Resolution of vectors — into components (any axes): Question bank

1.1.9 · D5 · Physics › Measurement, Vectors & Kinematics › Resolution of vectors — into components (any axes)

Yeh page sirf Resolution of vectors — into components (any axes) ke ideas assume karta hai: ek component kisi chosen direction mein vector ki shadow hoti hai, $A_{x} = A cos θ$ aur $A_{y} = A sin θ$ jab $θ$ ko x-axis se measure kiya jaata hai, aur projection ek unit direction $\overset{n}{^}$ pe dot product hai $A \cdot \overset{n}{^} = A cos ϕ$ .

Figure — Resolution of vectors — into components (any axes)

Figure dekho: coral vector $A$ ki lavender ruler $\overset{n}{^}$ pe shadow exactly $A cos ϕ$ length ki hai, jahan $ϕ$ unke beech ka opening angle hai.

True or false — justify

Ek vector ko resolve karna us vector ko khud badal deta hai

False. Resolution sirf us hi arrow ko chosen axes ke saath do (ya zyada) pieces ke roop mein re-describe karta hai; physical quantity — uski length aur direction — bilkul nahi badlti. Aapne description badli, cheez nahi.

Ek vector ke hamesha exactly do components hote hain

False. Components ki sankhya un axes ki sankhya ke barabar hoti hai jo aap choose karte hain. Ek plane mein hum usually do chunte hain, lekin usi vector ke alag-alag axis pairs pe alag-alag component values hote hain — aur koi bhi "real wala" nahi hota.

A_{x} = A cos θ

ek universal law hai jise aap hamesha quote kar sakte hain

False. Yeh tabhi kaam karta hai jab

θ

us axis se measure kiya jaaye jis pe aap project kar rahe hain. Dusri axis se angle measure karo to cosine aur sine swap ho jaate hain. Yeh formula ek picture encode karta hai, koi magic string nahi.

Ek vector ke do rectangular components hamesha positive hote hain

False. Sign axis ke saath direction batata hai. Second quadrant mein point karne wale vector ka

A_{x} < 0, A_{y} > 0

hota hai. Signs drop kar ke sirf

A cos θ

A sin θ

ko positive numbers ke roop mein rakhna ek classic error hai.

Ek vector ki magnitude

A_{x} + A_{y}

hoti hai

False. Components perpendicular directions mein hote hain, isliye woh Pythagoras se combine hote hain:

A = A_{x}^{2} + A_{y}^{2}

, kabhi bhi simple addition se nahi. Unhe seedha add karne se over-count ho jaata.

Agar aap oblique (non-90°) axes pe resolve karein, to magnitude phir bhi

a^{2} + b^{2}

hogi

False. Pythagoras ko axes ke beech right angle chahiye. Oblique axes ke liye components

a, b

aur

\overset{u}{^}

aur

\overset{v}{^}

ke beech angle

ψ

ke saath, law of cosines deta hai

∣ A ∣^{2} = a^{2} + b^{2} - 2 ab cos (18 0^{\circ} - ψ) = a^{2} + b^{2} + 2 ab cos ψ

; sirf jab

ψ = 9 0^{\circ}

ho tabhi yeh Pythagoras mein collapse hota hai.

A

\overset{n}{^}

ke along component

∣ A ∣

se bada ho sakta hai

False. Kyunki

A_{\overset{n}{^}} = A cos ϕ

aur

∣ cos ϕ ∣ \leq 1

, ek shadow kabhi bhi use daalne wali cheez se lambi nahi ho sakti. Equality (

= A

) sirf tab hoti hai jab

\overset{n}{^}

exactly

A

ke along point kare (yaani

ϕ = 0

Ek component zero ho sakta hai chahe vector zero na ho

True. Agar

A

\overset{n}{^}

ke perpendicular hai to

ϕ = 9 0^{\circ}

, isliye

A_{\overset{n}{^}} = A cos 9 0^{\circ} = 0

. Vector us direction pe koi shadow nahi daalta jo uske right angles pe ho.

Ek component negative ho sakta hai chahe length kabhi negative nahi hoti

True. Ek component ek signed number hai: yeh record karta hai kitna aur axis ke along kaun si taraf.

A cos ϕ < 0

jab

ϕ > 9 0^{\circ}

ho, matlab shadow axis ke negative side pe padti hai.

Perpendicular axes ke liye, projection parallelogram component ke barabar hoti hai

True. Orthogonality exactly woh condition hai jiske under "shadow" aur "woh piece jo wapas add hoti hai" ek saath milti hain — ek axis ka dusri mein koi leak nahi hota.

Spot the error

θ

angle ke ek incline pe, weight along the slope

W cos θ

hai."

Error: yeh

W sin θ

hai. Jaise figure dikhata hai, vertical weight

W

aur normal

N

(plane ke perpendicular) ke beech ka angle exactly

θ

hai, isliye along-slope part us triangle ka opposite side hai → sine. Sin/cos swap karna sabse common incline mistake hai.

"Oblique axis

\overset{u}{^}

ke along

A

ka component paane ke liye, bas

A \cdot \overset{u}{^}

compute karo."

Error: non-perpendicular axes ke liye projection

A \cdot \overset{u}{^}

mein

\overset{v}{^}

-part se ek "leak" aa jaata hai. Sahi parallelogram component

A = a \overset{u}{^} + b \overset{v}{^}

solve karne se aata hai, na ki ek single dot product se.

θ = tan^{- 1} (A_{y} / A_{x})

kisi bhi vector ki direction seedha deta hai."

Error:

tan

har

18 0^{\circ}

pe repeat karta hai, isliye bare arctan quadrant II aur IV mein fark nahi kar sakta. Aapko

A_{x}, A_{y}

ke signs inspect karne chahiye aur

18 0^{\circ}

tab add karna chahiye jab

A_{x} < 0

ho, warna calculator arrow ki opposite direction return karta hai.

"Ek smooth incline pe ruke huye block ko in-place rehne ke liye slope ke upar

m g

force chahiye."

Error: sirf gravity ka along-slope pull,

m g sin θ

, balance hona chahiye.

m g cos θ

part surface ke push se cancel ho jaata hai.

m g

up-slope supply karna block ko over-push kar dega.

"Kyunki

cos 9 0^{\circ} = 0

hai, ek vertical vector ka koi vertical component nahi hota."

Error: reference ki confusion hai. Ek vertical vector, vertical axis se

0^{\circ}

banata hai, isliye uska vertical component

A cos 0^{\circ} = A

hai (poori length).

cos 9 0^{\circ} = 0

uske horizontal component pe apply hota hai.

A

ko resolve karna aur phir components ko wapas add karna cos aur sin ki rounding ki wajah se slightly different vector deta hai."

Principle mein error: exactly,

A_{x} \overset{x}{^} + A_{y} \overset{y}{^} = A cos θ \overset{x}{^} + A sin θ \overset{y}{^}

A

ko perfectly reconstruct karta hai. Koi bhi difference aapki arithmetic rounding hai, resolution mein koi flaw nahi.

Why questions

Adjacent side angle ke saath cosine kyu use karta hai, sine kyun nahi?

Kyunki right triangle mein definition se

cos θ = adjacent / hypotenuse

hota hai. Cosine literally angle ko touch karne wali direction pe hypotenuse ki shadow measure karta hai. Mnemonic: "COS hugs the Angle."

Projection ke liye dot product sahi tool kyun hai, sirf magnitudes multiply karne ki jagah?

Kyunki projection poochti hai "kitna

A

\overset{n}{^}

ke along point karta hai", aur

A \cdot \overset{n}{^} = A cos ϕ

unke beech ka angle

ϕ

build in karta hai. Plain

A \times 1

yeh ignore karta hai ki woh aligned nahi bhi ho sakte.

Inclined plane pe horizontal–vertical axes ki jagah axes ko tilt kyun karte hain?

Kyunki motion slope ke along hoti hai, isliye tilted axes ek component ko pure driving force (

m g sin θ

) aur dusre ko fully balanced force (

m g cos θ

, normal force

N

se cancel — surface ka perpendicular push) banate hain. Horizontal–vertical dono effects ko dono axes mein smear kar deta.

Ek fixed axis ke along numbers ordinary numbers ki tarah add kyun hote hain, lekin poore vectors nahi?

Jab ek direction fix ho jaati hai, "us mein kitna" ek single signed scalar hai, aur scalars seedhe add hote hain. Poore vectors direction bhi carry karte hain, isliye unhe parallelogram/triangle rule chahiye.

Cos ya sin likhne se pehle angle ki reference direction jaanna zaroori kyun hai?

Kyunki cosine us side ke saath jaata hai jo angle ko touch karti hai. Agar aapko nahi pata ki angle kis axis se measure kiya gaya hai, to aapko nahi pata kaunsi axis "adjacent" hai, aur cos/sin galat components pe ja sakte hain.

Projection

A cos ϕ

kyun hai aur

A tan ϕ

A sin ϕ

kyun nahi?

Cosine along- $\overset{n}{^}$ shadow deta hai (projection triangle ki adjacent side). Sine perpendicular leftover deta, aur tangent koi shadow length hai hi nahi — woh ek side ko dusri se compare karta hai, hypotenuse se nahi.

Edge cases

Kisi vector ka us direction ke along component kya hoga jo uske perpendicular ho?

Exactly zero:

ϕ = 9 0^{\circ}

deta hai

A cos 9 0^{\circ} = 0

. Physically vector us screen pe koi shadow nahi daalta jiske saath woh parallel ho.

Kisi bhi axes pe zero vector ke components kya hote hain?

Dono zero, axes ki har choice pe. Zero vector ki na length hai na direction, isliye yeh har jagah zero shadow daalta hai — ek rare case jahan answer axis-independent hota hai.

θ \to 9 0^{\circ}

hone par

A_{x} = A cos θ

aur

A_{y} = A sin θ

ka kya hoga?

A_{x} \to 0

aur

A_{y} \to A

: vector purely vertical ho jaata hai, sab kuch y-component mein. Yeh limiting behaviour kisi bhi resolution ka quick sanity check hai.

Third quadrant mein ek vector ke liye (

A_{x} < 0, A_{y} < 0

), naive

tan^{- 1} (A_{y} / A_{x})

kya return karta hai?

Do negatives ka ratio positive hota hai, isliye arctan ek first-quadrant angle return karta hai — exact opposite direction. Aapko

18 0^{\circ}

add karna hoga kyunki tangent kisi direction aur uski reverse mein distinguish nahi kar sakta.

Agar

\overset{n}{^}

exactly

A

ke opposite point kare, to

A_{\overset{n}{^}}

kya hoga?

ϕ = 18 0^{\circ}

, isliye

A_{\overset{n}{^}} = A cos 18 0^{\circ} = - A

: poori magnitude lekin negative, yeh signal karta hai ki shadow

\overset{n}{^}

ke negative side pe completely land hoti hai.

Ek horizontal surface (

θ = 0

) pe, weight ke along-slope aur normal components kya hain?

Along-slope

= W sin 0^{\circ} = 0

aur normal

= W cos 0^{\circ} = W

. Koi slope nahi matlab koi driving component nahi, aur surface poora weight support karti hai — flat-ground intuition se match karta hai.

Ek vertical wall (

θ = 9 0^{\circ}

) pe, weight ke woh components kya hain?

Along-slope

= W sin 9 0^{\circ} = W

aur normal

= W cos 9 0^{\circ} = 0

. Gravity poori tarah "slope" ke neeche act karti hai aur wall koi normal support nahi deta — object freely vertically fall kar raha hai.

Connections

Dot product & scalar projection
Unit vectors and Cartesian coordinates
Inclined plane dynamics
Vectors — addition (parallelogram & triangle law)
Projectile motion
Equilibrium of concurrent forces