Introduction to Three Dimensional Geometry: Class 11 Maths Chapter 11 Hinglish mein

Space ke har point ke liye teen numbers, aur kinhi bhi do points ke beech ki distance

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

Published 18 Oct 2026 · Updated 18 Oct 2026

55 min
1 simulation
20 practice questions
17 MCQs

Is chapter mein aap kya seekhenge

Space ke rectangular coordinate system ko describe karna: origin, teen mutually perpendicular axes aur teen coordinate planes.
Aath octants ke naam batana aur kisi point ke coordinates ke signs se tay karna ki woh kisi octant mein hai, kisi coordinate plane par hai ya kisi axis par.
Perpendiculars drop karke space mein kisi point ke coordinates nikaalna, aur diye gaye ordered triplet wale point ko locate karna.
Samjhaana ki x, y aur z kisi point ki YZ-, ZX- aur XY-planes se signed distances kyun hain, aur har axis aur har coordinate plane ke points ka general form likhna.
Pythagoras' theorem ko do baar lagaakar space mein distance formula derive karna, aur usse points ke beech ki distances nikaalna.
Distances se collinearity test karna, triangles aur parallelograms ko classify karna, aur kisi distance condition se define hue points ke set ki equation nikaalna.
Space mein points ke midpoint aur centroid se missing vertices aur medians ki lengths nikaalna.

NCERT Mathematics Class 11, Chapter 11, pp. 208–216 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

11.1 Introduction

Cricket match mein kisi dost se poochho ki woh kahan baithi hai, to do baatein sab tay kar deti hain: stand aur seat number. Lekin usi match ko film karne wala drone alag kahani hai. Sirf yeh kehna kaafi nahi ki woh pitch ke upar hai; yeh bhi batana padega ki woh kitni oonchai par hover kar raha hai. Flat surface ke points ke liye do numbers chahiye, lekin space ke points ke liye teen.

Do numbers wali kahani tum pehle se jaante ho. Ek plane ko do perpendicular number lines milti hain, yaani axes, aur uske har point ko unse uski signed distances ke pair $(x, y)$ se describe kiya jaata hai. Is chapter mein hum ek teesra axis jodenge, jo dono ke perpendicular hai, aur space ke liye wahi tools dobara banayenge: coordinates, signs, regions aur distances.

Ek chhota example dikhata hai ki teen numbers bilkul kaafi kyun hain. Diwali par ek paper lantern (कंदील) kamre ke ek corner ke paas latka hai. Floor aur us corner par milne wali do deewarein teen flat surfaces hain, aur har ek baaki do ke perpendicular hai. Lantern ki in teeno se distance naapo, jaise Fig. 11.1 mein, to uski position pakki ho gayi: kamre ke kisi aur point ki yahi teen distances nahi hongi.

4/4Yeh floor se 2 m upar latka hai. Triplet $(1.5, 2.5, 2)$ iski position fix kar deta hai.

Fig. 11.1Teen perpendicular surfaces, teen distances, ek position.

Kamre ka ek corner tircha banaya gaya hai: ek back wall, left mein ek side wall aur floor, teeno ek corner par milte hain. Ek diamond-shaped lantern dhaage se latka hai. Step 1 mein sirf kamra aur lantern hai. Step 2 mein lantern se seedhe back wall tak ek dashed line judti hai, jis par 1.5 m likha hai. Step 3 mein seedhe side wall tak ek dotted line judti hai, jis par 2.5 m likha hai. Step 4 mein seedhe neeche floor tak ek solid line judti hai, jis par 2 m likha hai, aur upar triplet (1.5, 2.5, 2) dikhta hai.

Mathematics deewaron aur floor ki jagah teen mutually perpendicular planes leti hai aur teen distances ko point ke coordinates kehti hai. Baaki chapter isi ko precise banata hai: axes aur planes set karna (Section 11.2), coordinates padhna aur likhna (Section 11.3) aur do points ke beech ki distance naapna (Section 11.4). Class 12 mein tum inhi coordinates se space mein lines ko describe karoge.

Quick check

2 mein se 1

Ek drone playground ke upar hover kar raha hai. Kaun si information uski position poori tarah fix karti hai?

Ek corner post se uski east aur north ki distancesPost se uski east aur north ki distances, aur uski heightZameen se uski height aur uski speedCorner post se uski straight-line distance

Answer dekho

Answer (B) Post se uski east aur north ki distances, aur uski height

Do horizontal distances sirf drone ke neeche zameen wala spot fix karti hain; height teesra number hai. Post se sirf ek distance possible positions ka poora sphere chhod deti hai, aur speed position ke baare mein kuch nahi batati.

11.2 Three Dimensional Space mein Coordinate Axes aur Coordinate Planes

Ek plane se shuru karo, use floor samjho, aur usmein do perpendicular lines kheencho jo ek point $O$ par kaatti hain. Ab $O$ par ek teesri line seedhi khadi karo, floor ke perpendicular. Yeh teen lines, jinmein har ek baaki do se right angle banati hai, coordinate axes hain: $X^{'} O X$ $x$ -axis hai, $Y^{'} O Y$ $y$ -axis aur $Z^{'} O Z$ $z$ -axis. Inka common point $O$ origin hai (Fig. 11.2).

4/4 $O X$ , $O Y$ , $O Z$ ke along distances positive hain; $O X^{'}$ , $O Y^{'}$ , $O Z^{'}$ ke along negative.

Fig. 11.2Axes aur teen coordinate planes; har axis ka positive half solid hai, negative half dashed.NCERT Fig. 11.1 · p. 209

Step 1 mein ek horizontal plane hai, shaded parallelogram ki tarah bana, jismein point O se do perpendicular lines guzarti hain: X′OX peeche right se aage left ki taraf aur Y′OY left se right ki taraf. Step 2 mein O se vertical line Z′OZ judti hai, chhote right-angle marks ke saath. Step 3 mein do vertical planes shade hote hain, YZ-plane aur ZX-plane, horizontal XY-plane ke saath. Step 4 mein har axis ke solid end OX, OY aur OZ par plus sign aur dashed ends OX′, OY′ aur OZ′ par minus sign lagta hai.

Koi bhi do axes ek plane mein hote hain, isliye axes ke saath teen planes aate hain, har ek mein do axes hain aur woh teesre axis ke perpendicular hai. Ulti taraf se dekho to yeh teen planes ek doosre ko pairs mein theek axes ke along kaatte hain. Origin aur uske axes aur planes milkar space ke liye ek rectangular coordinate system banate hain.

Definition

Coordinate planes

Woh teen planes jinmein se har ek mein do axes hain: XY-plane (jismein $x$ - aur $y$ -axes hain), YZ-plane aur ZX-plane. Har ek bache hue axis ke perpendicular hai.

ExampleKamre mein floor ko XY-plane lo aur corner par usse milne wali do deewaron ko YZ- aur ZX-planes.

Paper par hum XY-plane ko tircha dikhte floor ki tarah banate hain, taaki $z$ -axis vertical rahe. Har axis ko ek direction bhi chahiye, kyunki is chapter ka har formula signs par depend karta hai. Convention hamesha ke liye fix hai:Isse system right-handed banta hai: apne right hand ki ungliyan $O X$ se $O Y$ ki taraf modo, to angootha $O Z$ ke along point karta hai. Class 12 ke vectors isi par chalte hain.

Aage ya peeche ( $x$ ). YZ-plane ke tumhari taraf wala point, yaani $O X$ side, positive $x$ paata hai; uske peeche wala point, $O X^{'}$ side par, negative.
Right ya left ( $y$ ). ZX-plane ki $O Y$ side par $y$ positive hota hai, $O Y^{'}$ side par $y$ negative.
Upar ya neeche ( $z$ ). Floor, yaani XY-plane, ke upar wale points ka $z > 0$ hota hai; uske neeche, $O Z^{'}$ ki taraf, wale points ka $z < 0$ .

Try karke dekho

Carton ke andar axes

Tumhe chahiye

ek khaali carton ya shoebox jiski ek side khuli ho
ek marker
ek ruler
ek chhota eraser

Pehle guess karo

Eraser carton ke bottom par rakha hai. Uska kaun sa coordinate

0

hona chahiye?

x

y

z

Khuli side apni taraf karo. Peeche, neeche aur left wale andar ke corner ko $O$ kaho.
$O$ se tumhari taraf aane wale bottom edge ko $O X$ , right ki taraf jaane wale bottom edge ko $O Y$ aur seedhe khade back edge ko $O Z$ mark karo.
Bottom panel par XY, back panel par YZ aur left panel par ZX likho.
Eraser ko andar daalo. Har panel se uski distance naapo aur teeno numbers ko triplet ki tarah likho.

Teen coordinate planes space ko aath tukdon mein kaat dete hain, waise hi jaise do khadi cuts aur ek seedhi cut ek cake ko aath hisson mein baant deti hain. Har tukda ek octant (अष्टांश) hai. Octant ka naam uske edges wale teen half-axes se rakha jaata hai: $X O Y Z$ ke edges $O X$ , $O Y$ aur $O Z$ hain, jabki $X^{'} O Y Z^{'}$ ke edges $O X^{'}$ , $O Y$ aur $O Z^{'}$ hain. Octants par Roman numerals bhi hote hain (Fig. 11.3). Octants I se IV XY-plane ke upar hain aur us plane ke chaar quadrants ke usual order mein chalte hain. Octants V se VIII neeche hain, har ek apne se chaar kam number wale octant ke theek neeche.

3/3Har octant ka $(x, y, z)$ ke signs ka apna pattern hai.

Fig. 11.3Teen coordinate planes space ko aath octants mein kaatte hain: I–IV XY-plane ke upar, V–VIII neeche.

Aath cubes ka ek block, teen coordinate planes ke along gaps se alag, jismein x-axis dekhne wale ki taraf, y-axis right ki taraf aur z-axis upar point karta hai. Step 1 mein upar ke chaar cubes saaf dikhte hain, I, II, III aur IV labelled, jabki neeche ke chaar cubes halke hain. Step 2 mein neeche ke cubes bhi dikhte hain, V, VI, VII aur VIII labelled, har ek apne se chaar kam number wale cube ke theek neeche. Step 3 mein har cube par x, y aur z ke signs judte hain, I par (+,+,+) se lekar VIII par (+,−,−) tak.

Definition

Octant

Woh aath regions jinmein teen coordinate planes space ko baantte hain. Octants I–IV XY-plane ke upar hain aur V–VIII uske neeche, aur octant $n + 4$ octant $n$ ke theek neeche hai.

11.3 Space mein Point ke Coordinates

Is section mein do sawaal hain. Diye gaye point ke kaun se teen numbers hain? Aur diye gaye teen numbers ka kaun sa point hai? Fig. 11.4 ek point $P$ ke liye dono ka jawaab deta hai.

Maan lo $M$ floor (XY-plane) ka woh point hai jo $P$ ke theek neeche ya upar hai, taaki $P M$ floor ke perpendicular ho, aur $L$ $x$ -axis ka woh point hai jo $M$ ke sabse paas hai, taaki $M L$ axis se right angle par mile. Teen signed lengths padho: $O L$ , phir $L M$ , phir $M P$ . Yeh $P$ ke coordinates hain, jinhe $P (x, y, z)$ likhte hain. Signed ka matlab hai ki har length apni direction ka sign rakhti hai: jab $P$ floor ke neeche ho to $M P$ negative hai, jab $M$ $O Y^{'}$ side par ho to $L M$ negative hai, aur jab $L$ $O$ ke peeche ho to $O L$ negative hai.

4/4 $O L = 3$ , $L M = 4$ , $M P = 5$ , isliye $P = (3, 4, 5)$ . $O \to L \to M \to P$ chalne se process ulta ho jaata hai.

Fig. 11.4Neeche floor tak, phir across

x

-axis tak: teen legs

x

y

aur

z

hain.NCERT Fig. 11.2 · p. 209

Axes OX, OY aur OZ, jismein XY-plane tiled floor ki tarah bana hai. Step 1 mein point P floor ke upar hawa mein dikhta hai. Step 2 mein P se floor par uske foot M tak ek vertical line girti hai, right-angle mark ke saath. Step 3 mein M se, y-axis ke parallel, x-axis par point L tak ek dashed line banti hai, phir se right-angle mark ke saath. Step 4 mein segment OL judta hai aur teen legs par x = 3, y = 4 aur z = 5 likha jaata hai; M par (3, 4, 0) mark hai aur heading P(3, 4, 5) hai.

Numbers se point tak jaana teen legs ki walk hai. $O$ se shuru karke $x$ -axis ke along $x$ units chalo ( $x < 0$ ho to peeche) aur $L$ tak pahuncho; mudo aur $y$ -axis ke parallel $y$ units chalkar $M$ tak pahuncho; phir $z$ units upar chadho, ya $z < 0$ ho to neeche utro, aur $P$ tak pahuncho. Space ke har point ka theek ek triplet hota hai aur har triplet theek ek point tak le jaata hai. Mathematicians kehte hain ki space ke points aur real numbers ke ordered triplets mein one-to-one correspondence hai.

Ordered shabd important hai. $(2, 5, - 1)$ aur $(5, 2, - 1)$ mein wahi teen numbers hain lekin yeh alag points hain: pehla YZ-plane ke 2 units aage aur ZX-plane ke 5 units right mein hai, doosra iska ulta.

Coordinates planes se distances ki tarah

Ek doosri picture teeno coordinates ko ek jaisa treat karti hai. Woh box imagine karo, ek rectangular parallelepiped, jiske opposite corners $O$ aur $P$ hain aur jiske edges axes ke along hain (Fig. 11.5). Iske teen faces coordinate planes mein hain; baaki teen $P$ se guzarte hain, har ek kisi ek coordinate plane ke parallel. Box ke teen corners axes par hain; unhe $x$ -axis par $A$ , $y$ -axis par $B$ aur $z$ -axis par $C$ kaho. $O$ se unke edges bilkul $x$ , $y$ aur $z$ hain: $O A = x$ , $O B = y$ , $O C = z$ .

4/4 $O A = 3$ , $O B = 5$ , $O C = 4$ . Har corner $P$ hi hai jiske kuch coordinates $0$ se badal diye gaye.

Fig. 11.5

P

se guzarte teen planes ek box banate hain; axes ke along uske edges

x

y

aur

z

hain.NCERT Fig. 11.3 · p. 210

Axes OX, OY aur OZ ke saath point P(3, 5, 4). Step 1 mein P se guzarta YZ-plane ke parallel ek plane shade hota hai; woh x-axis ko A par milta hai. Step 2 mein P se guzarta ZX-plane ke parallel plane judta hai, jo y-axis ko B par milta hai. Step 3 mein P se guzarta XY-plane ke parallel plane judta hai, jo z-axis ko C par milta hai. Step 4 mein OA par x = 3, OB par y = 5 aur OC par z = 4 likha jaata hai, aur box ke teen corners ke naam aate hain: P ke neeche floor par (3, 5, 0), ZX-plane par (3, 0, 4) aur YZ-plane par (0, 5, 4).

$P$ se guzarta box ka har face kisi coordinate plane ke parallel hai, aur do parallel planes ke beech ka gap kisi bhi common perpendicular ke along naapa ja sakta hai. Isliye $∣ x ∣$ YZ-plane se $P$ ki distance hai, $∣ y ∣$ ZX-plane se uski distance aur $∣ z ∣$ XY-plane se uski distance. Box ke corners bhi utne hi aasaan hain: $P$ ke kuch coordinates ko zero se badal do.

Zeros batate hain ki point kahan baitha hai (Fig. 11.6). Origin $(0, 0, 0)$ hai. $x$ -axis ka point ZX- aur XY-planes se zero distance par hai, isliye uska form $(x, 0, 0)$ hai; baaki axes ke points $(0, y, 0)$ aur $(0, 0, z)$ hain. XY-plane ke point ka form $(x, y, 0)$ hai, YZ-plane ke point ka $(0, y, z)$ aur ZX-plane ke point ka $(x, 0, z)$ .

3/3Ek coordinate zero hai: woh jiska axis plane mein nahi hai.

Fig. 11.6Ek coordinate zero: point kisi coordinate plane par hai. Do zeros: woh kisi axis par hai.

Axes OX, OY aur OZ. Step 1 mein origin O(0, 0, 0) mark hota hai. Step 2 mein har axis par ek point judta hai: x-axis par (x, 0, 0), y-axis par (0, y, 0) aur z-axis par (0, 0, z). Step 3 mein teen coordinate planes shade hote hain aur har ek mein ek point judta hai: XY-plane mein (x, y, 0), YZ-plane mein (0, y, z) aur ZX-plane mein (x, 0, z).

Planes se door, signs octant tay karte hain; Table 11.1 unki list deta hai. Yeh plane ke chaar quadrants ko do baar padhne jaisa hai: octants I–IV $(x, y)$ ke quadrant signs ko $z > 0$ ke saath dohraate hain, aur V–VIII unhe $z < 0$ ke saath. Jis point ka koi coordinate zero ho, woh kisi coordinate plane par hai aur isliye kisi octant mein nahi aata.

Table 11.1Aath octants mein coordinates ke signsNCERT Table 11.1 · p. 210

Octant	Edges	$x$	$y$	$z$	Sample point
I	$X O Y Z$	$+$	$+$	$+$	$(2, 1, 3)$
II	$X^{'} O Y Z$	$-$	$+$	$+$	$(- 2, 1, 3)$
III	$X^{'} O Y^{'} Z$	$-$	$-$	$+$	$(- 2, - 1, 3)$
IV	$X O Y^{'} Z$	$+$	$-$	$+$	$(2, - 1, 3)$
V	$X O Y Z^{'}$	$+$	$+$	$-$	$(2, 1, - 3)$
VI	$X^{'} O Y Z^{'}$	$-$	$+$	$-$	$(- 2, 1, - 3)$
VII	$X^{'} O Y^{'} Z^{'}$	$-$	$-$	$-$	$(- 2, - 1, - 3)$
VIII	$X O Y^{'} Z^{'}$	$+$	$-$	$-$	$(2, - 1, - 3)$

Example 11.1NCERT Example 1 · p. 211

Almirah ke corners

Ek steel almirah kamre ke corner mein khadi hai. Floor ke corner ko $O$ , floor ko XY-plane aur almirah ke peeche ki do deewaron ko YZ- aur ZX-planes lo, lengths metres mein. Almirah ek box hai jiska ek corner $O$ par hai aur opposite corner $P (1.2, 0.6, 2)$ par. Coordinates nikaalo: (a) $P$ ke theek neeche floor wala corner, (b) upar ka woh corner jo dono deewaron ko chhoota hai, (c) floor ka woh corner jo $y$ -axis par hai.

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

Example 11.2NCERT Example 2 · p. 211

Octant padhna

Points $A (4, - 2, - 3)$ , $B (- 5, - 1, 6)$ aur $C (0, 2, - 1)$ kahan hain?

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

InteractivePoint-in-space lab

Axes ka three-dimensional view, positive halves solid aur negative halves dashed. Point mode mein teen sliders x, y aur z ko −5 se 5 tak set karte hain; point P apne coordinate box ke saath dikhta hai, aur O se x-axis ke along L tak, y-axis ke parallel foot M tak, aur upar ya neeche P tak ka raasta bhi. Ek panel octant ka naam batata hai, ya woh plane ya axis jis par point hai, aur teen coordinate planes, teen axes aur origin se distances ki list deta hai. Two points mode mein ek toggle chunta hai ki sliders P ko move karein ya Q ko; diagonal PQ wala box banta hai jiski teen edge directions colour hoti hain, aur panel coordinates ke differences, unke squares, unka sum aur distance PQ dikhata hai. Picture ko drag karne ya do view sliders se woh ghoomti aur jhukti hai. Jab tak learner kuch na kare, kuch move nahi hota.

Fig. 11.7Sliders se

x

y

aur

z

set karo aur picture ko drag karke ghumao. Section 11.4 ka distance formula kaam karte dekhne ke liye Two points par switch karo.

Practice · 6 questions

Practice 11.1: answers ke saath

Short answerEasyNCERT Exercise 11.1, Q1
Ek point $z$ -axis par hai, XY-plane ke 4 units neeche. Uske $x$ - aur $y$ -coordinates kya hain, aur point kya hai?
Solution dikhao
Answer $x = 0$ , $y = 0$ ; point $(0, 0, - 4)$ hai.
$z$ -axis YZ-plane aur ZX-plane dono mein hai, isliye point dono se zero distance par hai: $x = 0$ aur $y = 0$ . XY-plane ke neeche $z$ -coordinate negative hai, isliye $z = - 4$ aur point $(0, 0, - 4)$ hai.
2
Short answerEasyNCERT Exercise 11.1, Q2
Tum kya keh sakte ho
1. YZ-plane ke har point ke $x$ -coordinate ke baare mein?
  Solution dikhao
  AnswerYeh $0$ hai.
  $x$ -coordinate YZ-plane se signed distance hai, aur us plane ka point usse $0$ distance par hai.
2. XY-plane ke har point ke $z$ -coordinate ke baare mein?
  Solution dikhao
  AnswerYeh $0$ hai.
  $z$ -coordinate XY-plane se signed distance hai, jo uske apne points ke liye $0$ hai. Aise points ka form $(x, y, 0)$ hai.
3
Quick answerEasyNCERT Exercise 11.1, Q3
Har point ko us octant se match karo jismein woh hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$(2, 5, 1)$ : I
$(- 3, 4, 7)$ : II
$(- 1, - 6, 2)$ : III
$(3, 1, - 4)$ : V
$(- 4, - 4, - 4)$ : VII
$(6, - 3, - 2)$ : VIII
Sign patterns yeh hain: $(+, +, +)$ I, $(-, +, +)$ II, $(-, -, +)$ III, $(+, +, -)$ V, $(-, -, -)$ VII, $(+, -, -)$ VIII.
Quick answerEasyNCERT Exercise 11.1, Q4
Blanks bharo. $y$ -axis aur $z$ -axis milkar -plane tay karte hain. ZX-plane ke har point ka form hai. Aath octants mein se mein $x$ -coordinate negative hai.
Answer dekho
Answer (1) YZ, (2) (x, 0, z), (3) 4
$y$ - aur $z$ -axes YZ-plane banate hain. ZX-plane par $y = 0$ , isliye points $(x, 0, z)$ hain. Octants II, III, VI aur VII mein $x < 0$ : unmein se chaar.
Quick answerMedium
Point $(5, - 12, 7)$ $z$ -axis se kitni door hai?
units
Answer dekho
Answer 13 units
$z$ -axis ka sabse paas wala point $(0, 0, 7)$ hai, isliye distance $\sqrt{5^{2} + 12^{2}} = 13$ hai. XY-plane se uski distance, $7$ , ek alag quantity hai.
Quick answerMedium
Ek point ka $x \neq 0$ , $y = 0$ aur $z < 0$ hai. Kaun sa statement sahi hai?
Yeh octant V ya VIII mein haiYeh ZX-plane par hai, $x$ -axis ke neecheYeh $z$ -axis par haiYeh XY-plane par hai
Answer dekho
Answer (B) Yeh ZX-plane par hai, $x$ -axis ke neeche
$y = 0$ point ko ZX-plane par rakhta hai, aur $z < 0$ use XY-plane ke neeche rakhta hai, jiska ZX-plane mein matlab hai $x$ -axis ke neeche. Coordinate plane par pada point kisi octant mein nahi hota.

11.4 Do Points ke beech Distance

Plane mein $(x_{1}, y_{1})$ aur $(x_{2}, y_{2})$ ke beech ki distance $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ hai, jo asal mein Pythagoras' theorem hi hai. Space mein ek right triangle kaafi nahi, lekin do kaafi hain: pehla ek horizontal plane mein leta hai, aur doosra uske upar seedha khada hai.

DerivationSpace mein distance formula

Koi bhi do points fix karo: $P$ jiske coordinates $(x_{1}, y_{1}, z_{1})$ hain aur $Q$ jiske coordinates $(x_{2}, y_{2}, z_{2})$ hain. $P$ se $Q$ tak jaane ke liye ek baar mein ek coordinate badlo, to teen moves ek aise box ke teen edges banate hain jiske edges axes ke parallel hain aur $P Q$ uska long diagonal hai (Fig. 11.8). Edge lengths $∣ x_{2} - x_{1} ∣$ , $∣ y_{2} - y_{1} ∣$ aur $∣ z_{2} - z_{1} ∣$ hain.

4/4 $S$ par right angle: $P Q^{2} = P S^{2} + S Q^{2}$ .

Fig. 11.8Pythagoras do baar: ek baar box ke floor ke across, ek baar uski side par upar.NCERT Fig. 11.4 · p. 211

Step 1 mein do points P aur Q ek dashed segment se jude hain, top left par dono points ke coordinates likhe hain aur sawaal PQ = ? hai. Step 2 mein axes ke parallel edges wala woh box banta hai jiska diagonal PQ hai. Step 3 mein bottom face mein triangle PRS shade hota hai: PR, x-axis ke parallel, ki length |x2 − x1| hai, RS, y-axis ke parallel, ki length |y2 − y1| hai, R par right angle hai aur note mein PS² = PR² + RS² likha hai. Step 4 mein khada triangle PSQ shade hota hai: vertical edge SQ ki length |z2 − z1| hai, S par right angle hai aur note mein PQ² = PS² + SQ² likha hai.

Kram se, moves $x$ -axis ke parallel $R (x_{2}, y_{1}, z_{1})$ tak jaate hain, $y$ -axis ke parallel $S (x_{2}, y_{2}, z_{1})$ tak, aur $z$ -axis ke parallel upar $Q$ tak. Points $P$ , $R$ aur $S$ sabki height $z_{1}$ hai, isliye triangle $P R S$ ek horizontal plane mein hai. Iska $R$ par right angle hai, kyunki $P R$ aur $R S$ do perpendicular axes ke parallel hain, aur $P R = ∣ x_{2} - x_{1} ∣$ aur $R S = ∣ y_{2} - y_{1} ∣$ :

\begin{aligned} P S^{2} & = P R^{2} + R S^{2} \\ = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \end{aligned}

(11.1)

Ab, $S Q$ $z$ -axis ke parallel hai, isliye woh triangle $P R S$ ke horizontal plane ke perpendicular hai aur khaas taur par $P S$ ke bhi. Isliye triangle $P S Q$ ka $S$ par right angle hai, aur $S Q = ∣ z_{2} - z_{1} ∣$ :

P Q^{2} = P S^{2} + S Q^{2}

(11.2)

(11.1) ko (11.2) mein rakho. Kyunki $∣ a ∣^{2} = a^{2}$ , absolute values hat jaati hain aur hamein distance formula milta hai; $P Q$ right side ka positive square root hai.

Distance formula

\begin{aligned} P Q^{2} = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \\ + (z_{2} - z_{1})^{2} \end{aligned}

(11.3)

Pehle point ko origin par lene se, aur doosre point ko $P (x, y, z)$ likhne se, $O$ se kisi bhi point ki distance milti hai:

Origin se distance

O P = \sqrt{x^{2} + y^{2} + z^{2}}

(11.4)

Teen baatein zyaadatar galtiyon se bachati hain. Points ka order matter nahi karta, kyunki $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ . Matching coordinates ghatao aur unke signs rakho: $- 3$ se $4$ tak change $7$ hai, $1$ nahi. Aur plane wala formula naye formula ke andar hi hai: agar dono points XY-plane mein hain, to $z_{1} = z_{2} = 0$ aur teesra term gayab ho jaata hai.

Example 11.3NCERT Example 3 · p. 212

Do points, ek distance

$P (2, - 1, 5)$ aur $Q (- 4, 1, 2)$ ke beech ki distance nikaalo.

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

SimulationExample 11.3 ka floor plan

Dono axes par −6 se 6 tak ek square grid, jise upar se dekhe XY-plane ki tarah padhna hai. Do draggable points, P′ (2, −1) par aur Q′ (−4, 1) par, Example 11.3 ke P(2, −1, 5) aur Q(−4, 1, 2) se perpendiculars ke feet hain. Segment P′Q′ bana hai aur uski length naapi jaati hai; yeh distance ka horizontal part hai, aur vertical part heights ka difference, 3, hai.

Fig. 11.9

P^{'}

aur

Q^{'}

XY-plane par

P (2, - 1, 5)

aur

Q (- 4, 1, 2)

ke feet hain, upar se dekhe hue. Inki distance derivation ka floor diagonal

P S

hai.

Try karo

$P^{'} Q^{'}$ ki length padho aur uska square karo. $(5 - 2)^{2} = 9$ jodo aur square root lo: tumhe $7$ milna chahiye, jo Example 11.3 ka answer hai.
$Q^{'}$ ko drag karke $P^{'}$ ke upar rakho. Ab $P$ aur $Q$ ek vertical line par hain: distance formula kya ban jaata hai?
$Q^{'}$ ki aisi position dhoondho jahan $P^{'} Q^{'} = 5$ ho. Agar heights $5$ aur $2$ hi rahein, to ab $P Q$ kya hai?

Distance formula se kai geometric sawaal arithmetic ban jaate hain:

Collinear points. Teen points tabhi collinear hote hain jab unke beech ki teen distances mein se sabse lambi baaki do ke sum ke barabar ho; warna woh triangle banate hain.
Triangle ke prakaar. Isosceles ya equilateral triangles pehchaanne ke liye side lengths compare karo, aur right angle pakadne ke liye sabse lambi side $c$ ke liye $a^{2} + b^{2} = c^{2}$ test karo (Pythagoras' theorem ka converse).
Points ke sets. Distances par koi condition, jaise do fixed points se barabar door hona, distances ko square karne par $x$ , $y$ , $z$ mein ek equation ban jaati hai. Jo set of points yeh describe karti hai use locus kehte hain.

Example 11.4NCERT Example 4 · p. 212

Ek line par teen points

Dikhao ki $A (- 1, 2, 0)$ , $B (1, 5, 6)$ aur $C (5, 11, 18)$ collinear hain.

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

Example 11.5NCERT Example 5 · p. 212

Kya right angle hai?

Kya corners $A (1, 2, 3)$ , $B (3, 3, 5)$ aur $C (3, 0, 2)$ wale triangle mein right angle hai?

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

Example 11.6NCERT Example 6 · p. 213

Ek condition sphere ban jaati hai

$A$ aur $B$ points $(2, - 1, 3)$ aur $(0, 3, - 1)$ hain. Ek point $P$ aise move karta hai ki $P A^{2} + P B^{2} = 30$ . Woh equation nikaalo jo yeh satisfy karta hai, aur uske path ko describe karo.

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

Space mein distance conditions surfaces describe karti hain. Fig. 11.10 woh teen dikhata hai jo is chapter ke examples aur exercises mein aate hain.

2/4 $P A = P B$ : $M$ se guzarta $A B$ ke right angle par plane.

Fig. 11.10Space mein fixed distance conditions surfaces deti hain, curves nahi.

Step 1 mein do points A aur B ek segment se jude hain, uske midpoint M ke saath. Step 2 mein segment ke across right angles par, M se guzarta ek plane khada hai, jis par ek point P dashed lines se A aur B se juda hai aur label PA = PB hai. Step 3 mein plane ki jagah M centre wala sphere aata hai, jis par ek point P A aur B se juda hai aur label PA² + PB² = constant hai. Step 4 mein sphere ki jagah A aur B ke around ek lamba, rugby-ball jaisa surface aata hai, jis par ek point P hai aur label PA + PB = constant hai.

Key idea

Space mein midpoint aur centroid

Class 10 ke do results space mein bhi chalte hain, ek baar mein ek coordinate. Current NCERT text inhe apne miscellaneous examples mein bina proof ke use karta hai, isliye yahan dekho ki yeh sahi kyun hain. $A (x_{1}, y_{1}, z_{1})$ aur $B (x_{2}, y_{2}, z_{2})$ ka midpoint $M$ hai

\begin{aligned} M = ( & \frac{x_{1} + x_{2}}{2}, \\ \frac{y_{1} + y_{2}}{2}, \\ \frac{z_{1} + z_{2}}{2}) \end{aligned}

(11.5)

Kyun: $A$ se $M$ tak har coordinate difference $A$ se $B$ tak ke difference ka aadha hai, aur $M$ se $B$ tak ka har difference bhi. (11.3) se, $A M = M B = \frac{1}{2} A B$ . Phir $A M + M B = A B$ , isliye $M$ segment $A B$ par hai, theek beech mein.

Teesre vertex $C (x_{3}, y_{3}, z_{3})$ wale triangle ka centroid $G$ hai

\begin{aligned} G = ( & \frac{x_{1} + x_{2} + x_{3}}{3}, \\ \frac{y_{1} + y_{2} + y_{3}}{3}, \\ \frac{z_{1} + z_{2} + z_{3}}{3}) \end{aligned}

(11.6)

Kyun: $G$ $C$ se nikalne wale median par hai, $C$ se $A B$ ke midpoint tak do-tihaai raaste par. $z$ -coordinate mein isse $z_{3} + \frac{2}{3} (\frac{z_{1} + z_{2}}{2} - z_{3}) = \frac{z_{1} + z_{2} + z_{3}}{3}$ milta hai, aur $x$ aur $y$ bhi aise hi chalte hain.

3/3 $G$ har median ko vertex se $2 : 1$ ratio mein divide karta hai.

Fig. 11.11Midpoint do points ka average hai; centroid teen ka.

Step 1 mein segment AB hai jiske dono halves par barabar tick marks hain aur uska midpoint M, top par midpoint formula likha hai. Step 2 mein teesra point C judta hai, triangle ABC shade hota hai aur C se M tak median dashed line ki tarah banta hai. Step 3 mein baaki do medians bante hain, unka common point G mark hota hai aur median CM ke do parts par 2 aur 1 likha hota hai, note CG : GM = 2 : 1 ke saath.

Practice · 7 questions

Practice 11.2: answers ke saath

1
NumericalEasyNCERT Exercise 11.2, Q1
Har pair of points ke beech ki distance nikaalo.
1. $(3, - 1, 2)$ aur $(5, 2, 8)$
  Answer dekho
  Answer 7
  Differences $2, 3, 6$ : $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .
2. $(- 2, 0, 5)$ aur $(1, - 4, 5)$
  Answer dekho
  Answer 5
  Differences $3, - 4, 0$ : $\sqrt{9 + 16 + 0} = 5$ . Dono points ki height $5$ hai, isliye yeh asal mein plane wala problem hai.
3. $(0, - 3, 1)$ aur $(2, 1, - 3)$
  Answer dekho
  Answer 6
  Differences $2, 4, - 4$ : $\sqrt{4 + 16 + 16} = 6$ .
4. $(\sqrt{2}, 1, - 1)$ aur $(0, 2, 1)$
  Answer dekho
  Answer 2.6458
  Differences $- \sqrt{2}, 1, 2$ : $\sqrt{2 + 1 + 4} = \sqrt{7} \approx 2.646$ .
ProofEasyNCERT Exercise 11.2, Q2
Dikhao ki $(2, - 1, 4)$ , $(4, 0, 2)$ aur $(8, 2, - 2)$ collinear hain.
Solution dikhao
AnswerDistances $3$ , $6$ aur $9$ hain, aur $3 + 6 = 9$ .
Points ko $A$ , $B$ , $C$ kaho. To $A B = \sqrt{2^{2} + 1^{2} + (- 2)^{2}} = 3$ , $B C = \sqrt{4^{2} + 2^{2} + (- 4)^{2}} = 6$ aur $A C = \sqrt{6^{2} + 3^{2} + (- 6)^{2}} = 9$ .
Kyunki $A B + B C = A C$ , points collinear hain, aur $B$ $A$ aur $C$ ke beech mein hai.
3
ProofMediumNCERT Exercise 11.2, Q3
Har statement verify karo.
1. $A (2, 0, 1)$ , $B (4, 1, 3)$ aur $C (2, 3, 1)$ ek isosceles triangle ke vertices hain.
  Solution dikhao
  $A B = \sqrt{4 + 1 + 4} = 3$ , $A C = \sqrt{0 + 9 + 0} = 3$ aur $B C = \sqrt{4 + 4 + 4} = 2 \sqrt{3}$ . Do sides barabar hain, aur $3 + 3 > 2 \sqrt{3}$ , isliye teeno points sach mein triangle banate hain aur woh isosceles hai.
2. $P (1, - 1, 2)$ , $Q (3, 1, 1)$ aur $R (7, - 1, 5)$ wale triangle $P Q R$ mein right angle hai.
  Solution dikhao
  $P Q^{2} = 4 + 4 + 1 = 9$ , $Q R^{2} = 16 + 4 + 16 = 36$ aur $P R^{2} = 36 + 0 + 9 = 45$ . Kyunki $P Q^{2} + Q R^{2} = 45 = P R^{2}$ , $Q$ par angle right angle hai.
3. $A (0, 1, 2)$ , $B (1, 3, 4)$ , $C (4, 2, 5)$ aur $D (3, 0, 3)$ ek parallelogram $A B C D$ ke vertices hain.
  Solution dikhao
  $A C$ ka midpoint $(2, \frac{3}{2}, \frac{7}{2})$ hai aur $B D$ ka midpoint bhi $(2, \frac{3}{2}, \frac{7}{2})$ hai. Diagonals ek doosre ko bisect karte hain, isliye $A B C D$ ek parallelogram hai.
  Check ke taur par, $A B = C D = 3$ aur $B C = D A = \sqrt{11}$ .
Short answerMediumNCERT Exercise 11.2, Q4
$(2, - 1, 1)$ aur $(0, 3, - 3)$ se barabar door points ke locus ki equation likho.
Solution dikhao
Answer $x - 2 y + 2 z + 3 = 0$
$P (x, y, z)$ ke liye, barabar distances ka matlab barabar squared distances: $(x - 2)^{2} + (y + 1)^{2} + (z - 1)^{2} = x^{2} + (y - 3)^{2} + (z + 3)^{2}$ .
Expand karne par left side $x^{2} + y^{2} + z^{2} - 4 x + 2 y - 2 z + 6$ hai aur right side $x^{2} + y^{2} + z^{2} - 6 y + 6 z + 18$ hai.
Cancel karke aur terms ikattha karke: $- 4 x + 8 y - 8 z - 12 = 0$ , yaani $x - 2 y + 2 z + 3 = 0$ . Check: midpoint $(1, 1, - 1)$ se $1 - 2 - 2 + 3 = 0$ milta hai.
Long answerHardNCERT Exercise 11.2, Q5
Ek point $P$ aise move karta hai ki $A (0, 0, 2)$ aur $B (0, 0, - 2)$ se uski distances ka sum hamesha $6$ rahe. Uske locus ki equation nikaalo.
Solution dikhao
Answer $9 x^{2} + 9 y^{2} + 5 z^{2} = 45$
$P A^{2} = x^{2} + y^{2} + (z - 2)^{2}$ aur $P B^{2} = x^{2} + y^{2} + (z + 2)^{2}$ , isliye $P B^{2} - P A^{2} = 8 z$ .
Kyunki $P B^{2} - P A^{2} = (P B + P A) (P B - P A) = 6 (P B - P A)$ , hamein $P B - P A = \frac{4 z}{3}$ milta hai. Ise $P B + P A = 6$ mein jodne se $P B = 3 + \frac{2 z}{3}$ milta hai.
Square karo: $x^{2} + y^{2} + z^{2} + 4 z + 4 = 9 + 4 z + \frac{4 z^{2}}{9}$ , isliye $x^{2} + y^{2} + \frac{5}{9} z^{2} = 5$ .
$9$ se multiply karke: $9 x^{2} + 9 y^{2} + 5 z^{2} = 45$ , $z$ -axis ke along khincha hua ek ellipsoid, jo us par $z = \pm 3$ tak pahunchta hai.
Quick answerMedium
$x$ -axis ka woh point jo $(1, 2, 3)$ aur $(3, 5, - 2)$ se barabar door hai, hai
$(6, 0, 0)$ $(- 6, 0, 0)$ $(4, 0, 0)$ $(0, 6, 0)$
Answer dekho
Answer (A) $(6, 0, 0)$
$(a, 0, 0)$ ke saath: $(a - 1)^{2} + 4 + 9 = (a - 3)^{2} + 25 + 4$ , isliye $- 2 a + 14 = - 6 a + 38$ aur $a = 6$ . $(0, 6, 0)$ to $x$ -axis par hai hi nahi.
Quick answerMedium
$(2, k, 1)$ aur $(- 1, 3, 5)$ ke beech ki distance $\sqrt{34}$ hai. $k$ ki saari possible values ka sum nikaalo.
Answer dekho
Answer 6
$9 + (k - 3)^{2} + 16 = 34$ se $(k - 3)^{2} = 9$ milta hai, isliye $k = 0$ ya $k = 6$ . Inka sum $6$ hai.

Miscellaneous Examples

Example 11.7NCERT Example 7 · p. 213

Aisa parallelogram jo rectangle nahi hai

Dikhao ki $A (1, 0, 2)$ , $B (3, 3, 3)$ , $C (4, 5, 1)$ aur $D (2, 2, 0)$ ek parallelogram $A B C D$ ke vertices hain, aur yeh rectangle nahi hai.

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

2/4Parallelogram mein diagonals ek doosre ko bisect karte hain.

Fig. 11.12Space mein parallelogram ko uske diagonals se test karo: barabar opposite sides kaafi nahi hain.

Step 1 mein parallelogram ABCD hai jismein AB aur DC par single tick marks aur BC aur AD par double tick marks hain. Step 2 mein diagonals AC aur BD bante hain, jo ek point par kaatte hain jis par same midpoint likha hai. Step 3 mein note judta hai ki AC BD ke barabar nahi hai, isliye figure rectangle nahi hai. Step 4 mein iski jagah ek tetrahedron aata hai jiske edges AB, BC, CD aur DA chaar barabar sides wala closed path banate hain; dashed diagonals AC aur BD ke midpoints alag hain jo ek chhote dotted segment se jude hain, aur note mein likha hai ki chaar barabar sides hain, par flat nahi.

Example 11.8NCERT Example 8 · p. 214

Do points se barabar door

Kaun se points $P$ $A (1, - 2, 4)$ se utni hi door hain jitni $B (3, 2, 0)$ se? Woh equation nikaalo jo yeh satisfy karte hain.

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

Example 11.9NCERT Example 9 · p. 214

Missing vertex

Triangle $A B C$ ka centroid $G (2, - 1, 3)$ hai. Agar $A = (4, 1, - 2)$ aur $B = (- 3, 0, 5)$ , to $C$ nikaalo.

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

Quick check

3 mein se 1

M (3, - 1, 4)

A B

ka midpoint hai, aur

A = (1, 2, 5)

B

z

-coordinate kya hai?

Answer dekho

Answer 3

\frac{5 + z_{B}}{2} = 4

z_{B} = 3

milta hai. Poora,

B = (5, - 4, 3)

Recap

Summary aur key points

Space ko teen coordinates chahiye: teen mutually perpendicular axes $X^{'} O X$ , $Y^{'} O Y$ aur $Z^{'} O Z$ origin $O$ par milte hain (Section 11.2).
XY-, YZ- aur ZX-planes mein se har ek mein do axes hain aur woh teesre ke right angle par khada hai (Section 11.2).
Coordinate planes ke along space aath octants mein bant jaata hai, aur $(x, y, z)$ ke signs batate hain ki point kis octant mein hai (Table 11.1).
Point $P (x, y, z)$ tak pahunchne ke liye $x$ -axis ke along $x$ chalo, phir $y$ -axis ke parallel $y$ , phir $z$ -axis ke parallel $z$ ; points aur ordered triplets mein one-to-one correspondence hai (Section 11.3).
$∣ x ∣$ , $∣ y ∣$ aur $∣ z ∣$ YZ-, ZX- aur XY-planes se $P$ ki distances hain (Coordinates planes se distances ki tarah).
Axes ke points ke forms $(x, 0, 0)$ , $(0, y, 0)$ , $(0, 0, z)$ hain; coordinate planes ke points ke forms $(x, y, 0)$ , $(0, y, z)$ , $(x, 0, z)$ hain (Fig. 11.6).
$P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ ((11.3)), aur $O P = \sqrt{x^{2} + y^{2} + z^{2}}$ ((11.4)).
Distances collinearity, triangle ke prakaar aur parallelograms tay karti hain, aur distance conditions ko surfaces ki equations mein badal deti hain (Section 11.4).
Midpoints aur centroids coordinates ka average lete hain; space mein quadrilateral tab parallelogram hota hai jab uske diagonals ka midpoint common ho (midpoint and centroid).

Flash cards · 7

Key ideas

1 / 7

Chapter 11 par Miscellaneous Exercise

Practice · 7 questions

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

Short answerMediumNCERT Miscellaneous Exercise, Q1
$A (2, - 3, 1)$ , $B (4, 1, - 2)$ aur $C (1, 2, 3)$ ek parallelogram $A B C D$ ke lagaatar corners hain. Chautha corner $D$ nikaalo.
Solution dikhao
Answer $D (- 1, - 2, 6)$
$A C$ ka midpoint $(\frac{3}{2}, - \frac{1}{2}, 2)$ hai. Yahi $B D$ ka midpoint bhi hona chahiye.
$D = (x, y, z)$ ke saath: $\frac{4 + x}{2} = \frac{3}{2}$ , $\frac{1 + y}{2} = - \frac{1}{2}$ aur $\frac{- 2 + z}{2} = 2$ , isliye $x = - 1$ , $y = - 2$ , $z = 6$ .
2
NumericalMediumNCERT Miscellaneous Exercise, Q2
Triangle ke vertices $A (4, 0, 0)$ , $B (0, 2, 0)$ aur $C (0, 0, 6)$ hain. Length nikaalo
1. $A$ se median ki
  Answer dekho
  Answer 5.099
  $B C$ ka midpoint $(0, 1, 3)$ hai; median $\sqrt{16 + 1 + 9} = \sqrt{26} \approx 5.10$ hai.
2. $B$ se median ki
  Answer dekho
  Answer 4.123
  $C A$ ka midpoint $(2, 0, 3)$ hai; median $\sqrt{4 + 4 + 9} = \sqrt{17} \approx 4.12$ hai.
3. $C$ se median ki
  Answer dekho
  Answer 6.403
  $A B$ ka midpoint $(2, 1, 0)$ hai; median $\sqrt{4 + 1 + 36} = \sqrt{41} \approx 6.40$ hai. Check: $26 + 17 + 41 = 84 = \frac{3}{4} (20 + 40 + 52)$ , jaisa squared medians ka sum hona hi chahiye.
Short answerMediumNCERT Miscellaneous Exercise, Q3
Triangle $P Q R$ ka centroid $(1, 2, - 1)$ hai, jahan $P = (3 a, - 2, 5)$ , $Q = (1, 4 b, - 8)$ aur $R = (- 7, 6, c + 1)$ . $a$ , $b$ aur $c$ nikaalo.
Solution dikhao
Answer $a = 3$ , $b = \frac{1}{2}$ , $c = - 1$
$x$ : $3 a + 1 - 7 = 3$ , isliye $a = 3$ . $y$ : $- 2 + 4 b + 6 = 6$ , isliye $b = \frac{1}{2}$ . $z$ : $5 - 8 + c + 1 = - 3$ , isliye $c = - 1$ .
Long answerHardNCERT Miscellaneous Exercise, Q4
$A$ aur $B$ $(1, 0, - 2)$ aur $(- 3, 2, 4)$ hain, aur $k$ ek constant hai. Us point $P$ ke locus ki equation nikaalo jo aise move karta hai ki $P A^{2} + P B^{2} = k^{2}$ . $k$ ki kin values ke liye aisa point exist karta hai?
Solution dikhao
Answer $2 (x^{2} + y^{2} + z^{2}) + 4 x - 4 y - 4 z + 34 - k^{2} = 0$ ; points tabhi exist karte hain jab $k^{2} \geq 28$ .
$P A^{2} = (x - 1)^{2} + y^{2} + (z + 2)^{2}$ aur $P B^{2} = (x + 3)^{2} + (y - 2)^{2} + (z - 4)^{2}$ .
Jodne par: $2 x^{2} + 2 y^{2} + 2 z^{2} + 4 x - 4 y - 4 z + 34 = k^{2}$ , kyunki constants $1 + 4 + 9 + 4 + 16 = 34$ hain.
Aadha karo aur squares complete karo: $(x + 1)^{2} + (y - 1)^{2} + (z - 1)^{2} = \frac{k^{2}}{2} - 14$ . Left side kabhi negative nahi hota, isliye points tabhi exist karte hain jab $k^{2} \geq 28$ .
$k^{2} > 28$ ke liye set ek sphere hai jiska centre $(- 1, 1, 1)$ hai, yaani $A B$ ka midpoint. $k^{2} = 28$ ke liye yeh sirf wahi ek point hai.
Quick answerMedium
Point $(6, - 2, 3)$ ki $y$ -axis se distance hai
$7$ $3 \sqrt{5}$ $\sqrt{13}$ $2 \sqrt{10}$
Answer dekho
Answer (B) $3 \sqrt{5}$
$\sqrt{6^{2} + 3^{2}} = \sqrt{45} = 3 \sqrt{5}$ . Baaki options origin ( $7$ ), $x$ -axis ( $\sqrt{13}$ ) aur $z$ -axis ( $2 \sqrt{10}$ ) se distances hain.
Quick answerMedium
Assertion (A): Point $(- 2, 0, 5)$ octant II mein hai. Reason (R): Octant II mein $x$ -coordinate negative hota hai aur $y$ - aur $z$ -coordinates positive.
A aur R dono true hain, aur R A ko explain karta haiA aur R dono true hain, lekin R A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (D) A false hai lekin R true hai
R octant II ka sahi sign pattern hai. Lekin point ka $y = 0$ hai, isliye woh ZX-plane par hai aur kisi octant mein nahi: A false hai.
Quick answerHard
$k$ ki kis value ke liye $(1, 2, 3)$ , $(3, 6, 7)$ aur $(k, 10, 11)$ collinear hain?
Answer dekho
Answer 5
Pehle point se doosre tak, $y$ aur $z$ $4$ badhte hain jabki $x$ $2$ badhta hai. Teesre point tak $y$ aur $z$ phir $4$ badhte hain, isliye $x$ ko bhi phir $2$ badhna chahiye: $k = 5$ . Distances bhi yahi confirm karti hain: $6 + 6 = 12$ .

Timeline

Space mein coordinates

1637
René Descartes
La Géométrie plane curves ko equations mein badal deti hai.
1700
Antoine Parent
Ek surface ko teen coordinates mein equation ki tarah likhte hain.
1731
Alexis Clairaut
Space mein curves par ek book publish karte hain, jo unhone lagbhag solah saal ki umar mein likhi thi.
1748
Leonhard Euler
Unki Introductio ke volume 2 ka appendix surfaces ko systematically treat karta hai.
1854
Bernhard Riemann
Ek famous lecture kisi bhi number of dimensions wale spaces ki ijaazat deta hai.
1908
Hermann Minkowski
Time ko chauthe coordinate ki tarah jodte hain: Einstein ki relativity ka space-time.

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 11.1, Q1	Practice 11.1 · Q1
NCERT Exercise 11.1, Q2	Practice 11.1 · Q2
NCERT Exercise 11.1, Q3	Practice 11.1 · Q3
NCERT Exercise 11.1, Q4	Practice 11.1 · Q4
NCERT Exercise 11.2, Q1	Practice 11.2 · Q1
NCERT Exercise 11.2, Q2	Practice 11.2 · Q2
NCERT Exercise 11.2, Q3	Practice 11.2 · Q3
NCERT Exercise 11.2, Q4	Practice 11.2 · Q4
NCERT Exercise 11.2, Q5	Practice 11.2 · Q5
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4

Quick check: MCQs

11.1 Introduction2 questions
11.2 Three Dimensional Space mein Coordinate Axes aur Coordinate Planes4 questions
Coordinates planes se distances ki tarah4 questions
11.4 Do Points ke beech Distance4 questions
Miscellaneous Examples3 questions

Glossary · 12

Key terms (English + Hindi)

centroidकेंद्रकkendrak: Triangle ke teen medians ka common point; iske coordinates vertices ke coordinates ka average hain.Chapter mein pehli baar yahan aaya →
collinearसंरेखsanrekh: Ek hi straight line par hona.Chapter mein pehli baar yahan aaya →
coordinate axesनिर्देशांक अक्षnirdeshank aksh: Origin se guzarne wali teen mutually perpendicular lines, $x$ -, $y$ - aur $z$ -axes, jinke along coordinates naape jaate hain.Chapter mein pehli baar yahan aaya →
coordinate planeनिर्देशांक तलnirdeshank tal: Do axes wala plane: XY-, YZ- ya ZX-plane.
coordinates of a pointनिर्देशांकnirdeshank: YZ-, ZX- aur XY-planes se kisi point ki signed distances ka ordered triplet $(x, y, z)$ .Chapter mein pehli baar yahan aaya →

distance formulaदूरी सूत्रdoori sootra: Rule $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ , jahan $P$ aur $Q$ ke coordinates $(x_{1}, y_{1}, z_{1})$ aur $(x_{2}, y_{2}, z_{2})$ hain.Chapter mein pehli baar yahan aaya →

locusबिंदुपथbindupath: Kisi di gayi condition ko satisfy karne wale saare points ka set; space mein yeh aksar ek surface hota hai.Chapter mein pehli baar yahan aaya →

midpointमध्य-बिंदुmadhya-bindu: Segment ka woh point jo uske dono ends se barabar door hai; iske coordinates unke coordinates ka average hain.Chapter mein pehli baar yahan aaya →

octantअष्टांशashtansh: Un aath regions mein se ek jinmein teen coordinate planes space ko baantte hain.Chapter mein pehli baar yahan aaya →
originमूल बिंदुmool bindu: Point $O$ jahan teeno axes milte hain; iske coordinates $(0, 0, 0)$ hain.Chapter mein pehli baar yahan aaya →

rectangular coordinate system: Origin, teen mutually perpendicular axes aur unse bane teen coordinate planes, sab milkar.Chapter mein pehli baar yahan aaya →
rectangular parallelepipedघनाभghanabh: Chhe rectangular faces se ghira solid, jaise eent ya band carton.Chapter mein pehli baar yahan aaya →

11.1 Introduction

11.2 Three Dimensional Space mein Coordinate Axes aur Coordinate Planes

11.3 Space mein Point ke Coordinates

Coordinates planes se distances ki tarah

11.4 Do Points ke beech Distance

Miscellaneous Examples

Chapter 11 par Miscellaneous Exercise

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye