Vectors in ℝⁿ — operations, geometric interpretation: Exercises

4.5.1 · D4 · Maths › Linear Algebra (Full) › Vectors in ℝⁿ — operations, geometric interpretation

Ek reminder ki har notation ka matlab kya hai, taaki tumhe koi unexplained symbol na mile:

Definition Symbol cheat-sheet

$v = (v_{1}, \dots, v_{n})$ — ek vector: numbers ki ek ordered list, saath hi origin se ek arrow bhi.
$0 = (0, \dots, 0)$ — zero vector: har component $0$ hai; arrow ke roop mein iska length zero hai aur yeh kisi direction mein point nahi karta.
$u + v$ — component-wise add karo (ek waqt mein ek axis).
$- v = (- 1) v$ — negation: har component ka sign flip karo (same length, opposite direction).
$u - v$ — subtraction: yeh defined hai negation add karne ke roop mein, $u + (- v) = (u_{1} - v_{1}, \dots, u_{n} - v_{n})$ . Toh yeh ab bhi component-wise hai; geometrically yeh $v$ ki tip se $u$ ki tip tak ka arrow hai.
$c v$ — scale: har component ko number $c$ se multiply karo ( $∣ c ∣$ se stretch karo, agar $c < 0$ ho toh flip karo).
$∥ v ∥$ — norm / length: origin se arrow ki tip ki straight-line distance.
$\hat{v} = v /∥ v ∥$ — unit vector: same direction, length exactly $1$ .
$u \cdot v$ — dot product: ek single number jo measure karta hai ki do arrows kitna same direction mein point karte hain.
$θ$ — do arrows ke beech ka angle, $cos θ = \frac{u \cdot v}{∥ u ∥∥ v ∥}$ se recover kiya gaya.
$u ⊥ v$ — "perpendicular", exactly tab sach hota hai jab $u \cdot v = 0$ .

Level 1 — Recognition

Yeh check karte hain ki tum dono operations ko geometry ke baare mein soche bina naam de sako aur execute kar sako.

Recall Solution 1.1

KYA: matching components add karo. KYUN: har axis independent hai, isliye hum unhe ek ek karke combine karte hain. $u + v = (2 + (- 3), - 1 + 5, 4 + 1) = (- 1, 4, 5) .$

Recall Solution 1.2

KYA: har component ko $3$ se multiply karo. KYUN: scaling poore arrow ko uniformly stretch karta hai. $3 u = (3 \cdot 2, 3 \cdot (- 1), 3 \cdot 4) = (6, - 3, 12) .$

Recall Solution 1.3

KYA: Pythagoras apply karo. KYUN: arrow ek right triangle ka hypotenuse hai jiske legs $6$ aur $8$ hain. $∥ w ∥ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10.$

Level 2 — Application

Ab operations ko combine karo aur woh quantities produce karo jo tum actually use karoge.

Recall Solution 2.1

KYA: har vector ko scale karo, phir subtract karo. KYUN: subtraction sirf negative scaling add karna hai ( $u - v = u + (- v)$ , cheat-sheet se). $2 u = (2, 4), \frac{1}{2} v = (2, - 1) .$ $2 u - \frac{1}{2} v = (2 - 2, 4 - (- 1)) = (0, 5) .$

Recall Solution 2.2

KYA: length se divide karo. KYUN: $∥ a ∥$ se divide karna rotate kiye bina length $1$ par rescale karta hai. Pehle check karo $a \neq = 0$ taaki division legal ho: yahan $a = (3, - 4) \neq = 0$ . ✓ $∥ a ∥ = 3^{2} + (- 4)^{2} = 25 = 5, \hat{a} = \frac{1}{5} (3, - 4) = (0.6, - 0.8) .$ Check: $0. 6^{2} + (- 0.8)^{2} = 0.36 + 0.64 = 1 = 1.$ ✓

Recall Solution 2.3

KYA: matching components multiply karo aur sum karo. KYUN: dot product weigh karta hai ki do arrows kitna align karte hain. $u \cdot v = 2 \cdot 4 + (- 1) \cdot 0 + 3 \cdot (- 2) = 8 + 0 - 6 = 2.$

Level 3 — Analysis

Yahan numbers ko interpret karna hai: angles, perpendicularity, aur dot product ka sign tumhe kya batata hai.

Neeche ki figure Exercise 3.1 dikhati hai. Kya observe karna hai: kala arrow $b = (1, 0)$ horizontal $x$ -axis ke saath flat padi hai, jabki laal arrow $a = (1, 1)$ diagonal climb karti hai. Unke beech ka chhota laal arc angle $θ$ hai. Dekho ki laal arrow $x$ -axis aur seedha upar ke bilkul beech mein baithti hai — woh visual "halfway" exactly woh $4 5^{\circ}$ hai jo hum compute karte hain.

Figure — Vectors in ℝⁿ — operations, geometric interpretation

Recall Solution 3.1

KYA: $cos θ = \frac{a \cdot b}{∥ a ∥∥ b ∥}$ use karo. KYUN yeh tool: dot product woh ek maatra operation hai jo angle ko arithmetic mein pack karta hai — hum $θ$ isolate karne ke liye iske geometric form ko rearrange karte hain. $a \cdot b = 1 \cdot 1 + 1 \cdot 0 = 1, ∥ a ∥ = 2, ∥ b ∥ = 1.$ $cos θ = \frac{1}{2 \cdot 1} = \frac{1}{2} \Rightarrow θ = 4 5^{\circ} .$ KAISA DIKHTA HAI: exactly upar wali figure ka laal arrow — flat aur vertical ke beech mein, isliye $4 5^{\circ}$ .

Recall Solution 3.2

KYA: dot product compute karo aur sign padho. KYUN: perpendicularity exactly $u \cdot v = 0$ ka statement hai, aur sign baaki cases mein batata hai ki angle acute hai ya obtuse. $u \cdot v = 3 \cdot 4 + (- 2) \cdot 6 = 12 - 12 = 0.$ Zero $\Rightarrow$ perpendicular ( $u ⊥ v$ ). Kyunki $cos θ = \frac{u \cdot v}{∥ u ∥∥ v ∥}$ aur norms positive hain, dot product ka sign $cos θ$ ka sign hai:

Dot $> 0$ : $cos θ > 0$ , toh $θ < 9 0^{\circ}$ — arrows same taraf jhuke hain.
Dot $= 0$ : $θ = 9 0^{\circ}$ — right angle.
Dot $< 0$ : $cos θ < 0$ , toh $θ > 9 0^{\circ}$ — arrows alag taraf jhuke hain (obtuse).

Recall Solution 3.3

KYA: same $cos θ$ formula apply karo, ab negative cosine expect karte hue. KYUN: dot product negative aaya, jo 3.2 ke sign rule se ek obtuse angle force karta hai — hum $θ$ ke liye same tarike se solve karte hain. $p \cdot q = 1 \cdot (- 1) + 0 + 0 = - 1, ∥ p ∥ = 1, ∥ q ∥ = (- 1)^{2} + 1^{2} = 2 .$ $cos θ = \frac{- 1}{1 \cdot 2} = - \frac{1}{2} \Rightarrow θ = 13 5^{\circ} .$ Negative dot product ek obtuse angle force karta hai — arrows alag half-planes mein point karte hain.

Level 4 — Synthesis

Addition, norms, aur dot products ko ek reasoning chain mein combine karo.

Neeche ki figure Exercise 4.1 illustrate karti hai. Kya observe karna hai: chaar kaale arrows ek parallelogram banate hain sides $u$ aur $v$ ke saath. Laal solid arrow ek diagonal $u + v$ hai; laal dashed segment doosra diagonal $u - v$ hai ( $u$ aur $v$ ki tips ke beech). Parallelogram law kehta hai: un do laal diagonals ko square karo, unhe add karo, aur tumhe exactly twice the squared sides milenge — laal objects dekho, wahi saara point hain.

Recall Solution 4.1

KYA: parallelogram ke dono diagonals aur dono sides compute karo. KYUN: law kehta hai ki do diagonals ki squared lengths mila kar do sides ki squared lengths se do guna ke barabar hoti hain. $u + v = (4, 1) \Rightarrow ∥ u + v ∥^{2} = 16 + 1 = 17.$ $u - v = (- 2, 3) \Rightarrow ∥ u - v ∥^{2} = 4 + 9 = 13.$ Left side: $17 + 13 = 30$ . $∥ u ∥^{2} = 1 + 4 = 5, ∥ v ∥^{2} = 9 + 1 = 10 \Rightarrow 2 (5) + 2 (10) = 30.$ Dono sides $30$ ke barabar hain. ✓ KAISA DIKHTA HAI: upar wali figure mein parallelogram ke do laal diagonals, $u + v$ (chhota, solid) aur $u - v$ (dashed), uske chaar barabar kaale sides ke against balance karte hain.

Recall Solution 4.2

KYA: dot product ko zero set karo. KYUN: perpendicular $⟺$ dot $= 0$ . $a \cdot b = 3 t + 2 t = 5 t = 0 \Rightarrow t = 0.$ Toh $a = (0, 2)$ (seedha upar) aur $b = (3, 0)$ (seedha daayein) — waqai ek right angle. Sirf $t = 0$ kaam karta hai.

Recall Solution 4.3

KYA: $∥ c u ∥ = ∣ c ∣ ∥ u ∥$ use karo. KYUN: $c$ se scaling length ko $∣ c ∣$ se multiply karta hai (absolute value, kyunki flipping length nahi badlata). $∥ u ∥ = 5, ∣ c ∣ \cdot 5 = 15 \Rightarrow ∣ c ∣ = 3 \Rightarrow c = 3 or c = - 3.$ Dono valid hain: $c = 3$ direction rakhta hai, $c = - 3$ use reverse karta hai — dono mein same length.

Level 5 — Mastery

Ek extended problem jo poore chapter ko ek saath weave karta hai.

Recall Solution 5.1

KYA kar rahe hain: $u$ ko $v$ ke saath wale piece aur uske square wale piece mein split karna — Projections and Orthogonal Decomposition ka pehla taste. KYUN yeh kaam karta hai: parallel piece $v$ ki ek scaling hai, aur sahi scaling isliye choose ki jaati hai taaki bacha hua perpendicular ho. Dot product exactly woh tool hai jo measure karta hai "kitna $u$ , $v$ ke along hai."

Step 0 — legality check. Hum $v \cdot v = ∥ v ∥^{2}$ se divide karne wale hain, isliye hume $v \neq = 0$ chahiye. Yahan $v = (1, 3) \neq = 0$ , toh projection well-defined hai. ( $0$ par project karna meaningless hota — $0$ kisi taraf point nahi karta.)

Step 1 — parallel part. Likho $p = k v$ kisi scalar $k$ ke liye (parallel matlab "ek scaled copy"). Woh formula jo remainder ko perpendicular banata hai woh hai $k = \frac{u \cdot v}{v \cdot v} .$ Compute karo: $u \cdot v = 2 \cdot 1 + 1 \cdot 3 = 5, v \cdot v = 1^{2} + 3^{2} = 10.$ $k = \frac{5}{10} = \frac{1}{2}, p = \frac{1}{2} (1, 3) = (\frac{1}{2}, \frac{3}{2}) .$

Step 2 — perpendicular part. Jo bhi bacha hua hai (subtraction $u - p = u + (- p)$ use karte hue): $q = u - p = (2 - \frac{1}{2}, 1 - \frac{3}{2}) = (\frac{3}{2}, - \frac{1}{2}) .$

Step 3 — (b) verify perpendicularity. $q \cdot v = \frac{3}{2} \cdot 1 + (- \frac{1}{2}) \cdot 3 = \frac{3}{2} - \frac{3}{2} = 0. ✓$ Aur $p + q = (\frac{1}{2} + \frac{3}{2}, \frac{3}{2} - \frac{1}{2}) = (2, 1) = u .$ ✓

KYUN $k$ ki choice forced thi: humne insist kiya ki bacha hua $q = u - k v$ , $v$ se perpendicular ho, yaani $q \cdot v = 0$ . Us requirement ko dot product ke distributive rule se expand karo: $(u - k v) \cdot v = u \cdot v - k (v \cdot v) = 0.$ Single unknown $k$ solve karne par $k = \frac{u \cdot v}{v \cdot v}$ milta hai — yeh ek maatra value hai jo bacha hua $v$ ke square mein banati hai. Koi bhi doosra $k$ nonzero dot product chhodta, toh $q$ perpendicular nahi hota. (Note karo $v \cdot v \neq = 0$ kyunki $v \neq = 0$ — wahi legality check jaise Step 0 mein.)

Recall Solution 5.2

KYUN yeh hold hona chahiye: $p$ aur $q$ right angle par milte hain, isliye woh ek right triangle banate hain $u$ as hypotenuse ke saath — Pythagoras apply hota hai. $∥ u ∥^{2} = 2^{2} + 1^{2} = 5.$ $∥ p ∥^{2} = (\frac{1}{2})^{2} + (\frac{3}{2})^{2} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = 2.5.$ $∥ q ∥^{2} = (\frac{3}{2})^{2} + (- \frac{1}{2})^{2} = \frac{9}{4} + \frac{1}{4} = 2.5.$ $∥ p ∥^{2} + ∥ q ∥^{2} = 2.5 + 2.5 = 5 = ∥ u ∥^{2} . ✓$

Connections

Projections and Orthogonal Decomposition — Exercise 5 poore projection formula ka seed hai.
Dot Product and Orthogonality — Levels 3–5 mein perpendicularity tests.
Norms and Distance in Rn — length aur parallelogram computations.
Vectors in ℝⁿ — operations, geometric interpretation — parent note jise yeh exercises drill karte hain.