Exercises — Vector addition — triangle law, parallelogram law

Kaun sa tool? Same direction matlab $\theta =0^{\circ }$. $\theta =0$ kyun? Unke beech ka angle zero hai — woh parallel arrows ek hi taraf point kar rahe hain. $R=\sqrt{A^2+B^2+2AB\cos 0^{\circ }}=\sqrt{A^2+B^2+2AB}=\sqrt{(A+B{)}^2}=A+B.$ $R=6+9=\boxed{15\text{m/s}}.$ Sanity check: parallel vectors ordinary numbers ki tarah add ho jaate hain — yeh maximum possible resultant hai.

Kaun sa tool? Opposite matlab $\theta =180^{\circ }$, aur $\cos 180^{\circ }=-1$. $R=\sqrt{A^2+B^2-2AB}=\sqrt{(A-B{)}^2}=|A-B|.$ $R=|6-9|=\boxed{3\text{m/s}}.$ Direction: resultant bade vector ki taraf point karta hai — yahaan $9\text{m/s}$ wala. Kyun? $9$ wale opposition se $6$ cancel hone ke baad, sirf $3$ bachta hai bade vector ki direction mein. Yeh minimum possible resultant hai.

Kaun sa tool? $\theta =90^{\circ }$, aur $\cos 90^{\circ }=0$ cross term ko khatam kar deta hai. $R=\sqrt{A^2+B^2+0}=\sqrt{25+144}=\sqrt{169}=\boxed{13\text{N}}.$ Yeh Pythagoras mein kyun collapse ho jaata hai: do forces perpendicular hain, isliye woh ek right triangle ki do legs banaati hain aur $R$ hypotenuse hai.

Magnitude. $\cos 60^{\circ }=\frac{1}{2}$: $R=\sqrt{4^2+3^2+2(4)(3)(0.5)}=\sqrt{16+9+12}=\sqrt{37}\approx \boxed{6.08}.$ Direction. $\sin 60^{\circ }=0.8660$: $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\frac{3(0.8660)}{4+3(0.5)}=\frac{2.598}{5.5}=\mathrm{0.4724.}$ $\alpha =\arctan (0.4724)\approx \boxed{25.3^{\circ }}.$ Arctan kyun? $\tan \alpha$ hume $\vec{R}$ ki tip se vertical drop karke bane right triangle ka ratio (opposite over adjacent) deta hai; $\arctan$ poochta hai "kaun sa angle woh tangent rakhta hai?" aur $\alpha$ wapas deta hai.

Magnitude. $\cos 120^{\circ }=-0.5$ (cross term ab subtract karta hai): $R=\sqrt{49+25+2(7)(5)(-0.5)}=\sqrt{49+25-35}=\sqrt{39}\approx \boxed{6.24}.$ Direction. $\sin 120^{\circ }=0.8660$: $\tan \alpha =\frac{5(0.8660)}{7+5(-0.5)}=\frac{4.330}{4.5}=0.9623\Rightarrow \alpha =\arctan (0.9623)\approx \boxed{43.9^{\circ }}.$ Denominator kyun chhhota hua: $B\cos \theta =-2.5$, $\vec{A}$ ke along backwards point karta hai, isliye horizontal reach $R_x=A+B\cos \theta$ sirf $4.5$ hai, $7+5$ nahi. Resultant zyada steep jhukta hai → bada $\alpha$.

$R=\sqrt{8^2+8^2+0}=\sqrt{128}=8\sqrt{2}\approx \boxed{11.31}.$ $\tan \alpha =\frac{8\sin 90^{\circ }}{8+8\cos 90^{\circ }}=\frac{8}{8}=1\Rightarrow \alpha =\boxed{45^{\circ }}.$ $45^{\circ }$ kyun? Equal vectors → resultant unke beech ke angle ko bisect karta hai, aur $90^{\circ }$ ka aadha $45^{\circ }$ hota hai. Symmetry ka tohfa.

Hum kya karenge: magnitude formula ko unknown $\theta$ ke liye ek equation maano. $R^2=A^2+B^2+2AB\cos \theta \Rightarrow 144=36+64+96\cos \theta .$ $144=100+96\cos \theta \Rightarrow \cos \theta =\frac{44}{96}=\mathrm{0.4583.}$ $\theta =\arccos (0.4583)\approx \boxed{62.7^{\circ }}.$ $\arccos$ kyun? Hum $\cos \theta$ jaante hain aur angle wapas chahiye — $\arccos$ cosine ka "undo" button hai. Kyunki $R=12$ minimum $|8-6|=2$ aur maximum $8+6=14$ ke beech mein hai, ek valid angle zaroor exist karta hai. ✔

"$\vec{A}$ ke perpendicular" ka matlab: resultant ka $\vec{A}$ ke along koi component nahi hai, yaani $R_x=0$. $R_x=A+B\cos \theta =0\Rightarrow \cos \theta =-\frac{A}{B}=-\frac{10}{6}=-\mathrm{1.667.}$ Lekin $\cos \theta$ kabhi $-1$ se kam nahi ho sakta! Toh aisa koi angle exist nahi karta — ek $6$-unit vector ke liye $10$-unit vector ko pure right angle tak swing karna impossible hai. $\boxed{\text{Impossible}}$. Geometrically kyun impossible hai: $\vec{B}$ ki poori length sirf $6$ hai, lekin $10$ units ki horizontal reach cancel karne ke liye use $10$ units ka backward component dena hoga. Neeche wali picture dekho — $\vec{B}$ ki tip itni door left nahi ja sakti.

Is baar $\vec{B}$ ke along setup karo. $\vec{B}$ ko x-axis par rakho; tab $\vec{A}$ usse $\theta$ angle banata hai. Resultant ka $\vec{B}$ ke along component zero hona chahiye: $B+A\cos \theta =0\Rightarrow \cos \theta =-\frac{B}{A}.$ $A=5,B=4$ ke liye: $\cos \theta =-0.8\Rightarrow \theta =\arccos (-0.8)\approx \boxed{143.1^{\circ }}.$ Yahaan possible kyun hai: humein $B=4$ ka backward reach chahiye, aur $\vec{A}$ ki length $5\ge 4$ hai, toh woh supply kar sakta hai. (L3.2 se compare karo, jahaan roles ne ise impossible bana diya tha.)

Components kyun, pair formula kyun nahi? Two-vector formula ek baar mein sirf do arrows handle karta hai. Teen ke saath, clean route har ek ko x aur y mein resolve karke alag-alag sum karna hai. $R_x=5+4\cos 90^{\circ }+3\cos 180^{\circ }=5+0-3=2\text{N}.$ $R_y=5\sin 0^{\circ }+4\sin 90^{\circ }+3\sin 180^{\circ }=0+4+0=4\text{N}.$ $R=\sqrt{R_x^2+R_y^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\approx \boxed{4.47\text{N}}.$ $\tan \alpha =\frac{R_y}{R_x}=\frac{4}{2}=2\Rightarrow \alpha =\arctan (2)\approx \boxed{63.4^{\circ }}\text{ above }+x.$ Sign check: $R_x>0,R_y>0$ → resultant first quadrant mein hai, isliye $+x$ se anticlockwise measure kiya hua $\alpha$ bilkul sahi hai.

Yeh vector addition disguise mein hai: true velocity $=$ boat velocity $+$ river velocity, aur woh $90^{\circ }$ par hain. $R=\sqrt{4^2+3^2}=\sqrt{25}=\boxed{5\text{km/h}}.$ $\tan \alpha =\frac{3}{4}=0.75\Rightarrow \alpha =\arctan (0.75)\approx \boxed{36.9^{\circ }}\text{ downstream of straight-across}.$ $90^{\circ }$ kyun: boat bilkul across aim karti hai jabki current bank ke along side mein push karta hai — do velocities perpendicular hain, isliye phir wahi $3$–$4$–$5$ right triangle hai.

Key idea Subtraction of vectors and the difference vector se: $\vec{B}$ subtract karna matlab $-\vec{B}$ add karna, jo $180^{\circ }$ opposite point karta hai. Toh magnitude formula ke andar effective angle $180^{\circ }-60^{\circ }=120^{\circ }$ ho jaata hai, jisse $\cos 120^{\circ }=-0.5$ milta hai: $|\vec{A}-\vec{B}|=\sqrt{25+25+2(25)(-0.5)}=\sqrt{50-25}=\sqrt{25}=\boxed{5}.$ Sundar check: length $5$ ki do equal sides aur $60^{\circ }$ ka gap hone se, teen points ek equilateral triangle banate hain, toh teesri side (difference) bhi $5$ hai. ✔

Magnitude formula mein $R=A$ aur $B=A$ set karo: $A^2=A^2+A^2+2A^2\cos \theta \Rightarrow 0=A^2+2A^2\cos \theta .$ $\Rightarrow \cos \theta =-\frac{1}{2}\Rightarrow \theta =\boxed{120^{\circ }}.$ Kyun samajh mein aata hai: $120^{\circ }$ apart teen equal-length arrows ek closed loop banate hain (jaise Mercedes star). Unmein se do tip-to-tail add kiye jaayein toh teesre ka reverse banana padega — same length $A$. ✔

Step 1 — known resultant ko components mein resolve karo. $R_x=R\cos 30^{\circ }=10(0.8660)=8.660,R_y=R\sin 30^{\circ }=10(0.5)=5.$ Step 2 — $\vec{A}$ hatao $\vec{B}$ ke components expose karne ke liye ($\vec{B}=\vec{R}-\vec{A}$): $B_x=R_x-A=8.660-8=0.660,B_y=R_y-0=5.$ Step 3 — magnitude aur angle mein wapas combine karo. $B=\sqrt{0.660^2+5^2}=\sqrt{0.4356+25}=\sqrt{25.44}\approx \boxed{5.04}.$ $\theta =\arctan \frac{B_y}{B_x}=\arctan \frac{5}{0.660}=\arctan (7.576)\approx \boxed{82.5^{\circ }}\text{ from }\vec{A}.$ Sign check: $B_x>0$ aur $B_y>0$ → $\vec{B}$ quadrant I mein hai, isliye $\theta$ ek seedha first-quadrant angle hai jo $90^{\circ }$ se thoda kam hai. Figure dekho — $\vec{B}$ ek chhota arrow hai jo almost seedha upar point kar raha hai.

(a) Maximum $\theta =0^{\circ }$ par: $R_{\max }=9+5=\boxed{14}$. Minimum $\theta =180^{\circ }$ par: $R_{\min }=|9-5|=\boxed{4}$. Har achievable resultant $[4,14]$ mein hoga. (b) Angle ke liye solve karo: $8^2=9^2+5^2+2(9)(5)\cos \theta \Rightarrow 64=106+90\cos \theta .$ $\cos \theta =\frac{64-106}{90}=\frac{-42}{90}=-0.4667\Rightarrow \theta =\arccos (-0.4667)\approx \boxed{117.8^{\circ }}.$ Check: $8$, $[4,14]$ ke andar hai, isliye ek valid $\theta$ exist karta hai — aur woh obtuse hai, negative cosine ke saath consistent. ✔