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

4.5.3 · D4 · Maths › Linear Algebra (Full) › Cross product — formula, geometric meaning (area), right-han

Quick reminders jinpe tum lean karoge (sab parent mein prove hain):

Yahan " $\times$ " ka matlab cross product hai (output ek vector hai), kabhi ordinary multiplication nahi. " $\cdot$ " ka matlab Dot Product hai (output ek number hai).

Level 1 — Recognition

L1·Q1

Components compute kiye bina, $a \times b$ ka type (scalar ya vector) batao aur ek guaranteed geometric fact batao, jahan $a = (2, 1, 0)$ , $b = (0, 1, 3)$ .

Recall Solution — L1·Q1

Kya hai: cross product output ek vector (teen numbers) deta hai, scalar nahi. Ek guaranteed fact: woh vector perpendicular to both $a$ aur $b$ hota hai. Yahi uski definition hai — yeh jaanne ke liye kuch compute karna zaroori nahi. Uski length aage jaake un dono vectors se bane parallelogram ka area bhi hogi, lekin "perpendicular vector" wala fact sabse pakka instant answer hai.

L1·Q2

Standard-basis products fill karo: $\hat{i} \times \hat{j} = ?$ , $\hat{j} \times \hat{k} = ?$ , $\hat{k} \times \hat{i} = ?$ , aur $\hat{j} \times \hat{i} = ?$

Recall Solution — L1·Q2

Cyclic order $i \to j \to k \to i$ (figure dekho): cycle mein forward jaane par plus sign aata hai.

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

\hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{k} = \hat{i}, \hat{k} \times \hat{i} = \hat{j} .

Cycle mein backward jaane par sign flip ho jaata hai, isliye

\hat{j} \times \hat{i} = - (\hat{i} \times \hat{j}) = - \hat{k} .

L1·Q3

$a = (3, - 1, 2)$ ke liye, $a \times a$ kya hai? Kyun?

Recall Solution — L1·Q3

Kya: $a \times a = 0 = (0, 0, 0)$ . Kyun: ek vector aur khud uske beech ka angle $θ = 0$ hota hai, isliye $sin θ = sin 0 = 0$ , jo magnitude $∣ a ∣^{2} \cdot 0 = 0$ deta hai. Geometrically, do identical arrows ek zero-width parallelogram banate hain — koi area nahi, koi flap nahi, kuch bahar nahi aata.

Level 2 — Application

L2·Q1

$a = (1, 3, - 2)$ , $b = (2, - 1, 4)$ ke liye $a \times b$ compute karo, aur $a$ ke saath perpendicularity verify karo.

Recall Solution — L2·Q1

Kya: component formula apply karo, ek slot at a time.

Pehla slot: $a_{2} b_{3} - a_{3} b_{2} = (3) (4) - (- 2) (- 1) = 12 - 2 = 10$ .
Beech wala slot (sign dhyan se): $a_{3} b_{1} - a_{1} b_{3} = (- 2) (2) - (1) (4) = - 4 - 4 = - 8$ .
Teesra slot: $a_{1} b_{2} - a_{2} b_{1} = (1) (- 1) - (3) (2) = - 1 - 6 = - 7$ . $a \times b = (10, - 8, - 7) .$ Verify (kyun?): ek genuine cross product $a$ ke saath perpendicular hona chahiye, yani dot product zero ho: $a \cdot (10, - 8, - 7) = (1) (10) + (3) (- 8) + (- 2) (- 7) = 10 - 24 + 14 = 0. ✓$

L2·Q2

$u = (2, 0, 1)$ aur $v = (1, 2, 0)$ se bane parallelogram ka area nikalo.

Recall Solution — L2·Q2

Kya/Kyun: area $= ∣ u \times v ∣$ , isliye pehle cross karo, phir length lo.

Pehla: $(0) (0) - (1) (2) = - 2$ .
Beech wala: $(1) (1) - (2) (0) = 1$ .
Teesra: $(2) (2) - (0) (1) = 4$ . $u \times v = (- 2, 1, 4), ∣ u \times v ∣ = (- 2)^{2} + 1^{2} + 4^{2} = 4 + 1 + 16 = 21 .$ Area $= 21 \approx 4.583$ .

L2·Q3

Vertices $A (1, 1, 0)$ , $B (3, 1, 2)$ , $C (1, 4, 1)$ wale triangle ka area nikalo.

Recall Solution — L2·Q3

Ek vertex se edges kyun? Ek triangle exactly apni do edges par bane parallelogram ka aadha hota hai (figure dekho). $A$ ko corner lo.

Edge vectors:

A B = (2, 0, 2)

A C = (0, 3, 1)

A B \times A C = ((0) (1) - (2) (3), (2) (0) - (2) (1), (2) (3) - (0) (0)) = (- 6, - 2, 6) .

∣ A B \times A C ∣ = 36 + 4 + 36 = 76 = 219 .

Triangle area $= \frac{1}{2} \cdot 219 = 19 \approx 4.359$ .

Level 3 — Analysis

L3·Q1

Do vectors satisfy karte hain $∣ a ∣ = 5$ , $∣ b ∣ = 4$ , aur $a \cdot b = 10$ . Components jaane bina $∣ a \times b ∣$ nikalo.

Recall Solution — L3·Q1

Kaun sa tool aur kyun? Components nahi diye, isliye component formula kaam ka nahi. Lekin lengths aur ek dot product diya hai — exactly Lagrange's identity ke ingredients, jo dono products ko link karta hai: $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} .$ Yeh tool isliye kaam aata hai kyunki yeh "kitna parallel" (dot) ko "kitna perpendicular" (cross) mein convert karta hai, sirf magnitudes use karke. $∣ a \times b ∣^{2} = (5^{2}) (4^{2}) - 1 0^{2} = 400 - 100 = 300.$ $∣ a \times b ∣ = 300 = 103 \approx 17.32.$ Sanity: $cos θ = \frac{10}{5 \cdot 4} = \frac{1}{2}$ , isliye $θ = 60°$ , $sin 60° = \frac{3}{2}$ , aur $∣ a ∣∣ b ∣ sin θ = 20 \cdot \frac{3}{2} = 103 .$ ✓

L3·Q2

Dikhao ki sabhi $a, b$ ke liye $(a + b) \times b = a \times b$ hota hai. Area ke baare mein yeh geometrically kya kehta hai?

Recall Solution — L3·Q2

Hum kya use karte hain: cross product addition par distribute karta hai (yeh har slot mein linear hai), aur $b \times b = 0$ . $(a + b) \times b = (a \times b) + (b \times b) = (a \times b) + 0 = a \times b .$ Geometric meaning: ek edge ki tip ko doosri edge ki direction mein slide karna (yani $a$ ki jagah $a + b$ lena) parallelogram ka area nahi badalta — yeh ek shear hai. Base aur height unchanged rehte hain (figure dekho), isliye flap ka area same hai, hence same cross product.

L3·Q3

Agar $a \times b = 0$ ho lekin na $a$ zero vector ho na $b$ , toh $a$ aur $b$ ke baare mein kya true hona chahiye? Sabhi cases cover karo.

Recall Solution — L3·Q3

Kyun: $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ = 0$ . Kyunki $∣ a ∣ \neq = 0$ aur $∣ b ∣ \neq = 0$ , product zero hone ka ek hi tarika hai: $sin θ = 0$ . $sin θ = 0$ hota hai $θ = 0°$ aur $θ = 180°$ par — dono cases:

$θ = 0°$ : vectors same direction mein point karte hain (parallel).
$θ = 180°$ : woh opposite directions mein point karte hain (antiparallel). Dono cases mein dono arrows ek line par hote hain — woh parallel (collinear) hain aur koi area span nahi karte, isliye koi flap nahi aur perpendicular arrow ki length zero hoti hai.

Level 4 — Synthesis

L4·Q1

Scalar Triple Product $a \cdot (b \times c)$ use karke $a = (1, 0, 0)$ , $b = (1, 1, 0)$ , $c = (1, 1, 1)$ par bane parallelepiped ka volume compute karo.

Recall Solution — L4·Q1

Kaun sa tool aur kyun? Teen edges par bane box ka volume $∣ a \cdot (b \times c) ∣$ hai: cross $b \times c$ ek aisa vector deta hai jiska length base area hai aur direction base ke perpendicular hai; $a$ ke saath dot karna phir measure karta hai ki $a$ us perpendicular ke along kitna bahar nikla hua hai — yani height. Base area × height = volume. $b \times c = ((1) (1) - (0) (1), (0) (1) - (1) (1), (1) (1) - (1) (1)) = (1, - 1, 0) .$ $a \cdot (b \times c) = (1) (1) + (0) (- 1) + (0) (0) = 1.$ Volume $= ∣1∣ = 1$ . (Sahi lagta hai: yeh teen unit "staircase" edges hain jo ek unit box enclose karte hain.)

L4·Q2

$a = (2, - 1, 1)$ aur $b = (1, 0, - 1)$ dono ke perpendicular ek unit vector nikalo.

Recall Solution — L4·Q2

Plan: cross ek perpendicular vector deta hai; use unit-long banane ke liye uski length se divide karo (Orthogonality). $a \times b = ((- 1) (- 1) - (1) (0), (1) (1) - (2) (- 1), (2) (0) - (- 1) (1)) = (1, 3, 1) .$ $∣ a \times b ∣ = 1 + 9 + 1 = 11 .$ $\overset{n}{^} = \frac{1}{11} (1, 3, 1) .$ Note: $- \overset{n}{^}$ bhi perpendicular aur unit-length hai — exactly do answers hain, plane ke har side ke liye ek. Cross product ka right-hand rule upar wala pick karta hai.

L4·Q3

Ek force $F = (0, 3, 0)$ N ek pivot se $r = (2, 0, 0)$ m door kaam karta hai. Torque $τ = r \times F$ nikalo aur uski direction interpret karo.

Recall Solution — L4·Q3

Cross product kyun? Torque and Angular Momentum: torque twisting effect measure karta hai, jo force ke perpendicular part aur rotation axis ki direction par depend karta hai — precisely wahi jo cross product encode karta hai. $τ = r \times F = ((0) (0) - (0) (3), (0) (0) - (2) (0), (2) (3) - (0) (0)) = (0, 0, 6) N\cdotm .$ Direction: $+ \hat{k}$ ke along ( $x y$ -plane se bahar). Right-hand rule se, $r$ ( $+ x$ ke along) se $F$ ( $+ y$ ke along) ki taraf curl karne par thumb upar point karta hai — object $x y$ -plane mein counterclockwise spin karta hai. Magnitude $6$ N·m $= ∣ r ∣∣ F ∣ sin 90° = 2 \cdot 3 \cdot 1$ .

Level 5 — Mastery

L5·Q1

Anticommutative law $b \times a = - (a \times b)$ component formula se prove karo (right-hand picture se nahi).

Recall Solution — L5·Q1

Hum kya karte hain: $b \times a$ compute karo har slot mein $a$ aur $b$ ki roles swap karke, aur compare karo. Formula se, $b \times a$ ka pehla component hai $b_{2} a_{3} - b_{3} a_{2} = - (a_{2} b_{3} - a_{3} b_{2}),$ jo exactly $a \times b$ ke pehle component ka $- 1$ times hai. Wahi swap middle slot ko bhi flip karta hai: $b_{3} a_{1} - b_{1} a_{3} = - (a_{3} b_{1} - a_{1} b_{3})$ , aur teesra: $b_{1} a_{2} - b_{2} a_{1} = - (a_{1} b_{2} - a_{2} b_{1})$ . Har component negate hota hai, isliye $b \times a = - (a \times b) . ■$ Geometry se match kyun karta hai: do vectors swap karna right hand ke curl direction ko reverse karta hai, isliye thumb flip ho jaata hai — same area, opposite direction. Algebra aur picture agree karte hain.

L5·Q2

$a = (1, 2, 2)$ , $b = (2, - 1, 0)$ ke liye, Lagrange's identity $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$ numerically verify karo, phir unke beech ka $sin θ$ padho.

Recall Solution — L5·Q2

Left side: $a \times b = ((2) (0) - (2) (- 1), (2) (2) - (1) (0), (1) (- 1) - (2) (2)) = (2, 4, - 5) .$ $∣ a \times b ∣^{2} = 4 + 16 + 25 = 45.$ Right side: $∣ a ∣^{2} = 1 + 4 + 4 = 9$ , $∣ b ∣^{2} = 4 + 1 + 0 = 5$ , $a \cdot b = 2 - 2 + 0 = 0$ . $∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = 9 \cdot 5 - 0 = 45. ✓$ Angle: $a \cdot b = 0$ ka matlab $cos θ = 0$ hai, isliye $θ = 90°$ aur $sin θ = 1$ . Check: $∣ a ∣∣ b ∣ sin θ = 3 \cdot 5 \cdot 1 = 45 = ∣ a \times b ∣$ . ✓

L5·Q3

Prove karo ki kisi bhi $a, b, c$ ke liye vector $a \times (b \times c)$ hamesha $b$ aur $c$ ke plane mein hota hai — aur "BAC–CAB" identity se confirm karo: $a \times (b \times c) = b (a \cdot c) - c (a \cdot b) .$ Ise $a = (1, 0, 0)$ , $b = (0, 1, 0)$ , $c = (0, 0, 1)$ par test karo.

Recall Solution — L5·Q3

Woh us plane mein kyun hai: identity $a \times (b \times c)$ ko $(number) b - (number) c$ ke roop mein likhti hai — sirf $b$ aur $c$ ka combination. Aisa koi bhi combination un dono ke span kiye plane mein rehta hai. Isliye triple product $b$ aur $c$ ke saath coplanar hai, kabhi unke plane se bahar nahi jaata. Numeric test:

Inner cross: $b \times c = (0, 1, 0) \times (0, 0, 1) = \hat{j} \times \hat{k} = \hat{i} = (1, 0, 0)$ .
Outer cross: $a \times (b \times c) = (1, 0, 0) \times (1, 0, 0) = 0$ (parallel!).
Right side: $a \cdot c = 0$ , $a \cdot b = 0$ , isliye $b (0) - c (0) = 0 . ✓$ Dono sides $0$ dete hain, aur actually $0$ trivially $b, c$ plane mein rehta hai.

Recap ladder

Grade yourself: kya tum har level par ek problem cleanly kar sakte ho?

L1 output type & basis products

Ek vector;

\hat{i} \times \hat{j} = \hat{k}

cyclically, reverse sign flip karta hai.

L2 compute & area

Component formula (middle minus dhyan se); area

= ∣ a \times b ∣

, triangle half.

L3 dot se relate karo

Lagrange

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}

; zero cross ⇒ parallel.

L4 isse build karo

Triple product = volume; unit normal ke liye normalize karo;

r \times F

= torque.

L5 structure prove karo

Anticommutative from components; not associative; BAC–CAB expansion.

Connections

Dot Product — $cos θ$ partner; Lagrange's identity ke zariye cross se juda hua.
Determinants — component formula hai hi ek $3 \times 3$ determinant expansion.
Scalar Triple Product — $a \cdot (b \times c)$ L4·Q1 mein volume ke liye use hua.
Area and Volume — parallelogram, triangle, parallelepiped measures.
Orthogonality — unit normals build karna (L4·Q2) aur cross perpendicularity kyun deta hai.
Torque and Angular Momentum — $r \times F$ L4·Q3 mein.