Vector Algebra: Class 12 Maths Chapter 10 Hinglish mein

Arrows jinhe tum add, stretch aur multiply kar sakte ho

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

125 min
1 simulation
73 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

Scalars aur vectors mein farq pehchano, aur vector ko uske magnitude, direction cosines aur direction ratios se describe karo
Zero, unit, coinitial, collinear, equal aur negative vectors pehchano
Triangle aur parallelogram laws se vectors ko add aur subtract karo, aur unki properties use karo
Vector ko scalar se multiply karo, unit vectors nikaalo, aur component form mein vectors ke saath kaam karo
Do points ko jodne wala vector nikaalo, aur segment ko divide karne wale point ka position vector nikaalo
Scalar product se angles nikaalo, perpendicularity test karo aur projections calculate karo
Vector product se perpendicular vectors aur parallelograms aur triangles ke areas nikaalo

NCERT Mathematics Class 12, Chapter 10, pp. 338–376 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

10.1 Introduction

Ek parcel drone Bengaluru ke ek warehouse se udta hai. Poocho ki woh kitna door uda, to ek number jawab de deta hai: 5 km. Lekin poocho ki woh kahan utra, to ek number kaafi nahi hai, kyunki 5 km north-east aur 5 km south-west do alag jagahen hain (Fig. 10.1).

2/2Direction jodne se landing point fix ho jaata hai: yahi vector hai.

Fig. 10.1Sirf distance se end points ka poora circle ban sakta hai; distance ke saath direction ho to ek hi point chuna jaata hai.

Ek start point ke around 5 km radius ka dashed circle bana hai aur ek chhota compass north aur east dikhata hai. Pehle step mein circle par chhe possible landing points hain, har ek par question mark, aur ek radius par 5 km likha hai. Doosre step mein woh points fade ho jaate hain aur ek blue arrow start se circle tak north-east direction mein jaata hai, jis par 5 km NE likha hai.

Kuch quantities ek real number aur unit se poori tarah describe ho jaati hain: ek kaddu ka mass, cricket pitch ki length, Jaipur mein dopahar ka temperature, bus ticket ka price. Aisi quantities ko scalars (अदिश) kehte hain. Kuch quantities ko size aur direction dono chahiye: drone ka displacement, train ki velocity, ankle se pakde gaye kabaddi raider par lagne wala khinchaav. Yeh vectors (सदिश) hain.

Temperature, energy, speed, volume aur price scalars hain; displacement, force, velocity aur momentum vectors hain. Ek quick test: agar kisi measurement ke saath "north ki taraf" jodne par bhi uska matlab banta hai, to woh shayad vector hai.

Is chapter mein hum vectors ko algebra ke objects ki tarah treat karte hain. Hum seekhenge ki unhe add kaise karte hain, numbers se stretch kaise karte hain, aur do vectors ko do alag tarikon se multiply kaise karte hain. Har rule ke saath ek picture hai, aur yahi pictures rules par bharosa karna aasaan banati hain.

Quick check

2 mein se 1

Inmein se kaun si vector quantities hain?

stumps ke beech ke 22 yardswest se 20 km/h se chalti hawaek tarbooz ka 1.5 kg massdarwaze ki taraf 40 N ka push

Saare sahi options chuno

Answer dekho

Answer

(B) west se 20 km/h se chalti hawa
(D) darwaze ki taraf 40 N ka push

Hawa aur push dono ke saath direction hai. 22 yards ki length aur 1.5 kg ka mass ek number se hi poore ho jaate hain, isliye woh scalars hain.

10.2 Kuch Basic Concepts

Vector ko aisi picture chahiye jo size aur sense dono saath mein dikhaye. Ek straight line $l$ par arrowhead lagao, jisse uske saath sirf ek taraf chalna allowed ho: ab humare paas ek directed line hai. A se B tak ka hissa kaat lo aur arrowhead rakho. Jo bachta hai, ek directed line segment, uski ek fixed length bhi hai aur direction bhi, aur yahi information vector mein hoti hai (Fig. 10.2).

Fig. 10.2Ek line, ek directed line aur ek directed line segment: aakhri wale mein length aur direction dono hain.

Teen rows. Pehli row mein ek saadi straight line hai jis par l likha hai. Doosri row mein wahi line hai jiske right end par arrowhead hai, jo ek chuni hui direction dikhata hai. Teesri row mein ek halki line hai jiska A se B tak ka hissa blue mein mota bana hai, B par arrowhead hai, A par initial point aur B par terminal point likha hai, aur upar vector AB label hai.

Definition

Vector

Aisi quantity jiska magnitude aur direction dono ho. Geometry mein, vector ek directed line segment hota hai.

Example $A$ se $B$ tak ka segment, jise $\vec{A B}$ likhte hain.

Pehle naam. A se B tak ke arrow ko $\vec{A B}$ likhte hain; jab end points matter nahi karte, to hum use bas $\vec{a}$ kehte hain. A, jahan se woh shuru hota hai, initial point (tail) hai, aur B terminal point (head) hai. Arrow kitna lamba hai, yaani distance AB, woh uska magnitude hai: $∣ \vec{A B} ∣$ , $∣ \vec{a} ∣$ , ya short mein sirf $a$ .

Kyunki yeh ek length hai, magnitude 0 ho sakta hai lekin kabhi negative nahi. $∣ \vec{a} ∣ = - 3$ jaisi equation ka koi solution hi nahi hai.

Position vector

Har point ka apna ek vector hota hai. Class 11 ke right-handed axes ke origin O ko tail maan lo; P(x, y, z) par khatam hone wala arrow $\vec{O P}$ tab P ka position vector hai, jise aksar $\vec{r}$ kehte hain. Iski length distance OP hai. Aadat se, points A, B, C, … ke position vectors $\vec{a}, \vec{b}, \vec{c}$ , … liye jaate hain.

Position vector ki length

∣ \vec{r} ∣ = \sqrt{x^{2} + y^{2} + z^{2}}

(10.1)

Example

Origin se kitna door?

Ek drone point P(1, 4, 8) par hover kar raha hai, distances sau metres mein hain. Woh origin O se kitna door hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Direction cosines

$\vec{r} = \vec{O P}$ kis taraf point karta hai? Ek poora jawab hai woh angle jo yeh har positive axis ke saath banata hai: x-axis ke saath α, y-axis ke saath β aur z-axis ke saath γ. Yeh direction angles 0 aur π ke beech hote hain. Practice mein hum unke cosines ke saath kaam karte hain, $l = \cos α$ , $m = \cos β$ aur $n = \cos γ$ , jo $\vec{r}$ ke direction cosines hain.

P se x-axis par perpendicular giraao, jo use A par mile. Triangle OAP ka right angle A par hai, uska hypotenuse $O P = r$ hai aur axis ke saath wali side $O A = x$ hai, isliye $\cos α = x / r$ (Fig. 10.3). Baaki do axes par perpendiculars isi tarah do aur right triangles dete hain.

2/4Triangle $O A P$ mein $A$ par right angle hai, isliye $\cos α = x / r$ .

Fig. 10.3Har direction cosine ek right triangle mein adjacent upon hypotenuse hai, jiska hypotenuse

r

hai.NCERT Fig. 10.3 · p. 340

Left mein teen axes O par milte hain aur ek blue arrow r O se P(x, y, z) tak jaata hai, coordinate planes tak halki guide lines ke saath. Right mein, pehle step mein r ki length ka formula dikhaya gaya hai. Doosre step mein axis ka hissa OA highlight hota hai, P se A tak dashed line girti hai, aur right mein asli shape ka right triangle OAP aata hai jiska hypotenuse r, base x, A par right angle, O par angle alpha aur label cos alpha equals x over r hai. Teesra aur chautha step yahi y-axis ke liye B aur beta ke saath, aur z-axis ke liye C aur gamma ke saath karte hain.

Direction cosines

l = \frac{x}{r}, m = \frac{y}{r}, n = \frac{z}{r}

(10.2)

Ulta padho, to (10.2) kehta hai $P = (l r, m r, n r)$ : coordinates cosines ko r se scale karke milte hain. Cosines ko kisi non-zero number se scale karke mili koi bhi triple, $a = k l$ , $b = k m$ , $c = k n$ , vector ke direction ratios ka ek set kehlaati hai. Coordinates $x, y, z$ choice $k = r$ hain.

Teeno cosines ko square karke add karo to hamesha $l^{2} + m^{2} + n^{2} = (x^{2} + y^{2} + z^{2}) / r^{2} = 1$ milta hai. Direction ratios aisa koi promise nahi karte: 2, 4, 4 aur 1, 2, 2 same direction batate hain, aur dono mein se kisi set ke squares ka sum 1 nahi hai.

Example

Point se cosines

P(1, 4, 8) ke position vector ke direction cosines nikaalo, aur woh angle jo yeh z-axis ke saath banata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek point P (2, −3, 6) par hai.

∣ \vec{O P} ∣

nikaalo.

Answer dekho

Answer 7

\sqrt{4 + 9 + 36} = \sqrt{49} = 7

10.3 Vectors ke Types

Kuch tarah ke vectors itni baar aate hain ki unke naam rakh diye gaye hain. List padhte waqt Fig. 10.4 ko saamne rakho.

Zero vector, jise $\vec{0}$ likhte hain, ek hi point par shuru aur khatam hota hai, jaise $\vec{A A}$ . Kisi taraf point karne ko kuch nahi hai, isliye iski koi khaas direction nahi hoti, aur suvidha ke liye hum ise jo direction chahiye woh de sakte hain.
Unit vector koi bhi vector hai jiski length exactly 1 ho; yeh sirf direction record karta hai, aur kuch nahi. Non-zero $\vec{a}$ ke liye, usi taraf point karne wale unit vector ko $\hat{a}$ kehte hain (padho “a hat”).
Coinitial vectors ki tail common hoti hai: sab ek hi point se shuru hote hain.
Collinear vectors sab ek line par ya parallel lines par hote hain. Unki lengths alag ho sakti hain, aur kuch doosron ke ulte point kar sakte hain.
Equal vectors length aur direction dono mein match karte hain; unhe kahan banaya gaya hai isse farq nahi padta. Hum likhte hain $\vec{a} = \vec{b}$ .
Vector ko ulta karne se uska negative milta hai: same length, ulta sense. B se A tak chalna A se B tak chalne ko undo kar deta hai, isliye $\vec{B A} = - \vec{A B}$ .

1/5Grid par bane chhe vectors.

Fig. 10.4Arrows compare karne ke liye grid squares gino: equal vectors same moves karte hain; collinear vectors proportional moves karte hain.NCERT Fig. 10.5 · p. 342

Ek square grid par chhe arrows hain jinke naam p, q, r, s, t aur u hain. Pehle step mein sab dikhte hain. Doosre step mein sirf p aur q bright rehte hain: dono 3 squares right aur 1 upar jaate hain, isliye equal hain. Teesre step mein r bhi judta hai: yeh 6 left aur 2 neeche jaata hai, same line direction mein, isliye p, q aur r collinear hain. Chauthe step mein s aur t bright hain: dono same grid point se shuru hote hain, isliye coinitial hain. Paanchve step mein t aur u bright hain: u 2 left aur 3 neeche jaata hai, bilkul t ka ulta, isliye u, t ka negative hai.

Example 1NCERT Example 1 · p. 341

Displacement banana

Ek survey drone north se 40° west ki direction mein 30 km udta hai. Is displacement ko diagram par dikhao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 10.5Example 1 ka displacement, scale ke saath bana hua.NCERT Fig. 10.4 · p. 341

Point O se guzarti lines north, south, east aur west dikhati hain. Ek blue arrow OD, O se north aur west ke beech, north se 40 degrees door nikalta hai, aur us par 30 km likha hai. Kone mein ek scale bar woh length dikhata hai jo 10 km ke barabar hai.

Example 2NCERT Example 2 · p. 341

Scalar ya vector?

Har measure ko scalars aur vectors mein baanto: (i) 90 minutes, (ii) 250 mL, (iii) 3 kg, (iv) seedha east ki taraf 15 m/s, (v) seedha neeche 9.8 m/s² ka acceleration, (vi) ₹500, (vii) 60 °C, (viii) upar ki taraf 50 N ka pull.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 342

Picture se types padhna

Fig. 10.4 mein kaun se vectors (i) equal, (ii) collinear, (iii) coinitial hain? Kya koi vector kisi doosre ka negative hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\vec{P Q}

aur

\vec{R S}

ki length same hai aur dono same taraf point karte hain, lekin P aur R alag points hain. To yeh do vectors

equal haincoinitial hainek doosre ke negative hainbilkul related nahi hain

Answer dekho

Answer (A) equal hain

Equality ke liye sirf same magnitude aur same direction chahiye; starting points alag ho sakte hain (free vectors). Yeh coinitial nahi hain, kyunki P ≠ R.

10.4 Vectors ka Addition

Ek auto-rickshaw tumhe ghar H se market M le jaata hai, aur phir market se school S. Galiyan kitni bhi ghoomein, tumhari position ka total change H se S tak ka seedha arrow hai. Displacements head to tail add hote hain: $\vec{H M} + \vec{M S} = \vec{H S}$ .

Yahi rule kinhi bhi do vectors ko add karta hai. $\vec{a} + \vec{b}$ nikaalne ke liye $\vec{b}$ ko, bina ghumaye ya stretch kiye, tab tak khiskao jab tak uski tail $\vec{a}$ ke head par na aa jaaye. $\vec{a}$ ki tail se khiskaye gaye $\vec{b}$ ke head tak ka arrow sum hai, jise resultant bhi kehte hain. Yeh triangle law of vector addition hai (Fig. 10.6).

Triangle law

\vec{A B} + \vec{B C} = \vec{A C}

(10.3)

3/5 $\vec{A C} = \vec{a} + \vec{b}$ triangle ko band karta hai.

Fig. 10.6Head to tail: triangle ki teesri side sum hai.NCERT Fig. 10.8 · p. 343

Pehle step mein A se B tak ek blue vector a aur alag se right mein ek orange vector b dikhta hai. Doosre step mein b bina ghoome tab tak khiskta hai jab tak uski tail B par na aa jaaye; ab uska head C hai. Teesre step mein A se C tak ek purple arrow par a plus b likha hai. Chauthe step mein C se wapas A tak ka arrow dashed hai aur label kehta hai AB plus BC plus CA equals zero vector. Paanchve step mein purple sum hat jaata hai; ek dashed red arrow minus b B se neeche D tak jaata hai, aur A se D tak ek purple arrow par a minus b likha hai.

Triangle ka ek chakkar lagane par tum wapas start par aa jaate ho, isliye order mein li gayi teeno sides ka sum zero vector hai: $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$ . Subtraction matlab negative ko add karna: $\vec{a} - \vec{b} = \vec{a} + (- \vec{b})$ .

Parallelogram law

Ek delivery drone ko seedha north 40 km/h se udne ke liye set kiya gaya hai, jabki ek crosswind use 30 km/h se east le jaati hai. Dono velocities same point par same time par act karti hain, isliye hum unhe coinitial arrows ki tarah banate hain. Drone asal mein unke banaye parallelogram ke diagonal ke saath chalta hai: 50 km/h se, north se lagbhag 37° east.

LawVector addition ka parallelogram law

Ek point se do vectors banao aur unka parallelogram poora karo. Us point se shuru hone wala diagonal unka sum hai.

Dono laws ek hi baat kehte hain. Parallelogram OACB mein, $\vec{O A} = \vec{a}$ aur $\vec{O B} = \vec{b}$ lekar, side AC, OB ke equal aur parallel hai, isliye $\vec{A C} = \vec{b}$ . Ab triangle OAC mein triangle law se $\vec{O C} = \vec{a} + \vec{b}$ milta hai (Fig. 10.7).

3/4Diagonal $\vec{O C}$ , $\vec{a} + \vec{b}$ hai.

Fig. 10.7Common tail se guzarne wala diagonal sum hai.NCERT Fig. 10.9 · p. 344

Point O se ek blue vector a, A tak aur ek orange vector b, B tak jaata hai. Doosre step mein A aur B se dashed lines ek shaded parallelogram poora karti hain jiska chautha corner C hai. Teesre step mein O se C tak purple diagonal par a plus b likha hai. Chauthe step mein dashed sides arrows ban jaati hain: A se C tak b aur B se C tak a, jo dikhata hai ki dono raaste, pehle a phir b aur pehle b phir a, C par khatam hote hain.

InteractiveVector addition lab

Dono axes par minus 5 se 5 tak ek square grid, origin se do vectors ke saath: blue mein a, jo 3i + j se shuru hota hai, aur orange mein b, jo i + 2j se shuru hota hai. Unke tips round handles hain jinhe whole-number points tak drag kar sakte ho, ya chaar number boxes se set kar sakte ho. Chaar buttons view chunte hain: triangle law (a ke head se dobara bana b aur purple sum), parallelogram (shaded parallelogram aur uska diagonal), a minus b (a ke head se bana minus b aur purple difference), aur lambda a (lambda ke liye minus 3 se 3 tak slider aur purple multiple). Ek status panel result ko components mein lengths ke saath deta hai. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 10.8

\vec{a}

aur

\vec{b}

ke tips drag karo (ya components type karo). Triangle law, parallelogram,

\vec{a} - \vec{b}

aur

λ \vec{a}

ke beech switch karo, aur components aur lengths padho.

Vector addition ki properties

Property 1Commutative property

Kinhi bhi do vectors ke liye, $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ .

Property 2Associative property

Kinhi bhi teen vectors ke liye, $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$ .

1/2 $\vec{P R} + \vec{R S} = \vec{P S}$ .

Fig. 10.9Teeno vectors ko kaise bhi group karo, sum

\vec{P S}

hi aata hai.NCERT Fig. 10.11 · p. 345

Chaar points P, Q, R, S head to tail jude hain: P se Q tak blue vector a, Q se R tak orange vector b aur R se S tak green vector c, aur P se S tak ek purple arrow par a plus b plus c likha hai. Pehle step mein P se R tak ek dashed red diagonal par a plus b likha hai, yeh pehli grouping hai. Doosre step mein dashed diagonal Q se S tak hai, jis par b plus c likha hai, yeh doosri grouping hai. Dono groupings S tak pahunchti hain.

Kyunki grouping se farq nahi padta, hum bas $\vec{a} + \vec{b} + \vec{c}$ likhte hain. Zero vector add karne se kuch nahi badalta, $\vec{a} + \vec{0} = \vec{a}$ , isliye $\vec{0}$ vector addition ki additive identity hai.

Quick check

4 mein se 1

Triangle XYZ mein,

\vec{X Y} + \vec{Y Z}

barabar hai

\vec{Z X}

\vec{X Z}

\vec{Y X}

\vec{0}

Answer dekho

Answer (B) $\vec{X Z}$

Head to tail: X se shuru karo, Y jao, phir Z. Net move X se Z tak hai.

\vec{Z X}

ki length sahi hai lekin direction galat.

10.5 Vector ka Scalar se Multiplication

Displacement ko double karne ka matlab hai same taraf do guna door jaana; use negate karne ka matlab hai wapas jaana. General mein, real number λ ke liye, vector $λ \vec{a}$ woh $\vec{a}$ hai jo factor |λ| se stretch hua ho, aur λ negative ho to ulta ghoom gaya ho (Fig. 10.10). Dono cases mein yeh $\vec{a}$ ke parallel rehta hai: ek vector aur uske multiples hamesha collinear hote hain.

Scalar multiple ki length

∣ λ \vec{a} ∣ = ∣ λ ∣ ∣ \vec{a} ∣

(10.4)

4/4 $\hat{a} = \frac{1}{2.5} \vec{a}$ ki length 1 hai.

Fig. 10.10λ se multiply karne par |λ| guna stretch hota hai, aur λ < 0 ho to direction ulti ho jaati hai.NCERT Fig. 10.12 · p. 346

Saare arrows ek dashed vertical line se shuru hote hain, units mein ruled background par. Pehle step mein ek blue arrow a right ki taraf point karta hai, length 2.5. Doosre step mein doosra arrow, 2a, right ki taraf, length 5. Teesre step mein ek red arrow, minus 1.5 a, left ki taraf, length 3.75. Chauthe step mein ek chhota purple arrow, a hat, right ki taraf, length 1. Right mein ek column lengths 2.5, 5, 3.75 aur 1 list karta hai.

Teen factors ke naam yaad rakhne laayak hain. λ = −1, $\vec{a}$ ko ulta karke $- \vec{a}$ bana deta hai, jo use cancel karta hai, $\vec{a} + (- \vec{a}) = \vec{0}$ ; isliye $- \vec{a}$ , $\vec{a}$ ka additive inverse hai. λ = 0 har vector ko daba kar $\vec{0}$ bana deta hai, aur koi bhi factor $\vec{0}$ ko lamba nahi kar sakta: $k \vec{0} = \vec{0}$ .

Teesra hai $λ = 1 / ∣ \vec{a} ∣$ , jahan $\vec{a} \neq \vec{0}$ . Yeh $\vec{a}$ ko bina ghumaye exactly length 1 par rescale karta hai, jisse $\hat{a}$ milta hai, yaani $\vec{a}$ ki taraf point karne wala unit vector:

a ke along unit vector

\hat{a} = \frac{1}{∣ \vec{a} ∣} \vec{a}

(10.5)

10.5.1 Vector ke components

Har positive axis ke along ek unit vector rakho: $\hat{i}$ , $\hat{j}$ aur $\hat{k}$ , O se (1, 0, 0), (0, 1, 0) aur (0, 0, 1) tak jaate hain. Space ka har vector in teeno se banaya ja sakta hai.

Point P(x, y, z) ke liye, A ko (x, 0, 0) aur Q ko (x, y, 0) maano, jo xy-plane ka woh point hai jo P ke theek neeche ya upar hai. Raasta O → A → Q → P axes ke parallel teen moves karta hai: $x \hat{i}$ , phir $y \hat{j}$ , phir $z \hat{k}$ (Fig. 10.11). Triangle law se P ka position vector inka sum hai.

Component form

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

(10.6)

4/4Teesra move: $\vec{Q P} = z \hat{k}$ ; Q par right angle.

Fig. 10.11Axes ke parallel teen moves milkar

\vec{r}

banate hain; do baar Pythagoras se uski length milti hai.NCERT Fig. 10.14 · p. 347

Teen axes O par milte hain. Pehla step x-, y- aur z-axes ke along chhote unit arrows i, j aur k dikhata hai. Doosra step O se x-axis par A tak ek blue arrow x i jodta hai. Teesra step A se Q tak y-axis ke parallel ek orange arrow y j, ek dashed line OQ, A par right-angle mark aur formula OQ squared equals x squared plus y squared jodta hai. Chautha step Q se upar P tak ek green arrow z k, O se P tak ek purple arrow r, Q par right-angle mark aur formula OP squared equals OQ squared plus z squared jodta hai.

Is tarah likha vector component form mein hota hai. Teen numbers x, y, z, uske scalar components, batate hain ki har axis ke along kitna jaana hai; teeno moves khud, $x \hat{i}$ , $y \hat{j}$ aur $z \hat{k}$ , uske vector components hain (kuch books inhe rectangular components kehti hain).

Iski length Pythagoras se aati hai, do baar use karke. Triangle OAQ ka right angle A par hai, isliye $O Q^{2} = x^{2} + y^{2}$ . Triangle OQP ka right angle Q par hai, isliye $O P^{2} = O Q^{2} + z^{2} = x^{2} + y^{2} + z^{2}$ , jo phir se (10.1) hai.

Components mein, vectors ka algebra matching entries par arithmetic ban jaata hai. Maano $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ aur $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ .

Sum: matching components add karo, to $\vec{a} + \vec{b}$ ke components $a_{1} + b_{1}$ , $a_{2} + b_{2}$ aur $a_{3} + b_{3}$ hain.
Difference: matching components subtract karo: $a_{1} - b_{1}$ , $a_{2} - b_{2}$ aur $a_{3} - b_{3}$ .
Equality: $\vec{a} = \vec{b}$ theek tab hota hai jab $a_{1} = b_{1}$ , $a_{2} = b_{2}$ aur $a_{3} = b_{3}$ .
Scalar multiple: $λ \vec{a}$ ke components $λ a_{1}$ , $λ a_{2}$ aur $λ a_{3}$ hain.

PropertyDistributive laws

Vectors $\vec{a}, \vec{b}$ aur scalars $k, m$ ke liye: (i) $k \vec{a} + m \vec{a} = (k + m) \vec{a}$ ; (ii) $k (m \vec{a}) = (k m) \vec{a}$ ; (iii) $k (\vec{a} + \vec{b}) = k \vec{a} + k \vec{b}$ .

Example 4NCERT Example 4 · p. 349

Equal vectors

p, q aur r aise nikaalo ki $2 p \hat{i} + (q - 3) \hat{j} + 5 \hat{k}$ aur $6 \hat{i} + 2 \hat{j} + (r + 1) \hat{k}$ equal hon.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 350

Same length, alag vectors

Maano $\vec{a} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k}$ aur $\vec{b} = - 8 \hat{i} + \hat{j} + 4 \hat{k}$ . Kya $∣ \vec{a} ∣ = ∣ \vec{b} ∣$ ? Kya vectors equal hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6NCERT Example 6 · p. 350

Ek unit vector

Kaun sa unit vector $\vec{a} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ ki taraf point karta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 350

Di gayi length ka vector

$2 \hat{i} + \hat{j} - 2 \hat{k}$ ki direction mein magnitude 12 ka vector nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 350

Sum ke along unit vector

$\vec{a} = 5 \hat{i} + \hat{j} - 3 \hat{k}$ aur $\vec{b} = - 3 \hat{i} + \hat{j} + 4 \hat{k}$ add karo, phir woh length 1 ka vector do jo unke sum ke along point karta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 351

Ratios aur cosines

$\vec{v} = 2 \hat{i} - 6 \hat{j} + 3 \hat{k}$ ke direction ratios likho aur uske direction cosines nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\vec{a}

ke comparison mein, vector

- 3 \vec{a}

3 guna lamba hai, same direction3 guna lamba hai, ulti directionek tihaai lamba hai, ulti directionutna hi lamba hai, ulti direction

Answer dekho

Answer (B) 3 guna lamba hai, ulti direction

∣ - 3 \vec{a} ∣ = 3 ∣ \vec{a} ∣

, aur negative factor direction ulti kar deta hai. Ek tihaai lamba

- \frac{1}{3} \vec{a}

hota.

10.5.2 Do points ko jodne wala vector

Maano $P_{1} (x_{1}, y_{1}, z_{1})$ aur $P_{2} (x_{2}, y_{2}, z_{2})$ do points hain. Triangle $O P_{1} P_{2}$ mein triangle law kehta hai $\vec{O P_{1}} + \vec{P_{1} P_{2}} = \vec{O P_{2}}$ (Fig. 10.12). Isliye $P_{1}$ se $P_{2}$ tak ka vector hai "head ka position vector minus tail ka position vector".

Fig. 10.12

\vec{P_{1} P_{2}} = \vec{O P_{2}} - \vec{O P_{1}}

: head minus tail.NCERT Fig. 10.15 · p. 351

Origin O se ek blue arrow P1 tak aur ek orange arrow P2 tak jaata hai. Ek purple arrow P1 ko P2 se jodta hai. Picture dikhati hai OP1 plus P1P2 equals OP2, isliye P1P2, OP2 minus OP1 hai.

Do points ko jodne wala vector

\begin{aligned} \vec{P_{1} P_{2}} & = (x_{2} - x_{1}) \hat{i} \\ + (y_{2} - y_{1}) \hat{j} \\ + (z_{2} - z_{1}) \hat{k} \end{aligned}

(10.7)

Iska magnitude distance $P_{1} P_{2}$ hai:

\begin{aligned} ∣ \vec{P_{1} P_{2}} ∣^{2} & = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \\ + (z_{2} - z_{1})^{2} \end{aligned}

(10.8)

Example 10NCERT Example 10 · p. 352

Head minus tail

A(4, −1, 2) se B(1, 3, −10) tak ka vector aur uski length nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

10.5.3 Section formula

Maano P aur Q ke position vectors $\vec{a}$ aur $\vec{b}$ hain, aur R line PQ par ek point hai. Agar R, P aur Q ke beech hai aur $\frac{P R}{R Q} = \frac{m}{n}$ (m, n > 0), to hum kehte hain ki R, PQ ko m : n ratio mein internally divide karta hai. Agar R line par segment ke bahar hai aur $\frac{P R}{Q R} = \frac{m}{n}$ , to woh PQ ko us ratio mein externally divide karta hai (Fig. 10.13).

1/2R, P aur Q ke beech, $P R : R Q = m : n$ ke saath.

Fig. 10.13Internal division mein R, P aur Q ke beech hota hai; external division mein R line par segment ke bahar hota hai.NCERT Fig. 10.16 · p. 352

Origin O se ek blue vector a, P tak aur ek orange vector b, Q tak jaata hai, aur P aur Q jude hain. Pehle step mein R, P aur Q ke beech hai, P se do-tihaai raaste par; segment ke upar braces PR ko m aur RQ ko n mark karte hain, aur ek purple vector r, O se R tak jaata hai. Doosre step mein R badhayi gayi line par Q ke aage hai; upar ek brace PR ko m aur neeche ek brace QR ko n mark karta hai, aur purple vector r, O se is naye R tak jaata hai.

DerivationDivide karne wale point ka position vector

Maano R, PQ ko m : n ratio mein internally divide karta hai aur uska position vector $\vec{r}$ hai. Head minus tail se, $\vec{P R} = \vec{r} - \vec{a}$ aur $\vec{R Q} = \vec{b} - \vec{r}$ .

Yeh dono vectors P se Q ki taraf point karte hain, aur unki lengths m : n ratio mein hain. Isliye $\vec{P R}$ ki n copies, $\vec{R Q}$ ki m copies ke barabar hain:

\begin{aligned} n (\vec{r} - \vec{a}) \\ = m (\vec{b} - \vec{r}) \end{aligned}

$\vec{r}$ wale terms ikatthe karo: $(m + n) \vec{r} = m \vec{b} + n \vec{a}$ . $m + n$ se divide karne par internal division ka section formula milta hai.

Internal division

\vec{r} = \frac{m \vec{b} + n \vec{a}}{m + n}

(10.9)

External division (m ≠ n) ke liye, $\vec{P R}$ aur $\vec{Q R}$ same taraf point karte hain aur $n \vec{P R} = m \vec{Q R}$ . $n (\vec{r} - \vec{a}) = m (\vec{r} - \vec{b})$ ko isi tarah solve karne par doosra formula milta hai.

External division

\vec{r} = \frac{m \vec{b} - n \vec{a}}{m - n}

(10.10)

Midpoint ke liye, (10.9) mein m = n rakho:

Midpoint

\vec{r} = \frac{\vec{a} + \vec{b}}{2}

(10.11)

SimulationR, PQ ko kahan divide karta hai?

Dono axes par minus 10 se 10 tak ek coordinate plane, do fixed points ke saath, P (−4, −5) par aur Q (2, 3) par, aur ek draggable point R jo midpoint (−1, −1) se shuru hota hai. R hilne par distances PR aur RQ bante aur measure hote hain, taaki learner check kar sake ki R, PQ ko kis ratio mein divide karta hai, internally ya externally.

Fig. 10.14P(−4, −5) aur Q(2, 3) fixed hain, 10 units door. R ko drag karo aur distances PR aur RQ compare karo.

Try karo

(10.9) se woh point predict karo jo PQ ko 2 : 3 ratio mein internally divide karta hai. R ko wahan drag karo. Kya PR aur RQ, 4 aur 6 dikhate hain?
R ko (5, 7) par drag karo. Yeh segment ke bahar hai. Yeh PQ ko kis ratio mein externally divide karta hai?
PQ ko 1 : 3 ratio mein externally divide karne wala point predict karo, phir check karne ke liye R ko drag karo.

Example 11NCERT Example 11 · p. 353

Dono tarah se divide karna

Points P aur Q ke position vectors $\vec{a} = 2 \hat{i} + \hat{j} - 3 \hat{k}$ aur $\vec{b} = 7 \hat{i} - 4 \hat{j} + 2 \hat{k}$ hain. PQ ko 2 : 3 ratio mein divide karne wale point ka position vector nikaalo, (i) internally, (ii) externally.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 353

Ek right-angled triangle

Dikhao ki corners A(2, 0, 1), B(3, 2, 3) aur C(5, 3, 1) wale triangle mein ek right angle hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

- 4 \hat{i} + 4 \hat{j} + 2 \hat{k}

ka magnitude nikaalo.

Answer dekho

Answer 6

\sqrt{16 + 16 + 4} = \sqrt{36} = 6

Practice · 19 questions

Practice 10.2: answers ke saath

1
Quick answerEasyNCERT Exercise 10.2, Q1
Har vector ka magnitude nikaalo.
1. $\vec{p} = 2 \hat{i} - \hat{j} + 2 \hat{k}$
  Answer dekho
  Answer 3
  $\sqrt{4 + 1 + 4} = 3$ .
2. $\vec{q} = 6 \hat{i} - 2 \hat{j} + 3 \hat{k}$
  Answer dekho
  Answer 7
  $\sqrt{36 + 4 + 9} = 7$ .
3. $\vec{s} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} - \frac{1}{\sqrt{2}} \hat{k}$
  Answer dekho
  Answer 1
  $\sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = 1$ : ek unit vector.
Short answerEasyNCERT Exercise 10.2, Q2
Do aise unequal vectors ka example do jinki lengths equal hon.
Solution dikhao
AnswerJaise $3 \hat{i} + 4 \hat{j}$ aur $5 \hat{k}$ .
$∣ 3 \hat{i} + 4 \hat{j} ∣ = \sqrt{9 + 16} = 5$ aur $∣ 5 \hat{k} ∣ = 5$ . Vectors alag directions mein point karte hain, isliye yeh same magnitude wale alag vectors hain.
Short answerEasyNCERT Exercise 10.2, Q3
Do aise unequal vectors ka example do jo bilkul same direction mein point karein.
Solution dikhao
AnswerJaise $2 \hat{i} - \hat{j} + \hat{k}$ aur $6 \hat{i} - 3 \hat{j} + 3 \hat{k}$ .
Doosra, pehle ka 3 guna hai. Positive scalar se multiply karne par direction wahi rehti hai aur length badal jaati hai ( $\sqrt{6}$ se $3 \sqrt{6}$ ), isliye vectors alag hain.
4
Quick answerEasyNCERT Exercise 10.2, Q4
x aur y aise nikaalo ki $(x + 2) \hat{i} - 4 \hat{j} = 5 \hat{i} + 2 y \hat{j}$ .
1. x = ?
  Answer dekho
  Answer 3
  $x + 2 = 5$ .
2. y = ?
  Answer dekho
  Answer -2
  $2 y = - 4$ .
Short answerEasyNCERT Exercise 10.2, Q5
Ek vector (−3, 4) se shuru hota hai aur (5, −2) par khatam hota hai. Uske scalar aur vector components nikaalo.
Solution dikhao
AnswerScalar: 8 aur −6. Vector: $8 \hat{i}$ aur $- 6 \hat{j}$ .
Vector hai $(5 + 3) \hat{i} + (- 2 - 4) \hat{j} = 8 \hat{i} - 6 \hat{j}$ .
Iske scalar components 8 aur −6 hain; iske vector components $8 \hat{i}$ aur $- 6 \hat{j}$ hain.
Short answerEasyNCERT Exercise 10.2, Q6
$\vec{a} = 3 \hat{i} + \hat{j} - 4 \hat{k}$ , $\vec{b} = - \hat{i} + 5 \hat{j} + 2 \hat{k}$ aur $\vec{c} = - 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ ka sum nikaalo.
Solution dikhao
Answer $3 \hat{j} + 4 \hat{k}$
$\hat{i}$ : $3 - 1 - 2 = 0$ . $\hat{j}$ : $1 + 5 - 3 = 3$ . $\hat{k}$ : $- 4 + 2 + 6 = 4$ .
To sum $3 \hat{j} + 4 \hat{k}$ hai, length 5 ka vector.
Short answerEasyNCERT Exercise 10.2, Q7
Kaun sa unit vector $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ ki taraf point karta hai?
Solution dikhao
Answer $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$
$∣ \vec{a} ∣ = \sqrt{4 + 4 + 1} = 3$ , isliye $\hat{a} = \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}$ .
Short answerEasyNCERT Exercise 10.2, Q8
P aur Q points (−1, 0, 2) aur (1, 6, −1) hain. $\vec{P Q}$ ke along unit vector do.
Solution dikhao
Answer $\frac{1}{7} (2 \hat{i} + 6 \hat{j} - 3 \hat{k})$
$\vec{P Q} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$ aur $∣ \vec{P Q} ∣ = \sqrt{4 + 36 + 9} = 7$ .
Unit vector $\frac{2}{7} \hat{i} + \frac{6}{7} \hat{j} - \frac{3}{7} \hat{k}$ hai.
Short answerMediumNCERT Exercise 10.2, Q9
$\vec{a} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ aur $\vec{b} = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ ke liye, $\vec{a} + \vec{b}$ ke along unit vector do.
Solution dikhao
Answer $\frac{1}{\sqrt{6}} (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{a} + \vec{b} = 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$ , length $\sqrt{24} = 2 \sqrt{6}$ .
Divide karne par: $\frac{1}{2 \sqrt{6}} (2 \hat{i} + 4 \hat{j} + 2 \hat{k}) = \frac{1}{\sqrt{6}} (\hat{i} + 2 \hat{j} + \hat{k})$ .
Short answerEasyNCERT Exercise 10.2, Q10
$4 \hat{i} + 4 \hat{j} - 7 \hat{k}$ ki direction mein magnitude 18 ka vector nikaalo.
Solution dikhao
Answer $8 \hat{i} + 8 \hat{j} - 14 \hat{k}$
$∣ 4 \hat{i} + 4 \hat{j} - 7 \hat{k} ∣ = \sqrt{16 + 16 + 49} = 9$ .
Required vector $\frac{18}{9} (4 \hat{i} + 4 \hat{j} - 7 \hat{k}) = 8 \hat{i} + 8 \hat{j} - 14 \hat{k}$ hai.
Quick answerEasyNCERT Exercise 10.2, Q11
Vectors $- 3 \hat{i} + \hat{j} + 5 \hat{k}$ aur $6 \hat{i} - 2 \hat{j} - 10 \hat{k}$ collinear hain: doosra, pehle ka λ guna hai. λ nikaalo.
Answer dekho
Answer -2
$\frac{6}{- 3} = \frac{- 2}{1} = \frac{- 10}{5} = - 2$ . Teeno ratios match karte hain, aur yahi collinear test hai.
Short answerEasyNCERT Exercise 10.2, Q12
$\hat{i} - 2 \hat{j} + 2 \hat{k}$ ke direction cosines nikaalo.
Solution dikhao
Answer $\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$
Length $\sqrt{1 + 4 + 4} = 3$ hai, isliye $l, m, n = \frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$ .
Short answerMediumNCERT Exercise 10.2, Q13
A(3, −1, 4) se B(−3, 2, 2) tak ke vector ke direction cosines nikaalo.
Solution dikhao
Answer $- \frac{6}{7}, \frac{3}{7}, - \frac{2}{7}$
$\vec{A B} = - 6 \hat{i} + 3 \hat{j} - 2 \hat{k}$ , length $\sqrt{36 + 9 + 4} = 7$ .
Isliye $l, m, n = - \frac{6}{7}, \frac{3}{7}, - \frac{2}{7}$ .
14
Short answerMediumNCERT Exercise 10.2, Q14
Axes ke saath angles.
1. Dikhao ki $2 \hat{i} + 2 \hat{j} + 2 \hat{k}$ , OX, OY aur OZ ke saath equal angles banata hai, aur woh angle nikaalo.
  Solution dikhao
  Answer $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$
  Length $\sqrt{12} = 2 \sqrt{3}$ hai, isliye har direction cosine $\frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}$ hai.
  Equal cosines equal angles dete hain, har ek $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$ .
2. Kya $\hat{i} - \hat{j} + \hat{k}$ teeno axes ki positive directions ke saath equal angles banata hai?
  Solution dikhao
  AnswerNahi
  Iske direction cosines $\frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ hain. OY ke saath angle obtuse hai (lagbhag ${125.3}^{\circ}$ ), jabki baaki do lagbhag ${54.7}^{\circ}$ hain.
15
Short answerMediumNCERT Exercise 10.2, Q15
P aur Q ke position vectors $3 \hat{i} - \hat{j} + 2 \hat{k}$ aur $- \hat{i} + 3 \hat{j} + 6 \hat{k}$ hain. PQ ko 3 : 1 ratio mein divide karne wale point R ka position vector nikaalo, (a) internally, (b) externally.
1. internally
  Solution dikhao
  Answer $2 \hat{j} + 5 \hat{k}$
  $\vec{r} = \frac{3 \vec{q} + \vec{p}}{4} = \frac{1}{4} (0 \hat{i} + 8 \hat{j} + 20 \hat{k}) = 2 \hat{j} + 5 \hat{k}$ .
2. externally
  Solution dikhao
  Answer $- 3 \hat{i} + 5 \hat{j} + 8 \hat{k}$
  $\vec{r} = \frac{3 \vec{q} - \vec{p}}{2} = \frac{1}{2} (- 6 \hat{i} + 10 \hat{j} + 16 \hat{k}) = - 3 \hat{i} + 5 \hat{j} + 8 \hat{k}$ .
Short answerEasyNCERT Exercise 10.2, Q16
P(−3, 5, 1) aur Q(7, −1, 4) ko jodne wale segment ke midpoint ka position vector nikaalo.
Solution dikhao
Answer $2 \hat{i} + 2 \hat{j} + \frac{5}{2} \hat{k}$
$\frac{1}{2} [(- 3 + 7) \hat{i} + (5 - 1) \hat{j} + (1 + 4) \hat{k}] = 2 \hat{i} + 2 \hat{j} + \frac{5}{2} \hat{k}$ .
ProofMediumNCERT Exercise 10.2, Q17
A, B, C ke position vectors $4 \hat{i} + \hat{k}$ , $5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ aur $9 \hat{i} + 2 \hat{j} + \hat{k}$ hain. Dikhao ki ABC ek right-angled triangle hai.
Solution dikhao
AnswerB par right-angled.
$\vec{A B} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , $\vec{B C} = 4 \hat{i} - 2 \hat{k}$ aur $\vec{A C} = 5 \hat{i} + 2 \hat{j}$ .
$A B^{2} = 9$ , $B C^{2} = 20$ aur $A C^{2} = 29$ .
Kyunki $9 + 20 = 29$ , B par angle (AC ke saamne) right angle hai.
Quick answerEasyNCERT Exercise 10.2, Q18
Triangle PQR mein kaun sa statement false hai?
$\vec{P Q} + \vec{Q R} = \vec{P R}$ $\vec{Q R} + \vec{R P} + \vec{P Q} = \vec{0}$ $\vec{P Q} - \vec{R Q} = \vec{P R}$ $\vec{P R} + \vec{R Q} = \vec{Q P}$
Answer dekho
Answer (D) $\vec{P R} + \vec{R Q} = \vec{Q P}$
P→R→Q, Q par khatam hota hai, isliye $\vec{P R} + \vec{R Q} = \vec{P Q}$ , $\vec{Q P}$ nahi. C sahi hai kyunki $- \vec{R Q} = \vec{Q R}$ .
19
Quick answerMediumNCERT Exercise 10.2, Q19
$\vec{a}$ aur $\vec{b}$ non-zero collinear vectors hain. Kaun se statements zaroor sahi hain?
kisi scalar λ ke liye $\vec{b} = λ \vec{a}$ $∣ \vec{a} ∣ = ∣ \vec{b} ∣$ unke components proportional hainwoh same taraf point karte hain
Saare sahi options chuno
Answer dekho
Answer
(A) kisi scalar λ ke liye $\vec{b} = λ \vec{a}$
(C) unke components proportional hain
Collinearity se $\vec{b} = λ \vec{a}$ milta hai, isliye components proportional hain. λ ko kuch fix nahi karta: yeh negative ho sakta hai (ulti directions) ya $∣ λ ∣ \neq 1$ ho sakta hai (alag lengths).

10.6 Do Vectors ka Product

Chalti cart par force jo work karta hai, woh sirf force ke us hisse par depend karta hai jo motion ke along hai. Spanner ka turning effect push ke us hisse par depend karta hai jo handle ke across hai. Yeh do zarooraton se vectors ko multiply karne ke do alag tarike nikalte hain: scalar (ya dot) product, jiski value ek number hai, aur vector (ya cross) product, jiski value ek vector hai.

10.6.1 Do vectors ka scalar (ya dot) product

Definition

Scalar (dot) product

$\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ$ , jahan $0 \leq θ \leq π$ non-zero vectors $\vec{a}$ aur $\vec{b}$ ke beech ka angle hai. Jab inmein se ek $\vec{0}$ ho, to koi angle hota hi nahi, aur product ko 0 define kiya jaata hai.

Scalar product

\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ

(10.12)

Angle θ common point se bane do vectors ke beech naapa jaata hai, isliye yeh kabhi π se zyada nahi hota. Magnitudes positive hain, isliye $\vec{a} \cdot \vec{b}$ ka sign wahi hai jo cos θ ka (Fig. 10.15).

1/3cos θ > 0, isliye $\vec{a} \cdot \vec{b} > 0$ .

Fig. 10.15Acute:

\vec{a} \cdot \vec{b} > 0

. Right angle:

\vec{a} \cdot \vec{b} = 0

. Obtuse:

\vec{a} \cdot \vec{b} < 0

.NCERT Fig. 10.19 · p. 356

Ek blue vector a, O se right ki taraf ek line ke along point karta hai jo O ke peeche dashed hai. Ek orange vector b, O se shuru hota hai. Pehle step mein b, a ke saath acute angle banata hai; a ki line par uski shadow a ke along padti hai aur plus sign ke saath green shaded hai, aur label kehta hai a dot b greater than 0. Doosre step mein b right angle par hai, square corner se mark kiya, koi shadow nahi hai, aur label kehta hai a dot b equals 0. Teesre step mein angle obtuse hai; shadow O ke peeche girti hai, minus sign ke saath red hatched hai, aur label kehta hai a dot b less than 0.

Definition ke kuch nateeje, jo yaad hone chahiye:

$\vec{a} \cdot \vec{b}$ ek real number hai, vector nahi.
Non-zero vectors ke liye, $\vec{a} \cdot \vec{b} = 0$ theek tab hota hai jab $\vec{a}$ aur $\vec{b}$ perpendicular hon.
Agar θ = 0, to $\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣$ . Khaas taur par $\vec{a} \cdot \vec{a} = ∣ \vec{a} ∣^{2}$ .
Agar θ = π, to $\vec{a} \cdot \vec{b} = - ∣ \vec{a} ∣ ∣ \vec{b} ∣$ . Khaas taur par $\vec{a} \cdot (- \vec{a}) = - ∣ \vec{a} ∣^{2}$ .
Axis unit vectors ke liye, $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ , jabki $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ .
Order se farq nahi padta: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ , kyunki angle aur uska cosine dono tarah same hain.

Definition ko rearrange karne par do non-zero vectors ke beech ka angle milta hai:

Do vectors ke beech ka angle

\cos θ = \frac{\vec{a} \cdot \vec{b}}{∣ \vec{a} ∣ ∣ \vec{b} ∣}

(10.13)

PropertyScalar product ke do rules

(1) Distributivity: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$ . (2) Scalars aaram se idhar-udhar ho sakte hain: $(λ \vec{a}) \cdot \vec{b} = λ (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (λ \vec{b})$ .

DerivationComponents se scalar product

Maano $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ aur $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ . Do rules se $\vec{a} \cdot \vec{b}$ ko expand karne par nau terms milte hain, har ek koi number times $\hat{i} \cdot \hat{j}$ jaisa product.

$\hat{i} \cdot \hat{j}$ jaise chhe mixed products 0 hain, aur $\hat{i} \cdot \hat{i}$ jaise teen squares 1 hain. Sirf teen terms bachte hain.

Components mein scalar product

\begin{aligned} \vec{a} \cdot \vec{b} = a_{1} b_{1} \\ + a_{2} b_{2} + a_{3} b_{3} \end{aligned}

(10.14)

InteractiveVector product lab

Dono axes par minus 5 se 5 tak ek square grid, origin se do vectors ke saath: blue mein a, jo 4i + j se shuru hota hai, aur orange mein b, jo i + 3j se shuru hota hai. Unke tips whole-number points tak drag kiye ja sakte hain ya number boxes se set kiye ja sakte hain. Teen buttons view chunte hain. Dot product angle ka arc aur ek panel dikhata hai jismein components se nikala a dot b, cos theta, degrees mein angle aur yeh ki angle acute, right ya obtuse hai. Projection b ki line banata hai, a ke tip se dashed perpendicular giraata hai aur projection vector ko purple mein uski signed length ke saath dikhata hai. Cross product a aur b par bane parallelogram ko shade karta hai, a se b tak ka turn dikhata hai, aur a cross b ko k ke multiple ki tarah, uska size area ki tarah, aur yeh ki woh screen se bahar ya andar point karta hai, deta hai. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 10.16

\vec{a}

aur

\vec{b}

ke tips drag karo. Dot product

\vec{a} \cdot \vec{b}

aur angle dikhata hai; Projection

\vec{a}

\vec{b}

ki line par giraata hai; Cross product (10.6.3 ke liye) woh parallelogram shade karta hai jiska area

∣ \vec{a} \times \vec{b} ∣

hai.

10.6.2 Line par vector ka projection

Ek directed line l ke along rakhe ruler ke upar ek pencil $\vec{A B}$ pakdo, aur roshni ko seedha neeche ruler par girne do. Pencil ki shadow l par uska projection hai. Agar θ, $\vec{a} = \vec{A B}$ aur l ke beech ka angle hai, to shadow $∣ \vec{a} ∣ ∣ \cos θ ∣$ lambi hai. Woh l ke saath chalti hai ya uske ulte, yeh cos θ ka sign tay karta hai (Fig. 10.17).

Definition

Line par projection

$\vec{a} = \vec{A B}$ ko aise khiskao ki A directed line l par aa jaaye, aur B se l par perpendicular giraao, jo use C par mile. Vector $\vec{A C}$ , l par $\vec{a}$ ka projection vector hai, aur signed number $∣ \vec{a} ∣ \cos θ$ , l par $\vec{a}$ ka projection hai.

ExampleLength 10 ka vector jo l se $60^{\circ}$ par hai, uska projection $10 \cos 60^{\circ} = 5$ hai.

3/3 $\vec{A C}$ projection vector hai; iski signed length $∣ \vec{a} ∣ \cos θ$ hai.

Fig. 10.17

\vec{b}

ki line par

\vec{a}

ka projection

∣ \vec{a} ∣ \cos θ = \vec{a} \cdot \hat{b}

hai.NCERT Fig. 10.20 · p. 357

Ek directed line l left se right jaati hai. Us par ek point A se, ek orange vector b line ke along hai aur ek blue vector a = AB angle theta par upar uthta hai. Doosre step mein B se line par C tak ek dashed perpendicular girta hai, right-angle mark ke saath. Teesre step mein line ke theek neeche ek purple arrow A se C tak jaata hai: yeh projection vector hai, jis par a-length times cos theta likha hai, jo a dot b-hat ke equal hai.

Agar $\hat{p}$ , l ke along unit vector hai, to l par $\vec{a}$ ka projection $\vec{a} \cdot \hat{p}$ hai, kyunki $∣ \hat{p} ∣ = 1$ .
Kisi doosre vector $\vec{b}$ par $\vec{a}$ ke projection mein $\hat{p} = \hat{b}$ lete hain, jisse (10.15) milta hai. Projection vector woh number times $\hat{b}$ hai.
Jab θ = 0 ya θ = π, to $\vec{a}$ , l ke along hai aur khud apna projection vector hai; projection $∣ \vec{a} ∣$ ya $- ∣ \vec{a} ∣$ hai.
Jab θ = π/2, to projection vector $\vec{0}$ hai.

b par a ka projection

\vec{a} \cdot \hat{b} = \frac{\vec{a} \cdot \vec{b}}{∣ \vec{b} ∣}

(10.15)

Example 13NCERT Example 13 · p. 358

Dot product se angle

Vectors $\vec{a}$ aur $\vec{b}$ ke magnitudes 4 aur 3 hain, aur $\vec{a} \cdot \vec{b} = - 6$ . Unke beech ka angle nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 359

Components se angle

$\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ aur $\vec{b} = 6 \hat{i} - 3 \hat{j} + 2 \hat{k}$ ke beech ka angle nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 15NCERT Example 15 · p. 359

Rhombus ke diagonals

$\vec{a} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ aur $\vec{b} = \hat{i} - 3 \hat{j} + 2 \hat{k}$ ke liye dikhao ki $\vec{a}$ aur $\vec{b}$ par bane parallelogram ke diagonals $\vec{a} + \vec{b}$ aur $\vec{a} - \vec{b}$ perpendicular hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 359

Ek projection

$\vec{a} = 4 \hat{i} - \hat{j} + 3 \hat{k}$ ka $\vec{b} = 2 \hat{i} + 2 \hat{j} - \hat{k}$ par projection, aur projection vector nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 359

Sum ki length

Agar $∣ \vec{a} ∣ = 3$ , $∣ \vec{b} ∣ = 5$ aur $\vec{a} \cdot \vec{b} = 7$ , to $∣ \vec{a} + \vec{b} ∣$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 360

Ek unknown length

Agar $∣ \vec{a} ∣ = 2$ aur $(\vec{x} + 2 \vec{a}) \cdot (\vec{x} - 2 \vec{a}) = 9$ , to $∣ \vec{x} ∣$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 360

Cauchy–Schwarz inequality

Dikhao ki kinhi bhi do vectors ke liye $∣ \vec{a} \cdot \vec{b} ∣ \leq ∣ \vec{a} ∣ ∣ \vec{b} ∣$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 360

Triangle inequality

Dikhao ki kinhi bhi do vectors ke liye $∣ \vec{a} + \vec{b} ∣ \leq ∣ \vec{a} ∣ + ∣ \vec{b} ∣$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 361

Ek line par teen points

Dikhao ki A(1, −2, 3), B(3, 1, 2) aur C(7, 7, 0) collinear hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

(2 \hat{i} - \hat{j} + 3 \hat{k}) \cdot (\hat{i} + 4 \hat{j} + \hat{k})

nikaalo.

Answer dekho

Answer 1

2 \cdot 1 + (- 1) \cdot 4 + 3 \cdot 1 = 2 - 4 + 3 = 1

Practice · 18 questions

Practice 10.3: answers ke saath

Short answerEasyNCERT Exercise 10.3, Q1
Magnitudes 2 aur $\sqrt{2}$ wale do vectors jinka scalar product −2 hai, unke beech ka angle nikaalo.
Solution dikhao
Answer $\frac{3 π}{4}$
$\cos θ = \frac{- 2}{2 \sqrt{2}} = - \frac{1}{\sqrt{2}}$ , isliye $θ = \frac{3 π}{4}$ .
Short answerEasyNCERT Exercise 10.3, Q2
$2 \hat{i} + \hat{j} - \hat{k}$ aur $\hat{i} - \hat{j} + 2 \hat{k}$ ke beech ka angle nikaalo.
Solution dikhao
Answer $\cos^{- 1} (- \frac{1}{6}) \approx {99.6}^{\circ}$
Dot product: $2 - 1 - 2 = - 1$ . Dono lengths $\sqrt{6}$ hain.
$\cos θ = \frac{- 1}{6}$ , isliye $θ = \cos^{- 1} (- \frac{1}{6}) \approx {99.6}^{\circ}$ .
Quick answerEasyNCERT Exercise 10.3, Q3
$\hat{i} - \hat{j}$ par $3 \hat{i} + \hat{j}$ ka projection nikaalo.
Answer dekho
Answer 1.4142
$\frac{3 - 1}{\sqrt{2}} = \sqrt{2} \approx 1.414$ .
Short answerEasyNCERT Exercise 10.3, Q4
$\hat{i} + 2 \hat{j} + 2 \hat{k}$ par $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ ka projection nikaalo.
Solution dikhao
Answer $\frac{8}{3}$
$\frac{2 - 6 + 12}{3} = \frac{8}{3}$ .
ProofMediumNCERT Exercise 10.3, Q5
Dikhao ki $\frac{1}{9} (\hat{i} + 4 \hat{j} + 8 \hat{k})$ , $\frac{1}{9} (4 \hat{i} + 7 \hat{j} - 4 \hat{k})$ aur $\frac{1}{9} (8 \hat{i} - 4 \hat{j} + \hat{k})$ unit vectors hain, aur har ek baaki do ke perpendicular hai.
Solution dikhao
AnswerTeeno ki length 1 hai aur teeno dot products 0 hain.
Lengths: $\frac{1}{9} \sqrt{1 + 16 + 64} = \frac{1}{9} \sqrt{16 + 49 + 16} = \frac{1}{9} \sqrt{64 + 16 + 1} = 1$ .
Dot products, factor $\frac{1}{81}$ ko chhod kar: $4 + 28 - 32 = 0$ , $32 - 28 - 4 = 0$ aur $8 - 16 + 8 = 0$ .
Short answerMediumNCERT Exercise 10.3, Q6
Agar $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 27$ aur $∣ \vec{a} ∣ = 2 ∣ \vec{b} ∣$ , to $∣ \vec{a} ∣$ aur $∣ \vec{b} ∣$ nikaalo.
Solution dikhao
Answer $∣ \vec{a} ∣ = 6$ , $∣ \vec{b} ∣ = 3$
$∣ \vec{a} ∣^{2} - ∣ \vec{b} ∣^{2} = 27$ aur $∣ \vec{a} ∣^{2} = 4 ∣ \vec{b} ∣^{2}$ , isliye $3 ∣ \vec{b} ∣^{2} = 27$ .
Isliye $∣ \vec{b} ∣ = 3$ aur $∣ \vec{a} ∣ = 6$ .
Short answerMediumNCERT Exercise 10.3, Q7
$(2 \vec{a} + 3 \vec{b}) \cdot (4 \vec{a} - \vec{b})$ ko expand karo. Phir $∣ \vec{a} ∣ = 1$ , $∣ \vec{b} ∣ = 2$ aur $\vec{a} \cdot \vec{b} = 1$ hone par iski value nikaalo.
Solution dikhao
Answer $8 ∣ \vec{a} ∣^{2} + 10 \vec{a} \cdot \vec{b} - 3 ∣ \vec{b} ∣^{2} = 6$
$8 \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + 12 \vec{b} \cdot \vec{a} - 3 \vec{b} \cdot \vec{b} = 8 ∣ \vec{a} ∣^{2} + 10 \vec{a} \cdot \vec{b} - 3 ∣ \vec{b} ∣^{2}$ .
Di gayi values ke saath: $8 + 10 - 12 = 6$ .
Quick answerEasyNCERT Exercise 10.3, Q8
Do vectors ke magnitudes equal hain, unke beech ka angle $120^{\circ}$ hai, aur unka scalar product −8 hai. Unka common magnitude nikaalo.
Answer dekho
Answer 4
$∣ \vec{a} ∣^{2} \cdot (- \frac{1}{2}) = - 8$ , isliye $∣ \vec{a} ∣^{2} = 16$ aur $∣ \vec{a} ∣ = 4$ .
Quick answerMediumNCERT Exercise 10.3, Q9
$\vec{a}$ ek unit vector hai aur $(2 \vec{x} - \vec{a}) \cdot (2 \vec{x} + \vec{a}) = 15$ . $∣ \vec{x} ∣$ nikaalo.
Answer dekho
Answer 2
$4 ∣ \vec{x} ∣^{2} - ∣ \vec{a} ∣^{2} = 15$ , isliye $4 ∣ \vec{x} ∣^{2} = 16$ aur $∣ \vec{x} ∣ = 2$ .
Quick answerMediumNCERT Exercise 10.3, Q10
Maano $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ aur $\vec{c} = \hat{i} + 3 \hat{j} + \hat{k}$ . Kis λ ke liye $\vec{a} + λ \vec{b}$ , $\vec{c}$ ke perpendicular hai?
Answer dekho
Answer 0.5
$\vec{a} + λ \vec{b} = (1 + 2 λ) \hat{i} + (λ - 2) \hat{j} + (3 - λ) \hat{k}$ . $\vec{c}$ ke saath dot karne par: $(1 + 2 λ) + 3 (λ - 2) + (3 - λ) = 4 λ - 2 = 0$ , isliye $λ = \frac{1}{2}$ .
ProofMediumNCERT Exercise 10.3, Q11
Non-zero vectors $\vec{a}$ aur $\vec{b}$ ke liye dikhao ki $∣ \vec{b} ∣^{2} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ , $\vec{b}$ ke perpendicular hai.
Solution dikhao
Answer $\vec{b}$ ke saath iska dot product 0 hai.
$(∣ \vec{b} ∣^{2} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot \vec{b} = ∣ \vec{b} ∣^{2} (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{b}) ∣ \vec{b} ∣^{2} = 0$ .
Isliye vector $\vec{b}$ ke perpendicular hai (ya $\vec{0}$ hai, jo tab hota hai jab $\vec{a}$ aur $\vec{b}$ collinear hon). $∣ \vec{b} ∣^{2}$ se divide karne par dikhta hai ki yeh $\vec{a}$ minus $\vec{b}$ par uska projection vector hai.
Short answerEasyNCERT Exercise 10.3, Q12
Maano $\vec{u} \cdot \vec{u} = 0$ . $\vec{u}$ kya hai? Tab $\vec{u} \cdot \vec{v} = 0$ tumhe $\vec{v}$ ke baare mein kya batata hai?
Solution dikhao
Answer $\vec{u} = \vec{0}$ ; $\vec{v}$ ke baare mein kuch nahi.
$∣ \vec{u} ∣^{2} = 0$ hone se $\vec{u} = \vec{0}$ hi hoga.
Tab har $\vec{v}$ ke liye $\vec{u} \cdot \vec{v} = 0$ , isliye $\vec{v}$ koi bhi vector ho sakta hai.
Quick answerMediumNCERT Exercise 10.3, Q13
$\vec{a}$ , $\vec{b}$ , $\vec{c}$ har ek ka magnitude 2 hai aur $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ nikaalo.
Answer dekho
Answer -6
$0 = ∣ \vec{a} + \vec{b} + \vec{c} ∣^{2} = 4 + 4 + 4 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ , isliye sum $- 6$ hai. (Har pair $120^{\circ}$ par hai.)
Short answerEasyNCERT Exercise 10.3, Q14
Do vectors dhoondho, dono mein se koi bhi $\vec{0}$ nahi, jinka scalar product 0 ho. Tumhara example is claim ke baare mein kya kehta hai ki $\vec{a} \cdot \vec{b} = 0$ hone par ek vector $\vec{0}$ hona hi chahiye?
Solution dikhao
Answer $\vec{a} = 2 \hat{i} - \hat{j}$ , $\vec{b} = \hat{i} + 2 \hat{j}$
$(2 \hat{i} - \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = 2 - 2 = 0$ , phir bhi koi vector zero nahi hai: woh perpendicular hain.
Short answerMediumNCERT Exercise 10.3, Q15
Ek triangle ke vertices A(3, 1, −1), B(1, 2, 1) aur C(2, 4, 3) hain. ∠ABC nikaalo, yaani $\vec{B A}$ aur $\vec{B C}$ ke beech ka angle.
Solution dikhao
Answer $\cos^{- 1} (- \frac{4}{9}) \approx {116.4}^{\circ}$
$\vec{B A} = 2 \hat{i} - \hat{j} - 2 \hat{k}$ aur $\vec{B C} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , dono ki length 3.
Unka dot product $2 - 2 - 4 = - 4$ hai, isliye $\cos ∠ A B C = - \frac{4}{9}$ aur $∠ A B C \approx {116.4}^{\circ}$ .
ProofEasyNCERT Exercise 10.3, Q16
Dikhao ki A(−1, 4, 2), B(0, 2, 4) aur C(2, −2, 8) collinear hain.
Solution dikhao
Answer $∣ A B ∣ + ∣ B C ∣ = 3 + 6 = 9 = ∣ A C ∣$ .
$\vec{A B} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ (length 3), $\vec{B C} = 2 \hat{i} - 4 \hat{j} + 4 \hat{k}$ (length 6), $\vec{A C} = 3 \hat{i} - 6 \hat{j} + 6 \hat{k}$ (length 9).
Kyunki 3 + 6 = 9, B segment AC par hai, isliye points collinear hain.
ProofMediumNCERT Exercise 10.3, Q17
Scalar product use karke dikhao ki A(2, −1, 3), B(3, 1, 1) aur C(4, 1, 6) ek right-angled triangle banate hain.
Solution dikhao
Answer $\vec{A B} \cdot \vec{A C} = 0$ : A par right angle.
$\vec{A B} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ aur $\vec{A C} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ .
$\vec{A B} \cdot \vec{A C} = 2 + 4 - 6 = 0$ , isliye A par angle right angle hai. (Check: $A B^{2} + A C^{2} = 9 + 17 = 26 = B C^{2}$ .)
Quick answerEasyNCERT Exercise 10.3, Q18
$\vec{a}$ ka magnitude 5 hai. Kis λ ke liye $λ \vec{a}$ ek unit vector hai?
λ = 5λ = ±1/5λ = 1/25λ = −5
Answer dekho
Answer (B) λ = ±1/5
$∣ λ ∣ \cdot 5 = 1$ se $∣ λ ∣ = \frac{1}{5}$ milta hai; dono signs chalte hain, kyunki negative λ sirf direction ulti karta hai.

10.6.3 Do vectors ka vector (ya cross) product

Yaad karo ki coordinate axes ko right-handed kya banata hai. Apne right hand ki ungliyon ko positive x-axis se positive y-axis ki taraf mod lo, to tumhara thumb positive z-axis ki taraf point karta hai. Isi tarah, x se y ki taraf ghumaya gaya right-handed screw z ke along chalta hai.

Definition

Vector (cross) product

Non-zero vectors $\vec{a}$ , $\vec{b}$ ke liye, jinke beech angle θ hai ( $0 \leq θ \leq π$ ), $\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ . Yahan $\hat{n}$ dono ke perpendicular woh unit vector hai jiske liye $\vec{a}$ , $\vec{b}$ , $\hat{n}$ ek right-handed system banate hain. Agar koi bhi vector $\vec{0}$ hai, to hum $\vec{a} \times \vec{b} = \vec{0}$ rakhte hain.

Vector product

\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}

(10.16)

"Right-handed system" ka matlab hai: $\vec{a}$ ko θ se $\vec{b}$ ki taraf ghumane par right-handed screw $\hat{n}$ ke along chalta hai (Fig. 10.18).

2/3 $\vec{a}$ ko $\vec{b}$ ki taraf ghumana: $\vec{a} \times \vec{b}$ normal ke along upar point karta hai.

Fig. 10.18Pehle vector se doosre tak ka turn tay karta hai ki

\hat{n}

kis taraf point karega.NCERT Fig. 10.23 · p. 363

Ek tilted, shaded plane mein ek blue vector a aur ek orange vector b common point O se shuru hote hain, unke beech angle theta hai. Doosre step mein ek curved arrow a se b ki taraf ghoomta hai aur ek purple arrow n-hat O se seedha upar uthta hai, jis par a cross b likha hai. Teesre step mein curved arrow ulti taraf ghoomta hai, b se a ki taraf, aur ek dashed red arrow O se plane ke paar seedha neeche point karta hai, jis par b cross a likha hai.

$\vec{a} \times \vec{b}$ ek vector hai, jo $\vec{a}$ aur $\vec{b}$ dono ke perpendicular hai.
Non-zero vectors ke liye, $\vec{a} \times \vec{b} = \vec{0}$ theek tab hota hai jab $\vec{a}$ aur $\vec{b}$ parallel hon (θ = 0 ya π). Khaas taur par $\vec{a} \times \vec{a} = \vec{0}$ aur $\vec{a} \times (- \vec{a}) = \vec{0}$ .
Agar θ = π/2, to $∣ \vec{a} \times \vec{b} ∣ = ∣ \vec{a} ∣ ∣ \vec{b} ∣$ .
Axis vectors ke liye, $\hat{i} \times \hat{j} = \hat{k}$ , aur cycle mein aage badhte hue, $\hat{j} \times \hat{k} = \hat{i}$ aur phir $\hat{k} \times \hat{i} = \hat{j}$ . $\hat{i} \times \hat{i}$ , $\hat{j} \times \hat{j}$ aur $\hat{k} \times \hat{k}$ mein se har ek $\vec{0}$ hai.
Angle length se padha ja sakta hai: $\sin θ = \frac{∣ \vec{a} \times \vec{b} ∣}{∣ \vec{a} ∣ ∣ \vec{b} ∣}$ .
Order matter karta hai: $\vec{b} \times \vec{a} = - (\vec{a} \times \vec{b})$ . $\vec{b}$ ko $\vec{a}$ ki taraf ghumana ulta turn hai, isliye screw doosri taraf chalta hai.
Isliye $\hat{j} \times \hat{i} = - \hat{k}$ , $\hat{k} \times \hat{j} = - \hat{i}$ aur $\hat{i} \times \hat{k} = - \hat{j}$ (Fig. 10.19).

Fig. 10.19Cycle

\hat{i} \to \hat{j} \to \hat{k}

ke around aage chalne par + milta hai; peeche chalne par −.NCERT Fig. 10.24 · p. 364

Ek ring ke around teen circles hain jin par i, j aur k likha hai, purple arrows clockwise i se j, j se k aur k se i jaate hain, aur beech mein clockwise plus likha hai. Right mein aage wale products list hain: i cross j equals k, j cross k equals i, k cross i equals j. Left mein ulte products minus signs ke saath list hain: j cross i equals minus k, k cross j equals minus i, i cross k equals minus j.

Cross product se areas

Size $∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ$ ki ek picture hai. $\vec{a}$ ko ek parallelogram ke base ke along rakho aur $\vec{b}$ ko uski tirchi side. Height $∣ \vec{b} ∣ \sin θ$ hai, isliye area base times height hai (Fig. 10.20).

Parallelogram ka area

area = ∣ \vec{a} \times \vec{b} ∣

(10.17)

Ek diagonal parallelogram ko do congruent triangles mein kaat deta hai. Isliye ek vertex se bani sides $\vec{a}$ aur $\vec{b}$ wale triangle ka area uska aadha hai.

Triangle ka area

area = \frac{1}{2} ∣ \vec{a} \times \vec{b} ∣

(10.18)

2/3Iski height $∣ \vec{b} ∣ \sin θ$ hai, isliye iska area $∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ$ hai.

Fig. 10.20Area base

∣ \vec{a} ∣

times height

∣ \vec{b} ∣ \sin θ

hai, jo parallelogram ke liye

∣ \vec{a} \times \vec{b} ∣

ke equal hai aur triangle ke liye uska aadha.NCERT Fig. 10.27 · p. 365

Ek shaded parallelogram OACB mein O se A tak ek blue base vector a aur O se B tak ek orange side vector b hai, O par angle theta. Doosre step mein B se base par E tak ek dashed height girti hai, right-angle mark ke saath, jis par length of b times sine theta likha hai. Teesre step mein diagonal AB banta hai aur triangle OAB ko aur gehra shade kiya jaata hai, jis par one half of the length of a cross b likha hai.

Cross product calculate karna

Property 3Vector product ki distributivity

Vectors $\vec{a}, \vec{b}, \vec{c}$ aur ek scalar λ ke liye: (i) $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ ; (ii) $λ (\vec{a} \times \vec{b}) = (λ \vec{a}) \times \vec{b} = \vec{a} \times (λ \vec{b})$ .

DerivationComponents se cross product

$(a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k})$ ko term by term expand karo. Repeated unit vector wale teen terms zero ho jaate hain. Baaki chhe mein se har ek $\hat{i}, \hat{j}, \hat{k}$ mein se kisi ek ka plus ya minus hai, jo Fig. 10.19 ke cycle se padha jaata hai:

\begin{aligned} \vec{a} \times \vec{b} \\ = a_{1} b_{2} \hat{k} - a_{1} b_{3} \hat{j} \\ - a_{2} b_{1} \hat{k} + a_{2} b_{3} \hat{i} \\ + a_{3} b_{1} \hat{j} - a_{3} b_{2} \hat{i} \end{aligned}

Terms ko unit vector ke hisaab se group karne par ek pattern milta hai jo determinant ki tarah yaad rakhna sabse aasaan hai, pehli row ke along expand karke:

Components mein cross product

\begin{aligned} \vec{a} \times \vec{b} \\ = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array} ∣ \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} \\ - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \end{aligned}

(10.19)

Product lab (Fig. 10.16) mein xy-plane ke vectors ke liye ek Cross product mode hai. Wahan $\vec{a} \times \vec{b} = (a_{1} b_{2} - a_{2} b_{1}) \hat{k}$ : jab $\vec{a}$ se $\vec{b}$ ka turn anticlockwise ho to yeh screen se bahar point karta hai, aur iski length shaded area hai.

Example 22NCERT Example 22 · p. 366

Cross product ki length

$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ aur $\vec{b} = \hat{i} + 3 \hat{j} - 2 \hat{k}$ ke liye $∣ \vec{a} \times \vec{b} ∣$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 23NCERT Example 23 · p. 367

Ek unit normal

$\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ aur $\vec{b} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ dono ke perpendicular ek unit vector nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 24NCERT Example 24 · p. 367

Triangle ka area

Ek triangle ke corners A(1, 1, −1), B(3, 0, 1) aur C(2, 3, −1) hain. Iska area kitna hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 25NCERT Example 25 · p. 368

Parallelogram ka area

Adjacent sides $\vec{a} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ aur $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ wale parallelogram ka area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 12 questions

Practice 10.4: answers ke saath

Short answerEasyNCERT Exercise 10.4, Q1
$\vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ aur $\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$ ke liye $∣ \vec{a} \times \vec{b} ∣$ nikaalo.
Solution dikhao
Answer $5 \sqrt{3}$
$\vec{a} \times \vec{b} = (6 - 1) \hat{i} - (4 + 1) \hat{j} + (- 2 - 3) \hat{k} = 5 \hat{i} - 5 \hat{j} - 5 \hat{k}$ .
Iski length $\sqrt{75} = 5 \sqrt{3}$ hai.
Short answerMediumNCERT Exercise 10.4, Q2
$\vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ aur $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ ke liye, $\vec{a} + \vec{b}$ aur $\vec{a} - \vec{b}$ dono ke perpendicular ek unit vector nikaalo.
Solution dikhao
Answer $\pm \frac{1}{\sqrt{42}} (4 \hat{i} - 5 \hat{j} - \hat{k})$
$\vec{a} + \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ aur $\vec{a} - \vec{b} = \hat{i} + 4 \hat{k}$ .
Unka cross product $(8 - 0) \hat{i} - (12 - 2) \hat{j} + (0 - 2) \hat{k} = 8 \hat{i} - 10 \hat{j} - 2 \hat{k}$ hai, length $\sqrt{168} = 2 \sqrt{42}$ .
Isliye unit vectors $\pm \frac{1}{\sqrt{42}} (4 \hat{i} - 5 \hat{j} - \hat{k})$ hain. (Shortcut: $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = - 2 (\vec{a} \times \vec{b})$ .)
Short answerMediumNCERT Exercise 10.4, Q3
Ek unit vector $\vec{a}$ , $\hat{i}$ ke saath angle $\frac{2 π}{3}$ , $\hat{j}$ ke saath angle $\frac{π}{3}$ aur $\hat{k}$ ke saath ek acute angle θ banata hai. θ aur $\vec{a}$ ke components nikaalo.
Solution dikhao
Answer $θ = \frac{π}{4}$ ; $\vec{a} = - \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$
$\cos^{2} θ = 1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}$ . Kyunki θ acute hai, $\cos θ = \frac{1}{\sqrt{2}}$ aur $θ = \frac{π}{4}$ .
Component form mein unit vector apne direction cosines ke equal hota hai, isliye $\vec{a} = - \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ .
ProofMediumNCERT Exercise 10.4, Q4
Dikhao ki $(\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) = - 5 (\vec{a} \times \vec{b})$ .
Solution dikhao
AnswerExpand karo aur simplify karo.
$(\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) = 2 \vec{a} \times \vec{a} - \vec{a} \times \vec{b} + 4 \vec{b} \times \vec{a} - 2 \vec{b} \times \vec{b}$ .
Pehla aur aakhri term $\vec{0}$ hain, aur $4 \vec{b} \times \vec{a} = - 4 \vec{a} \times \vec{b}$ . Total $- 5 (\vec{a} \times \vec{b})$ hai.
Short answerMediumNCERT Exercise 10.4, Q5
λ aur μ nikaalo agar $(3 \hat{i} - 6 \hat{j} + 9 \hat{k}) \times (2 \hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0}$ .
Solution dikhao
Answer $λ = - 4$ , $μ = 6$
Vectors parallel hone chahiye, isliye unke components proportional hain: $\frac{2}{3} = \frac{λ}{- 6} = \frac{μ}{9}$ .
Isliye $λ = - 4$ aur $μ = 6$ . (Sach mein doosra vector pehle ka $\frac{2}{3}$ hai.)
Short answerEasyNCERT Exercise 10.4, Q6
Kya do non-zero vectors ka scalar product aur vector product ek saath zero ho sakta hai? Jab $\vec{a} \cdot \vec{b} = 0$ aur $\vec{a} \times \vec{b} = \vec{0}$ dono saath mein hon, to kya sahi hona chahiye?
Solution dikhao
Answer $\vec{a} = \vec{0}$ ya $\vec{b} = \vec{0}$
Agar dono non-zero hote, to pehli condition unhe perpendicular banati (cos θ = 0) aur doosri parallel (sin θ = 0). Kisi bhi angle ka sine aur cosine dono zero nahi hote.
Isliye kam se kam ek vector zero vector hai.
Short answerMediumNCERT Exercise 10.4, Q7
$\vec{a} = \hat{i} + 2 \hat{j}$ , $\vec{b} = 3 \hat{j} - \hat{k}$ aur $\vec{c} = 2 \hat{i} + \hat{k}$ ke liye distributive law $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ check karo. Phir samjhao ki yahi working sabhi vectors ke liye ise kyun prove karti hai.
Solution dikhao
AnswerDono sides $- \hat{k}$ ke equal hain.
$\vec{b} + \vec{c} = 2 \hat{i} + 3 \hat{j}$ , isliye $\vec{a} \times (\vec{b} + \vec{c}) = (1 \cdot 3 - 2 \cdot 2) \hat{k} = - \hat{k}$ .
$\vec{a} \times \vec{b} = - 2 \hat{i} + \hat{j} + 3 \hat{k}$ aur $\vec{a} \times \vec{c} = 2 \hat{i} - \hat{j} - 4 \hat{k}$ ; inka sum $- \hat{k}$ hai.
General mein, (10.19) ka har component $b_{1}, b_{2}, b_{3}$ mein linear hai: jaise $a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2}) = (a_{2} b_{3} - a_{3} b_{2}) + (a_{2} c_{3} - a_{3} c_{2})$ . Isliye law har component par sahi hai.
Short answerEasyNCERT Exercise 10.4, Q8
Kya yeh sach hai ki $\vec{a} \times \vec{b} = \vec{0}$ sirf tab hota hai jab ek vector $\vec{0}$ ho? Ek example se tay karo.
Solution dikhao
AnswerNahi: jaise $\hat{i} - 2 \hat{j} + 3 \hat{k}$ aur $- 2 \hat{i} + 4 \hat{j} - 6 \hat{k}$ .
Doosra vector pehle ka $- 2$ guna hai, isliye woh parallel hain aur unka cross product $\vec{0}$ hai, halaanki koi bhi $\vec{0}$ nahi hai.
Short answerMediumNCERT Exercise 10.4, Q9
Triangle PQR ka area kya hai, jahan P (2, −1, 1), Q (3, 1, 0) aur R (1, 0, 3) hai?
Solution dikhao
Answer $\frac{\sqrt{35}}{2} \approx 2.96$
$\vec{P Q} = \hat{i} + 2 \hat{j} - \hat{k}$ aur $\vec{P R} = - \hat{i} + \hat{j} + 2 \hat{k}$ .
$\vec{P Q} \times \vec{P R} = (4 + 1) \hat{i} - (2 - 1) \hat{j} + (1 + 2) \hat{k} = 5 \hat{i} - \hat{j} + 3 \hat{k}$ , length $\sqrt{35}$ .
Area $\frac{\sqrt{35}}{2}$ square units hai.
Short answerEasyNCERT Exercise 10.4, Q10
Adjacent sides $3 \hat{i} + \hat{j} - 2 \hat{k}$ aur $\hat{i} - 3 \hat{j} + 4 \hat{k}$ wale parallelogram ka area nikaalo.
Solution dikhao
Answer $10 \sqrt{3} \approx 17.3$
Cross product $(4 - 6) \hat{i} - (12 + 2) \hat{j} + (- 9 - 1) \hat{k} = - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$ hai.
Iski length $\sqrt{4 + 196 + 100} = \sqrt{300} = 10 \sqrt{3}$ square units hai.
Quick answerMediumNCERT Exercise 10.4, Q11
$∣ \vec{a} ∣ = 4$ aur $∣ \vec{b} ∣ = \frac{1}{2}$ . Agar $\vec{a} \times \vec{b}$ ek unit vector hai aur $\vec{a}$ aur $\vec{b}$ ke beech ka angle θ acute hai, to θ hai
$\frac{π}{3}$ $\frac{π}{6}$ $\frac{π}{2}$ $\frac{π}{4}$
Answer dekho
Answer (B) $\frac{π}{6}$
$4 \cdot \frac{1}{2} \sin θ = 1$ se $\sin θ = \frac{1}{2}$ milta hai, aur acute solution $\frac{π}{6}$ hai.
Quick answerMediumNCERT Exercise 10.4, Q12
Ek rectangle ke vertices A(−2, 1, 3), B(2, 1, 3), C(2, −1, 3) aur D(−2, −1, 3) hain. Iska area hai
164 $2 \sqrt{5}$ 8
Answer dekho
Answer (D) 8
$\vec{A B} = 4 \hat{i}$ aur $\vec{A D} = - 2 \hat{j}$ , isliye $\vec{A B} \times \vec{A D} = - 8 \hat{k}$ aur area 8 hai. (Sides 4 aur 2.)

Miscellaneous Examples

Example 26NCERT Example 26 · p. 369

Plane ka har unit vector

Ek flat floor (ise XY-plane maano) par ek robot cleaner kisi bhi direction mein face kar sakta hai. Uski heading ke along unit vector ka formula likho, aur dikhao ki formula har possible heading ko cover karta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1/3 $\vec{r} = \cos θ \hat{i} + \sin θ \hat{j}$ .

Fig. 10.21Plane ke unit vectors ke heads unit circle par hote hain.NCERT Fig. 10.28 · p. 369

Axes O par cross karte hain, radius 1 ka ek dashed circle hai. Pehle step mein O se P tak ek purple unit vector x-axis ke saath angle theta banata hai; x-axis ke along ek blue arrow cos theta i aur upar ki taraf ek orange arrow sin theta j uske components dikhate hain, aur P par cos theta, sin theta likha hai. Doosre step mein aath blue unit vectors alag directions mein point karte hain, har ek circle par khatam hota hai. Teesre step mein circle solid ban jaata hai aur us par x squared plus y squared equals 1 likha hai: plane ke saare unit vectors ke heads.

Example 27NCERT Example 27 · p. 370

Angle jo collinearity dikhata hai

A, B, C, D ke position vectors $2 \hat{i} + \hat{j} - \hat{k}$ , $4 \hat{i} + 2 \hat{k}$ , $3 \hat{j} + \hat{k}$ aur $- 4 \hat{i} + 5 \hat{j} - 5 \hat{k}$ hain. $\vec{A B}$ aur $\vec{C D}$ ke beech ka angle nikaalo, aur isse dikhao ki woh collinear hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 28NCERT Example 28 · p. 370

Har ek baaki do ke sum ke perpendicular

Vectors $\vec{a}, \vec{b}, \vec{c}$ ke magnitudes 2, 3 aur 6 hain, aur har ek baaki do vectors ke sum ke saath right angle par hai. $∣ \vec{a} + \vec{b} + \vec{c} ∣$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 29NCERT Example 29 · p. 371

Dot products ka sum

$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ wale vectors ke liye $∣ \vec{a} ∣ = 5$ , $∣ \vec{b} ∣ = 12$ aur $∣ \vec{c} ∣ = 13$ . $μ = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ ki value nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 30NCERT Example 30 · p. 371

Vector ko doosre ke along aur across todna

$\vec{α} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ aur $\vec{β} = 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ ke saath, $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$ likho jahan ${\vec{β}}_{1}$ , $\vec{α}$ ke parallel hai aur ${\vec{β}}_{2}$ , $\vec{α}$ ke perpendicular hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

XY-plane mein positive x-axis ke saath 60° banane wala unit vector kaun sa hai?

\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}

\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}

\hat{i} + \sqrt{3} \hat{j}

\frac{1}{2} \hat{i} + \frac{1}{2} \hat{j}

Answer dekho

Answer (A) $\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$

\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}

. Option B 30° wala unit vector hai; C sahi taraf point karta hai lekin uski length 2 hai; D unit vector nahi hai.

History seNCERT Historical Note · p. 375

Parallelogram of forces se vector analysis tak

Vector shabd aane se bahut pehle, engineers do pushes ya pulls ko parallelogram rule se jodte the. Flemish engineer Simon Stevin ne 1586 mein statics par apni book De Beghinselen der Weeghconst ("The Principles of the Art of Weighing") mein forces ke liye iska use kiya.

Lagbhag 1800 mein, Caspar Wessel aur Jean-Robert Argand ne complex number a + ib ko plane mein ek arrow ki tarah socha, to complex numbers add karna arrows add karna ban gaya. William Rowan Hamilton space ke arrows ke liye bhi yahi chahte the. 1843 mein unhone quaternions invent kiye, a + bi + cj + dk form ke numbers, aur humein vector shabd diya, jo Latin mein "carrier" (le jaane wala) hai.

1880s mein United States mein Josiah Willard Gibbs aur England mein Oliver Heaviside ne, alag-alag kaam karte hue, quaternion product ko is chapter ke do products mein tod diya, dot aur cross product. Gibbs ne 1881 aur 1884 mein apne students ke liye Elements of Vector Analysis print kiya, aur unka vector analysis aaj physics aur engineering ki bhasha hai.

Recap

Summary aur key points

Scalar ek number hai; vector ka magnitude aur direction hota hai (Section 10.1).
P(x, y, z) ka position vector $\vec{O P} = x \hat{i} + y \hat{j} + z \hat{k}$ hai, length $\sqrt{x^{2} + y^{2} + z^{2}}$ ((10.6)).
Direction cosines $l = \frac{x}{r}$ , $m = \frac{y}{r}$ , $n = \frac{z}{r}$ hain, jahan $l^{2} + m^{2} + n^{2} = 1$ ; koi bhi proportional triple direction ratios ka ek set hai ((10.2)).
Zero, unit, coinitial, collinear, equal aur negative vectors; vectors khiskne ke liye free hain (Section 10.3).
Vectors head to tail (triangle law) ya parallelogram ke diagonal ki tarah add hote hain; triangle ki sides order mein add karne par $\vec{0}$ milta hai (Section 10.4).
Addition commutative aur associative hai, $\vec{0}$ identity hai (Vector addition ki properties).
$λ \vec{a}$ ki length $∣ λ ∣ ∣ \vec{a} ∣$ hai aur λ < 0 hone par direction ulti ho jaati hai; $\hat{a} = \vec{a} / ∣ \vec{a} ∣$ (Section 10.5).
Components mein, entry by entry add, subtract, compare aur scale karo; collinear vectors ke components proportional hote hain (Section 10.5.1).
$P_{1}$ se $P_{2}$ tak ka vector head minus tail hai ((10.7)).
Section formula: internally $\frac{m \vec{b} + n \vec{a}}{m + n}$ , externally $\frac{m \vec{b} - n \vec{a}}{m - n}$ (Derivation).
$\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ ; perpendicular vectors ke liye zero ((10.14)).
$\vec{b}$ par $\vec{a}$ ka projection $\frac{\vec{a} \cdot \vec{b}}{∣ \vec{b} ∣}$ hai ((10.15)).
$\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ , determinant (10.19); iski length parallelogram ka area hai aur iska aadha triangle ka area ((10.19)).

Chapter 10 par Miscellaneous Exercise

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 10.1, Q1	Practice 10.1 · Q1
NCERT Exercise 10.1, Q2	Practice 10.1 · Q2
NCERT Exercise 10.1, Q3	Practice 10.1 · Q3
NCERT Exercise 10.1, Q4	Practice 10.1 · Q4
NCERT Exercise 10.1, Q5	Practice 10.1 · Q5
NCERT Exercise 10.2, Q1	Practice 10.2 · Q1
NCERT Exercise 10.2, Q2	Practice 10.2 · Q2
NCERT Exercise 10.2, Q3	Practice 10.2 · Q3
NCERT Exercise 10.2, Q4	Practice 10.2 · Q4
NCERT Exercise 10.2, Q5	Practice 10.2 · Q5
NCERT Exercise 10.2, Q6	Practice 10.2 · Q6
NCERT Exercise 10.2, Q7	Practice 10.2 · Q7
NCERT Exercise 10.2, Q8	Practice 10.2 · Q8
NCERT Exercise 10.2, Q9	Practice 10.2 · Q9
NCERT Exercise 10.2, Q10	Practice 10.2 · Q10
NCERT Exercise 10.2, Q11	Practice 10.2 · Q11
NCERT Exercise 10.2, Q12	Practice 10.2 · Q12
NCERT Exercise 10.2, Q13	Practice 10.2 · Q13
NCERT Exercise 10.2, Q14	Practice 10.2 · Q14
NCERT Exercise 10.2, Q15	Practice 10.2 · Q15
NCERT Exercise 10.2, Q16	Practice 10.2 · Q16
NCERT Exercise 10.2, Q17	Practice 10.2 · Q17
NCERT Exercise 10.2, Q18	Practice 10.2 · Q18
NCERT Exercise 10.2, Q19	Practice 10.2 · Q19
NCERT Exercise 10.3, Q1	Practice 10.3 · Q1
NCERT Exercise 10.3, Q2	Practice 10.3 · Q2
NCERT Exercise 10.3, Q3	Practice 10.3 · Q3
NCERT Exercise 10.3, Q4	Practice 10.3 · Q4
NCERT Exercise 10.3, Q5	Practice 10.3 · Q5
NCERT Exercise 10.3, Q6	Practice 10.3 · Q6
NCERT Exercise 10.3, Q7	Practice 10.3 · Q7
NCERT Exercise 10.3, Q8	Practice 10.3 · Q8
NCERT Exercise 10.3, Q9	Practice 10.3 · Q9
NCERT Exercise 10.3, Q10	Practice 10.3 · Q10
NCERT Exercise 10.3, Q11	Practice 10.3 · Q11
NCERT Exercise 10.3, Q12	Practice 10.3 · Q12
NCERT Exercise 10.3, Q13	Practice 10.3 · Q13
NCERT Exercise 10.3, Q14	Practice 10.3 · Q14
NCERT Exercise 10.3, Q15	Practice 10.3 · Q15
NCERT Exercise 10.3, Q16	Practice 10.3 · Q16
NCERT Exercise 10.3, Q17	Practice 10.3 · Q17
NCERT Exercise 10.3, Q18	Practice 10.3 · Q18
NCERT Exercise 10.4, Q1	Practice 10.4 · Q1
NCERT Exercise 10.4, Q2	Practice 10.4 · Q2
NCERT Exercise 10.4, Q3	Practice 10.4 · Q3
NCERT Exercise 10.4, Q4	Practice 10.4 · Q4
NCERT Exercise 10.4, Q5	Practice 10.4 · Q5
NCERT Exercise 10.4, Q6	Practice 10.4 · Q6
NCERT Exercise 10.4, Q7	Practice 10.4 · Q7
NCERT Exercise 10.4, Q8	Practice 10.4 · Q8
NCERT Exercise 10.4, Q9	Practice 10.4 · Q9
NCERT Exercise 10.4, Q10	Practice 10.4 · Q10
NCERT Exercise 10.4, Q11	Practice 10.4 · Q11
NCERT Exercise 10.4, Q12	Practice 10.4 · Q12
NCERT Miscellaneous Exercise 10, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise 10, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise 10, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise 10, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise 10, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise 10, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise 10, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise 10, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise 10, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise 10, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise 10, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise 10, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise 10, Q13	Miscellaneous exercise · Q13
NCERT Miscellaneous Exercise 10, Q14	Miscellaneous exercise · Q14
NCERT Miscellaneous Exercise 10, Q15	Miscellaneous exercise · Q15
NCERT Miscellaneous Exercise 10, Q16	Miscellaneous exercise · Q16
NCERT Miscellaneous Exercise 10, Q17	Miscellaneous exercise · Q17
NCERT Miscellaneous Exercise 10, Q18	Miscellaneous exercise · Q18
NCERT Miscellaneous Exercise 10, Q19	Miscellaneous exercise · Q19

Quick check: MCQs

10.1 Introduction2 questions
Direction cosines3 questions
10.3 Vectors ke Types3 questions
Vector addition ki properties4 questions
10.5.1 Vector ke components3 questions
10.5.3 Section formula4 questions
10.6.2 Line par vector ka projection4 questions
Cross product calculate karna4 questions
Miscellaneous Examples3 questions

Glossary · 16

Key terms (English + Hindi)

coinitial vectorsसह-आदिम सदिशsah-aadim sadish: Same initial point wale vectors.Chapter mein pehli baar yahan aaya →
collinear vectorsसंरेख सदिशsanrekh sadish: Ek hi line ke parallel vectors; ek doosre ka scalar multiple hai.Chapter mein pehli baar yahan aaya →

direction cosinesदिक्-कोसाइनdik-cosine: Vector jo angles positive axes ke saath banata hai unke cosines $l, m, n$ ; $l^{2} + m^{2} + n^{2} = 1$ .Chapter mein pehli baar yahan aaya →
direction ratiosदिक्-अनुपातdik-anupaat: Koi bhi teen numbers jo vector ke direction cosines ke proportional hon.Chapter mein pehli baar yahan aaya →

equal vectorsसमान सदिशsamaan sadish: Same magnitude aur direction wale vectors, chahe kahin se bhi shuru hon.Chapter mein pehli baar yahan aaya →

magnitudeपरिमाणparimaan: Vector ki length, jise $∣ \vec{a} ∣$ likhte hain; kabhi negative nahi.Chapter mein pehli baar yahan aaya →

position vectorस्थिति सदिशsthiti sadish: Origin se point P tak ka vector $\vec{O P}$ .Chapter mein pehli baar yahan aaya →
projectionप्रक्षेपprakshep: Directed line par $\vec{a}$ ki shadow ki signed length $∣ \vec{a} ∣ \cos θ$ .

resultantपरिणामीparinaami: Do ya zyada vectors ka sum.Chapter mein pehli baar yahan aaya →

scalarअदिशadish: Aisi quantity jo ek real number se poori tarah describe ho jaaye, jaise mass ya temperature.Chapter mein pehli baar yahan aaya →
scalar componentsअदिश घटकadish ghatak: $x \hat{i} + y \hat{j} + z \hat{k}$ mein numbers $x, y, z$ .Chapter mein pehli baar yahan aaya →
scalar productअदिश गुणनफलadish gunanphal: $\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ$ , ek number.

unit vectorमात्रक सदिशmaatrak sadish: Magnitude 1 wala vector; $\hat{a} = \vec{a} / ∣ \vec{a} ∣$ .Chapter mein pehli baar yahan aaya →

vectorसदिशsadish: Aisi quantity jiska magnitude aur direction dono ho, jise directed line segment ki tarah banate hain.Chapter mein pehli baar yahan aaya →
vector productसदिश गुणनफलsadish gunanphal: $\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ , dono ke perpendicular ek vector.

zero vectorशून्य सदिशshunya sadish: Length 0 wala vector $\vec{0}$ , jiske initial aur terminal points ek hi hain.Chapter mein pehli baar yahan aaya →

Vector Algebra: Class 12 Maths Chapter 10 Hinglish mein

Is chapter mein aap kya seekhenge

10.1 Introduction

10.2 Kuch Basic Concepts

Position vector

Direction cosines

10.3 Vectors ke Types

Practice 10.1: answers ke saath

10.4 Vectors ka Addition

Parallelogram law

Vector addition ki properties

10.5 Vector ka Scalar se Multiplication

10.5.1 Vector ke components

10.5.2 Do points ko jodne wala vector

10.5.3 Section formula

Practice 10.2: answers ke saath

10.6 Do Vectors ka Product

10.6.1 Do vectors ka scalar (ya dot) product

10.6.2 Line par vector ka projection

Practice 10.3: answers ke saath

10.6.3 Do vectors ka vector (ya cross) product

Cross product se areas

Cross product calculate karna

Practice 10.4: answers ke saath

Miscellaneous Examples

Summary aur key points

Chapter 10 par Miscellaneous Exercise

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

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