Exercises — Unit vectors — î, ĵ, k̂; constructing unit vector

Unit vector ki magnitude exactly $1$ hoti hai. Har length ko $\sqrt{\text{sum of squares}}$ se check karo.

• (a) $|\vec{P}|=\sqrt{1^2+1^2}=\sqrt{2}\approx 1.414$ → unit nahi.
• (b) $|\vec{Q}|=\sqrt{0.6^2+(-0.8{)}^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$ → unit ✓ (minus sign sirf direction set karta hai, square hone par chala jaata hai).
• (c) $|\vec{R}|=\sqrt{0^2+0^2+1^2}=1$ → unit ✓ ($\hat{k}$ ek standard basis vector hai, definition se length $1$).
• (d) $|\vec{S}|=\sqrt{2^2}=2$ → unit nahi.

Answer: (b) aur (c).

Galat. Ek component hona length ke baare mein kuch nahi kehta. Length hai $|\vec{A}|=\sqrt{5^2}=5$, na ki $1$. Iska unit vector hoga $\hat{A}=\frac{5\hat{i}}{5}=\hat{i}$. Components ki sankhya aur magnitude do alag-alag baatein hain.

Step 1 (KYA): length nikalo. $|\vec{A}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. KYUN: hum length tabhi "divide out" kar sakte hain jab hum usse jaante hon. Step 2: har component ko $10$ se divide karo: $\hat{A}=0.6\hat{i}+0.8\hat{j}$. Check: $\sqrt{0.6^2+0.8^2}=\sqrt{1}=1$ ✓.

Step 1: $|\vec{B}|=\sqrt{2^2+(-3{)}^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$. KYUN squares minus ko khatam karte hain: squaring se har term positive ho jaata hai; direction component ke sign mein stored rehta hai, length mein nahi. Step 2: $\hat{B}=\frac{2}{7}\hat{i}-\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}$. $\hat{j}$ par minus rehta hai — yeh kehta hai "$-y$ ki taraf point karo".

Step 1: $\vec{d}$ ko pure direction mein strip karo. $|\vec{d}|=\sqrt{(-4{)}^2+3^2}=\sqrt{25}=5$, isliye $\hat{d}=-0.8\hat{i}+0.6\hat{j}$. KYUN: hum sirf $\vec{d}$ ka aim chahte hain, uska size nahi. Step 2: $\vec{v}=(20\text{m/s})\hat{d}=-16\hat{i}+12\hat{j}\text{m/s}$. Check: $|\vec{v}|=\sqrt{16^2+12^2}=\sqrt{400}=20\text{m/s}$ ✓.

Length equation set karo: $|\vec{A}|=\sqrt{3^2+c^2}=5$. Dono sides square karo (root undo karne ke liye): $9+c^2=25\Rightarrow c^2=16$. Root lo — DONO signs rakho: $c=+4$ ya $c=-4$. KYUN do answers: magnitude sirf $c^2$ fix karta hai. Geometrically, length $5$ ka ek arrow jiska horizontal reach $3$ hai, upar ($c=4$) ya neeche ($c=-4$) jhuk sakta hai — do mirror-image vectors, dono length $5$ ke. Negative wala mat chhodo.

Idea: do 2D vectors perpendicular hote hain jab unka dot product $0$ ho. $(A_x,A_y)$ ko $90^{\circ }$ rotate karne ka ek quick tarika hai components swap karo aur ek sign flip karo: $(A_x,A_y)\to (-A_y,A_x)$. Step 1: ek perpendicular vector hai $\vec{n}=-4\hat{i}+3\hat{j}$. Perpendicularity check: $\vec{A}\cdot \vec{n}=(3)(-4)+(4)(3)=-12+12=0$ ✓. Step 2: normalize karo. $|\vec{n}|=\sqrt{(-4{)}^2+3^2}=5$, isliye $\hat{n}=-0.8\hat{i}+0.6\hat{j}$. (Opposite direction $\hat{n}=0.8\hat{i}-0.6\hat{j}$ bhi utni hi valid hai — do perpendicular unit vectors hote hain.)

"Kisi axis ke along fraction" exactly unit vector ka corresponding component hota hai — kyunki $\hat{A}$, $\vec{A}$ hai jo scale karke length $1$ kar diya gaya hai, isliye uska har component us axis ka ek unit mein hissa hai. Step 1: $|\vec{A}|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$. Step 2: $\hat{A}=\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}$. Answer: $z$ ka share $\frac{1}{3}\approx 0.333$ hai, matlab length ka lagbhag $33.3\%$ $z$ ke along point karta hai.

Step 1 — displacement direction (displacement = end minus start): $\overrightarrow{PQ}={\vec{r}}_Q-{\vec{r}}_P=(4-1)\hat{i}+(6-2)\hat{j}=3\hat{i}+4\hat{j}$. Step 2 — pure direction: $|\overrightarrow{PQ}|=\sqrt{3^2+4^2}=5$, isliye $\hat{u}=0.6\hat{i}+0.8\hat{j}$. KYUN: "$Q$ ki taraf" ka matlab hai hum sirf $\overrightarrow{PQ}$ ka aim lete hain, uski poori length $5$ nahi. Step 3 — $10$-unit step: $\Delta \vec{r}=10\hat{u}=6\hat{i}+8\hat{j}$. Step 4 — start mein add karo (tip-to-tail): ${\vec{r}}_{\text{final}}={\vec{r}}_P+\Delta \vec{r}=(1+6)\hat{i}+(2+8)\hat{j}=7\hat{i}+10\hat{j}$. Note karo ki $Q$ khud sirf $5$ units door hai $P$ se, isliye $10$ move karna $Q$ se aage nikal jaata hai — jo hamaara answer confirm karta hai.

Step 1 — components add karo: $\vec{F}={\vec{F}}_1+{\vec{F}}_2=(3-1)\hat{i}+(4+2)\hat{j}=2\hat{i}+6\hat{j}\text{N}$. Step 2 — magnitude: $|\vec{F}|=\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}\approx 6.32\text{N}$. Step 3 — direction: $\hat{F}=\frac{2\hat{i}+6\hat{j}}{2\sqrt{10}}=\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}\approx 0.316\hat{i}+0.949\hat{j}$. KYUN sum ko normalize karo, parts ko nahi: net direction depend karta hai dono forces ke combine hone par, isliye pehle add karna zaroori hai, phir length strip karo.

Maano $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$ jahan $|\vec{A}|=m=\sqrt{A_x^2+A_y^2+A_z^2}$, aur $m\ne 0$. Tab $\hat{A}=\frac{A_x}{m}\hat{i}+\frac{A_y}{m}\hat{j}+\frac{A_z}{m}\hat{k}$. Iske magnitude compute karo: $|\hat{A}|=\sqrt{{\left(\frac{A_x}{m}\right)}^2+{\left(\frac{A_y}{m}\right)}^2+{\left(\frac{A_z}{m}\right)}^2}=\sqrt{\frac{A_x^2+A_y^2+A_z^2}{m^2}}=\sqrt{\frac{m^2}{m^2}}=\sqrt{1}=1.$ Dimension kabhi matter nahi kiya: root ke andar numerator exactly $m^2$ hai definition se. $■$

$m\ne 0$ kyun zaroori hai: $m=0$ se divide karna undefined hai — Problem 5.2 dekho.

$|\vec{0}|=\sqrt{0+0+0}=0$. Recipe demand karegi $\hat{0}=\frac{\vec{0}}{0}$, jo undefined hai — zero se divide nahi kar sakte. Geometric reason: zero vector ek point hai, arrow nahi — iska koi direction nahi jis taraf point kare. Unit vector ka kaam sirf direction encode karna hai, isliye represent karne ke liye kuch hai hi nahi. Conclusion: zero vector ka koi unit vector nahi hota. Yeh woh ek input hai jahan construction fail ho jaata hai, aur fail isliye hota hai kyunki concept hi (ek direction) absent hai.

$x$, $y$, $z$ ke saath equal angles aur sab-positive matlab equal positive components: $\vec{A}=\hat{i}+\hat{j}+\hat{k}$ lo (koi bhi positive multiple same direction dega). Step 1: $|\vec{A}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$. Step 2: $\hat{A}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\approx 0.577(\hat{i}+\hat{j}+\hat{k})$. Sanity check: $3\times {\left(\frac{1}{\sqrt{3}}\right)}^2=3\times \frac{1}{3}=1$ ✓. Yeh ek unit cube ki famous "body diagonal" direction hai.

Pehle length ($\sqrt{\text{sum of squares}}$), phir point karne ke liye divide karo ($\hat{A}=\vec{A}/|\vec{A}|$), har sign rakho, magnitude conditions se $\pm$ expect karo, normalize karne se pehle full vectors add karo — aur kabhi $\vec{0}$ ko normalize mat karo.