Exercises — Dot product — formula, geometric meaning, work calculation

KYA HAI: Dot product hamesha do arrows ko ek single plain number mein collapse kar deta hai — ek scalar. Compute karo component form $\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$ se: $\vec{A}\cdot \vec{B}=(1)(4)+(2)(5)+(3)(6)=4+10+18=32.$ Koi direction attached nahi — bas number $32$. Yeh ek scalar hai.

KYUN yeh test: do non-zero vectors perpendicular hote hain exactly tab jab unka dot product $0$ ho (kyunki $\cos 90^{\circ }=0$). $\vec{A}\cdot \vec{B}=(2)(3)+(3)(-2)=6-6=0.$ Dot product $0$ hai, isliye haan — woh perpendicular hain ($\theta =90^{\circ }$).

KYUN: ek vector khud se dot kiya jaaye to $|\vec{A}{|}^2$ milta hai (khud se angle $0^{\circ }$ hai, $\cos 0=1$). $\vec{A}\cdot \vec{A}=(3)(3)+(4)(4)=9+16=25,|\vec{A}|=\sqrt{25}=5.$

Like-with-like multiply karo, phir add karo: $\vec{A}\cdot \vec{B}=(5)(-3)+(-2)(4)+(1)(2)=-15-8+2=-21.$ SIGN ka matlab: negative ⇒ vectors mostly oppose kar rahe hain ek doosre ko (angle $90^{\circ }$ se zyada).

KYUN $\cos \theta$: force ka sirf woh part jo motion ke saath ho woh work karta hai — woh part hai $|\vec{F}|\cos \theta$ (figure mein amber segment, $\vec{F}$ ki shadow $\vec{d}$ par, exactly). $W=|\vec{F}||\vec{d}|\cos \theta =(20)(4)\cos 30^{\circ }=80\times 0.8660=69.28\text{ J}.$ Units: $\text{N⋅m}=\text{J}$.

KYUN angle skip karein: components haath mein hain to $W=\vec{F}\cdot \vec{d}$ bas multiply-and-add hai. $W=(6)(3)+(-8)(1)=18-8=10\text{ J}.$

KYUN magnitudes se divide karein: geometric form rearrange karne par milta hai $\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}$; denominator lengths strip out karta hai taaki sirf alignment bache. $\vec{A}\cdot \vec{B}=(1)(2)+(2)(0)+(2)(-1)=2+0-2=0.$ $\cos \theta =\frac{0}{|\vec{A}||\vec{B}|}=0\Rightarrow \theta =90^{\circ }.$ Dot product $0$ hai, isliye woh perpendicular hain — yahan magnitudes compute karne ki zaroorat bhi nahi.

$\vec{A}\cdot \vec{B}=(-2)(3)+(1)(1)=-6+1=-5.$ Interpret: negative ⇒ $\cos \theta <0$ ⇒ $\theta$ obtuse hai ($>90^{\circ }$). Vectors ek doosre ko partly oppose kar rahe hain. (Figure dekhao: agar $\vec{A}$ ki shadow $\vec{B}$ ke opposite side par pade, to shadow length — aur isliye dot product — negative ho jaata hai.)

KYUN yeh formula: projection length hai $|\vec{A}|\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{B}|}$ (dekho Projection of a vector). Kyunki $\vec{B}=(1,0)$ $x$-axis ke saath point karta hai, uski length $1$ hai. $\vec{A}\cdot \vec{B}=(4)(1)+(3)(0)=4,|\vec{B}|=1.$ $\text{projection}=\frac{4}{1}=4.$ Yeh exactly $\vec{A}$ ka $x$-component hai — $x$-axis par shadow hai hi $A_x$.

KYUN: perpendicular ⇔ dot product $=0$. Setup karo aur $k$ ke liye solve karo. $\vec{A}\cdot \vec{B}=(2)(3)+(k)(-2)+(1)(4)=6-2k+4=10-2k.$ $10-2k=0\Rightarrow k=5.$

KYUN add karein: work ek scalar hai, aur dot product distributive hai, isliye total work stages ka sum hai: $W=\vec{F}\cdot {\vec{d}}_1+\vec{F}\cdot {\vec{d}}_2=\vec{F}\cdot ({\vec{d}}_1+{\vec{d}}_2)$. $\vec{F}\cdot {\vec{d}}_1=(3)(2)+(4)(0)=6,\vec{F}\cdot {\vec{d}}_2=(3)(0)+(4)(3)=12.$ $W=6+12=18\text{ J}.$ Check combined displacement ${\vec{d}}_1+{\vec{d}}_2=(2,3)$ se: $(3)(2)+(4)(3)=6+12=18$ ✓.

Pehle $\vec{R}=(1,0)+(0,1)=(1,1)$ banao. $\vec{A}\cdot \vec{R}=(1)(1)+(0)(1)=1.$ $|\vec{A}|=1,|\vec{R}|=\sqrt{1^2+1^2}=\sqrt{2}.$ $\cos \theta =\frac{1}{1\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }.$ Sahi lagta hai: $\vec{R}$ ek unit square ka diagonal hai, $\vec{A}$ se $45^{\circ }$ tilted.

Displacement components mein: $\vec{d}=(5,0)\text{ m}$ (horizontal). Applied force: $W_{\text{app}}={\vec{F}}_{\text{app}}\cdot \vec{d}=(12)(5)+(5)(0)=60\text{ J}.$ Sirf horizontal $12\text{ N}$ part work karta hai; vertical $5\text{ N}$, $\vec{d}$ ke perpendicular hai aur $0$ contribute karta hai (uski shadow $\vec{d}$ par zero length ki hai). Friction: $W_{\text{fric}}=\vec{f}\cdot \vec{d}=(-4)(5)+(0)(0)=-20\text{ J}.$ Negative — friction energy remove karta hai, exactly jaisa hona chahiye. Net work: $W_{\text{net}}=60+(-20)=40\text{ J}.$ Work-Energy Theorem ke according, yeh $+40\text{ J}$ block ki kinetic energy ki gain ke barabar hai.

KYUN hum zero expect karte hain: Circular motion mein centripetal force hamesha centre ki taraf point karta hai, jo velocity ke (aur displacement ke, jo $\vec{v}$ follow karta hai) perpendicular hota hai. $W=\vec{F}\cdot \vec{d}=(5)(0)+(0)(0.03)=0\text{ J}.$ Perpendicular force koi work nahi karta, isliye speed (aur kinetic energy) constant rehti hai chahe direction badhalta rahe.

Setup: $\cos 60^{\circ }=\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}$ with $\cos 60^{\circ }=\frac{1}{2}$. $\vec{A}\cdot \vec{B}=(2)(1)+(1)(c)+(-2)(1)=2+c-2=c.$ $|\vec{A}|=\sqrt{4+1+4}=3,|\vec{B}|=\sqrt{1+c^2+1}=\sqrt{c^2+2}.$ $\frac{c}{3\sqrt{c^2+2}}=\frac{1}{2}\Rightarrow 2c=3\sqrt{c^2+2}.$ Dono sides square karo (valid tabhi jab $c>0$ ho, kyunki $60^{\circ }$ acute angle chahiye): $4c^2=9(c^2+2)\Rightarrow 4c^2=9c^2+18\Rightarrow -5c^2=18.$ Isse milta hai $c^2=-3.6$, jo impossible hai — koi real $c$ kaam nahi karta. ISKA MATLAB: geometry yahan $60^{\circ }$ angle forbid karta hai. Jaise $c$ badhta hai, $\vec{B}$ ki length dot product se zyada tez badhti hai, isliye $\cos \theta$ kabhi $\frac{1}{2}$ tak nahi pahunch sakta; attainable angle $60^{\circ }$ se upar rehta hai. Algebra par trust karne se pehle hamesha sanity-check karo ki solution exist kar sakta hai.