Dot product — formula, geometric meaning, work calculation: Question bank

1.1.11 · D5 · Physics › Measurement, Vectors & Kinematics › Dot product — formula, geometric meaning, work calculation

True or false — justify

A \cdot B

ek vector hai kyunki tumne do vectors se shuru kiya.

False — dot product collapses to a single scalar mein ho jaata hai; direction bilkul khatam ho jaati hai. Jo operation direction rakhta hai woh cross product hai.

Agar

A \cdot B = 0

toh dono mein se kam se kam ek vector zero vector hona chahiye.

False — do non-zero vectors tab zero dete hain jab woh perpendicular hon (

cos 9 0^{\circ} = 0

). Zero dot product ka matlab hai "koi alignment nahi," naa ki "koi length nahi."

A \cdot A

negative ho sakta hai agar

A

teesre quadrant mein point karta ho.

False —

A \cdot A = ∣ A ∣^{2} cos 0 = ∣ A ∣^{2}

, yeh ek length squared hai, isliye yeh hamesha

\geq 0

hota hai, chahe

A

kisi bhi direction mein point kare.

Order badal dene se,

B \cdot A

, result ka sign change ho sakta hai.

False — dot product commutative hota hai;

cos θ

same rehta hai chahe angle

A

B

ki taraf napo ya

B

A

ki taraf, isliye

A \cdot B = B \cdot A

Motion ke perpendicular force thoda kaam toh karta hi hai kyunki woh ek asli, strong force hai.

False — work hai

F d cos θ

, aur

cos 9 0^{\circ} = 0

, isliye ek perpendicular force exactly zero work karta hai chahe woh kitni bhi badi kyun na ho (dekho Circular motion).

Dono vectors ko do guna lamba karne se unke beech ka

cos θ

unchanged rehta hai.

True — scaling se

∣ A ∣

aur dot product change hota hai, lekin

cos θ = \frac{A \cdot B}{∣ A ∣∣ B ∣}

mein extra factors upar aur neeche cancel ho jaate hain, isliye angle unchanged rehta hai.

A \cdot B

∣ A ∣∣ B ∣

se bada ho sakta hai agar vectors strongly aligned hon.

False —

cos θ

zyada se zyada

+ 1

ho sakta hai, isliye

A \cdot B \leq ∣ A ∣∣ B ∣

hamesha. Isse bada number aana matlab tumne arithmetic mein koi galti ki hai.

Distributivity,

A \cdot (B + C) = A \cdot B + A \cdot C

, sirf tab kaam karti hai jab vectors perpendicular hon.

False — yeh kisi bhi vectors ke liye hold karta hai, kyunki projections add hote hain:

B + C

A

par shadow, dono alag-alag shadows ka sum hota hai.

Spot the error

cos θ = A \cdot B

, toh main dot product plug in karta hoon aur

cos^{- 1}

leta hoon."

Denominator missing hai — sahi relation hai

cos θ = \frac{A \cdot B}{∣ A ∣∣ B ∣}

. Magnitudes se divide karna bhool jaane par ek "

cos θ

" 1 se bada aa sakta hai, jo impossible hai.

"Work force times distance hai, isliye ek

6 0^{\circ}

par rope ke liye

W = (10 N) (6 m) = 60 J

Sirf force ka aligned part hi work karta hai. Tumhe

cos 6 0^{\circ}

zaroor include karna hoga:

W = 60 \cdot 0.5 = 30 J

. Plain

F d

sirf

θ = 0

ka special case hai.

"Friction ne

- 40 J

work kiya — yeh ek error hai, work negative nahi ho sakta."

Yeh error nahi hai — negative work ka matlab hai energy remove ho gayi object se (friction, braking). Minus sign physical hai, jo

cos θ < 0

se aata hai jab force motion ko oppose kare.

\hat{i} \cdot \hat{j} = 1

kyunki dono unit vectors hain jinka length 1 hai."

Magnitudes 1 hain, lekin woh perpendicular hain, isliye

\hat{i} \cdot \hat{j} = (1) (1) cos 9 0^{\circ} = 0

. Matched pairs 1 dete hain; mixed pairs 0 dete hain — yahi component formula ke peeche ka poora trick hai.

A \cdot B = (3) (2) + (4) (- 1) = A_{x} B_{x} - A_{y} B_{y}

Formula hai

A_{x} B_{x} + A_{y} B_{y}

— tum products ko add karte ho; yahan minus

B_{y} = - 1

se aaya, kisi subtraction rule se nahi. Sign component ke andar rehta hai, operation mein kabhi nahi.

"Dot product

A \cdot B

A

aur

B

ke beech ki direction mein point karta hai."

Ek scalar ki koi direction hi nahi hoti. Kuch bhi point karne ko nahi hai — tum ise ek vector operation jaise cross product se confuse kar rahe ho.

Yeh prove kiya ja chuka hai ki dono equal hain: har vector ko perpendicular unit-vector basis mein likho aur distributivity ke saath expand karo, toh har mixed term khatam ho jaata hai, aur projection picture component sum mein convert ho jaati hai (dekho Trigonometry — cosine and components).

A \cdot B

ka sign akela tumhe kyun bata sakta hai ki angle sharp hai ya wide?

Kyunki

∣ A ∣, ∣ B ∣ > 0

, isliye sign purely

cos θ

se aata hai: ==positive for

θ < 9 0^{\circ}

== (partly aligned),

9 0^{\circ}

par zero,

θ > 9 0^{\circ}

par negative (opposing).

Edge cases

Jab

A

zero vector ho toh

A \cdot B

kya hoga?

Zero — zero vector ki na koi length hai na koi direction, isliye kisi bhi cheez par uska shadow kuch nahi hota. (Note: yahan angle

θ

undefined hai, isliye safe rehne ke liye component form use karo.)

Do vectors antiparallel hain (bilkul opposite direction mein,

θ = 18 0^{\circ}

). Unka dot product kya hoga?

Jitna negative ho sakta hai utna:

A \cdot B = ∣ A ∣∣ B ∣ cos 18 0^{\circ} = - ∣ A ∣∣ B ∣

. Yahi sign milta hai jab friction directly motion ko oppose kare.

Ek puck constant speed par circle mein slide karta hai; centripetal force ek poore lap mein kitna work karta hai?

Zero — centripetal force hamesha velocity ke perpendicular point karta hai, isliye har instant par

cos 9 0^{\circ} = 0

hota hai, zero work milta hai aur speed unchanged rehti hai (dekho Circular motion).

Agar

θ

exactly

9 0^{\circ}

ho lekin vectors bahut bade hon, toh kya dot product bada hoga?

Nahi —

cos 9 0^{\circ} = 0

hone par magnitude matter nahi karta; vectors chahe kitne bhi lambe hon, product zero hi hoga. Perpendicularity, size ko beat karta hai.

Maximum dot product

A \cdot B = + ∣ A ∣∣ B ∣

vectors ke baare mein kya batata hai?

Woh parallel aur same direction mein hain (

θ = 0

cos 0 = 1

A

ka har bit

B

ke along hai, isliye shadow poori length ke barabar hai.

Kya

A \cdot B

positive ho sakta hai jabki vectors visibly alag directions mein point karein?

Haan —

0^{\circ}

aur

9 0^{\circ}

ke beech ka koi bhi angle positive (lekin maximum nahi) result deta hai; vectors partly aligned hain, identical nahi. Positive ka sirf matlab hai "sharp angle."

Agar tumhare paas sirf components hain aur koi angle nahi, toh kya tum phir bhi work find kar sakte ho?

Haan —

W = F \cdot d = F_{x} d_{x} + F_{y} d_{y}

directly use karo; component form ko koi angle chahiye hi nahi, isliye yeh aksar faster route hota hai.

Recall One-line self-test

Answers cover karo. Agar tum har "True/False" ko ek reason ke saath justify kar sako (coin-flip se nahi), aur tum bata sako kyun $cos θ$ , $sin θ$ ko yahan beat karta hai, toh yeh topic tumhara hai. Ready ::: Toh cross product par aage badho — woh " $sin θ$ " partner jo ek direction rakhta hai.

Question bank — Dot product — formula, geometric meaning, work calculation

True or false — justify

Spot the error

Why questions

Edge cases