Projection of vectors: Exercises — Hinglish deep dive

4.5.4 · D4 · Maths › Linear Algebra (Full) › Projection of vectors

Shuru karne se pehle, Projection of vectors ke yeh do engines dhyan mein rakho:

Upar ka blueprint woh picture hai jo har exercise ke peeche hai: $a$ seedha neeche shadow daalta hai $b$ ke through jaane wali line par. Amber segment hai $proj_{b} a$ ; aur cyan segment jo line se upar jaati hai woh bacha hua perpendicular part $a_{⊥}$ hai.

Level 1 — Recognition

Goal: definition padho aur plug in karo. Interpretation ke koi trap nahi, bas sahi arithmetic.

L1.1 $a = (6, 8)$ ka scalar projection compute karo $b = (1, 0)$ par.

Recall Solution

KYA karte hain: $comp_{b} a = \frac{a \cdot b}{∣ b ∣}$ apply karo. KYUN: $b = (1, 0)$ $x$ -axis direction hai aur already length $1$ hai, isliye shadow ki length hai hi $x$ -component. $a \cdot b = 6 (1) + 8 (0) = 6, ∣ b ∣ = 1^{2} + 0^{2} = 1.$ $comp_{b} a = \frac{6}{1} = 6 .$ KAISA DIKHTA HAI: $(6, 8)$ ka shadow horizontal axis par $6$ tak jaata hai — exactly uski horizontal reach.

L1.2 $a = (0, 5)$ ka vector projection compute karo $b = (0, 2)$ par.

Recall Solution

KYA/KYUN: $proj_{b} a = \frac{a \cdot b}{b \cdot b} b$ use karo kyunki hume vector chahiye, sirf length nahi. $a \cdot b = 0 (0) + 5 (2) = 10, b \cdot b = 0^{2} + 2^{2} = 4.$ $proj_{b} a = \frac{10}{4} (0, 2) = (0, 5) = (0, 5) .$ KAISA DIKHTA HAI: $a$ already seedha upar $b$ ke same line mein point karta hai, isliye uska shadow khud woh hi hai. Fraction $\frac{10}{4} = 2.5$ times $(0, 2)$ exactly $(0, 5)$ par land karta hai. Perfect self-projection.

L1.3 $a = (3, 4)$ , $b = (0, 1)$ ke liye, $a \cdot b$ aur scalar projection batao.

Recall Solution

$a \cdot b = 3 (0) + 4 (1) = 4, ∣ b ∣ = 1, comp_{b} a = \frac{4}{1} = 4 .$ Dot product ne vertical component $4$ pick out kiya kyunki $b$ $y$ -axis hai.

Level 2 — Application

Goal: non-unit $b$ handle karo, jahan denominator real kaam karta hai, aur signs sahi padho.

L2.1 $a = (4, 3)$ ka vector projection $b = (3, 0)$ par.

Recall Solution

$a \cdot b = 4 (3) + 3 (0) = 12, b \cdot b = 3^{2} + 0^{2} = 9.$ $proj_{b} a = \frac{12}{9} (3, 0) = (4, 0) = (4, 0) .$ KYUN $b \cdot b$ aur $∣ b ∣$ nahi? $\frac{1}{9}$ long vector $(3, 0)$ ki "extra length" cancel karta hai. Notice karo answer $(4, 0)$ ki length $4$ hai — $a$ ki horizontal reach — $12$ nahi. Agar hum sirf ek baar $∣ b ∣ = 3$ se divide karte, toh miltaa $(4, 0) \cdot 3 = (12, 0)$ , jo teen guna zyada lamba hai.

L2.2 $a = (2, 3)$ ka scalar projection $b = (3, 4)$ par.

Recall Solution

$a \cdot b = 2 (3) + 3 (4) = 6 + 12 = 18, ∣ b ∣ = 3^{2} + 4^{2} = 25 = 5.$ $comp_{b} a = \frac{18}{5} = 3.6 .$ KAISA DIKHTA HAI: $(3, 4)$ ke through jaane wali slanted line par $a$ ka shadow $3.6$ units lamba hai — positive, isliye woh usi side girti hai jis taraf $b$ point karta hai.

L2.3 Obtuse case. $a = (1, 1)$ ka scalar projection $b = (- 2, 0)$ par.

Recall Solution

$a \cdot b = 1 (- 2) + 1 (0) = - 2, ∣ b ∣ = (- 2)^{2} = 2.$ $comp_{b} a = \frac{- 2}{2} = - 1 .$ KYUN negative? $b$ left ki taraf point karta hai ( $- x$ ), lekin $a$ right aur up ki taraf jhuka hai — dono ke beech angle $9 0^{\circ}$ se zyada hai, isliye $cos θ < 0$ . Sign data hai: shadow $b$ ke peeche wali side par girti hai.

Level 3 — Analysis

Goal: projection use karo decompose karne ke liye, perpendicular parts dhundhne ke liye, aur distances measure karne ke liye.

L3.1 $a = (3, 3)$ ko ek part $b = (1, 0)$ ke along aur ek part uske perpendicular mein split karo. Perpendicular check verify karo.

Recall Solution

Parallel part: $a \cdot b = 3, b \cdot b = 1, proj_{b} a = \frac{3}{1} (1, 0) = (3, 0) .$ Perpendicular part (jo bacha hua hai — dekho Orthogonal decomposition): $a_{⊥} = a - proj_{b} a = (3, 3) - (3, 0) = (0, 3) .$ Orthogonality check: $a_{⊥} \cdot b = (0) (1) + (3) (0) = 0$ ✓. KAISA DIKHTA HAI: $(3, 3)$ = horizontal reach $(3, 0)$ + vertical climb $(0, 3)$ , ek clean right-angle split.

L3.2 $a = (2, 1, 2)$ , $b = (1, 2, 2)$ ke liye, vector projection aur perpendicular remainder dhundho.

Recall Solution

$a \cdot b = 2 (1) + 1 (2) + 2 (2) = 2 + 2 + 4 = 8, b \cdot b = 1 + 4 + 4 = 9.$ $proj_{b} a = \frac{8}{9} (1, 2, 2) = (\frac{8}{9}, \frac{16}{9}, \frac{16}{9}) .$ $a_{⊥} = (2 - \frac{8}{9}, 1 - \frac{16}{9}, 2 - \frac{16}{9}) = (\frac{10}{9}, - \frac{7}{9}, \frac{2}{9}) .$ Check: $a_{⊥} \cdot b = \frac{10}{9} (1) + (- \frac{7}{9}) (2) + \frac{2}{9} (2) = \frac{10 - 14 + 4}{9} = \frac{0}{9} = 0$ ✓.

L3.3 $a = (5, 2)$ ki tip se origin ke through $b = (1, 1)$ direction mein jaane wali line tak ki distance dhundho.

Recall Solution

KYUN projection distance deta hai: perpendicular part $a_{⊥}$ tip se line tak ka straight-line gap hai, aur uski length woh distance hai. $a \cdot b = 5 (1) + 2 (1) = 7, b \cdot b = 2, proj_{b} a = \frac{7}{2} (1, 1) = (3.5, 3.5) .$ $a_{⊥} = (5, 2) - (3.5, 3.5) = (1.5, - 1.5) .$ $distance = ∣ a_{⊥} ∣ = 1. 5^{2} + (- 1.5)^{2} = 4.5 = \frac{3}{2} \approx 2.1213 .$

Level 4 — Synthesis

Goal: projection ko doosre ideas ke saath combine karo — orthonormalising, unknowns, physics.

L4.1 (Gram-Schmidt process ka ek step.) $u = (1, 0)$ aur $v = (2, 3)$ diye hain, $v$ se $u$ ke along wala sab kuch hatao, phir result ko unit vector mein normalise karo.

Recall Solution

$u$ -component hatao: $v \cdot u = 2, u \cdot u = 1, proj_{u} v = (2, 0) .$ $w = v - proj_{u} v = (2, 3) - (2, 0) = (0, 3) .$ Normalise ( $∣ w ∣ = 3$ ): $\overset{w}{^} = \frac{( 0 , 3 )}{3} = (0, 1) .$ KAISA DIKHTA HAI: $u$ $x$ ke along point karta hai; Gram-Schmidt $v$ ka horizontal share strip karta hai aur sirf pure vertical rakhta hai, jo orthonormal partner $(0, 1)$ deta hai.

L4.2 $t$ ki kaunsi value par $a = (t, 4)$ perpendicular hoga $b = (2, 1)$ par? Projection ke through interpret karo.

Recall Solution

KYUN dot product: do vectors perpendicular hote hain exactly tab jab ek doosre par projection zero ho, yaani $a \cdot b = 0$ . $a \cdot b = 2 t + 4 = 0 \Rightarrow t = - 2 .$ Check: $(- 2, 4) \cdot (2, 1) = - 4 + 4 = 0$ ✓. $t = - 2$ ke saath, $proj_{b} a = 0$ — bilkul koi shadow nahi.

L4.3 (Work done by a force.) Ek force $F = (6, 8)$ N ek object ko displacement $d = (4, 0)$ m ke along push karta hai. Kiya gaya work dhundho, aur force ka woh component jo motion ke perpendicular mein waste hua.

Recall Solution

KYUN dot product work hai: $W = F \cdot d$ sirf force ka woh part count karta hai jo displacement ke along hai — projection hi disguise mein. $W = F \cdot d = 6 (4) + 8 (0) = 24 J .$ Force ka wasted (perpendicular) part: $F$ ko $d$ par project karo, phir subtract karo. $proj_{d} F = \frac{24}{16} (4, 0) = (6, 0), F_{⊥} = (6, 8) - (6, 0) = (0, 8) N .$ $(0, 8)$ N sideways push karta hai aur koi work nahi karta kyunki yeh $d$ ke perpendicular hai. $W = 24 J, F_{⊥} = (0, 8)$ .

Level 5 — Mastery

Goal: ek general property prove karo aur limiting / degenerate cases ke baare mein reason karo.

L5.1 Prove karo ki same $b$ par do baar project karne se kuch nahi badlta: $proj_{b} (proj_{b} a) = proj_{b} a .$ (Yeh "idempotence" isliye hai ki projection ek genuine shadow hai.)

Recall Solution

Maano $p = proj_{b} a = \frac{a \cdot b}{b \cdot b} b$ . Likho $k = \frac{a \cdot b}{b \cdot b}$ , toh $p = k b$ . Ab $p$ ko dobara project karo: $proj_{b} p = \frac{p \cdot b}{b \cdot b} b = \frac{( k b ) \cdot b}{b \cdot b} b = \frac{k ( b \cdot b )}{b \cdot b} b = k b = p .$ MATLAB: $p$ already $b$ ki line par hai, isliye uska shadow woh khud hi hai. Shadow ka shadow wahi shadow hota hai. $■$

L5.2 L5.1 claim ko $a = (2, 3)$ , $b = (1, 1)$ ke liye numerically verify karo.

Recall Solution

$a \cdot b = 5, b \cdot b = 2, p = \frac{5}{2} (1, 1) = (2.5, 2.5) .$ Re-project: $p \cdot b = 5, \frac{5}{2} (1, 1) = (2.5, 2.5)$ — unchanged ✓.

L5.3 Degenerate case. $comp_{b} a$ aur $proj_{b} a$ ka kya hota hai jab $b \to 0$ ? Words aur symbols mein explain karo.

Recall Solution

Scalar: $comp_{b} a = \frac{a \cdot b}{∣ b ∣}$ . Jab $b \to 0$ , numerator $a \cdot b \to 0$ lekin denominator $∣ b ∣ \to 0$ bhi — ek $\frac{0}{0}$ jiska koi limit nahi (yeh depend karta hai us direction par jis taraf se $b$ approach karta hai). Numerically, $a = (1, 0)$ lo: $b = (ε, 0)$ ke along milta hai $\frac{ε}{ε} = 1$ , lekin $b = (0, ε)$ ke along milta hai $\frac{0}{ε} = 0$ . Do alag answers ⇒ undefined. Vector: $proj_{b} a = \frac{a \cdot b}{b \cdot b} b$ . Yahan numerator scale karta hai $ε$ jaisa, denominator $ε^{2}$ jaisa, aur trailing $b$ $ε$ jaisa: overall $\frac{ε}{ε ^{2}} \cdot ε = O (1)$ lekin phir bhi direction-dependent — koi single limit nahi. KYUN fail hona zaroori hai: $0$ par project karna matlab poochna "kya $a$ ka kitna hissa us direction ke along hai jo exist hi nahi karta." Koi line hi nahi hai jis par shadow daali jaye. Rule: projection ke liye $b \neq = 0$ zaroori hai.

Recall Poori ladder ka ek-line recap

L1 plug in · L2 denominator aur sign dhyan rakho · L3 $a_{⊥}$ se decompose karo aur distance measure karo · L4 Gram-Schmidt, perpendicularity, work ke saath fuse karo · L5 idempotence prove karo aur $b \neq = 0$ rule respect karo.

Connections

Projection of vectors — woh parent jo yeh page drill karta hai.
Dot product — har solution ki pehli line.
Unit vectors — L4.1 mein normalising.
Orthogonal decomposition — L3 splits.
Gram-Schmidt process — L4.1 uska ek step hai.
Least squares regression — same "drop a perpendicular" idea higher dimensions mein.
Work done by a force — L4.3.