Worked examples — Scalars vs vectors — definition, examples

1.1.6 · D3 · Physics › Measurement, Vectors & Kinematics › Scalars vs vectors — definition, examples

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Scenario matrix

Kuch bhi work karne se pehle, aao har tarah ke cases list karein jo ek scalar/vector problem mein aa sakte hain. Baad ke har example mein uska cell tag hoga. Reader ko koi aisa scenario nahi milna chahiye jo hum skip kar dein.

#	Case class	Kya special hai isme	Example jo ise cover karta hai
C1	Scalar addition	direction ignore, plain $A+B$	Ex 1
C2	Aligned vectors ($\theta =0^{\circ }$)	$\cos \theta =+1$, vector max $=A+B$	Ex 2
C3	Anti-aligned vectors ($\theta =180^{\circ }$)	$\cos \theta =-1$, vector min $=	A-B
C4	Perpendicular ($\theta =90^{\circ }$)	$\cos \theta =0$, Pythagoras	Ex 3
C5	General acute angle ($0<\theta <90^{\circ }$)	cross term positive, poora formula chahiye	Ex 4
C6	General obtuse angle ($90^{\circ }<\theta <180^{\circ }$)	cross term negative	Ex 5
C7	Zero / degenerate input	ek magnitude $0$ hai, ya dono equal aur opposite	Ex 6
C8	Scalar ka sign vs vector ka sign	negative matlab "zero se neeche" vs "reversed"	Ex 7
C9	Direction-trap scalar	direction hai par arithmetically add hota hai	Ex 8
C10	Real-world word problem	tumhe khud decide karna hai scalar ya vector	Ex 9
C11	Exam-style twist	round trip, cancellation, mixed	Ex 10

Har figure mein accent colour red se us ek object ko mark karenge jo sabse zyada matter karta hai — usually resultant $\vec{R}$.

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Notation set karna (pehle samjho, phir use karo)

Figure s01 — Red curve hai $\cos \theta$, "helping dial". $\theta =0^{\circ }$ par yeh $+1$ read karta hai (arrows poori tarah help karte hain), $90^{\circ }$ par $0$ (neutral), $180^{\circ }$ par $-1$ (arrows poori tarah fight karte hain). Formula is helping term $2AB$ ko is dial se multiply karta hai.

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Worked Examples

Ex 1 — Cell C1: plain scalar addition

Figure s02 — Do black mass blocks ek pan par stacked hain; red total ($7\text{kg}$) sirf arithmetic sum hai — koi arrow nahi, koi direction nahi. Isko baad ke har figure se contrast karo jahan red object ek directed resultant hai.

1. Quantity identify karo. Mass. Yeh step kyun? Kyunki addition rule decide hota hai quantity kis type ki hai se, numbers se nahi.
2. Test check karo. Mass ki koi direction nahi; do masses kabhi "alag ways" mein point nahi karte. Toh plain arithmetic: $3+4=7\text{kg}$. Yeh step kyun? Yeh cell C1 hai — scalar case jahan $A+B$ hamesha answer hai.

Verify: Units: $\text{kg}+\text{kg}=\text{kg}$ ✓. Scale "7 kg north" nahi read kar sakta, confirm karta hai ki mass ek scalar hai. Answer $=7\text{kg}$ — forecast se match karta hai.

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Ex 2 — Cells C2 & C3: aligned aur anti-aligned vectors

Figure s03 — Top: do black force arrows same direction mein; red resultant $\vec{R}=14\text{N}$ unka poora sum hai (max). Bottom: arrows opposite directions mein point karte hain; red resultant $\vec{R}=2\text{N}$ chhota bacha hua hai (min).

1. (a) Same direction, $\theta =0^{\circ }$. $\cos 0^{\circ }=1$, toh $|\vec{R}|=\sqrt{6^2+8^2+2\cdot 6\cdot 8\cdot 1}=\sqrt{36+64+96}=\sqrt{196}=14\text{N}.$ Yeh step kyun? Jab $\cos \theta =+1$ toh formula collapse hota hai $\sqrt{(A+B{)}^2}=A+B$ mein. Yeh sabse bada resultant hai jo kabhi ho sakta hai. (Cell C2.) Direction: $\varphi =0^{\circ }$ (dono ke saath same direction mein).
2. (b) Opposite directions, $\theta =180^{\circ }$. $\cos 180^{\circ }=-1$, toh $|\vec{R}|=\sqrt{6^2+8^2-2\cdot 6\cdot 8}=\sqrt{36+64-96}=\sqrt{4}=2\text{N}.$ Yeh step kyun? Jab $\cos \theta =-1$ toh formula collapse hota hai $\sqrt{(A-B{)}^2}=|A-B|$ mein. Yeh sabse chhota possible magnitude hai. (Cell C3.) Yeh bade force yaani $8\text{N}$ wale ki direction mein point karta hai.

Verify: $A+B=14$ ✓, $|A-B|=|6-8|=2$ ✓ — dono forecast se match karte hain. In do forces ka har real resultant $2\text{N}$ aur $14\text{N}$ ke beech hi hoga — yeh rule yaad rakhne layak hai.

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Ex 3 — Cell C4: perpendicular (Pythagoras) — direction ke saath

Figure s04 — Black arrows: $3\text{km}$ East phir $4\text{km}$ North, right angle par milte hue. Red arrow resultant displacement hai $\vec{R}=5\text{km}$, East se $53^{\circ }$ upar jhuka hua.

1. Legs ke beech angle nikalo. East aur North $\theta =90^{\circ }$ apart hain. Yeh step kyun? Formula ko tail-to-tail draw kiye arrows ke beech angle chahiye; right-angle turn $90^{\circ }$ hota hai.
2. Magnitude. $\cos 90^{\circ }=0$, toh cross term khatam ho jaata hai: $|\vec{R}|=\sqrt{3^2+4^2+0}=\sqrt{9+16}=\sqrt{25}=5\text{km}.$ Yeh step kyun? $90^{\circ }$ par do arrows na help karte hain na fight, toh formula pure Pythagoras ban jaata hai. (Cell C4.)
3. Direction. Maan lo $\vec{A}=$ East ($3\text{km}$), $\vec{B}=$ North ($4\text{km}$), $\theta =90^{\circ }$ toh $\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$. Forward $=3+4\cdot 0=3>0$, toh hum safe first-quadrant case mein hain: $\tan \varphi =\frac{B\sin \theta }{A+B\cos \theta }=\frac{4\cdot 1}{3+4\cdot 0}=\frac{4}{3}.$ Toh $\varphi =\arctan \frac{4}{3}\approx 53^{\circ }$ North of East measure kiya gaya. Yeh step kyun? $\tan$ "sideways-over-forward" ratio $\frac{4}{3}$ ko actual tilt angle mein waapis convert karta hai, jo $\arctan$ ("kaun sa angle is tan ke saath hai?" sawal) recover karta hai.

Verify: $3$-$4$-$5$ right triangle: $3^2+4^2=25=5^2$ ✓. $\arctan \frac{4}{3}=53.13^{\circ }>45^{\circ }$, toh yeh East se zyada North ki taraf jhuka hai — forecast se match karta hai ✓. Distance $=3+4=7\text{km}$ (scalar) abhi bhi $5\text{km}$ displacement se alag hai. Dekho Distance vs Displacement.

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Ex 4 — Cell C5: general acute angle — direction ke saath

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Ex 5 — Cell C6: general obtuse angle — quadrant trap ke saath

Figure s05 — $\vec{A}$ (black, axis ke along) se dono obtuse cases. $\theta =120^{\circ }$ par red resultant abhi bhi forward jhukta hai ($\varphi \approx 79^{\circ }$). $\theta =160^{\circ }$ par red resultant $\vec{A}$ ke peeche jhukta hai ($\varphi \approx 129^{\circ }$) — woh case jahan blindly $\arctan$ par trust karna tumhe galat answer dega.

Verify: (a) Bounds $2\le 5.29\le 10$ ✓, $\sqrt{28}<\sqrt{52}$ ✓, $\varphi =79.1^{\circ }$. (b) Bounds $2\le 2.63\le 10$ ✓; forward negative ⇒ $\varphi >90^{\circ }$, aur $128.6^{\circ }$ sach mein $90^{\circ }$ se zyada hai — forecast confirm. "Resultant speed" ke liye Speed vs Velocity dekho.

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Ex 6 — Cell C7: zero aur degenerate inputs

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Ex 7 — Cell C8: minus sign ka matlab kya hai

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Ex 8 — Cell C9: direction-trap scalar

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Ex 9 — Cell C10: real-world word problem (tum decide karo)

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Ex 10 — Cell C11: exam twist (round trip)

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Recall Quick self-test (guess karne ke baad reveal karo)

Do forces $6\text{N}$ aur $8\text{N}$, resultant $90^{\circ }$ par? ::: $\sqrt{36+64}=10\text{N}$ Same forces, maximum possible resultant? ::: aligned, $6+8=14\text{N}$ Same forces, minimum possible resultant? ::: opposite, $|6-8|=2\text{N}$ $5\text{N}$ aur $8\text{N}$ at $60^{\circ }$ — magnitude? ::: $\sqrt{25+64+40}=\sqrt{129}\approx 11.36\text{N}$ $5\text{N}$ aur $8\text{N}$ at $60^{\circ }$ — $5\text{N}$ se direction? ::: $\arctan \frac{8\sin 60^{\circ }}{5+8\cos 60^{\circ }}\approx 37.6^{\circ }$ $\varphi$ ke liye plain $\arctan$ kab galat hota hai, aur fix kya hai? ::: jab forward $=A+B\cos \theta <0$; tab $180^{\circ }$ add karo correct (second) quadrant mein aane ke liye Junction par $2\text{A}$ + $3\text{A}$ = $5\text{A}$ kyun, $\sqrt{13}$ nahi? ::: current ek scalar hai; yeh arithmetically add hota hai (charge conservation), parallelogram se nahi

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Connections

• Scalars vs vectors — definition, examples — parent: definitions aur decision test.
• Vector Addition — Triangle & Parallelogram Law — jahan $\sqrt{A^2+B^2+2AB\cos \theta }$ aur $\tan \varphi =\frac{B\sin \theta }{A+B\cos \theta }$ derive hote hain.
• Distance vs Displacement — Ex 3 & Ex 9 detail mein.
• Speed vs Velocity — Ex 5, Ex 9, Ex 10.
• Components of a Vector — inhi answers ka doosra route (East/North parts mein split karo).
• Dot and Cross Products — jahan $\cos \theta$ dobara $\vec{A}\cdot \vec{B}$ ke roop mein aata hai.
• Units and Dimensions — upar ka har answer apna unit carry karta tha; yeh optional nahi hai.

Concept Map