Worked examples — Dot product — formula, geometric meaning, work calculation

1.1.11 · D3 · Physics › Measurement, Vectors & Kinematics › Dot product — formula, geometric meaning, work calculation

Yeh page ek tour hai un sabhi situations ka jo dot product tumhare saamne rakh sakta hai. Hum pehle ek table mein saare case-classes list karte hain, phir har ek ka example solve karte hain — saath mein ek Forecast taaki tum compute karne se pehle guess karo, aur ek Verify taaki tum kabhi bhi kisi number par aankhein band karke bharosa na karo.

Yahan sab kuch parent page par build karta hai: Dot product — formula, geometric meaning, work calculation. Agar koi bhi symbol unfamiliar lage, pehle Vectors — components and unit vectors aur Trigonometry — cosine and components dekh lo.

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The scenario matrix

Yeh har distinct case-class hai jisme ek dot-product question land kar sakta hai. Neeche ka har worked example apni cell letter ke saath tagged hai.

Ab hum har cell ko hit karte hain.

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Examples

C1 — Partly aligned (positive, acute angle)

1. Component form: $\vec{A}\cdot \vec{B}=(3)(2)+(1)(2)=6+2=8$. Yeh step kyun? Matching components multiply karo aur add karo — number tak pahunchne ka fastest route.
2. Magnitudes: $|\vec{A}|=\sqrt{3^2+1^2}=\sqrt{10}$, $|\vec{B}|=\sqrt{2^2+2^2}=\sqrt{8}$. Yeh step kyun? Geometric form ko $\cos \theta$ isolate karne ke liye dono lengths chahiye.
3. $\cos \theta =\frac{8}{\sqrt{10}\sqrt{8}}=\frac{8}{\sqrt{80}}\approx 0.894$, toh $\theta ={\cos }^{-1}(0.894)\approx 26.6^{\circ }$. Yeh step kyun? Rearranged geometric form number ko angle mein convert karta hai.

Verify: $8>0$ ✓ forecast ke anusaar positive. $26.6^{\circ }<90^{\circ }$ ✓ acute. Figure dekho — $\vec{A}$ ka blue shadow $\vec{B}$ par $\vec{B}$ ke same direction mein padta hai, positive sign confirm karta hai.

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C2 — Perpendicular (dot product exactly zero hai)

1. $\vec{A}\cdot \vec{B}=(4)(-1)+(2)(2)=-4+4=0$. Yeh step kyun? Seedha component form; yahan zero perpendicularity test hai.
2. Kyunki dono vectors non-zero hain aur dot product $0$ hai, $\cos \theta =0\Rightarrow \theta =90^{\circ }$. Yeh step kyun? Sirf $\cos 90^{\circ }=0$ hota hai, toh geometric form right angle force karta hai.

Verify: Slopes: $\vec{A}$ $2/4=0.5$ rise karta hai; $\vec{B}$ $2/(-1)=-2$ rise karta hai. Slopes ka product $=0.5\times (-2)=-1$, classic perpendicular test ✓. Figure mein arrows ek clean right angle par milte hain.

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C3 — Opposing (negative, obtuse angle)

1. $\vec{A}\cdot \vec{B}=(2)(-3)+(1)(1)=-6+1=-5$. Yeh step kyun? Component form; negative sign hi poori kahaani hai.
2. $|\vec{A}|=\sqrt{5}$, $|\vec{B}|=\sqrt{10}$, toh $\cos \theta =\frac{-5}{\sqrt{5}\sqrt{10}}=\frac{-5}{\sqrt{50}}\approx -0.707$. Yeh step kyun? $\cos \theta$ isolate karne ke liye magnitudes se divide karo (denominator kabhi mat chhodna!).
3. $\theta ={\cos }^{-1}(-0.707)\approx 135^{\circ }$. Yeh step kyun? Negative cosine second quadrant mein aata hai — obtuse, jaise predict kiya tha.

Verify: $-5<0$ ✓ negative. $135^{\circ }>90^{\circ }$ ✓ obtuse. Figure mein $\vec{A}$ ka shadow $\vec{B}$ ke opposite direction mein padta hai — woh peeche ki taraf wala shadow hi "negative" jaisa dikhta hai.

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C4 — Parallel aur anti-parallel (limiting cases)

1. $|\vec{A}|=\sqrt{3^2+4^2}=5$; $|\vec{B}|=\sqrt{6^2+8^2}=10$; $|\vec{C}|=10$. Yeh step kyun? Hum theoretical max $|\vec{A}||\vec{B}|$ se compare karna chahte hain.
2. (i) $\vec{A}\cdot \vec{B}=(3)(6)+(4)(8)=18+32=50$. Yeh step kyun? Component form. Compare karo: $|\vec{A}||\vec{B}|=5\times 10=50$ — ceiling hit kar li, toh $\cos \theta =1$, $\theta =0^{\circ }$.
3. (ii) $\vec{A}\cdot \vec{C}=(3)(-6)+(4)(-8)=-18-32=-50$. Yeh step kyun? Same magnitude but $=-|\vec{A}||\vec{C}|$, toh $\cos \theta =-1$, $\theta =180^{\circ }$.

Verify: $50=+|\vec{A}||\vec{B}|$ ✓ (max), $-50=-|\vec{A}||\vec{C}|$ ✓ (min). Dot product kabhi bhi $\pm |\vec{A}||\vec{B}|$ se exceed nahi kar sakta kyunki $\cos \theta$ $[-1,1]$ mein rehta hai — yeh do boundary hain. Dekho bhi Projection of a vector.

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C5 — Degenerate: ek zero vector

1. $\vec{A}\cdot \vec{B}=(0)(5)+(0)(-2)+(0)(7)=0$. Yeh step kyun? Har product mein ek zero factor hai, toh sum $0$ hai.

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C6 — 3-D vectors: angle find karo

1. $\vec{A}\cdot \vec{B}=(1)(2)+(2)(0)+(2)(-1)=2+0-2=0$. Yeh step kyun? Component form 3-D mein identically kaam karta hai — bas extra term add karo.
2. Kyunki dot product $0$ hai aur koi bhi vector zero nahi: $\cos \theta =0\Rightarrow \theta =90^{\circ }$. Yeh step kyun? 2-D jaisa hi perpendicularity test; dimension se logic nahi badalta.

Verify: $|\vec{A}|=\sqrt{1+4+4}=3$, $|\vec{B}|=\sqrt{4+0+1}=\sqrt{5}$, dono non-zero ✓, toh $90^{\circ }$ genuine hai. Dot product kitne bhi dimensions ki parwah nahi karta — shadow idea same hai.

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C7 — Real-world work: ek problem mein lift, drag, aur brake

1. Rope: $W_1=|{\vec{F}}_1||\vec{d}|\cos 30^{\circ }=(20)(8)(0.8660)\approx 138.6\text{ J}$. Yeh step kyun? Sirf horizontal component $20\cos 30^{\circ }$ $\vec{d}$ ke along hai; dot product use extract karta hai.
2. Friction: yeh motion ke opposite direction mein point karta hai, toh $\theta =180^{\circ }$: $W_2=(6)(8)\cos 180^{\circ }=(48)(-1)=-48\text{ J}$. Yeh step kyun? Anti-parallel ⇒ $\cos 180^{\circ }=-1$; friction energy remove karta hai (negative work physical hai, error nahi).
3. Gravity: horizontal path ke $\theta =90^{\circ }$ par, toh $W_g=(50)(8)\cos 90^{\circ }=0\text{ J}$. Yeh step kyun? Perpendicular force koi work nahi karta — same reason ek uthaya hua bag level ground par koi work nahi karta.

Verify: Signs forecast se match karte hain: $+138.6$, $-48$, $0$ ✓. Units: N·m = J ✓. Net work $=138.6-48+0=90.6\text{ J}$ kinetic energy mein jaata hai — dekho Work-Energy Theorem. Gravity-does-zero-work idea Circular motion ko bhi drive karta hai (centripetal force ⟂ velocity).

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C8 — Exam twist: unknown solve karo jo unhe perpendicular banata hai

1. Perpendicular condition: $\vec{A}\cdot \vec{B}=0$. Yeh step kyun? "Perpendicular" ko zero-dot-product test ke zariye algebra mein translate karo.
2. $\vec{A}\cdot \vec{B}=(3)(2)+(k)(-6)=6-6k=0$. Yeh step kyun? Component form solve karne ka equation deta hai.
3. $6-6k=0\Rightarrow k=1$. Yeh step kyun? $k$ isolate karo.

Verify: Plug back karo: $(3)(2)+(1)(-6)=6-6=0$ ✓. Toh $\vec{A}=(3,1)$ waakai $(2,-6)$ ke perpendicular hai.

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C9 — Self dot product = magnitude squared

1. $\vec{A}\cdot \vec{A}=(6)(6)+(-8)(-8)=36+64=100$. Yeh step kyun? $\vec{B}=\vec{A}$ ke saath component form.
2. $\vec{A}\cdot \vec{A}=|\vec{A}{|}^2$, toh $|\vec{A}|=\sqrt{100}=10$. Yeh step kyun? $\theta =0$ ke saath geometric form: $|\vec{A}||\vec{A}|\cos 0=|\vec{A}{|}^2$.

Verify: Direct magnitude: $\sqrt{6^2+(-8{)}^2}=\sqrt{100}=10$ ✓. Self-dot length-squared ka shortcut hai — aakhri step tak square root mat lo.

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Recall Kaun sa cell kaun sa tha?

C1 acute-positive ::: $(3,1)\cdot (2,2)=8$, $\theta \approx 26.6^{\circ }$ C2 perpendicular-zero ::: $(4,2)\cdot (-1,2)=0$, $\theta =90^{\circ }$ C3 obtuse-negative ::: $(2,1)\cdot (-3,1)=-5$, $\theta \approx 135^{\circ }$ C4 parallel / anti-parallel limits ::: $+50$ (max) aur $-50$ (min) C5 zero vector ::: $0$ degeneracy se, angle undefined C7 work signs ::: rope $+138.6$ J, friction $-48$ J, gravity $0$ J

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Connections

• Dot product — formula, geometric meaning, work calculation (parent)
• Vectors — components and unit vectors
• Trigonometry — cosine and components
• Projection of a vector
• Work-Energy Theorem
• Circular motion
• Cross product — area and torque