Worked examples — Vector representation — magnitude, direction, components

1.1.7 · D3 · Physics › Measurement, Vectors & Kinematics › Vector representation — magnitude, direction, components

Shuru karne se pehle, ek notation jo hum neeche har jagah use karte hain. Jab hum $\vec{A}=A_x\hat{i}+A_y\hat{j}$ likhte hain, to chote hats $\hat{i}$ aur $\hat{j}$ unit vectors hain: bilkul $1$ length ke chote arrows, ek $+x$ ki taraf ($\hat{i}$) aur ek $+y$ ki taraf ($\hat{j}$). Hat "$\hat{}$" ka matlab hamesha hota hai "is arrow ki length 1 hai" — pure direction, koi size nahi. Toh "$6\hat{i}+8\hat{j}$" literally matlab hai "$+x$ direction mein $6$ kadam jao, phir $+y$ direction mein $8$ kadam jao." (Iske baare mein aur Unit Vectors i, j, k mein padho.)

Iske saath: ek vector ek aisa arrow hai jisme ek size hoti hai (iska magnitude $A\ge 0$) aur ek heading hoti hai (iska direction angle $\theta$, $+x$ axis se anticlockwise measure kiya hua). Iske components $A_x,A_y$ wo shadows hain jo yeh horizontal aur vertical axes par dalta hai. Do recipes jo hum drill karte hain:

---

Scenario matrix

Is topic ke har problem ka ghar in cells mein se ek mein hai. Hum sabko cover karenge.

#	Cell (scenario class)	Kya mushkil karta hai	Covered by
1	Quadrant I ($A_x>0,A_y>0$)	easy baseline	Ex A
2	Quadrant II ($A_x<0,A_y>0$)	calculator jhooth bolta hai, $180°$ add karo	Ex B
3	Quadrant III ($A_x<0,A_y<0$)	dono negative, phir bhi $180°$ add karo	Ex C
4	Quadrant IV ($A_x>0,A_y<0$)	negative ya $+360°$	Ex D
5	Degenerate: purely axial ($A_x=0$ or $A_y=0$)	${\tan }^{-1}$ blow up karta hai / ratio $0$ hai	Ex E
6	Zero vector ($A_x=A_y=0$)	direction undefined hai	Ex E
7	Limiting angle ($\theta \to 90°$)	boundary par component behaviour	Ex F
8	Real-world word problem (do vectors add karo)	words $\to$ components translate karo	Ex G
9	Exam twist: angle galat axis se	$+y$ se measure kiya, $+x$ se nahi	Ex H
10	Exam twist: unit vector + reconstruction	size strip karo, phir rebuild karo	Ex I

---

Ex A — Quadrant I (baseline) cell 1

Figure mein orange arrow $\vec{A}$ hai; x-axis par iski plum shadow $A_x=6$ hai ($\theta$ ke adjacent) aur tip tak upar teal dashed shadow $A_y=8$ hai ($\theta$ ke opposite). Origin par chota ink arc wo angle $\theta$ hai jo hum solve kar rahe hain.

1. Magnitude. $A=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. Yeh step kyun? Plum aur teal shadows figure mein right triangle ki do legs hain; Pythagoras un legs ko hypotenuse mein convert karta hai, jo ki orange arrow ki length hai.
2. Raw angle. $\theta ={\tan }^{-1}(8/6)={\tan }^{-1}(1.333)=53.13°$. Yeh step kyun? $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{A_y}{A_x}$ measure karta hai ki orange arrow kitna steeply climb karta hai (teal over plum); ${\tan }^{-1}$ poochta hai "kis angle ki yeh steepness hai?"
3. Quadrant check. $A_x>0,A_y>0\Rightarrow$ Quadrant I $\Rightarrow$ koi fix nahi. Final $\theta =53.13°$. Yeh step kyun? Calculator $(-90°,90°)$ mein angles return karta hai, jo QI ke liye exactly sahi hai, toh figure ka ink arc ko koi correction ki zaroorat nahi.

Verify: components rebuild karo — $10\cos 53.13°=6.00$, $10\sin 53.13°=8.00$. ✓ Wapas start par. Units: components vector ke units share karte hain (yahan pure numbers hain).

---

Ex B — Quadrant II (calculator jhooth bolta hai) cell 2

Figure mein orange arrow true vector $\vec{A}$ hai jo up-left point kar raha hai; faint plum arrow down-right point karta hai jo calculator ka guess hai ($-67.4°$). Notice karo ki dono arrows origin se ek straight line par hain — isliye unka same $\tan$ ratio hai, aur exactly isliye calculator unhe confuse karta hai. Teal arc corrected angle hai.

1. Magnitude. $A=\sqrt{(-5{)}^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$. Yeh step kyun? Pythagoras legs ke squares use karta hai, toh $A_x$ ka sign gayab ho jaata hai — magnitude hamesha $\ge 0$ hota hai.
2. Raw angle. ${\theta }_{\text{raw}}={\tan }^{-1}\left(\frac{12}{-5}\right)={\tan }^{-1}(-2.4)=-67.38°$. Yeh step kyun? ${\tan }^{-1}$ sirf ratio $-2.4$ jaanta hai, toh yeh figure mein plum down-right arrow return karta hai — galat direction, lekin same line par.
3. Quadrant fix. $A_x<0\Rightarrow$ $180°$ add karo: $\theta =-67.38°+180°=112.62°$. Yeh step kyun? $\tan$ har $180°$ par repeat hota hai; plum guess ko half turn rotate karne se exactly orange arrow par land hote hain, bina ratio change kiye.

Verify: $13\cos 112.62°=-5.00$, $13\sin 112.62°=12.00$. ✓ Signs original $(-5,+12)$ se match karte hain.

---

Ex C — Quadrant III (dono negative) cell 3

Figure mein orange arrow down-left point karta hai (true $\vec{A}$); faint plum arrow up-right hai jo phir se calculator ka guess hai ($36.9°$). Ex B jaisi hi kahani — do arrows, ek line, ek shared ratio. Teal arc corrected angle tak sweep karta hai.

1. Magnitude. $A=\sqrt{(-8{)}^2+(-6{)}^2}=\sqrt{64+36}=\sqrt{100}=10$. Yeh step kyun? Har leg ko square karna dono minus signs erase karta hai, toh Pythagoras QI mein plum mirror arrow ke barabar length deta hai — magnitude kabhi direction ka sign nahi carry karta.
2. Raw angle. ${\theta }_{\text{raw}}={\tan }^{-1}\left(\frac{-6}{-8}\right)={\tan }^{-1}(0.75)=36.87°$. Yeh step kyun? Ratio ke andar dono minus signs cancel ho jaate hain, toh calculator plum up-right arrow return karta hai — yeh nahi dekh sakta ki dono legs actually backward point kar rahe hain.
3. Quadrant fix. $A_x<0\Rightarrow$ $180°$ add karo: $\theta =36.87°+180°=216.87°$. Yeh step kyun? Ex B jaisi hi rule: jab bhi $A_x<0$ ho (QII ya QIII) calculator half-turn off hota hai — plum arrow ko $180°$ rotate karne se orange par land hote hain.

Verify: $10\cos 216.87°=-8.00$, $10\sin 216.87°=-6.00$. ✓ Dono negative, jaise chahiye tha.

---

Ex D — Quadrant IV (negative ya +360°) cell 4

Figure mein orange arrow x-axis ke neeche QIV mein dip karta hai; teal arc $+x$ axis se clockwise curl karta hai, angle ko $-45°$ ke negative sweep ke roop mein dikhata hai. Plum dashed shadows $A_x=5$ (rightward) aur $A_y=-5$ (downward) mark karte hain.

1. Magnitude. $A=\sqrt{5^2+(-5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\approx 7.07$. Yeh step kyun? Pythagoras har leg ko square karta hai, toh negative $A_y$ (downward plum shadow) positive $A_x$ ki tarah ek positive $25$ contribute karta hai — length downward direction ke liye blind hai.
2. Raw angle. $\theta ={\tan }^{-1}\left(\frac{-5}{5}\right)={\tan }^{-1}(-1)=-45°$. Yeh step kyun? $A_x>0$ matlab arrow sach mein right half mein hai, exactly wahan jahan ${\tan }^{-1}$ rehta hai — teal clockwise arc ko koi fix ki zaroorat nahi.
3. Do valid names. $\theta =-45°$ ya equivalently $-45°+360°=315°$. Yeh step kyun? Angles cyclic hote hain; $-45°$ aur $315°$ identical direction point karte hain. Jo bhi tumhare problem ka convention ho use karo.

Verify: $5\sqrt{2}\cos (-45°)=5\sqrt{2}\cdot \frac{1}{\sqrt{2}}=5.00$, $5\sqrt{2}\sin (-45°)=-5.00$. ✓

Recall Ek-line quadrant rule

Calculator ka ${\tan }^{-1}$ kab $+180°$ maangta hai? Jab bhi $A_x<0$ ho (Quadrants II aur III). Agar $A_x>0$ ho toh already sahi hai.

---

Ex E — Degenerate & zero vectors cells 5, 6

Figure mein teal arrow seedha upar point karta hai ($\vec{P}$), orange arrow seedha left point karta hai ($\vec{Q}$), aur plum dot origin par $\vec{Z}$ hai — ek arrow jisme koi length nahi aur isliye point karne ke liye koi tip nahi.

1. (i) purely $+y$. $P=\sqrt{0^2+7^2}=7$. Teal arrow seedha upar point karta hai, toh $\theta =90°$. Yeh step kyun? $A_x=0$ se $A_y/A_x$ zero se division ban jaata hai — calculator error deta hai. Lekin picture bilkul clear hai: seedha upar $90°$ hai.
2. (ii) purely $-x$. $Q=\sqrt{(-4{)}^2+0^2}=4$. Orange arrow seedha left point karta hai, toh $\theta =180°$. Yeh step kyun? $A_y=0$ ratio $0$ deta hai, jiska ${\tan }^{-1}$ $0°$ hai — lekin $A_x<0$ matlab hum $180°$ add karte hain: $0°+180°=180°$. Quadrant rule phir bhi hume bachata hai.
3. (iii) zero vector. $Z=\sqrt{0^2+0^2}=0$. Direction undefined hai. Yeh step kyun? Plum dot ki length $0$ hai, toh iski koi tip nahi — yeh kahan bhi point nahi karta. Zero vector ke liye kabhi angle quote mat karo.

Verify: $7\cos 90°=0,7\sin 90°=7$ ✓; $4\cos 180°=-4,4\sin 180°=0$ ✓; zero vector mein $0=0$ hai aur koi angle nahi. ✓

---

Ex F — Limiting behaviour jab $\theta \to 90°$ cell 7

Figure mein equal length $10$ ke teen arrows upar fan karte hain — plum $0°$ par (flat along $+x$), teal $45°$ par (diagonal), orange $89°$ par (almost vertical). Dotted vertical drop-lines har arrow ka horizontal shadow $A_x$ hain: dekho ki jaise fan upar sweep karta hai yeh lines kitni chhoti hoti jaati hain, exactly wo trend jo forecast ne predict kiya tha.

1. $\theta =0°$: $A_x=10\cos 0°=10.00$, $A_y=10\sin 0°=0.00$. Yeh step kyun? $\cos 0°=1$ aur $\sin 0°=0$: plum arrow flat $+x$ ke along hai, toh uska shadow poori length horizontally hai aur vertically kuch nahi.
2. $\theta =45°$: $A_x=10\cos 45°=7.07$, $A_y=10\sin 45°=7.07$. Yeh step kyun? $\cos 45°=\sin 45°=\frac{1}{\sqrt{2}}\approx 0.707$: teal diagonal par dono shadows equal hain, length evenly split ho jaati hai.
3. $\theta =89°$: $A_x=10\cos 89°=0.175$, $A_y=10\sin 89°=9.998$. Yeh step kyun? $\cos 89°\approx 0.017$ almost mar chuka hai jabki $\sin 89°\approx 0.9998$ nearly $1$ hai: orange arrow ka horizontal shadow figure mein y-axis ke paas wala chota stub hai, aur vertical part nearly poora hai.
4. $\theta =90°$: $A_x=0.00$, $A_y=10.00$. Yeh step kyun? $\cos 90°=0$ aur $\sin 90°=1$ — wo limit jis taraf fan ja raha tha. Horizontal shadow bilkul kuch nahi reh gaya hai; poori length ab $y$ mein hai.

Verify: har angle par $A_x^2+A_y^2=100$ (magnitude rotation ke under conserved hai): $\theta =45°$ check karo: $7.07^2+7.07^2\approx 100$. ✓

---

Ex G — Word problem: do displacements cell 8

Figure mein plum arrow leg 1 hai ($30$ m at $40°$), teal arrow leg 2 hai ($20$ m straight up) plum tip se tip-to-tail draw kiya gaya hai, aur orange arrow resultant $\vec{R}$ hai start se final tip tak. Orange arrow plum se visibly steeper hai — northward leg ne total ko upar tilt kar diya, exactly jaise forecast tha.

1. Leg 1 components. $x_1=30\cos 40°=22.98$, $y_1=30\sin 40°=19.28$ (east, north). Yeh step kyun? East hamara $+x$ hai, north $+y$ hai; angle east ($+x$) se measure kiya gaya hai, toh east ko cosine milta hai.
2. Leg 2 components. Due north $\Rightarrow x_2=0,y_2=20$. Yeh step kyun? "Due north" seedha upar hai (teal arrow): koi eastward part nahi.
3. Componentwise add karo. $R_x=22.98+0=22.98$, $R_y=19.28+20=39.28$. Yeh step kyun? Same axis ke along components plain numbers ki tarah add hote hain — yahi reason hai humne resolve kiya (dekho Vector Addition — Triangle & Parallelogram Law).
4. Magnitude rebuild karo. $R=\sqrt{22.98^2+39.28^2}=\sqrt{528.3+1543.0}=\sqrt{2071.3}=45.51\text{ m}$. Yeh step kyun? Summed components $R_x,R_y$ resultant ke right triangle ki legs hain (figure mein orange); Pythagoras un legs ko single straight-line distance mein convert karta hai jo hiker start se end tak hai.
5. Angle rebuild karo. $\theta ={\tan }^{-1}(39.28/22.98)={\tan }^{-1}(1.709)=59.66°$. Dono components positive $\Rightarrow$ QI, koi fix nahi. Yeh step kyun? Resultant ke opposite-over-adjacent ratio ka ${\tan }^{-1}$ uski heading recover karta hai; dono components positive hain toh orange arrow QI mein hai aur koi $180°$ correction ki zaroorat nahi. (Rounding ke baare mein ek note: agar tum $x_1,y_1$ mein zyada decimals rakhte ho toh ratio $1.709$ aur angle $59.66°$ aata hai; components ko hard round karke $22.98$ aur $39.28$ karne se pehle divide karne se answer thoda $59.8°$ ke aas paas aa sakta hai — hamesha last line tak full precision carry karo.)

Verify: rebuild karo — $45.51\cos 59.66°=22.98$, $45.51\sin 59.66°=39.28$ ✓. Sanity: $45.51\text{ m}<30+20=50\text{ m}$ (arrows fully align nahi hue, toh total naive sum se kam hai) ✓, aur $59.66°>40°$ jaise forecast tha ✓. Units: poore mein metres.

---

Ex H — Exam twist: galat axis se angle cell 9

Figure mein orange arrow force hai; teal arc $+y$ axis ke paas hai, dikhata hai ki $25°$ vertical se measure kiya gaya hai, x-axis se nahi. Plum dashed shadows $F_x$ hain (short, horizontal — $25°$ ke opposite) aur $F_y$ (tall, vertical — $25°$ ke adjacent). Picture saaf kar deta hai ki kaunsi leg adjacent hai.

1. Adjacent side identify karo. Teal arc ka $25°$ force aur $+y$ axis ke beech hai, toh $y$-side $25°$ ke adjacent hai. Yeh step kyun? Cosine = adjacent/hypotenuse by definition; adjacent leg woh hai jis se angle khulta hai — yahan tall vertical plum shadow.
2. $y$ ko cosine milta hai. $F_y=50\cos 25°=45.32\text{ N}$. Yeh step kyun? $F_y$ $25°$ angle ke adjacent leg hai, aur adjacent/hypotenuse exactly cosine hai — toh yeh projection $\cos$ use karta hai, $\sin$ nahi.
3. $x$ ko sine milta hai. $F_x=50\sin 25°=21.13\text{ N}$. Yeh step kyun? Short horizontal plum shadow $F_x$ $25°$ ke opposite hai, aur opposite/hypotenuse sine hai.

Alternative (angle convert karo): $+x$ se angle $90°-25°=65°$ hai, jo deta hai $F_x=50\cos 65°=21.13$, $F_y=50\sin 65°=45.32$ — identical. ✓

Verify: $\sqrt{21.13^2+45.32^2}=\sqrt{446.5+2054}=\sqrt{2500}=50.00\text{ N}$. ✓ Magnitude recover ho gayi. Aur dekhna ho toh Resolution of Forces mein aur bhi hain.

---

Ex I — Exam twist: unit vector phir reconstruct cell 10

Figure mein lamba orange arrow $\vec{p}$ hai (length $15$) QIV mein down-right point karta hua; chota teal arrow exactly usi line par chadha hua hai jo $\hat{p}$ hai — identical direction, length $1$ tak shrunk. Same heading, alag size: yahi unit vector ka poora idea hai.

1. Magnitude. $p=\sqrt{9^2+(-12{)}^2}=\sqrt{81+144}=\sqrt{225}=15$. Yeh step kyun? Humein orange arrow ki length chahiye pehle use $1$ tak shrink karne se; do legs $9$ aur $-12$ par Pythagoras wo length deta hai (negative square ho jaata hai).
2. Unit vector. $\hat{p}=\frac{\vec{p}}{p}=\frac{9}{15}\hat{i}+\frac{-12}{15}\hat{j}=0.6\hat{i}-0.8\hat{j}$. Yeh step kyun? Har component ko same $p$ se divide karna length ko $1$ tak shrink karta hai (figure mein orange → teal) lekin ratio $A_y/A_x$ — direction — unchanged rehta hai (dekho Unit Vectors i, j, k).
3. Check karo ki length 1 hai. $\sqrt{0.6^2+(-0.8{)}^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$. ✓ Yeh step kyun? Ek "unit" vector ka magnitude exactly $1$ hona chahiye by definition; uske components par Pythagoras run karna confirm karta hai ki teal arrow sach mein length $1$ ka hai.
4. Reconstruct karo. $p\hat{p}=15(0.6\hat{i}-0.8\hat{j})=9\hat{i}-12\hat{j}=\vec{p}$. Yeh step kyun? magnitude $\times$ direction $=$ poora vector — teal unit arrow ko $15$ se wapas stretch karna exactly orange $\vec{p}$ par land karta hai. Do operations inverses hain.

Verify: direction check — $\vec{p}$ ka angle ${\tan }^{-1}(-12/9)=-53.13°$ hai (QIV, $A_x>0$ toh koi fix nahi); $\hat{p}$ ka angle ${\tan }^{-1}(-0.8/0.6)=-53.13°$ — same heading. ✓ $\hat{p}$ ke units: koi nahi (pure direction); reconstructed $\vec{p}$ wapas kg·m/s mein.

---

Active recall

Recall Reveal

Ex B mein $180°$ kyun add kiya lekin Ex D mein angle negative chhod diya? ::: Ex B mein $A_x<0$ hai (QII) toh calculator half-turn off hai; Ex D mein $A_x>0$ hai (QIV) toh ${\tan }^{-1}$ already sahi hai. Ex E(iii) mein koi angle kyun nahi? ::: Zero vector ki length $0$ hai aur isliye koi direction nahi — undefined. Ex G mein resultant $45.5\text{ m}$ kyun $50\text{ m}$ se kam hai? ::: Do legs parallel nahi theen, toh wo partly alag alag directions mein point karti hain; sirf parallel arrows poore sum tak add hote hain. Ex H mein jis axis se tum measure karte ho use kaun sa trig function milta hai? ::: Cosine (wo axis adjacent side hai).

Connections

• Vector representation — magnitude, direction, components — parent recipes yahan drill ki gayi hain.
• Scalars vs Vectors — kyun ek single number yeh kaam nahi kar sakta.
• Vector Addition — Triangle & Parallelogram Law — Ex G ke peeche ka engine.
• Resolution of Forces — Ex H apne natural mechanics habitat mein.
• Dot Product and Cross Product — components par build hone wale agle tools.
• Projectile Motion — Ex F ke limiting case ki tarah velocity split karna.
• Unit Vectors i, j, k — Ex I mein use kiya gaya basis.