Exercises — Vector representation — magnitude, direction, components

KYA: hum bas har formula ko uske word se match kar rahe hain.

• (i) $\sqrt{A_x^2+A_y^2}$ magnitude $A$ hai — yeh right triangle ki dono legs par Pythagoras hai, aur yeh ek length deta hai (ek single non-negative number).
• (ii) ${\tan }^{-1}(A_y/A_x)$ direction $\theta$ hai — woh angle jiska tangent (opposite÷adjacent) sides ke ratio ke barabar hai.

YEH KYUN KAAM KARTA HAI: length hypotenuse se aati hai; angle arrow ke slope $A_y/A_x$ se aata hai. Length angle nahi ho sakti aur vice-versa, toh koi ambiguity nahi hai.

Ek savdhani (isko skip mat karo): raw ${\tan }^{-1}(A_y/A_x)$ tabhi true direction hai jab $A_x>0$ ho. Jab bhi $A_x<0$ ho, tum $180°$ add karo (upar wala quadrant check), kyunki $\tan$ do opposite arrows ke liye same ratio deta hai. Hum yeh Level 3 se explicitly karenge — lekin calculator ka angle kabhi bina $A_x$ ka sign dekhe trust mat karo.

$A_x=A\cos \theta ,A_y=A\sin \theta .$ YEH KYUN: arrow ki tip se x-axis par ek perpendicular daalo. X-side uss axis ke saath lie karti hai jis se tumne $\theta$ measure kiya — yeh adjacent side hai, aur adjacent÷hypotenuse $=\cos \theta$. $A_x$ paane ke liye $A$ se multiply karo. Y-side opposite hai, isliye woh $\sin$ use karta hai.

False. $A=\sqrt{(-8{)}^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$. YEH KYUN: squaring se har minus sign mit jaata hai, isliye magnitude kabhi negative nahi aa sakti. "Kis taraf" wali information (ki arrow up-left point karta hai) components ke signs mein hoti hai, $A$ mein nahi.

KYA: polar → components, toh hum $A_x=A\cos \theta$, $A_y=A\sin \theta$ use karenge. $A_x=10\cos 30°=10\times 0.8660=8.66\text{ m}$ $A_y=10\sin 30°=10\times 0.5=5\text{ m}$ YEH FUNCTIONS KYUN: $30°$ x-axis se measure hota hai, isliye x-side adjacent hai (cosine), y-side opposite (sine). CHECK: $\sqrt{8.66^2+5^2}=\sqrt{75+25}=\sqrt{100}=10$ ✓ — magnitude wapas aa gayi.

$F=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13.$ YEH PYTHAGORAS KYUN: components right triangle ki do perpendicular legs hain; vector khud hypotenuse hai, aur hypotenuse² legs ke squares ke sum ke barabar hota hai.

KYA: unit vector woh arrow hai jo length 1 tak shrink kiya gaya ho: $\hat{v}=\vec{v}/|\vec{v}|$. Pehle length: $|\vec{v}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. Har component ko $10$ se divide karo: $\hat{v}=\frac{6}{10}\hat{i}+\frac{8}{10}\hat{j}=0.6\hat{i}+0.8\hat{j}.$ YEH DIVIDE KARNA DIRECTION KYUN RAKHTA HAI: dono components same factor $1/10$ se shrink hote hain, isliye unka ratio $A_y/A_x$ — jo angle set karta hai — unchanged rehta hai. Sirf length badlati hai. CHECK: $\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$ ✓.

Magnitude: $A=\sqrt{(-4{)}^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$. Raw angle: ${\tan }^{-1}(3/(-4))={\tan }^{-1}(-0.75)=-36.87°$. Quadrant check: $A_x<0$ aur $A_y>0$ ⇒ arrow upar aur baayein point karta hai ⇒ Quadrant II. Calculator ka $-36.87°$ neeche-daayein (Quadrant IV) point karta hai — galat direction. $180°$ add karo: $\theta =-36.87°+180°=143.13°.$ CALCULATOR KYUN JHOOTH BOLTA HAI: $\tan$ har $180°$ par repeat hota hai, isliye $\tan (143.13°)=\tan (-36.87°)$ — do opposite arrows ek ratio share karte hain. Sirf components ke signs hi unhe alag karte hain.

Neeche ki figure mein, green arrow true vector $\vec{A}=(-4,3)$ hai jo Quadrant II mein hai; dashed red arrow woh hai jahan calculator ka $-36.87°$ actually point karta hai — $(4,-3)$, bilkul opposite corner (Quadrant IV). Dono ke beech se guzarti white dotted line dikhati hai ki woh ek seedhi line par hain, isliye precisely woh same $\tan$ ratio share karte hain. Yellow label woh $+180°$ jump mark karta hai jo tumhe galat arrow se sahi tak le jaata hai.

Raw angle: ${\tan }^{-1}((-2)/(-2))={\tan }^{-1}(1)=45°$. Quadrant check: dono components negative ⇒ arrow neeche-baayein point karta hai ⇒ Quadrant III. $180°$ add karo: $\theta =45°+180°=225°.$ YEH KYUN: ratio $A_y/A_x=(-2)/(-2)=+1$ Quadrant I ke $\overrightarrow{(2,2)}$ jaisi hi hai, isliye calculator $45°$ deta hai. Do minus signs true direction ko diagonally-opposite, $225°$, mark karti hain.

"$X$-axis ke neeche" matlab angle clockwise measure hoti hai, isliye $\theta =-40°$. $F_x=50\cos (-40°)=50\times 0.7660=38.30\text{ N}$ $F_y=50\sin (-40°)=50\times (-0.6428)=-32.14\text{ N}$ $F_y$ NEGATIVE KYUN HAI: arrow axis ke neeche jaata hai, isliye uski vertical chhaaya neechey ($-y$) point karti hai. $\cos$ even hai ($\cos (-40°)=\cos 40°$) isliye $F_x$ positive rehta hai; $\sin$ odd hai ($\sin (-40°)=-\sin 40°$) isliye $F_y$ sign flip karta hai. Maths automatically "downward" ko minus mein record kar leta hai.

Step 1 — components (YEH KYUN: tum components add karte ho, kabhi magnitudes nahi). $F_{1x}=30\cos 0°=30,F_{1y}=30\sin 0°=0.$ $F_{2x}=40\cos 90°=0,F_{2y}=40\sin 90°=40.$ Step 2 — same type ke saath add karo: $R_x=30+0=30,R_y=0+40=40.$ Step 3 — wapas polar mein: $R=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\text{ N}.$ $\theta ={\tan }^{-1}(40/30)={\tan }^{-1}(1.333)=53.13°.$ Dono components positive ⇒ Quadrant I ⇒ koi correction nahi. Resultant: $50\text{ N}$ at $53.13°$. Geometric picture ke liye dekho Vector Addition — Triangle & Parallelogram Law.

Neeche ki figure mein, yellow arrow ${\vec{F}}_1$ hai (30 N along $+x$) aur green arrow ${\vec{F}}_2$ hai (40 N straight up), tip-to-tail draw kiya gaya hai: ${\vec{F}}_2$ wahan se shuru hota hai jahan ${\vec{F}}_1$ khatam hota hai. Origin se final tip tak jaane wala blue arrow resultant $\vec{R}$ hai — familiar 3-4-5 triangle jo das guna scale up hua hai, isliye uski hypotenuse exactly 50 hai. Origin par red arc woh $53.13°$ direction hai jo humne compute ki.

Vectors ka sum: $\vec{A}+\vec{B}=(5-5)\hat{i}+0\hat{j}=\vec{0}$, isliye $|\vec{A}+\vec{B}|=0$. Magnitudes ka sum: $|\vec{A}|+|\vec{B}|=5+5=10$. YEH CLASH KYUN KARTE HAIN: dono arrows bilkul opposite direction mein point karte hain, isliye woh cancel ho jaate hain — resultant zero-length arrow hai. Lekin magnitudes sirf lengths hain; do lengths add karne se kabhi cancel nahi hoga. Isliye hum components add karte hain (jo sign carry karte hain) na ki magnitudes. Equality sirf tabhi hoti jab arrows parallel aur same direction mein hon.

$v_x=20\cos 60°=20\times 0.5=10\text{ m/s}$ $v_y=20\sin 60°=20\times 0.8660=17.32\text{ m/s}$ $\frac{v_y}{v_x}=\frac{17.32}{10}=1.732=\tan 60°✓$ RATIO $\tan \theta$ KYUN HAI: dono component equations divide karne par, $20$ cancel ho jaata hai: $\frac{20\sin 60°}{20\cos 60°}=\tan 60°$. Launch magnitude drop out ho jaata hai — sirf velocity ki slope bachti hai, aur exactly yahi $\tan$ measure karta hai. Yeh split Projectile Motion ki neenv hai.

KYA GALAT JAATA HAI: $A_y/A_x=-7/0$ undefined hai — calculator refuse karta hai. Lekin vector bilkul real hai: yeh seedha neeche point karta hai. ARROW SEEDHA NEECHE KYUN HAI: $A_x=0$ matlab koi horizontal chhaaya nahi, aur $A_y=-7<0$ matlab poora arrow $-y$ mein point karta hai. Seedha neeche $\theta =270°$ hai (ya equivalently $-90°$). FORMULA KYUN TOOT JAATA HAI: jaise arrow vertical hota hai, uska slope $A_y/A_x\to \pm \infty$ hota hai; $\tan$ $90°$ aur $270°$ par infinity ki taraf jaata hai, isliye inverse ka wahan koi finite input nahi hai. Yeh limiting behaviour hai jise tum inspection se handle karo, formula mein plug in karke nahi. Safe rule: agar $A_x=0$ ho: $\theta =90°$ jab $A_y>0$, $\theta =270°$ (ya $-90°$) jab $A_y<0$, aur $\theta$ undefined hai agar $A_y=0$ bhi ho (zero vector ki koi direction nahi hoti).

Pehle, notation: kisi bhi letter ke upar hat "$\hat{}$" "us direction mein unit vector" ka standard sign hai. Isliye $\hat{A}$ ka $\vec{A}$ se wahi rishta hai jo $\hat{i}$ ka $+x$ axis se hai: same aim, length 1 tak trim. Definition $\hat{A}=\vec{A}/A$ se, uske components $(A_x/A,A_y/A)$ hain. Sidha magnitude compute karo: $|\hat{A}|=\sqrt{{\left(\frac{A_x}{A}\right)}^2+{\left(\frac{A_y}{A}\right)}^2}=\sqrt{\frac{A_x^2+A_y^2}{A^2}}=\frac{\sqrt{A_x^2+A_y^2}}{A}.$ Lekin $\sqrt{A_x^2+A_y^2}$ by definition magnitude $A$ hai. Isliye $|\hat{A}|=\frac{A}{A}=1.$ YEH HAMESHA 1 KYUN HAI: humne har component ko usi number se divide kiya jo length hai. Squaring aur re-summing se upar $A^2$ milta hai, jo neeche $A^2$ cancel karta hai — direction ($A_x:A_y$ ka ratio) preserve hoti hai, lekin size 1 par normalise ho jaata hai. Har quadrant ke liye kaam karta hai kyunki sirf squares appear hote hain.

Original: $A_x=A\cos \theta ,A_y=A\sin \theta$. $+90°$ rotate karne ke baad, new angle $\theta +90°$ hai: $A_x^{\prime }=A\cos (\theta +90°),A_y^{\prime }=A\sin (\theta +90°).$ Shift identities use karo $\cos (\theta +90°)=-\sin \theta$ aur $\sin (\theta +90°)=\cos \theta$: $A_x^{\prime }=-A\sin \theta =-A_y,A_y^{\prime }=A\cos \theta =A_x.$ Isliye $(A_x,A_y)\to (-A_y,A_x)$. YEH GEOMETRICALLY KYUN SAHI LAGTA HAI: arrow ko quarter-turn anticlockwise ghuma do toh "right" "up" ban jaata hai aur "up" "left" ban jaata hai. Purani x-shadow new y-shadow ban jaati hai ($A_y^{\prime }=A_x$), aur purani y-shadow new negative x ban jaati hai ($A_x^{\prime }=-A_y$). Yeh swap-and-negate cross product ki perpendicularity ka beej hai.

Hum components add kyun karte hain, magnitudes kyun nahi? ::: Kyunki arrows direction carry karte hain aur cancel ho sakte hain; components sign carry karte hain, magnitudes nahi. $A_x<0$ hone par ${\tan }^{-1}$ ko kya correction chahiye? ::: $180°$ add karo (vector Quadrant II ya III mein hai). $\vec{A}=(0,-7)$ ki direction? ::: Seedha neeche, $270°$ (formula fail hota hai — inspection se karo). Kisi bhi unit vector ki magnitude? ::: Exactly $1$, hamesha. $\hat{A}$ mein hat ka matlab kya hai? ::: "$\vec{A}$ ki direction mein unit vector", yani $\vec{A}/|\vec{A}|$. $(A_x,A_y)$ ko $+90°$ rotate karne par kya milta hai? ::: $(-A_y,A_x)$.