Unit vectors — î, ĵ, k̂; constructing unit vector: Worked examples

1.1.10 · D3 · Physics › Measurement, Vectors & Kinematics › Unit vectors — î, ĵ, k̂; constructing unit vector

Shuru karne se pehle, un do symbols ka ek reminder jo tumhe kabhi confuse nahi karne chahiye:

$A$ (upar arrow) = poora vector — iska ek size aur ek direction hota hai.
$∣ A ∣$ (vertical bars) = magnitude — ek single positive number, vector ki length.
$\hat{A}$ (hat) = unit vector — $A$ jaise hi direction, lekin length $1$ pe locked.

Recipe inhe jorti hai: $\hat{A} = \frac{A}{∣ A ∣}$ , aur iska reverse $A = ∣ A ∣ \hat{A}$ .

Scenario matrix

Is topic ke har problem ka ek (ya blend) in case classes mein se ek hota hai:

#	Case class	Isme tricky kya hai	Example
C1	Sab-positive 2D	Friendly baseline	Ex 1
C2	Mixed signs (pehle quadrant ke alawa koi aur)	Signs $\hat{A}$ mein survive karne chahiye	Ex 2
C3	Negative component ke saath 3D	Pythagoras do baar; sign rakho	Ex 3
C4	Pehle se unit vector	$1$ se divide karne par kuch nahi badlega	Ex 4
C5	Zero vector (degenerate)	$\hat{A}$ undefined hai — koi direction nahi	Ex 5
C6	Ugly / irrational magnitude	Answer exact rehna chahiye, round nahi karo	Ex 6
C7	Real-world word problem (units!)	Direction paane ke liye units strip karo, baad mein wapas lagao	Ex 7
C8	Exam twist: vector ek difference mein chhupa hai	Pehle vector banao, phir normalise karo	Ex 8
C9	Limiting behaviour: ek component bahut bada	$\hat{A}$ us axis ki taraf jhukne lagta hai	Ex 9

Neeche har example us cell ko fill karta hai jisme woh fit hota hai.

Worked Examples

Worked example Example 1 — C1: sab-positive 2D

$A = 6 \hat{i} + 8 \hat{j}$ ka unit vector nikalo.

Forecast: Dono components positive hain, toh $\hat{A}$ upar-aur-daayein point karna chahiye, pehle quadrant mein. Guess: kuch aisa $0.6 \hat{i} + 0.8 \hat{j}$ ?

Step 1. $∣ A ∣ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10$ . Yeh step kyun? Jo length tune measure nahi ki, usse divide nahi kar sakte. Squares-then-root isliye kyunki dono components ek right triangle ke perpendicular sides hain (figure dekho) aur sirf Pythagoras perpendicular lengths ko combine karta hai.

Step 2. $\hat{A} = \frac{6 i ^ + 8 j ^}{10} = 0.6 \hat{i} + 0.8 \hat{j}$ . Yeh step kyun? Positive number $10$ se divide karne par length $1$ ho jaati hai lekin kabhi rotate nahi hoti, toh direction untouched rehti hai.

Verify: $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1 = 1. ✓$ Dono components positive rahe — abhi bhi pehla quadrant. Forecast confirmed.

Worked example Example 2 — C2: mixed signs (third quadrant)

$A = - 3 \hat{i} - 4 \hat{j}$ ka unit vector nikalo.

Forecast: Dono components negative $\Rightarrow$ yeh arrow neeche-aur-baayein point karta hai (third quadrant). Magnitude yeh jaanti nahi — usse sirf size se matlab hai. Toh main expect karta hoon ki signs $\hat{A}$ mein survive karein.

Step 1. $∣ A ∣ = (- 3)^{2} + (- 4)^{2} = 9 + 16 = 25 = 5$ . Yeh step kyun? Squaring minus signs ko khatam kar deti hai — magnitude hamesha positive hoti hai. Yahi wajah hai ki magnitude akele kabhi quadrant nahi bata sakti.

Step 2. $\hat{A} = \frac{- 3 i ^ - 4 j ^}{5} = - 0.6 \hat{i} - 0.8 \hat{j}$ . Yeh step kyun? Hum poore vector ko divide karte hain, minus signs samet. Ek positive number se divide karne par har component ka sign preserve hota hai, toh pointer abhi bhi third quadrant mein aim karta hai.

Verify: $(- 0.6)^{2} + (- 0.8)^{2} = 0.36 + 0.64 = 1. ✓$ Ex 1 se compare karo: same length- $1$ pointer, lekin opposite direction mein flip ho gaya — signs ne apna kaam kiya.

Worked example Example 3 — C3: negative component ke saath 3D

$B = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ ke liye $\hat{B}$ nikalo.

Forecast: Teen components, ek negative. Magnitude teeno squares use karta hai, toh agar lucky raha toh whole number milega ( $2, 3, 6$ ek Pythagorean triple lagte hain). $\hat{B}$ ko $\hat{j}$ par minus rakhna chahiye.

Step 1. $∣ B ∣ = 2^{2} + (- 3)^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7$ . Yeh step kyun? 3D mein Pythagoras do baar apply hota hai — pehle base par ( $x, y$ ), phir $z$ tak — jo collapse hokar teeno squares ke sum ka ek square root ban jaata hai. Signs squaring ke neeche gayab ho jaate hain, toh $- 3$ contribute karta hai $+ 9$ .

Step 2. $\hat{B} = \frac{2 i ^ - 3 j ^ + 6 k ^}{7} = \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k}$ . Yeh step kyun? 2D jaisi hi recipe; teesra axis normalise karne ke tarike mein kuch nahi badalta.

Verify: $(\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{6}{7})^{2} = \frac{4 + 9 + 36}{49} = \frac{49}{49} = 1. ✓$ $\hat{j}$ par minus survive kiya — direction intact hai.

Worked example Example 4 — C4: pehle se unit vector

$C = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j}$ ko normalise karo.

Forecast: Yeh components suspiciously clean lagte hain. Agar $C$ pehle se hi length $1$ ka hai, toh uske magnitude ( $= 1$ ) se divide karne par woh seedha wapas aa jaana chahiye — ek "do-nothing" operation.

Step 1. $∣ C ∣ = (\frac{1}{2})^{2} + (\frac{1}{2})^{2} = \frac{1}{2} + \frac{1}{2} = 1 = 1$ . Yeh step kyun? Hamesha pehle measure karo. Yahaan measurement reveal karta hai ki hum pehle se unit length par the.

Step 2. $\hat{C} = \frac{C}{1} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j}$ (unchanged). Yeh step kyun? $1$ se divide karna identity operation hai — koi shrink nahi, koi stretch nahi, koi rotation nahi.

Verify: Wahi vector andar jaata hai aur bahar aata hai, aur uski length $1$ hai. $✓$ Lesson: ek unit vector ko normalise karna ek no-op hai, jo ek accha sanity check hai ki tumhari recipe self-consistent hai.

Worked example Example 5 — C5: zero vector (degenerate case)

$0 = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$ ka unit vector nikalne ki koshish karo.

Forecast: Length $0$ wale vector ki koi direction nahi hoti — woh ek single point hai, bina shaft ke ek arrow. "Point" karne ke liye kuch hai hi nahi. Toh main predict karta hoon ki recipe toot jaayegi.

Step 1. $∣ 0 ∣ = 0^{2} + 0^{2} + 0^{2} = 0$ . Yeh step kyun? Length hamesha ki tarah measure karo — aur woh zero aati hai.

Step 2. $\hat{0} = \frac{0}{0} = \frac{0 i ^ + 0 j ^ + 0 k ^}{0} \Rightarrow undefined$ . Yeh step kyun? Zero se division undefined hai. Geometrically: pointer banane ke liye tumhe ek real arrow ko shrink karna hoga, lekin zero-length arrow ke paas yeh kehne ka koi tarika nahi ki woh kis taraf jaata hai. Zero vector ka koi unit vector nahi hota.

Verify: Koi arithmetic plug back karne ke liye nahi — check conceptual hai: kisi bhi candidate $\hat{0}$ ka magnitude $1$ hona chahiye, lekin $\hat{0} = 0 /0$ produce karta hai $0/0$ , jo meaningless hai. $✓$ (Yeh woh ek case hai jisse tumhe code mein aur exams par guard karna chahiye.)

Worked example Example 7 — C7: real-world word problem (units matter)

Ek velocity hai $v = 30 \hat{i} + 40 \hat{j}$ measured in $m/s$ . Ek drone ko $100 m/s$ par usi direction mein fly karne ke liye kaha jaata hai. Drone ka velocity vector nikalo.

Forecast: "Same direction" matlab unit vector — $v$ ki length strip karo pure pointer paane ke liye, phir $100$ se re-stretch karo. Kyunki $30, 40$ scale hota hai $3, 4, 5$ par, ek tidy answer expect hai.

Step 1. $∣ v ∣ = 3 0^{2} + 4 0^{2} = 900 + 1600 = 2500 = 50 m/s$ . Yeh step kyun? Hume sirf $v$ ki direction chahiye, toh pehle uska size jaanna zaroori hai taaki hum use divide kar sakein. Units ( $m/s$ ) magnitude ke saath ride karte hain.

Step 2. $\overset{v}{^} = \frac{30 i ^ + 40 j ^}{50} = 0.6 \hat{i} + 0.8 \hat{j}$ (dimensionless). Yeh step kyun? $m/s$ ko $m/s$ se divide karne par units cancel ho jaati hain — ek unit vector dimensionless hota hai, pure direction. Yeh parent note ke "unit vector ke koi units nahi hote" point se match karta hai.

Step 3. $v_{drone} = (100 m/s) \overset{v}{^} = 60 \hat{i} + 80 \hat{j} m/s$ . Yeh step kyun? Vector = magnitude × direction. $100 m/s$ naya size supply karta hai aur units wapas attach karta hai.

Verify: $∣ v_{drone} ∣ = 6 0^{2} + 8 0^{2} = 3600 + 6400 = 10000 = 100 m/s . ✓$ Sahi speed, aur $\overset{v}{^}$ unchanged $\Rightarrow$ sahi direction.

Worked example Example 8 — C8: exam twist — vector ek difference mein chhupa hai

Points $P = (1, 2, 2)$ aur $Q = (4, 6, 2)$ . $P$ se $Q$ ki taraf point karne wala unit vector nikalo.

Forecast: Vector mujhe diya nahi gaya — mujhe pehle use $Q - P$ ke roop mein build karna hoga (yeh displacement hai). Tab hi normalise karoon. Agar subtraction bhool gaya toh galat cheez normalise karoon ga.

Step 1. $P Q = Q - P = (4 - 1) \hat{i} + (6 - 2) \hat{j} + (2 - 2) \hat{k} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k}$ . Yeh step kyun? " $P$ se $Q$ " ki direction displacement $Q - P$ hai — head minus tail. Dekho Position vector and displacement. Note karo $\hat{k}$ -component $0$ hai: dono points same height share karte hain, toh displacement ek plane mein flat rehta hai.

Step 2. $∣ P Q ∣ = 3^{2} + 4^{2} + 0^{2} = 9 + 16 + 0 = 25 = 5$ . Yeh step kyun? Zero component length mein kuch contribute nahi karta — lekin tum fir bhi use include karte ho taaki recipe uniform rahe.

Step 3. $P Q = \frac{3 i ^ + 4 j ^ + 0 k ^}{5} = 0.6 \hat{i} + 0.8 \hat{j}$ . Yeh step kyun? Same normalisation. $0 \hat{k}$ simply drop out ho jaata hai.

Verify: $0. 6^{2} + 0. 8^{2} + 0^{2} = 1 = 1. ✓$ Direction ka sanity check: $Q$ , $P$ ke daayein aur upar hai, aur sach mein dono non-zero components positive hain.

Worked example Example 9 — C9: limiting behaviour — ek component dominate karta hai

$E_{1} = 10 \hat{i} + 1 \hat{j}$ aur $E_{2} = 1000 \hat{i} + 1 \hat{j}$ ke unit vectors compare karo. Jab $x$ -component bina bound ke barhta jaata hai toh kya hota hai?

Forecast: Jaise $x$ , $y$ ko dwarf karta hai, arrow $+ x$ ke along almost flat ho jaata hai, toh $\hat{E}$ ko $\hat{i}$ hi approach karna chahiye. Main expect karta hoon ki $\hat{i}$ -coefficient $\to 1$ aur $\hat{j}$ -coefficient $\to 0$ .

Step 1. $E_{1}$ ke liye: $∣ E_{1} ∣ = 1 0^{2} + 1^{2} = 101 \approx 10.0499$ , toh $\hat{E}_{1} = \frac{10 i ^ + 1 j ^}{101} \approx 0.9950 \hat{i} + 0.0995 \hat{j} .$ Yeh step kyun? Bas recipe; trend dekhne ke liye kaafi decimals rakho.

Step 2. $E_{2}$ ke liye: $∣ E_{2} ∣ = 100 0^{2} + 1^{2} = 1000001 \approx 1000.0005$ , toh $\hat{E}_{2} \approx 0.9999995 \hat{i} + 0.0009999995 \hat{j} .$ Yeh step kyun? $x$ jitna bada hota hai, $\hat{i}$ -coefficient utna hi $1$ ke paas aata hai aur $\hat{j}$ -coefficient $0$ ki taraf shrink hota hai.

Step 3 (limit). Jab $x$ -component $\to \infty$ aur $y$ fixed ho, toh $\frac{x}{x ^{2} + 1} \to 1$ aur $\frac{1}{x ^{2} + 1} \to 0$ , isliye $\hat{E} \to \hat{i}$ . Yeh step kyun? Yeh dikhata hai ki $\hat{i}$ kisi bhi aisi vector ka limiting pointer hai jo $x$ -axis ki taraf zyada se zyada jhukti jaati hai — standard basis vectors normalisation ke extreme cases hain.

Verify: $0.995 0^{2} + 0.099 5^{2} = 0.99005 + 0.00990 \approx 1.000$ (rounding), aur exactly $\frac{1 0 ^{2} + 1 ^{2}}{101} = \frac{101}{101} = 1. ✓$

Recall Quick self-test across the matrix

Cover the answers and name the trap in each.

Which case gives an undefined unit vector, and why? ::: Zero vector (C5) — magnitude $0$ forces division by zero, and a zero-length arrow has no direction. Normalising an already-unit vector gives what? ::: The same vector back (C4) — dividing by $1$ is a no-op. When you're asked for the direction "from $P$ to $Q$ ", what must you build first? ::: The displacement $P Q = Q - P$ (C8), then normalise. As one component grows without bound, the unit vector approaches what? ::: The basis vector of that axis (C9), e.g. $\hat{i}$ . Do negative components survive into $\hat{A}$ ? ::: Yes — dividing by the positive magnitude keeps every sign (C2/C3).

Connections

Parent: Unit vectors
Magnitude and direction of a vector
Vectors — components and resolution
Pythagoras theorem
Position vector and displacement
Vector addition — triangle and parallelogram laws