Worked examples — Vector addition — triangle law, parallelogram law

1.1.8 · D3 · Physics › Measurement, Vectors & Kinematics › Vector addition — triangle law, parallelogram law

Shuru karne se pehle, teen words ko dobara clearly define karte hain taaki koi confused na rahe:

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Scenario matrix

Do arrows surprisingly kam genuinely alag tareekon se mil sakte hain. Yahan har wo cell hai jo hume cover karni hai, aur kaun sa worked example use cover karta hai.

#	Case class	Usmein kya khaas hai	Covered by
C1	Acute angle ($0^{\circ }<\theta <90^{\circ }$)	cross term $2AB\cos \theta$ positive hai	Ex 2
C2	Right angle ($\theta =90^{\circ }$)	cross term zero hai → pure Pythagoras	Ex 1
C3	Obtuse angle ($90^{\circ }<\theta <180^{\circ }$)	cross term negative hai → resultant shrink karta hai	Ex 3
C4	Degenerate: same direction ($\theta =0^{\circ }$)	arrows collinear, resultant maximum $A+B$ hai	Ex 4
C5	Degenerate: opposite ($\theta =180^{\circ }$)	arrows anti-parallel, resultant minimum $|A-B|$ hai	Ex 4
C6	Zero input (ek arrow ki length $0$ hai)	"kuch nahi" add karna — identity check	Ex 4
C7	Inverse problem ($R$ diya hai, $\theta$ find karo)	formula ko ulta solve karo	Ex 5
C8	Direction tricky quadrant mein jaati hai	$R_x<0$, toh naive $\arctan$ galat answer deta hai	Ex 6
C9	Real-world word problem	boat river cross kar rahi hai (velocity + velocity)	Ex 7
C10	Exam twist	"$\vec{A}$ ke perpendicular resultant hai" condition	Ex 8

Ab hum har cell ko cover karte hain.

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Example 1 — Right angle (cell C2)

Forecast: aage padhne se pehle resultant guess karo. Kya ye $6+8=14$ se bada hai ya chhota? (Chhota hona chahiye — ye dono fully ek doosre ki help nahi karte.)

1. Cross term khatam karo. Ye step kyun? $\theta =90^{\circ }$ par, $\cos 90^{\circ }=0$, isliye $2AB\cos \theta =0$. Formula simplify hokar $R=\sqrt{A^2+B^2}$ ban jaata hai — plain Pythagoras, kyunki do arrows bilkul ek right triangle ki do chhoti sides ki tarah right angle banaate hain. $R=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{N}.$
2. Direction. Ye step kyun? $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$; $\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$ rakhne par ye $\frac{8}{6}$ ho jaata hai. $\tan \alpha =\frac{8}{6}=1.333\Rightarrow \alpha =\arctan (1.333)\approx 53.13^{\circ }.$

Verify: $10$ allowed band $[|6-8|,6+8]=[2,14]$ ke andar hai. ✔ Units: N + N → N. ✔ Aur $53.13^{\circ }$ north ki taraf tilt karta hai — sensible hai, kyunki northward pull ($8$) dono mein zyada strong hai.

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Example 2 — Acute angle (cell C1)

Forecast: arrows equal hain aur ek doosre ki taraf jhuk rahe hain. Answer kahan point karna chahiye? (Bilkul beech mein — symmetry yahi demand karti hai.)

1. Magnitude. Ye step kyun? $\cos 60^{\circ }=0.5$, isliye cross term $2AB\cos \theta =2(5)(5)(0.5)=25$ positive hai aur length add karta hai. $R=\sqrt{25+25+25}=\sqrt{75}=5\sqrt{3}\approx 8.66\text{N}.$
2. Direction. Ye step kyun? Wahi formula; $\sin 60^{\circ }\approx 0.866$. $\tan \alpha =\frac{5(0.866)}{5+5(0.5)}=\frac{4.330}{7.5}=0.5774\Rightarrow \alpha =30^{\circ }.$

Verify: $30^{\circ }$ exactly $60^{\circ }$ ka aadha hai — resultant angle bisect karta hai, jo equal vectors ke saath hona hi chahiye. ✔ $8.66$ $[0,10]$ ke andar hai. ✔

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Example 3 — Obtuse angle (cell C3)

Forecast: Example 2 jaisi hi magnitudes hain lekin zyada door. Bada resultant hoga ya chhota? (Chhota — ab ye ek doosre se zyada fight kar rahe hain.)

1. Magnitude. Ye step kyun? $\cos 120^{\circ }=-0.5$, isliye cross term ab negative hai: $2(5)(5)(-0.5)=-25$. Obtuse angles ka yahi poora point hai — arrows partly cancel karte hain. $R=\sqrt{25+25-25}=\sqrt{25}=5\text{N}.$
2. Direction. $\sin 120^{\circ }\approx 0.866$, $\cos 120^{\circ }=-0.5$: $\tan \alpha =\frac{5(0.866)}{5+5(-0.5)}=\frac{4.330}{2.5}=1.732\Rightarrow \alpha =60^{\circ }.$

Verify: Ex 2 se compare karo: same lengths, wider angle → chhota resultant ($5<8.66$). ✔ Bisector logic abhi bhi hold karta hai: $\alpha =60^{\circ }=\frac{1}{2}(120^{\circ })$. ✔

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Example 4 — Degenerate trio (cells C4, C5, C6)

Forecast: collinear arrows ordinary numbers ki tarah behave karte hain. Algebra se pehle teeno predict karo.

1. (a) Same direction, $\theta =0^{\circ }$ (C4). Kyun? $\cos 0^{\circ }=1$, isliye $R^2=A^2+B^2+2AB=(A+B{)}^2$. $R=\sqrt{49+9+2(7)(3)}=\sqrt{100}=10=A+B.$ Ye maximum hai jo koi bhi resultant reach kar sakta hai.
2. (b) Opposite, $\theta =180^{\circ }$ (C5). Kyun? $\cos 180^{\circ }=-1$, isliye $R^2=A^2+B^2-2AB=(A-B{)}^2$. $R=\sqrt{49+9-42}=\sqrt{16}=4=|A-B|=|7-3|.$ Ye minimum hai. (Hum $|A-B|$ likhte hain taaki length positive rahe even if $B>A$.)
3. (c) Zero add karna, $B=0$ (C6). Kyun? Length $0$ ka arrow matlab hai "jagah par raho". Formula deta hai $R=\sqrt{A^2+0+0}=A$, aur $\tan \alpha =\frac{0}{A}=0\Rightarrow \alpha =0^{\circ }$. $R=7,\alpha =0^{\circ }.$ Zero vector identity hai: $\vec{A}+\vec{0}=\vec{A}$.

Verify: in do arrows ka har resultant $[|7-3|,7+3]=[4,10]$ mein hona chahiye. Cases (a) aur (b) us band ke do ends hain, aur (c) mein $7\in [4,10]$ milta hai. ✔

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Example 5 — Inverse problem: resultant diya hai, angle find karo (cell C7)

Forecast: $6$, $8$, $10$... kya ye trio koi Pythagorean bell bajata hai? Agar haan, toh Pythagoras kaunsa angle demand karta hai?

1. Formula likho, $\theta$ ko unknown maano. Ye step kyun? Hume $R$ aur dono magnitudes pata hain; sirf $\cos \theta$ missing hai, isliye use isolate karo. $R^2=A^2+B^2+2AB\cos \theta .$
2. Substitute karo. $100=36+64+2(6)(8)\cos \theta =100+96\cos \theta .$
3. Solve karo. Ye step kyun? Dono sides se $100$ subtract karo: $0=96\cos \theta \Rightarrow \cos \theta =0\Rightarrow \theta =90^{\circ }.$

Verify: Plug back karo: $\sqrt{36+64+0}=\sqrt{100}=10$. ✔ Diya hua resultant match karta hai, aur $6,8,10$ classic right triangle hai — $\theta =90^{\circ }$ ke saath consistent. ✔

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Example 6 — Tricky quadrant mein direction (cell C8)

Forecast: $B$ bahut lamba hai aur mostly backwards point karta hai (upar aur baayi taraf). Isliye resultant upper-left mein jaana chahiye — $90^{\circ }$ se aage. Dekho kaise naive arctan ise chhupane ki koshish karta hai.

1. Components mein toddo. Ye step kyun? Kyunki resultant clearly $\vec{A}$ ke paas nahi baithega; $R_x,R_y$ se jaana quadrant ko honestly padhne deta hai. $\cos 150^{\circ }=-0.8660$, $\sin 150^{\circ }=0.5$ ke saath: $R_x=A+B\cos \theta =2+6(-0.8660)=2-5.196=-3.196,$ $R_y=B\sin \theta =6(0.5)=3.$
2. Magnitude. Ye step kyun? Un components par Pythagoras. $R=\sqrt{(-3.196{)}^2+3^2}=\sqrt{10.215+9}=\sqrt{19.215}\approx \mathrm{4.384.}$ (Direct formula se cross-check: $R=\sqrt{4+36+2(2)(6)(-0.8660)}=\sqrt{40-20.78}=\sqrt{19.22}$ ✔.)
3. Direction trap. Ye step kyun? $\tan \alpha =\frac{R_y}{R_x}=\frac{3}{-3.196}=-0.9387$. Calculator ka $\arctan (-0.9387)$ return karta hai $-43.2^{\circ }$ — jo down-and-right point karega, galat quadrant! Kyunki $R_x<0$ aur $R_y>0$, arrow actually second quadrant mein hai (upper-left). Fix: $180^{\circ }$ add karo. $\alpha =-43.2^{\circ }+180^{\circ }\approx 136.8^{\circ }.$

Verify: $136.8^{\circ }$ $90^{\circ }$ se aage hai, hamare "upper-left" forecast se match karta hai. Aur $R\approx 4.38$ $[|2-6|,2+6]=[4,8]$ ke andar hai. ✔ Lesson ye hai: jab bhi $A+B\cos \theta <0$ ho, resultant backward lean karta hai aur tumhe bare arctangent mein $180^{\circ }$ add karna hoga.

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Example 7 — Real-world word problem (cell C9)

Forecast: boat seedha across aim karti hai lekin paani use sideways carry karta hai. Ye $3+4=7$ se tez nahi ja sakti. Guess karo true speed kahan land karegi.

1. Do vectors identify karo. Ye step kyun? Velocities exactly displacements ki tarah add hoti hain (ye "like quantities" hain). Yahan $A=3$ (across) aur $B=4$ (downstream) $90^{\circ }$ par milte hain.
2. Magnitude. Ye step kyun? Right angle → Pythagoras. $R=\sqrt{3^2+4^2+2(3)(4)\cos 90^{\circ }}=\sqrt{9+16}=5\text{m/s}.$
3. Direction. Ye step kyun? "Seedha across" ($3\text{m/s}$) vector se angle: $\tan \alpha =\frac{4\sin 90^{\circ }}{3+4\cos 90^{\circ }}=\frac{4}{3}=1.333\Rightarrow \alpha \approx 53.13^{\circ }\text{seedha-across se downstream ki taraf}.$

Verify: $5\text{m/s}$ $|3-4|=1$ aur $3+4=7$ ke beech hai. ✔ Units: poore mein m/s. ✔ Ye wahi $3$-$4$-$5$ geometry hai jaise Example 1 mein — ek reassuring echo (full boat-and-river treatment ke liye Relative velocity dekho).

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Example 8 — Exam twist: resultant $\vec{A}$ ke perpendicular (cell C10)

Forecast: "$\vec{R}\perp \vec{A}$" ek disguised equation hai. Agar $\vec{R}$ $\vec{A}$ se seedha upar point karta hai toh $\vec{R}$ ka horizontal part kya hona chahiye? (Wo vanish hona chahiye.)

1. Geometry ko algebra mein translate karo. Ye step kyun? Agar $\vec{R}$ $\vec{A}$ (hamare x-axis) ke perpendicular hai, toh $\vec{R}$ ka koi x-component nahi hoga: $R_x=0$. $R_x=A+B\cos \theta =4+B\cos 120^{\circ }=4-0.5B=0.$
2. $B$ solve karo. Ye step kyun? Ek equation, ek unknown. $4=0.5B\Rightarrow B=8.$
3. $R$ y-component se find karo. Ye step kyun? Kyunki $R_x=0$ hai, poora resultant hi uska y-component hai. $R=R_y=B\sin \theta =8\sin 120^{\circ }=8(0.8660)=6.928\approx 4\sqrt{3}.$

Verify: Full magnitude formula se check karo: $R=\sqrt{4^2+8^2+2(4)(8)\cos 120^{\circ }}=\sqrt{16+64-32}=\sqrt{48}=4\sqrt{3}\approx 6.928$. ✔ Aur $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\frac{6.928}{0}$ → undefined → $\alpha =90^{\circ }$, confirm karta hai $\vec{R}\perp \vec{A}$. ✔

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Recall Pure matrix par quick self-test

$6$ aur $8$ ka right-angle resultant ::: $10$, $\arctan (8/6)\approx 53.13^{\circ }$ par Equal vectors ($5,5$) $60^{\circ }$ par: magnitude aur direction ::: $5\sqrt{3}\approx 8.66$, $30^{\circ }$ par bisect karta hua Equal vectors ($5,5$) $120^{\circ }$ par: magnitude ::: $5$ $7$ aur $3$ ka maximum aur minimum resultant ::: $10$ ($0^{\circ }$ par) aur $4$ ($180^{\circ }$ par) $6$ aur $8$ se resultant $10$: $\theta$ kya hai? ::: $90^{\circ }$ (kyunki $\cos \theta =0$) Arctan direction mein $180^{\circ }$ kab add karna chahiye? ::: Jab $R_x=A+B\cos \theta <0$ ho "$\vec{A}$ ke perpendicular resultant" ek equation ke roop mein ::: $A+B\cos \theta =0$

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Connections

• Vectors — components and unit vectors — Examples 6 aur 8 poori tarah component picture par lean karte hain.
• Resolution of a vector into components — $R_x,R_y$ split jo poore mein use hua.
• Subtraction of vectors and the difference vector — Example 4(b) ($\theta =180^{\circ }$) subtraction ka seed hai.
• Relative velocity — Example 7 pehli jhalak hai; river problem yahan full mein hai.
• Law of cosines — magnitude formula disguise mein yahi rule hai; inverse problem (Ex 5) matlab hai "cosine rule ko angle ke liye solve karo".
• Scalar (dot) product — cross term mein chhupa $\cos \theta$ yahan se aata hai.
• Vector addition — triangle law, parallelogram law — parent note jiske liye ye examples serve karte hain.