Vectors in ℝⁿ — operations, geometric interpretation: Worked examples

4.5.1 · D3 · Maths › Linear Algebra (Full) › Vectors in ℝⁿ — operations, geometric interpretation

Do notation pieces jinhe hum baar baar use karenge

Do symbols almost har example mein aate hain, toh chaliye inhe yahan plain words mein pin karte hain, use karne se pehle.

Formula Angle identity jo hum reuse karenge (dot product = length × length × cosine)

Dot product $a \cdot b = \sum a_{i} b_{i}$ (matching components multiply karke sum karo) do arrows ke beech angle $θ$ se is tarah juda hai: $a \cdot b = ∥ a ∥ ∥ b ∥ cos θ$ Yeh sach kyun hai? Parent note ise Law of Cosines se derive karta hai: $∥ a - b ∥^{2}$ ko do alag tarike se expand karo — ek baar geometrically ( $∥ a ∥^{2} + ∥ b ∥^{2} - 2∥ a ∥∥ b ∥ cos θ$ ) aur ek baar algebraically ( $∥ a ∥^{2} - 2 a \cdot b + ∥ b ∥^{2}$ ); $∥ a ∥^{2}$ aur $∥ b ∥^{2}$ cancel ho jaate hain, aur boxed identity bach jaati hai. Rearrange karne par, yeh humein arithmetic se angle read off karne deti hai: $cos θ = \frac{a \cdot b}{∥ a ∥ ∥ b ∥} .$ Dot ka sign akela hi bata deta hai: positive ⟹ acute, zero ⟹ right angle, negative ⟹ obtuse. Dekho Dot Product and Orthogonality.

Scenario matrix

Kuch bhi kaam karne se pehle, chaliye har case class list karte hain jo yeh teen operations produce kar sakti hain. Har row ek "cell" hai — ek alag situation apne behaviour ke saath. Neeche ke examples mein har ek par cell(s) ka tag lagaa hai.

Cell	Operation	Scenario	Kya special hai / trap kya hai
A	add / scale	all-positive components	"friendly" baseline
B	add / scale	mixed signs, negative scalar	direction flip karna, sign bookkeeping
C1	scale	scalar $c = 0$	degenerate: $0 \cdot v = 0$ , direction vanish ho jaata hai
C2	add	zero vector $0$ add karna	identity: $v + 0 = v$ , kuch nahi badalata
D	norm	ek vector ko normalise karna; $0$ ki length	length se divide karna; $0$ ko normalise nahi kar sakte
E	dot	acute angle ( $θ < 9 0^{\circ}$ )	dot product positive
F	dot	right angle ( $θ = 9 0^{\circ}$ )	dot product exactly zero
G	dot	obtuse / opposite ( $θ > 9 0^{\circ}$ , ya $18 0^{\circ}$ )	dot product negative
H	dot	$R^{4}$ mein (koi picture possible nahi)	formula bina geometry ke bhi kaam karta hai
I	word problem	real displacement / force	units track karna, sahi operation choose karna
J	exam twist	unknown component dhundho jo $⊥$ banaye	$u \cdot v = 0$ se equation solve karo

Dhyan raho ki C1 (zero se scale karna) aur C2 (zero vector add karna) alag edge cases hain: C1 ek arrow ko destroy karta hai, C2 ek ko untouched chhodta hai. Example 3 dono ko handle karta hai, clearly alag karke. Teen examples follow karte hain; har header batata hai kaunsa cell hit hota hai.

Example 1 — Cell A (all-positive add & scale)

Worked example Do "friendly" arrows ko combine karo

Maano $u = (2, 1)$ aur $v = (1, 3)$ . $3 u + v$ compute karo.

Forecast: Dono arrows up-and-right point karte hain. $u$ ko triple karne se ek lamba right-leaning arrow banta hai, phir $v$ use thoda upar nudge karta hai. Guess: result kahin dur up-and-right mein land karega, pehla component doosre se bada. Apna guess likho.

Step 1. $3 u = (3 \cdot 2, 3 \cdot 1) = (6, 3)$ . Yeh step kyun? Scalar multiplication component-wise hai — har axis independent hai, toh hum har coordinate ko alag se $3$ se multiply karte hain.

Step 2. $3 u + v = (6 + 1, 3 + 3) = (7, 6)$ . Yeh step kyun? Addition bhi component-wise hai (east-parts add karo, north-parts add karo).

Verify: Dono components positive hain aur $7 > 6$ — forecast se match karta hai. Sanity check: $(7, 6)$ ko $(6, 3)$ ke baad tip-to-tail baithna chahiye; figure dekho — green arrow $v$ wahan se start hota hai jahan lamba lavender $3 u$ khatam hota hai, aur coral resultant path close karta hai. ✓

Figure — Vectors in ℝⁿ — operations, geometric interpretation

Example 2 — Cell B (mixed signs, negative scalar)

Worked example Subtraction matlab

$- 1$ se scale karna phir add karna Maano $u = (1, 2)$ , $v = (3, - 1)$ . $2 u - v$ compute karo.

Forecast: $v$ down-right point karta hai (uska doosra component negative hai). Ise subtract karne ka matlab hai uska flip add karna, jo up-left point karta hai. Toh answer $2 u$ ke comparison mein left aur upar drift karna chahiye. Dono components ke signs guess karo.

Step 1. $2 u = (2, 4)$ . Yeh step kyun? Component-wise scaling.

Step 2. $- v = (- 3, + 1)$ . Yeh step kyun? $v$ subtract karna waisa hi hai jaise $(- 1) v$ add karna. $- 1$ se multiply karne par direction flip ho jaati hai — har sign reverse ho jaata hai, toh $(3, - 1) \to (- 3, 1)$ .

Step 3. $2 u - v = (2 - 3, 4 + 1) = (- 1, 5)$ . Yeh step kyun? Component-wise add karo.

Verify: Pehla component negative (left drift hua), doosra positive aur bada (upar drift hua) — bilkul forecast jaisa. Figure mein, dashed coral arrow $- v$ hai jo up-left point karta hai, aur resultant upper-left region mein land karta hai. ✓

Example 3 — Cells C1 & C2 (do alag degenerate cases)

Worked example Zero se scale karna (C1) vs. zero vector add karna (C2)

Cell C1: (a) $0 \cdot (5, - 7)$ compute karo. Cell C2: (b) $(4, - 2) + 0$ compute karo jahan $0 = (0, 0)$ . Phir (c) $∥ 0 ∥$ kya hai?

Forecast: Yeh do edge cases opposite behave karte hain. Zero se scale karne par arrow destroy ho jaana chahiye (shrink hokar kuch nahi). "Koi movement nahi" add karne par arrow preserve hona chahiye (kuch nahi badalega). "Koi arrow nahi" ki length $0$ honi chahiye.

Step 1 (C1, part a). $0 \cdot (5, - 7) = (0 \cdot 5, 0 \cdot (- 7)) = (0, 0) = 0$ . Yeh step kyun? Component-wise scaling; har coordinate times $0$ equals $0$ . Yeh C1 behaviour hai: direction vanish ho jaati hai — zero-length arrow kahin point nahi karta.

Step 2 (C2, part b). $(4, - 2) + (0, 0) = (4 + 0, - 2 + 0) = (4, - 2)$ . Yeh step kyun? Yeh C2 behaviour hai, C1 se bilkul alag: $0$ additive identity hai. "Move zero east, zero north" add karne par tum exactly wahin raho jahan the — arrow untouched hai, destroy nahi hua.

Step 3 (part c). $∥ 0 ∥ = 0^{2} + 0^{2} = 0$ . Yeh step kyun? Norm formula (root-sum-of-squares se length) phir bhi apply hoti hai; saare squared components $0$ hain, toh length $0$ hai.

Verify: C1 ne arrow ko $0$ mein collapse kiya; C2 ne $(4, - 2)$ unchanged chhoda — dono cases genuinely alag hain, jaise forecast mein tha. ⚠️ Dhyan raho: kyunki $∥ 0 ∥ = 0$ hai, tum $0$ ko kabhi normalise nahi kar sakte — $0$ se divide karna undefined hai. Zero vector ki koi direction nahi hoti. ✓

Example 4 — Cell D (normalising, aur kyun yahan fail nahi ho sakta)

Worked example Ek vector ko unit arrow mein badlo

$w = (- 6, 8)$ ko normalise karo.

Forecast: $w$ up-left point karta hai. Unit version same direction mein point karega lekin length exactly $1$ hogi, toh uske components size mein chhote hone chahiye. Guess karo kya dono same sign rakheinge.

Step 1. $∥ w ∥ = (- 6)^{2} + 8^{2} = 36 + 64 = 100 = 10$ . Yeh step kyun? Humein current length (norm) chahiye $1$ par rescale karne se pehle. Note karo $(- 6)^{2} = 36$ — squaring ke neeche sign disappear ho jaata hai, isliye length hamesha positive hoti hai.

Step 2. $\hat{w} = \frac{1}{10} (- 6, 8) = (- 0.6, 0.8)$ . Yeh step kyun? Har component ko length $10$ se divide karna arrow ko length $1$ par shrink karta hai bina rotate kiye — direction (dono signs) untouched rehti hai. Hat ise unit version ke roop mein mark karta hai.

Verify: Nayi length check karo: $(- 0.6)^{2} + (0.8)^{2} = 0.36 + 0.64 = 1 = 1$ . ✓ Unit length confirmed, signs preserved (abhi bhi up-left) — forecast se match karta hai.

Example 5 — Cell E (acute angle → positive dot)

Worked example Do arrows jo mostly agree karte hain

$a = (3, 1)$ , $b = (2, 2)$ . Unke beech angle $θ$ dhundho.

Forecast: Dono up-right point karte hain, direction mein kaafi close. Toh $cos θ$ $1$ ke karib hona chahiye, yaani $θ$ chhota (well under $9 0^{\circ}$ ), aur dot product positive hona chahiye.

Step 1. $a \cdot b = 3 \cdot 2 + 1 \cdot 2 = 6 + 2 = 8$ . Yeh step kyun? Matching components ke products ka sum — yeh ek number alignment encode karta hai. Positive hona already bata deta hai ki angle acute hai.

Step 2. $∥ a ∥ = 9 + 1 = 10$ , $∥ b ∥ = 4 + 4 = 8 = 22$ . Yeh step kyun? Upar boxed identity, $a \cdot b = ∥ a ∥∥ b ∥ cos θ$ , $cos θ$ isolate karne ke liye dono lengths (norms) chahiye.

Step 3. $cos θ = \frac{8}{10 \cdot 2 2} = \frac{8}{2 20} = \frac{8}{4 5} = \frac{2}{5} \approx 0.894$ . Yeh step kyun? Boxed identity ko rearrange karke $cos θ$ solve kiya. Radicals rule $a \cdot b = ab$ se simplify hote hain: toh $10 \cdot 2 = 20$ , giving $10 \cdot 22 = 220$ denominator mein. Phir $20 = 4 \cdot 5 = 25$ (perfect square $4$ ko $2$ ke roop mein baahar nikalo), toh $220 = 45$ ; upar ka $8$ aur $4$ cancel hoke $\frac{2}{5}$ milta hai.

Step 4. $θ = arccos (0.894) \approx 26.5 7^{\circ}$ . Yeh step kyun? $arccos$ jawaab deta hai "kis angle ka yeh cosine hai?" — yeh cosine ko undo karta hai.

Verify: $θ \approx 26. 6^{\circ}$ chhota aur acute hai, dot product positive — forecast holds. Figure mein do almost-aligned arrows ke beech tight sliver of angle dikhta hai. ✓

Example 6 — Cell F (right angle → zero dot)

Example 7 — Cell G (obtuse aur fully opposite → negative dot)

Worked example Arrows jo disagree karte hain

(a) $c = (- 1, 2)$ , $d = (3, 1)$ : acute, right, ya obtuse? (b) $e = (2, 1)$ , $f = (- 2, - 1)$ : angle dhundho.

Forecast: (a) ke liye, $c$ left lean karta hai, $d$ right lean karta hai — yeh disagree karte lagte hain, toh obtuse guess karo (dot negative). (b) ke liye, $f$ exactly $- e$ hai, yaani flip, toh yeh opposite point karte hain — $18 0^{\circ}$ guess karo.

Step 1 (a). $c \cdot d = (- 1) (3) + (2) (1) = - 3 + 2 = - 1$ . Yeh step kyun? Negative dot product matlab $cos θ < 0$ , hence $θ > 9 0^{\circ}$ — obtuse. Classify karne ke liye humein full angle ki zaroorat nahi; sign akela hi bata deta hai.

Step 2 (b). $e \cdot f = (2) (- 2) + (1) (- 1) = - 4 - 1 = - 5$ . $∥ e ∥ = ∥ f ∥ = 5$ , toh $cos θ = \frac{- 5}{5 \cdot 5} = \frac{- 5}{5} = - 1$ . Yeh step kyun? $cos θ = - 1$ sabse zyada negative cosine ho sakta hai — matlab arrows exactly opposite directions mein point karte hain.

Step 3 (b). $θ = arccos (- 1) = 18 0^{\circ}$ . Yeh step kyun? $arccos (- 1) = 18 0^{\circ}$ , ek seedhi line — dono arrows anti-parallel hain.

Verify: (a) dot $= - 1 < 0$ ⟹ obtuse ✓. (b) $θ = 18 0^{\circ}$ " $f = - e$ " se match karta hai. Figure mein $c, d$ right angle se wide open dikhte hain, aur $e, f$ ek seedhi line par opposite directions mein. ✓

Example 8 — Cell H (dimension 4, koi picture nahi)

Worked example Jab draw nahi kar sakte, formula par trust karo

$R^{4}$ mein, maano $u = (1, 0, - 1, 2)$ aur $v = (2, 3, 0, 1)$ . $u \cdot v$ aur $∥ u ∥$ dhundho.

Forecast: Hum 4D picture nahi bana sakte, lekin algebra identical hai — matching components ke products sum karo, aur length squares ke sum ka square root hai. Guess karo kya dot positive hai.

Step 1. $u \cdot v = (1) (2) + (0) (3) + (- 1) (0) + (2) (1) = 2 + 0 + 0 + 2 = 4$ . Yeh step kyun? Dot product formula $\sum u_{i} v_{i}$ ki koi dimension limit nahi — yeh $R^{4}$ mein waisa hi kaam karta hai jaise $R^{2}$ mein. Iske liye kisi picture ki zaroorat nahi.

Step 2. $∥ u ∥ = 1^{2} + 0^{2} + (- 1)^{2} + 2^{2} = 1 + 0 + 1 + 4 = 6 \approx 2.449$ . Yeh step kyun? Norm repeated Pythagoras hai, ek perpendicular axis ek baar — char baar karo aur phir bhi bas squares add karte ho.

Verify: Dot $= 4 > 0$ , toh agar hum dekh paate toh angle acute hota. Length $6$ ek plain positive number hai. Formulas ko kabhi geometry ki zaroorat nahi thi — yahi $R^{n}$ ki power hai. ✓

Example 9 — Cell I (real-world word problem, units ke saath)

Worked example Hike ke do legs

Ek hiker $a = (3, 4)$ km (3 km east, 4 km north) chalta hai, phir $b = (- 6, 2)$ km. Dhundho (a) start se unki final position, aur (b) start se seedha-line distance.

Forecast: Pehla leg up-right jaata hai, doosra far up-left. Net result start se left mein khatam hona chahiye lekin upar. Seedha-line distance total path se kam honi chahiye. Final position ke east-component ka sign guess karo.

Step 1 (a). Final position $= a + b = (3 - 6, 4 + 2) = (- 3, 6)$ km. Yeh step kyun? Do displacements ek ke baad ek karna tip-to-tail addition hai — vectors add karne ka natural matlab. Units poore time km rehte hain.

Step 2 (b). Start se distance $= ∥ (- 3, 6) ∥ = (- 3)^{2} + 6^{2} = 9 + 36 = 45 = 35 \approx 6.708$ km. Yeh step kyun? Seedha-line distance resultant displacement ki length hai — norm. $- 3$ square karne par sign remove ho jaata hai, toh "west gaya" bhi distance mein positively add hota hai.

Verify: Final east-component $- 3$ (start se west mein khatam hua) forecast se match karta hai; north $+ 6$ (upar) sense banata hai. Units: $km^{2} = km$ ✓. Figure mein do legs tip-to-tail aur seedha coral resultant dikhta hai. ✓

Example 10 — Cell J (exam twist: missing component solve karo)

Worked example Ise perpendicular banao

$k$ ki kis value ke liye $s = (k, 2)$ , $t = (4, - 3)$ ke perpendicular hai?

Forecast: Perpendicular matlab dot product $0$ par force hota hai. Yeh ek equation, ek unknown hai — toh $k$ ki ek single value nikalni chahiye. Guess karo kya $k$ positive hai.

Step 1. Perpendicular $⟺ s \cdot t = 0$ . Yeh step kyun? Yeh right angle ki algebraic definition hai — geometric condition ko ek aisi equation mein baadlo jo hum solve kar sakein. (Zero dot matlab $cos θ = 0$ , yaani $θ = 9 0^{\circ}$ .)

Step 2. $s \cdot t = (k) (4) + (2) (- 3) = 4 k - 6$ . Yeh step kyun? Dot-product formula apply karo (matching components multiply karke sum karo), $k$ ko symbolic rakhte hue kyunki yeh wo unknown hai jo hum solve kar rahe hain.

Step 3. Dot product ko zero set karo aur solve karo: $4 k - 6 = 0 ⟹ 4 k = 6 ⟹ k = \frac{6}{4} = \frac{3}{2} = 1.5.$ Yeh step kyun? Step 1 ki perpendicular condition kehti hai ki Step 2 ka expression $0$ ke equal hona chahiye; yeh ek unknown mein ek linear equation hai, toh hum $k$ isolate karte hain dono sides mein $6$ add karke phir $4$ se divide karke.

Verify: $k = 1.5$ substitute karo: $s \cdot t = (1.5) (4) + (2) (- 3) = 6 - 6 = 0$ ✓. Dot product zero ho jaata hai, toh $s = (1.5, 2)$ sach mein $t$ ke perpendicular hai, aur $k = 1.5 > 0$ forecast se match karta hai. Final answer: $k = \frac{3}{2}$ .

Recall Dot product ka kaun sa sign kya matlab rakhta hai?

Positive dot ::: acute angle ( $θ < 9 0^{\circ}$ ) — arrows mostly agree karte hain. Zero dot ::: right angle ( $θ = 9 0^{\circ}$ ) — perpendicular. Negative dot ::: obtuse angle ( $θ > 9 0^{\circ}$ ); aur $cos θ = - 1$ (dot sabse zyada negative) matlab exactly opposite ( $18 0^{\circ}$ ). Kya tum zero vector ko normalise kar sakte ho? ::: Nahi — uski length $0$ hai aur $0$ se divide karna undefined hai; uski koi direction nahi hoti. Negative $c$ se scale karne par kya hota hai? ::: Length $∣ c ∣$ se stretch hoti hai aur arrow opposite direction mein flip ho jaata hai. Norm symbol $∥ v ∥$ ka kya matlab hai? ::: Arrow ki length, $v_{1}^{2} + \dots + v_{n}^{2}$ .

Connections

Dot Product and Orthogonality — perpendicular tests (Cells F, J) yahan generalize hote hain.
Norms and Distance in Rn — normalising (Cell D) aur hike distance (Cell I).
Linear Combinations and Span — add-and-scale examples (Cells A, B) building blocks hain.
Projections and Orthogonal Decomposition — Cells E–G ki acute/obtuse sign logic use karta hai.