Exercises — Scalars vs vectors — definition, examples

Hum kya karte hain: har ek ke liye poochho "kya mujhe ise poori tarah batane ke liye direction chahiye — AUR kya yeh parallelogram law se add hota hai?"

• (a) mass → scalar (sirf kilograms).
• (b) velocity → vector (speed aur direction).
• (c) temperature → scalar (sirf degrees).
• (d) force → vector (magnitude aur direction, parallelogram se add hota hai).
• (e) electric current → scalar — yeh ek trap hai; iska wire ke saath ek direction hota hai lekin currents arithmetically add hote hain (Kirchhoff), parallelogram se nahi.
• (f) displacement → vector.

Answer: speed. Distance aur speed scalar members hain; displacement aur velocity vector members hain. Distance total path length hai (no direction), speed distance-over-time hai (no direction). Dekho Distance vs Displacement aur Speed vs Velocity.

(a) Distance (scalar). Distance direction ignore karta hai, isliye yeh plain addition hai: $d=3+4=7\text{km}.$ (b) Displacement (vector). East aur North perpendicular hain, isliye $\theta =90^{\circ }$ aur $\cos 90^{\circ }=0$. Cross term $2AB\cos \theta$ (upar define kiya) vanish ho jaata hai: $|\vec{R}|=\sqrt{3^2+4^2+2\cdot 3\cdot 4\cdot 0}=\sqrt{9+16}=\sqrt{25}=5\text{km}.$ Yeh kaisa dikhta hai: ek right triangle jiske legs 3 aur 4 hain; diagonal (displacement) hypotenuse 5 hai. Isliye scalar 7 deta hai lekin vector 5 deta hai.

Hum kya karte hain: force ek vector hai, isliye resultant formula use karo $\theta =90^{\circ }$ ke saath, $\cos 90^{\circ }=0$: $|\vec{R}|=\sqrt{6^2+8^2+2\cdot 6\cdot 8\cdot 0}=\sqrt{36+64}=\sqrt{100}=10\text{N}.$

(a) Speed total path length (distance) use karta hai: $\text{speed}=\frac{400}{100}=4\text{m/s}.$ (b) Velocity displacement use karta hai. Car start pe waapis aati hai, isliye net displacement $=0$: $|\text{velocity}|=\frac{0}{100}=0\text{m/s}.$ Ek directed quantity apne aap cancel ho sakti hai; ek path length nahi ho sakti.

Hum kya karte hain: $A=B=5$, $\theta =120^{\circ }$, aur $\cos 120^{\circ }=-\frac{1}{2}$. Negative cosine sum ko shrink karta hai: $|\vec{R}|=\sqrt{5^2+5^2+2\cdot 5\cdot 5\cdot (-\frac{1}{2})}=\sqrt{25+25-25}=\sqrt{25}=5\text{N}.$ Yeh sahi kyun lagta hai: teen equal arrows of $5\text{N}$ at $120^{\circ }$ ek equilateral pattern mein close hote hain; unme se do combine hokar teesre ka size dete hain. Isliye resultant har force ke barabar hota hai.

Hum kya karte hain: current ek scalar hai, isliye alag alag directions ke bawajood hum plain addition use karte hain (charge conservation, Kirchhoff): $I_{\text{out}}=2+3=5\text{A}.$ Hum parallelogram nahi use karte — agar hum galti se karte, toh say $90^{\circ }$ par hume $\sqrt{4+9}\approx 3.6\text{A}$ milta, jo physically nonsense hai (charge disappear ho jaata). Yahi reason hai ki "direction hai ⇒ vector" fail karta hai.

$\theta =0^{\circ }$ rakho, toh $\cos 0^{\circ }=1$: $|\vec{R}|=\sqrt{A^2+B^2+2AB\cdot 1}=\sqrt{A^2+2AB+B^2}=\sqrt{(A+B{)}^2}=A+B.$ Interpretation: aligned arrows sirf head-to-tail stack hote hain, isliye lengths add hoti hain. "$+2AB\cos \theta$" term precisely yahi hai jo vectors ko scalars se alag banata hai jab angle zero nahi hota.

Step 1 — between-angle dhundho. Pehla leg East point karta hai. Doosra leg East se $60^{\circ }$ away turn karta hai. Tail-to-tail rakhne par, do arrows ke beech ka angle $\theta =60^{\circ }$ hai, aur $\cos 60^{\circ }=\frac{1}{2}$. Step 2 — formula apply karo $A=B=4$ ke saath: $|\vec{R}|=\sqrt{4^2+4^2+2\cdot 4\cdot 4\cdot \frac{1}{2}}=\sqrt{16+16+16}=\sqrt{48}=4\sqrt{3}\approx 6.93\text{km}.$ Yeh kaisa dikhta hai: do legs aur resultant ek triangle banate hain. Do extreme cases se compare karo taaki $6.93$ ka context ho: agar arrows opposite point karte ($\theta =180^{\circ }$) toh answer minimum $|4-4|=0$ hota; agar woh same way point karte ($\theta =0^{\circ }$) toh maximum $4+4=8$ hota. Kyunki turn gentle hai ($60^{\circ }$), hamaara $6.93$ in dono limits ke beech comfortably baitha hai, aligned maximum ke kareeb.

Maximum, $A+B$: chahiye $\cos \theta =+1$, isliye $\theta =0^{\circ }$ — arrows same direction mein point karte hain, lengths stack hoti hain. Minimum, $|A-B|$: chahiye $\cos \theta =-1$, isliye $\theta =180^{\circ }$ — arrows opposite point karte hain, woh partly cancel hote hain: $\sqrt{A^2+B^2-2AB}=\sqrt{(A-B{)}^2}=|A-B|.$ Takeaway: do vectors ke har possible resultant ka range $|A-B|\le |\vec{R}|\le A+B$ mein hota hai. Yeh range hi vector addition ki pehchaan hai.

Setup: $A=B=F$, chahiye $|\vec{R}|=F$: $F=\sqrt{F^2+F^2+2F^2\cos \theta }.$ Dono sides square karo (dono sides positive hain, square karna safe hai): $F^2=2F^2+2F^2\cos \theta .$ $F^2$ se divide karo (allowed kyunki $F\ne 0$): $1=2+2\cos \theta \Rightarrow 2\cos \theta =-1\Rightarrow \cos \theta =-\frac{1}{2}.$ Cosine undo karo: woh angle jiska cosine $-\frac{1}{2}$ hai woh hai $\theta =120^{\circ }.$ Check: yeh Exercise 3.1 se exactly match karta hai — do $5\text{N}$ forces $120^{\circ }$ par $5\text{N}$ deti hain. Algebra picture confirm karta hai.

Ulta assume karo: suppose current ek vector hota jo parallelogram law follow karta. Toh "out" current hota $\sqrt{2^2+3^2+2\cdot 2\cdot 3\cos 90^{\circ }}=\sqrt{4+9+0}=\sqrt{13}\approx 3.61\text{A}.$ Contradiction: $2\text{A}$ charge per second plus $3\text{A}$ charge per second junction mein har second $5$ coulombs deliver karte hain. Charge create ya destroy nahi ho sakta, isliye $5\text{A}$ zaroor bahar jaana chahiye. Lekin parallelogram ne $3.61\text{A}$ diya — charge ka loss, jo impossible hai. Conclusion: vector assumption charge conservation tod deta hai, isliye current ek scalar hai, arithmetically $5\text{A}$ mein add hota hai. Wire ke saath direction real hai, lekin yeh parallelogram law follow nahi karta — jo parent note ke "steel-man" trap ka poora point hai.

(a) Boat ki across velocity aur river ki along velocity do vectors hain $90^{\circ }$ par. Yeh parallelogram se add hote hain: $|\vec{v}|=\sqrt{3^2+4^2+2\cdot 3\cdot 4\cos 90^{\circ }}=\sqrt{9+16}=5\text{m/s}.$ (b) Velocities (boat, river, resultant) vectors hain — isliye woh $7$ nahi $5$ mein combine hue. Speeds $3$, $4$, $5$ (unke magnitudes) scalars hain. Time, agar involved hota, scalar hota. Yeh Speed vs Velocity action mein hai aur relative velocity ko preview karta hai.

Question ::: Answer hai — triple colon ke baad wala part reveal karo) Woh range jisme har do-vector resultant rehta hai ::: $|A-B|\le |\vec{R}|\le A+B$ Do equal forces kis angle par ek hi force ke barabar resultant deti hain? ::: $120^{\circ }$ Current direction hone ke bawajood scalar kyun hai? ::: yeh arithmetically add hota hai (Kirchhoff), parallelogram law se nahi