Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof: Worked examples

4.5.2 · D3 · Maths › Linear Algebra (Full) › Dot product — formula, cosine formula, Cauchy-Schwarz inequa

Scenario matrix

Neeche har worked example us cell ke saath tagged hai jo wo fill karta hai. Goal: poori table cover karna.

Cell	Case class	$a \cdot b$ ka sign	Angle $θ$	Example
C1	Acute — arrows ek hi taraf jhuke	positive	$0^{\circ} < θ < 9 0^{\circ}$	Ex 1
C2	Right angle — perpendicular	zero	$θ = 9 0^{\circ}$	Ex 2
C3	Obtuse — arrows alag taraf jhuke	negative	$9 0^{\circ} < θ < 18 0^{\circ}$	Ex 3
C4	Parallel same-way (C–S mein equality)	$= + ∥ a ∥∥ b ∥$	$θ = 0^{\circ}$	Ex 4
C5	Anti-parallel (equality, doosra sign)	$= - ∥ a ∥∥ b ∥$	$θ = 18 0^{\circ}$	Ex 4
C6	Degenerate — ek vector $0$ hai	$= 0$ lekin perpendicular nahi	undefined	Ex 5
C7	Higher dimension ( $R^{3} / R^{4}$ )	koi bhi	koi bhi	Ex 6
C8	Real-world word problem (work / teamwork)	koi bhi	koi bhi	Ex 7
C9	Exam twist — unknown ke liye solve karo	ek value pe forced	forced	Ex 8
C10	Cauchy–Schwarz sanity, strict vs equality	—	—	Ex 9

Example 1 — Acute angle (Cell C1)

Worked example Positive overlap

$a = (3, 1), b = (1, 2)$ . $a \cdot b$ aur angle $θ$ nikalo.

Forecast: dono arrows upar-aur-right point kar rahe hain, toh guess karo positive dot product aur angle $9 0^{\circ}$ se kam.

$a \cdot b = 3 \cdot 1 + 1 \cdot 2 = 3 + 2 = 5.$ Yeh step kyun? Component formula: matching slots multiply karo, add karo. Positive hai, jaise forecast kiya tha.
$∥ a ∥ = 3^{2} + 1^{2} = 10$ , $∥ b ∥ = 1^{2} + 2^{2} = 5$ . Yeh step kyun? Length $= a \cdot a$ ; $cos θ$ expose karne ke liye lengths chahiye.
$cos θ = \frac{5}{10 5} = \frac{5}{50} = \frac{5}{5 2} = \frac{1}{2} .$ Yeh step kyun? $cos θ$ ke liye geometric formula rearrange karne se angle isolate hota hai.
$θ = arccos (\frac{1}{2}) = 4 5^{\circ} .$ Yeh step kyun? $arccos$ jawaab deta hai "kis angle ka yeh cosine hai?" — yeh cosine ko undo karta hai.

Verify: $0^{\circ} < 4 5^{\circ} < 9 0^{\circ}$ ✔ acute, positive dot product se match karta hai. Figure dekho — arrows clearly ek hi taraf jhuke hain.

Figure — Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof

Example 2 — Right angle (Cell C2)

Example 3 — Obtuse angle (Cell C3)

Worked example Negative overlap

$a = (1, 2), b = (- 3, 1)$ . $θ$ nikalo.

Forecast: $b$ upar-aur-left point karta hai jabki $a$ upar-aur-right — yeh alag taraf jhuke hain. Guess karo negative dot product aur $θ > 9 0^{\circ}$ .

$a \cdot b = 1 \cdot (- 3) + 2 \cdot 1 = - 3 + 2 = - 1.$ Yeh step kyun? Negative sum ⇒ negative $cos θ$ ⇒ obtuse angle, jaise forecast kiya tha.
$∥ a ∥ = 5, ∥ b ∥ = 10 .$ Yeh step kyun? Cosine formula ke liye lengths chahiye.
$cos θ = \frac{- 1}{5 10} = \frac{- 1}{50} = \frac{- 1}{5 2} \approx - 0.1414.$ Yeh step kyun? Angle isolate karo. $cos θ$ ka sign "alag taraf jhukne" ki information carry karta hai.
$θ = arccos (- 0.1414) \approx 98.1 3^{\circ} .$ Yeh step kyun? Negative number ka $arccos$ automatically $(9 0^{\circ}, 18 0^{\circ})$ mein aata hai — koi quadrant fix nahi chahiye, kyunki $θ$ sirf $[0^{\circ}, 18 0^{\circ}]$ pe define hota hai.

Verify: $θ \approx 98.1 3^{\circ} > 9 0^{\circ}$ ✔ obtuse, negative dot product se match karta hai.

Example 4 — Parallel & anti-parallel (Cells C4, C5)

Worked example Cauchy–Schwarz ke do equality cases

Lo $a = (2, 4)$ . Case C4: $b = (1, 2)$ (same direction, $a = 2 b$ ). Case C5: $c = (- 1, - 2)$ (opposite direction, $a = - 2 c$ ).

Forecast: parallel vectors mein $∣ cos θ ∣ = 1$ hota hai, toh Cauchy–Schwarz ek equality hona chahiye, strict inequality nahi. Same-way → $+$ ; opposite → $-$ .

$a \cdot b = 2 \cdot 1 + 4 \cdot 2 = 2 + 8 = 10.$ Aur $∥ a ∥∥ b ∥ = 20 \cdot 5 = 100 = 10.$ Yeh step kyun? Hum equality test karne ke liye dot product ko length-product se compare karte hain.
Toh $cos θ = \frac{10}{10} = 1 \Rightarrow θ = 0^{\circ}$ — perfectly aligned (C4). Yeh step kyun? $cos θ = 1$ sirf ek value hai jo $θ = 0$ deti hai; arrows direction mein overlap karte hain.
$a \cdot c = 2 (- 1) + 4 (- 2) = - 2 - 8 = - 10.$ Aur $∥ a ∥∥ c ∥ = 20 \cdot 5 = 10.$ Yeh step kyun? Same magnitudes, lekin sign flip hota hai kyunki arrow reverse ho gaya hai.
$cos θ = \frac{- 10}{10} = - 1 \Rightarrow θ = 18 0^{\circ}$ — anti-parallel (C5). Yeh step kyun? $cos θ = - 1$ $θ = 18 0^{\circ}$ pe unique value hai.

Verify: Dono cases mein $∣ a \cdot b ∣ = ∥ a ∥∥ b ∥ = 10$ — Cauchy–Schwarz mein equality exactly tab jab parallel ho ✔. Ex 9 se compare karo, jahan yeh strict hai.

Example 5 — Degenerate: zero vector (Cell C6)

Worked example

$a \cdot b = 0$ hamesha "perpendicular" kyun nahi hota $a = (5, - 7), b = 0 = (0, 0)$ . $a \cdot b$ kya hai, aur angle kya hai?

Forecast: log reflexively kehte hain " $0$ dot product ⇒ perpendicular." Lekin zero-length arrow ki koi direction nahi hoti — toh baat karne ke liye koi angle hi nahi. Guess: dot product $0$ , lekin "perpendicular" yahan meaningless hai.

$a \cdot b = 5 \cdot 0 + (- 7) \cdot 0 = 0.$ Yeh step kyun? Har term mein $0$ ka factor hai, toh sum collapse hokar $0$ ho jaata hai.
Angle formula try karo: $cos θ = \frac{0}{∥ a ∥ \cdot 0} = \frac{0}{0}$ — undefined. Yeh step kyun? $∥ b ∥ = 0$ denominator mein aata hai. Zero se division ka matlab hai koi angle exist nahi karta.
Conclusion: $a \cdot 0 = 0$ kisi bhi $a$ ke liye hota hai, lekin yeh ek degenerate zero hai, perpendicularity nahi. Yeh step kyun? Asli perpendicularity (Ex 2) ke liye dono vectors nonzero hone chahiye.

Verify: Koi doosra $a$ lo, jaise $(100, 3)$ : $a \cdot 0 = 0$ phir bhi. Zero output $b = 0$ se forced hai, direction se independent ✔.

Example 6 — Higher dimensions (Cell C7)

Worked example Wahi machine

$R^{4}$ mein $a = (1, - 1, 2, 0), b = (3, 1, 1, 4)$ $R^{4}$ mein. $a \cdot b$ aur $cos θ$ nikalo.

Forecast: Hum 4D draw nahi kar sakte, lekin formula ko koi fark nahi padta — yeh sirf char products sum karta hai. Yahan algebraic definition apna kamal dikhaati hai (parent ne yeh stress kiya tha).

$a \cdot b = 1 \cdot 3 + (- 1) \cdot 1 + 2 \cdot 1 + 0 \cdot 4 = 3 - 1 + 2 + 0 = 4.$ Yeh step kyun? Component formula kisi bhi dimension tak extend hota hai — har coordinate ke liye ek slot.
$∥ a ∥ = 1 + 1 + 4 + 0 = 6, ∥ b ∥ = 9 + 1 + 1 + 16 = 27 = 33 .$ Yeh step kyun? $∥ \cdot ∥ = x \cdot x$ har dimension mein hold karta hai.
$cos θ = \frac{4}{6 \cdot 3 3} = \frac{4}{3 18} = \frac{4}{9 2} \approx 0.3143.$ Yeh step kyun? Geometric formula abhi bhi $R^{4}$ mein ek angle define karta hai — aur Cauchy–Schwarz (parent §3) exactly yahi guarantee karta hai ki yeh ratio $[- 1, 1]$ mein rahe.
$θ = arccos (0.3143) \approx 71.6 8^{\circ} .$

Verify: Cauchy–Schwarz check: $∣ a \cdot b ∣ = 4$ aur $∥ a ∥∥ b ∥ = 318 \approx 12.73$ , toh $4 \leq 12.73$ ✔ — cosine safely $[- 1, 1]$ ke andar hai. Abstract versions ke liye Inner product spaces bhi dekho.

Example 7 — Real-world word problem (Cell C8)

Worked example Force dwara kiya gaya work

Ek box par force $F = (6, 2)$ newtons lagti hai jabki yeh displacement $d = (4, 0)$ metres ke along slide karta hai. Work define hota hai $W = F \cdot d$ (joules). $W$ compute karo, aur dekho kitni force actually helpful rahi.

Forecast: sirf force ka woh part jo motion ke along ho, work karta hai. Force thoda upar jhuki hui hai, toh kuch "waste" ho raha hai. Guess karo $W$ thoda kam hoga $∣ F ∣ \cdot ∣ d ∣$ se.

$W = F \cdot d = 6 \cdot 4 + 2 \cdot 0 = 24$ joules. Yeh step kyun? Work hai hi ek dot product — yeh force-in-the-direction-of-motion measure karta hai, exactly wahi jo dot product karta hai.
$∥ F ∥ = 36 + 4 = 40 = 210, ∥ d ∥ = 4.$ Yeh step kyun? Angle aur "useful fraction" nikalne ke liye yeh chahiye.
$cos θ = \frac{24}{2 10 \cdot 4} = \frac{24}{8 10} = \frac{3}{10} \approx 0.9487, θ \approx 18.4 3^{\circ} .$ Yeh step kyun? Angle batata hai ki push kitna well aligned hai slide ke saath.
Force component along $d$ : $∥ F ∥ cos θ = 210 \cdot \frac{3}{10} = 6$ N. Yeh $F$ ka $d$ pe projection hai. Yeh step kyun? $W = (useful force) \times (distance) = 6 \times 4 = 24$ J — step 1 se match karta hai, ek achha consistency check.

Verify: Units: N·m $=$ J ✔. Agar force seedha upar hoti ( $F = (0, 2)$ ), toh $W = (0) (4) + (2) (0) = 0$ — koi work nahi, sahi hai, kyunki motion ke sideways push karne se kuch nahi hota ✔.

Example 8 — Exam twist: unknown ke liye solve karo (Cell C9)

Worked example

$k$ ki value nikalo jo vectors ko perpendicular banaye $k$ ki kaun si value(s) ke liye $a = (k, 3)$ aur $b = (k, - 12)$ perpendicular hain?

Forecast: perpendicular ⇔ dot product $= 0$ . Yeh $k$ mein ek equation hai — likely quadratic, toh do answers expect karo.

Perpendicular condition: $a \cdot b = 0$ . Yeh step kyun? Yeh right angle ka defining test hai (humari solutions ke liye dono vectors nonzero honge).
$a \cdot b = k \cdot k + 3 \cdot (- 12) = k^{2} - 36.$ Yeh step kyun? Component formula symbolically apply karo, $k$ ko unknown rakhte hue.
Set karo $k^{2} - 36 = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6.$ Yeh step kyun? Quadratic dono signs deta hai — common exam trap hai $k = - 6$ drop karna.

Verify:

$k = 6$ : $a = (6, 3), b = (6, - 12)$ , dot $= 36 - 36 = 0$ ✔.
$k = - 6$ : $a = (- 6, 3), b = (- 6, - 12)$ , dot $= 36 - 36 = 0$ ✔. Dono mein nonzero vectors hain, toh dono genuine right angles (orthogonal) hain ✔.

Example 9 — Cauchy–Schwarz: strict vs equality (Cell C10)

Worked example Inequality ko tight hote dekhna

Do pairs ko $∣ a \cdot b ∣ \leq ∥ a ∥∥ b ∥$ ke against compare karo: Pair A (parallel nahi): $a = (3, 4), b = (1, 2)$ . Pair B (parallel): $a = (3, 4), b = (6, 8)$ .

Forecast: Pair A strict hona chahiye ( $<$ ); Pair B, parallel hone ki wajah se, equality hona chahiye ( $=$ ).

Pair A: $∣ a \cdot b ∣ = ∣3 + 8∣ = 11.$ RHS $= ∥ a ∥∥ b ∥ = 5 \cdot 5 \approx 11.180.$ Yeh step kyun? Inequality ke dono sides mein directly plug karo.
$11 < 11.180$ — strict, kyunki koi scalar $t$ nahi hai jo $a = t b$ de (check: $3/1 \neq = 4/2$ ). Yeh step kyun? Strictness ⇔ non-parallel, parent ki equality condition ke anusaar.
Pair B: $∣ a \cdot b ∣ = ∣3 \cdot 6 + 4 \cdot 8∣ = ∣18 + 32∣ = 50.$ RHS $= 5 \cdot 100 = 5 \cdot 10 = 50.$ Yeh step kyun? Yahan $b = 2 a$ exactly hai, toh "leftover perpendicular part" zero hai aur bound hit hota hai.
$50 = 50$ — equality, confirm karta hai Pair B parallel hai.

Verify: $11 \leq 11.180$ ✔ (strict) aur $50 \leq 50$ ✔ (equality). Discriminant proof (parent §3) exactly yahi predict karta hai: equality iff kisi $t$ ke liye $∥ a - t b ∥ = 0$ ✔.

Recap

Recall Kaun sa cell kaun sa sign hai?

Positive dot product ka matlab angle hai... ::: acute ( $0^{\circ} < θ < 9 0^{\circ}$ ) Dono vectors nonzero hone par zero dot product ka matlab... ::: perpendicular ( $θ = 9 0^{\circ}$ ) Negative dot product ka matlab angle hai... ::: obtuse ( $9 0^{\circ} < θ < 18 0^{\circ}$ ) $a \cdot 0 = 0$ batata hai angle... ::: undefined hai ( $0$ ki koi direction nahi), perpendicular NAHI Cauchy–Schwarz exactly tab equality hota hai jab... ::: vectors parallel hon

Related: Vector projection, Triangle inequality, Law of cosines, Cross product (vector-valued cousin).