Exercises — Cross product — formula, direction (right-hand rule), torque - area calculation

Cycle $\hat{i}\to \hat{j}\to \hat{k}\to \hat{i}$ yaad karo. Is loop par aage jaana $+$ hai; peeche jaana $-$ hai; ek vector ka khud se cross product $\vec{0}$ hota hai.

• (a) $\hat{j}\times \hat{k}=\hat{k}$? Nahi — loop check karo. $\hat{j}\to \hat{k}$ forward hai, aur $\hat{k}$ ke baad ka agla letter $\hat{i}$ hai. Toh $\hat{j}\times \hat{k}=+\hat{i}$.
• (b) $\hat{k}\times \hat{j}$: yeh $\hat{j}\times \hat{k}$ hai order swap kiya hua, toh sign flip hoti hai: $-\hat{i}$.
• (c) $\hat{i}\times \hat{i}=\vec{0}$: ek vector aur khud ke beech angle $0$ hota hai, aur $\sin 0=0$. Koi spread nahi, koi area nahi.
• (d) $\hat{k}\times \hat{i}$: kya $\hat{k}\to \hat{i}$ loop par forward hai? $\dots \hat{k}\to \hat{i}$ haan (loop wrap around hota hai). Agla letter $\hat{j}$ hai. Toh $\hat{k}\times \hat{i}=+\hat{j}$.

KYA: seedha tool 1 mein plug karo. KYUN: humein lengths aur angle diya gaya hai, jo exactly $AB\sin \theta$ ko chahiye. $|\vec{A}\times \vec{B}|=AB\sin \theta =(6)(5)\sin 30^{\circ }=30\times \frac{1}{2}=15$ Woh $15$ do arrows jo tilaa hua patch span karte hain uska area hai.

Components padho: $A_x=1,A_y=2,A_z=3$ aur $B_x=4,B_y=5,B_z=6$.

• $\hat{i}$: $A_y{B}_z-A_z{B}_y=(2)(6)-(3)(5)=12-15=-3$
• $\hat{j}$: $-(A_x{B}_z-A_z{B}_x)=-[(1)(6)-(3)(4)]=-(6-12)=+6$. Middle term hamesha minus carry karta hai — yeh $+,-,+$ cofactor pattern hai.
• $\hat{k}$: $A_x{B}_y-A_y{B}_x=(1)(5)-(2)(4)=5-8=-3$ $\vec{A}\times \vec{B}=-3\hat{i}+6\hat{j}-3\hat{k}$ Quick sanity check: yeh result dono se perpendicular hona chahiye. $\vec{A}$ ke saath dot karo: $(-3)(1)+(6)(2)+(-3)(3)=-3+12-9=0$. ✓

$\vec{r}$ $x$ ke along point karta hai, $\vec{F}$ $y$ ke along. Sirf $\hat{k}$ term survive karta hai kyunki $\hat{i}\times \hat{j}=\hat{k}$. $\vec{\tau }=(r_y{F}_z-r_z{F}_y)\hat{i}+(r_z{F}_x-r_x{F}_z)\hat{j}+(r_x{F}_y-r_y{F}_x)\hat{k}$ $=(0)\hat{i}+(0)\hat{j}+(3\cdot 4-0)\hat{k}=12\hat{k}\text{N⋅m}$. Magnitude $=12$ N·m, $+\hat{k}$ direction mein point kar raha hai (page se bahar). Right-hand rule se iska matlab hai counter-clockwise turning. Torque and rotational equilibrium dekho.

KYA: same vertex $P$ se do edge vectors banao. KYUN: cross product ko dono arrows ka tail share karna padta hai (figure dekho — dono green edges $P$ par start hote hain). $\overrightarrow{PQ}=Q-P=(3,0,0),\overrightarrow{PR}=R-P=(0,4,0)$ Dono $z=0$ plane mein hain, toh sirf $\hat{k}$ component survive kar sakta hai: $\overrightarrow{PQ}\times \overrightarrow{PR}=(0)\hat{i}+(0)\hat{j}+(3\cdot 4-0\cdot 0)\hat{k}=12\hat{k}$ Parallelogram ka Area $=|12\hat{k}|=12$. Triangle uska aadha hota hai: $\text{Area}=\frac{1}{2}(12)=6\text{ square units}$ Base × height se check: $PQ$ $x$ ke along hai aur length $3$ hai (base); $R$ us base line se $4$ ki height par baith hai. Triangle area $=\frac{1}{2}(3)(4)=6$. ✓

Angle ke liye cross product KYUN? Kyunki $|\vec{A}\times \vec{B}|=AB\sin \theta$, toh jab hum teeno lengths jaante hain toh $\sin \theta$ solve kar sakte hain. Cross product (plane mein sirf $\hat{k}$ survive karta hai): $A_x{B}_y-A_y{B}_x=(2)(3)-(1)(1)=5$, toh $|\vec{A}\times \vec{B}|=5$. Lengths: $A=\sqrt{2^2+1^2}=\sqrt{5}$, $B=\sqrt{1^2+3^2}=\sqrt{10}$. $\sin \theta =\frac{|\vec{A}\times \vec{B}|}{AB}=\frac{5}{\sqrt{5}\sqrt{10}}=\frac{5}{\sqrt{50}}=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$ Toh $\theta =45^{\circ }$. Note: $\sin$ akele $45^{\circ }$ aur $135^{\circ }$ mein distinguish nahi kar sakta. Yahan ek quick dot-product check ($\vec{A}\cdot \vec{B}=2+3=5>0$) confirm karta hai ki angle acute hai, toh $45^{\circ }$ correct hai. Dot product — formula, projection, work calculation dekho.

$\vec{r}$ $y$ ke along, $\vec{p}$ $z$ ke along. Cross terms:

• $\hat{i}$: $r_y{p}_z-r_z{p}_y=(2)(3)-(0)(0)=6$
• $\hat{j}$: $-(r_x{p}_z-r_z{p}_x)=-(0-0)=0$
• $\hat{k}$: $r_x{p}_y-r_y{p}_x=0-0=0$ $\vec{L}=6\hat{i}{\text{kg⋅m}}^2/\text{s}$ $\hat{j}\times \hat{k}=\hat{i}$ ke consistent hai. Angular momentum — L = r × p dekho.

Pehle cross product $\vec{v}\times \vec{B}$: $\vec{v}$ $x$ ke along, $\vec{B}$ $z$ ke along.

• $\hat{i}$: $v_y{B}_z-v_z{B}_y=0-0=0$
• $\hat{j}$: $-(v_x{B}_z-v_z{B}_x)=-[(1)(2)-0]=-2$
• $\hat{k}$: $v_x{B}_y-v_y{B}_x=0-0=0$ Toh $\vec{v}\times \vec{B}=-2\hat{j}$. $q$ se multiply karo: $\vec{F}=q(\vec{v}\times \vec{B})=2(-2\hat{j})=-4\hat{j}\text{N}$ Force dono motion aur field ke perpendicular hai — isliye magnetic forces paths curve karti hain rather than cheezein speed up karne ke. Magnetic force — F = qv × B dekho.

KYA: component formula mein $\vec{B}=\vec{A}$ set karo, toh $B_x=A_x$, $B_y=A_y$, $B_z=A_z$.

• $\hat{i}$: $A_y{A}_z-A_z{A}_y=0$
• $\hat{j}$: $A_z{A}_x-A_x{A}_z=0$
• $\hat{k}$: $A_x{A}_y-A_y{A}_x=0$ Har component ek quantity minus khud hai. Toh $\vec{A}\times \vec{A}=\vec{0}$ sabhi $\vec{A}$ ke liye. KYUN aisa hona chahiye: ek vector khud se angle $0$ banata hai, $\sin 0=0$ — dono pictures agree karti hain.

Dono vectors plane mein hain, toh cross product ka sirf $\hat{k}$ component nonzero hai: $\vec{A}\times \vec{B}=(A_x{B}_y-A_y{B}_x)\hat{k}=(a\cdot b\sin \varphi -0\cdot b\cos \varphi )\hat{k}=ab\sin \varphi \hat{k}$ Iska magnitude hai $S=|ab\sin \varphi |=ab\sin \varphi$ ($0\le \varphi \le 180^{\circ }$ ke liye, $\sin \varphi \ge 0$). Yeh exactly tool 1 hai: doosre vector ki height base $\vec{A}$ ke upar $b\sin \varphi$ hai (figure mein pink dashed height dekho), base $a$ hai, toh area $=a\cdot b\sin \varphi$. Determinant aur geometry ek hi statement hain. $a=4,b=3,\varphi =90^{\circ }$ ke liye: $S=(4)(3)\sin 90^{\circ }=12$ square units.

$\vec{A}\times \vec{B}$ ek vector return karta hai ya scalar? ::: Ek vector (magnitude $AB\sin \theta$, direction by right-hand rule). Agar dono vectors $xy$-plane mein hain, toh kaun sa component survive karta hai? ::: Sirf $\hat{k}$ component, $A_x{B}_y-A_y{B}_x$ ke equal. Do triangle edges ko kya share karna chahiye? ::: Ek common tail (same vertex se start karna).