \cdot a = 0$ aur $(a \times b) \cdot b = 0$ ; agar koi bhi nonzero hai toh galti hui hai (usually middle sign).

Connections

Determinants — upar har computation $3 \times 3$ determinant disguise mein hai.
Area and Volume — Ex 5 (triangle area) aur parallelogram picture.
Orthogonality — Ex 2, Ex 6 mein perpendicularity checks.
Dot Product — Ex 4 mein "coS saath mein, Sin sideways" contrast.
Scalar Triple Product — agla step: $a \cdot (b \times c)$ signed volume ke liye.
Torque and Angular Momentum — Ex 7 ki physics.