Projection of vectors: Question bank — Hinglish deep dive

4.5.4 · D5 · Maths › Linear Algebra (Full) › Projection of vectors

Upar: $cos θ$ ke peeche ka right triangle (left) aur idempotence ke peeche ki "already-on-the-line" picture (right). Drills karte waqt inhe dobara dekhte rehna.

True ya false — justify karo

Har line mein: true/false decide karo, phir reason do — sirf yes/no se kuch nahi milega.

The vector projection

proj_{b} a

hamesha

b

ki same direction mein point karta hai.

False. Jab angle

θ

obtuse hota hai (

a \cdot b < 0

) toh scalar factor negative hota hai, isliye projection

b

ke opposite point karta hai. Yeh

b

ki line par rehta hai, lekin hamesha same direction mein nahi.

Agar

b

ki length double kar do, toh vector projection

proj_{b} a

same rehta hai.

True.

b

2 b

se replace karo: numerator

a \cdot (2 b) = 2 (a \cdot b)

, denominator

(2 b) \cdot (2 b) = 4 (b \cdot b)

, aur vector factor

2 b

ho jaata hai. Toh

\frac{2 ( a \cdot b )}{4 ( b \cdot b )} (2 b) = \frac{4}{4} \cdot \frac{a \cdot b}{b \cdot b} b = \frac{a \cdot b}{b \cdot b} b

— unchanged. Sirf

b

ki direction matter karti hai, uski length nahi.

Agar

b

ki length double kar do, toh scalar projection

comp_{b} a

same rehta hai.

True.

comp_{b} a = a \cdot \hat{b}

sirf unit direction

\hat{b}

par depend karta hai, aur

b

ko double karne se

\hat{b}

(uski direction) unchanged rehti hai.

proj_{b} a = proj_{a} b

jab bhi

∣ a ∣ = ∣ b ∣

False. Equal lengths direction fix nahi karte: ek result

b

ke along hota hai, doosra

a

ke along. Ye sirf tab agree karte hain jab

a

aur

b

already same direction mein point kar rahe hon.

Scalar projection

∣ a ∣

se bada ho sakta hai.

False.

comp_{b} a = ∣ a ∣ cos θ

aur

∣ cos θ ∣ \leq 1

, toh iska magnitude kabhi

∣ a ∣

se zyada nahi hoga. Shadow kabhi laathi se lamba nahi hota.

Agar

proj_{b} a = 0

, toh

a = 0

False. Projection tab zero hota hai jab

a ⊥ b

(unka dot product zero hai), chahe

a

kitna bhi lamba kyun na ho. Ek vertical laathi seedhi upar sun ke neeche floor par koi shadow nahi dalti.

Perpendicular part

a_{⊥} = a - proj_{b} a

construction ke hisaab se hamesha

b

ke perpendicular hota hai.

True.

a_{⊥} \cdot b = a \cdot b - \frac{a \cdot b}{b \cdot b} (b \cdot b) = 0

algebraically, toh orthogonality guaranteed hai, lucky nahi. Yahi Orthogonal decomposition ki neenv hai.

a

b

par project karo, phir us result ko dobara

b

par project karo — doosri baar kuch nahi badalta.

True.

p = proj_{b} a = \frac{a \cdot b}{b \cdot b} b

maan lo.

p

ko project karne par

\frac{p \cdot b}{b \cdot b} b

milta hai; lekin

p \cdot b = \frac{a \cdot b}{b \cdot b} (b \cdot b) = a \cdot b

, toh doosra projection exactly

p

return karta hai. Do baar karna ek baar karne ke barabar hai — yahi idempotent ka matlab hai (right-hand figure dekho).

Ek unit vector

\hat{b}

ke liye (toh

∣ \hat{b} ∣ = 1

), scalar projection aur vector projection ki magnitude equal hoti hain.

True. Scalar number

a \cdot \hat{b}

hai; vector

(a \cdot \hat{b}) \hat{b}

hai, jiska magnitude

∣ a \cdot \hat{b} ∣ \cdot 1

hai. Toh vector ki length scalar ki absolute value ke equal hai — same size, alag type ka object.

Error dhundho

Har ek mein ek "result" bataya gaya hai. Bolo kya galat hai aur sahi reasoning do.

proj_{b} a = \frac{a \cdot b}{∣ b ∣} b

Galat denominator. Yeh ek

∣ b ∣

se divide karta hai lekin poore

b

se multiply karta hai, jisse answer

∣ b ∣

factor se zyada lamba ho jaata hai. Sahi:

\frac{a \cdot b}{b \cdot b} b

— ek length shadow ke liye, ek direction ko unitise karne ke liye.

"Kyunki angle

θ

12 0^{\circ}

hai, scalar projection

∣ a ∣ cos 12 0^{\circ} = ∣ a ∣ \cdot \frac{1}{2}

hai."

cos 12 0^{\circ} = - \frac{1}{2}

hai,

+ \frac{1}{2}

nahi. Sign drop karne se yeh chhup jaata hai ki shadow

b

ke back side par padta hai; scalar projection yahan negative hai.

a

mein se kitna

b

ke along point karta hai yeh jaanne ke liye, main

∣ comp_{b} a ∣

compute karta hoon toh hamesha clean positive answer milega."

Absolute value lene se direction information destroy ho jaati hai. Sign hi yeh answer hai ki "

b

ke saath kaunsi direction mein" — ise phenkne se obtuse aur acute cases ek ho jaate hain.

(0, 5)

(1, 0)

par projection

(0, 5)

hai kyunki yahi woh vector hai jisse humne shuru kiya."

x

-axis par projection sirf

x

-component rakhta hai, jo yahan

0

hai. Sahi answer

(0, 0)

hai: ek purely vertical vector ka koi horizontal shadow nahi hota.

comp_{b} a = a \cdot b

— dot product hi scalar projection hai."

Yeh sirf tab true hai jab

∣ b ∣ = 1

ho. Generally

∣ b ∣

se divide karna zaroori hai:

comp_{b} a = \frac{a \cdot b}{∣ b ∣}

. Raw dot product mein

b

ki length bhi bundle hoti hai.

"Projection linear hai, isliye dot product ki symmetry se

proj_{b} (a) = proj_{a} (b)

Dot product symmetric hai, lekin projection mein ek direction factor

b

(ya

a

) bhi hota hai. Roles badalne se shadow kisi alag line par padta hai; operation symmetric nahi hai.

a = proj_{b} a + a_{⊥}

sirf 2D mein hold karta hai."

Yeh kisi bhi dimension mein hold karta hai. Identity pure algebra hai:

a_{⊥}

defined hi

a - proj_{b} a

ke roop mein hai, toh unhe wapas add karne par hamesha

a

milta hai, chahe vectors mein kitne bhi coordinates hon.

Why questions

Scalar projection mein

cos θ

kyun use hota hai,

sin θ

kyun nahi?

Hum

b

ke along component chahte hain (figure mein right triangle ki adjacent side), aur

cos θ = \frac{adjacent}{hypotenuse}

exactly wahi deta hai.

sin θ

perpendicular (opposite) part deta, yaani

∣ a_{⊥} ∣

Vector projection

∣ b ∣

se divide kyun nahi karta,

b \cdot b

se kyun karta hai?

∣ b ∣

ka ek factor dot product ko true shadow length mein convert karta hai; doosra factor

b

ko unit direction

\hat{b}

mein badalta hai. Dono milke

∣ b ∣^{2} = b \cdot b

dete hain.

Projection ka "engine" dot product kyun hai, angle directly measure karne ke bajaye?

Dot product secretly

cos θ

contain karta hai via

a \cdot b = ∣ a ∣∣ b ∣ cos θ

, toh yeh bina protractor ke coordinates se shadow extract karne deta hai. Dekho Dot product.

proj_{b} a = (a \cdot \hat{b}) \hat{b}

likhte waqt

\hat{b}

unit vector kyun hona chahiye?

Formula "(length)×(direction)" tabhi kaam karta hai jab direction ki length exactly 1 ho; warna direction factor answer mein extra length ghusa dega. Dekho Unit vectors.

Gram-Schmidt process mein projection subtract karne se perpendicular kyun milta hai?

a

ka woh part hatana jo

b

ke along hai, sirf wahi bachta hai jo

b

ke orthogonal hai. Har pehle vector ke against yahi repeat karna exactly woh hai jo Gram–Schmidt orthonormal set banane ke liye karta hai.

Work done by a force mein work

W = F \cdot d

secretly projection kyun use karta hai?

Sirf force ka woh component jo displacement ke along hai, kaam karta hai; woh component scalar projection

∣ F ∣ cos θ

hai, aur

∣ d ∣

se multiply karne par

F \cdot d

milta hai.

Scalar projection negative kyun ho sakta hai jabki physical "length" nahi ho sakti?

Scalar projection ek signed length hai jo

b

-line ke saath direction bhi record karta hai. Minus sign kehta hai "shadow peeche ki taraf point karta hai," jo ek plain length chhupa leti.

Edge cases

proj_{b} a

kya hoga jab

a

b

ke perpendicular hai (

θ = 9 0^{\circ}

Zero vector. Kyunki

a \cdot b = 0

, numerator zero ho jaata hai —

a

ka koi bhi part

b

ke along nahi hai.

proj_{b} a

kya hoga jab

a

already

b

ke parallel hai (same direction,

θ = 0^{\circ}

Yeh

a

khud ke barabar hoga.

a

ka poora part

b

ke along hai, toh shadow poore vector ko capture kar leta hai aur

a_{⊥} = 0

Jab

b = 0

ho toh projection formulas ka kya hoga?

Dono undefined hain: scalar

\frac{a \cdot b}{∣ b ∣}

aur vector

\frac{a \cdot b}{b \cdot b} b

dono zero se divide karte hain (kyunki

∣ 0 ∣ = 0

aur

0 \cdot 0 = 0

), aur project karne ke liye koi direction hi nahi hai.

proj_{b} a

kya hoga jab

a = 0

Zero vector. Zero vector ka koi shadow nahi hota — numerator

a \cdot b = 0

use khatam kar deta hai, aur

a_{⊥} = 0

bhi.

Exactly

θ = 9 0^{\circ}

par, kya projection ek "chhota" vector hai ya exactly zero?

Exactly zero, kyunki

cos 9 0^{\circ} = 0

. Yeh positive (acute) aur negative (obtuse) shadows ke beech ki clean boundary hai.

Jaise-jaise

θ \to 18 0^{\circ}

, scalar projection kya approach karta hai?

Yeh

- ∣ a ∣

approach karta hai, apni most negative value, kyunki

cos 18 0^{\circ} = - 1

. Shadow sabse lamba hota hai jab

a

seedha

b

ke peeche point kare.

Agar

a

aur

b

exact opposite direction mein point kar rahe hain (

θ = 18 0^{\circ}

), toh

a_{⊥}

kya hai?

Zero vector. Anti-parallel vectors phir bhi same line par hote hain, toh projection

a

poora capture kar leta hai aur kuch bhi perpendicular nahi bachta.

Recall Jaane se pehle ek one-line self-test

Inhe cover karo aur apne dimag mein jawab do. $θ = 9 1^{\circ}$ par scalar projection ka sign? ::: Negative — $9 0^{\circ}$ se thoda aage, $cos θ < 0$ . Vector projection ka denominator? ::: $b \cdot b = ∣ b ∣^{2}$ . Kya $b$ ko lamba karne se $proj_{b} a$ badalta hai? ::: Nahi — sirf $b$ ki direction matter karti hai. $b = 0$ par projection? ::: Undefined — dono scalar aur vector forms zero se divide karte hain. "Idempotent" yahan kya matlab hai? ::: Doosri baar project karne se kuch nahi badalta: $P (P (a)) = P (a)$ .