Class 11 Maths · Chapter 11
Short answer:
Class 11 Maths Chapter 11 — Introduction to Three Dimensional Geometry teen naye ideas deta hai: 3D me coordinate axes/planes, kisi point ke 3D coordinates (x, y, z), do points ke beech distance formula, aur section formula (internal division). Yeh chapter chhota hai (sirf 2 exercises + Miscellaneous) par 2D coordinate geometry (Straight Lines, Conic Sections) ka natural 3D extension hai — is liye agar tum class 11 maths ncert solutions chapter-by-chapter revise kar rahe ho to Straight Lines aur Conic Sections ke baad isse padhna sabse logical order hai. Full class 11 maths formulas pdf download karke bhi is chapter ke 4-5 formulas ek page pe revise ho jaate hain — utna hi content hai.
Class 10 tak humne sirf 2D coordinate geometry padhi — ek point define karne ke liye do coordinates (x, y) kaafi the, aur woh ek flat plane pe rehta tha. Class 11 Maths Chapter 11 "Introduction to Three Dimensional Geometry" us plane ko real duniya jaisa banata hai — ab har point ek room ke corner jaisa hai, jisko locate karne ke liye teen numbers chahiye: length, breadth, height — yaani (x, y, z).
Yeh chapter foundation-level hai — Class 12 me jo "Three Dimensional Geometry" (direction cosines, lines, planes) padhoge uski poori base yahi chapter deta hai. Isliye is chapter ko halka mat lo sirf isliye ki isme sirf 2 exercises hain. Yahan jo distance formula aur section formula seekhoge, woh exactly wahi logic hai jo Class 11 me hi Straight Lines chapter (2D) me padha tha — bas ek extra z-coordinate add ho gaya hai. Agar tumne class 11 maths straight lines ncert solutions already revise kiye hain, to yeh chapter bahut fast lagega.
Chapter 4 sections cover karta hai: coordinate axes/planes in 3D, kisi bhi point ke coordinates likhna, distance between two points, aur section formula (ek line segment ko given ratio me divide karna). Miscellaneous Exercise me thoda tricky application-based questions hain — parallelogram ka 4th vertex, triangle ki medians, centroid se unknowns nikalna, aur locus (set of points) wale questions, jo application-based thinking test karte hain.
Chapter 11 Summary — 5 Minute Revision
Is chapter ka poora core sirf 4 cheezon pe tika hai: (1) 3D space me x, y, z axes ek doosre ko perpendicular origin pe cut karte hain aur 3 coordinate planes (XY, YZ, ZX) banate hain jo space ko 8 octants me divide karte hain; (2) kisi point P ka coordinate (x, y, z) uski teeno axes se signed distance batata hai; (3) distance formula — d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²] — origin se distance ke liye sirf √(x²+y²+z²); (4) section formula — kisi point ko m:n ratio me internally divide karne wale point ke coordinates nikalna, jiska special case midpoint aur centroid formula hai.
Exam ke liye priority: distance formula ke sawaal (collinearity check, isosceles/right-triangle verify) sabse zyada aate hain, uske baad octant naming aur locus/equidistant-point wale application questions (Miscellaneous Exercise). Formula yaad rakhna kaafi nahi — sign convention (kaunsa octant kaunse (±,±,±) combination se match karta hai) bhi clearly yaad rakho, warna MCQ ya short-answer type questions me galti ho jaati hai.
In-Text Questions — Solutions
1. Ek point x-axis par hai. Uske y-coordinate aur z-coordinate kya honge?
x-axis par jo bhi point hoga, uska y aur z dono coordinates hamesha zero hote hain, kyunki x-axis wahi line hai jahan sirf x badalta hai, baaki dono planes (y aur z) ko point chhoo raha hota hai.
Point ka form: (x, 0, 0)
2. Ek point XZ-plane me hai. Uske y-coordinate ke baare me kya keh sakte ho?
XZ-plane wo plane hai jaha y-axis ki direction me koi displacement nahi hota — is liye XZ-plane ke har point ka y-coordinate = 0 hota hai.
Point ka form: (x, 0, z)
3. In points ke octant batao: (1, 2, 3), (4, −2, 3), (4, −2, −5), (4, 2, −5), (−4, 2, −5), (−4, 2, 5), (−3, −1, 6), (2, −4, −7)
Octant sign convention yaad rakho — pehle 4 octants me z positive (+), agle 4 me z negative (−); har group me x, y ka sign cyclic order me change hota hai.
- (1, 2, 3) → (+,+,+) → Octant I
- (4, −2, 3) → (+,−,+) → Octant IV
- (4, −2, −5) → (+,−,−) → Octant VIII
- (4, 2, −5) → (+,+,−) → Octant V
- (−4, 2, −5) → (−,+,−) → Octant VI
- (−4, 2, 5) → (−,+,+) → Octant II
- (−3, −1, 6) → (−,−,+) → Octant III
- (2, −4, −7) → (+,−,−) → Octant VIII
4. Blanks fill karo: (i) x-axis aur y-axis milkar jo plane banate hain use ______ kehte hain. (ii) XY-plane ke points ka form ______ hota hai. (iii) Coordinate planes space ko ______ octants me divide karte hain.
(i) XY-plane (kyunki x-axis aur y-axis dono XY-plane me lete hain).
(ii) XY-plane ka har point z = 0 pe hota hai, isliye form: (x, y, 0).
(iii) Teen coordinate planes (XY, YZ, ZX) space ko eight (8) octants me divide karte hain.
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q11)
Ex 11.2, Q1. P(1, −3, 4) aur Q(−4, 1, 2) ke beech distance nikalo.
Distance formula lagate hain:
d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
d = √[(−4−1)² + (1−(−3))² + (2−4)²]
d = √[(−5)² + (4)² + (−2)²] = √[25 + 16 + 4] = √45
d = 3√5 units
Ex 11.2, Q2. Dikhao ki (−2, 3, 5), (1, 2, 3), (7, 0, −1) collinear hain.
Let A(−2,3,5), B(1,2,3), C(7,0,−1). Teeno pairwise distances nikalte hain:
AB = √[(1+2)² + (2−3)² + (3−5)²] = √[9+1+4] = √14
BC = √[(7−1)² + (0−2)² + (−1−3)²] = √[36+4+16] = √56 = 2√14
AC = √[(7+2)² + (0−3)² + (−1−5)²] = √[81+9+36] = √126 = 3√14
Ab check karo: AB + BC = √14 + 2√14 = 3√14 = AC
Chunki sum of two smaller distances = largest distance, teeno points ek hi line par hain — collinear proved.
Ex 11.2, Q3. Verify karo ki (0, 7, −10), (1, 6, −6), (4, 9, −6) ek isosceles triangle ke vertices hain.
Let A(0,7,−10), B(1,6,−6), C(4,9,−6). Calculation aasan rakhne ke liye squares nikalte hain:
AB² = (1−0)² + (6−7)² + (−6+10)² = 1 + 1 + 16 = 18 → AB = √18
BC² = (4−1)² + (9−6)² + (−6+6)² = 9 + 9 + 0 = 18 → BC = √18
CA² = (4−0)² + (9−7)² + (−6+10)² = 16 + 4 + 16 = 36 → CA = 6
Chunki AB = BC = √18, do sides equal hain — isosceles triangle verified.
Ex 11.2, Q4. Verify karo ki (0, 7, 10), (−1, 6, 6), (−4, 9, 6) ek right-angled triangle ke vertices hain.
Let A(0,7,10), B(−1,6,6), C(−4,9,6). Squares of sides nikalte hain:
AB² = (−1)² + (−1)² + (−4)² = 1+1+16 = 18
BC² = (−3)² + (3)² + (0)² = 9+9+0 = 18
CA² = (−4)² + (2)² + (−4)² = 16+4+16 = 36
Check Pythagoras: AB² + BC² = 18 + 18 = 36 = CA²
Yeh condition satisfy karta hai, is liye triangle right-angled hai (right angle vertex B par hai).
Ex 11.2, Q5. Un points ka set (equation) nikalo jo (1, 2, 3) aur (3, 2, −1) se equidistant hain.
Let P(x, y, z) ek aisa point ho jo A(1,2,3) aur B(3,2,−1) se equidistant hai, matlab PA = PB, ya PA² = PB².
(x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)²
(y−2)² dono side cancel ho jata hai. Baaki expand karte hain:
(x²−2x+1) + (z²−6z+9) = (x²−6x+9) + (z²+2z+1)
−2x + 1 − 6z + 9 = −6x + 9 + 2z + 1
−2x − 6z + 10 = −6x + 2z + 10
4x − 8z = 0 ⟹ x − 2z = 0
Required set of points ka equation: x − 2z = 0.
Misc, Q1. Parallelogram ABCD ke teen vertices A(3, −1, 2), B(1, 2, −4), C(−1, 1, 2) diye hain. Chautha vertex D nikalo.
Parallelogram me diagonals ek doosre ko bisect karte hain — matlab AC ka midpoint = BD ka midpoint.
Midpoint of AC = ((3+(−1))/2, (−1+1)/2, (2+2)/2) = (1, 0, 2)
Let D = (x, y, z). Midpoint of BD bhi (1, 0, 2) hona chahiye:
(1+x)/2 = 1 ⟹ x = 1
(2+y)/2 = 0 ⟹ y = −2
(−4+z)/2 = 2 ⟹ z = 8
Fourth vertex: D(1, −2, 8).
Misc, Q2. Triangle A(0,0,6), B(0,4,0), C(6,0,0) ki medians ki lengths nikalo.
Median A se BC ke midpoint tak, median B se AC ke midpoint tak, aur median C se AB ke midpoint tak jaati hai.
Midpoint of BC = (3, 2, 0); AD (median from A) = √[(3−0)²+(2−0)²+(0−6)²] = √[9+4+36] = √49 = 7
Midpoint of AC = (3, 0, 3); BE (median from B) = √[(3−0)²+(0−4)²+(3−0)²] = √[9+16+9] = √34
Midpoint of AB = (0, 2, 3); CF (median from C) = √[(0−6)²+(2−0)²+(3−0)²] = √[36+4+9] = √49 = 7
Medians ki lengths: 7, √34, 7 units.
Misc, Q3. Agar origin, triangle PQR ka centroid hai jiske vertices P(2a, 2, 6), Q(−4, 3b, −10), R(8, 14, 2c) hain, to a, b, c nikalo.
Centroid formula: centroid ke coordinates = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3). Yahan centroid origin (0,0,0) hai.
(2a − 4 + 8)/3 = 0 ⟹ 2a + 4 = 0 ⟹ a = −2
(2 + 3b + 14)/3 = 0 ⟹ 3b + 16 = 0 ⟹ b = −16/3
(6 − 10 + 2c)/3 = 0 ⟹ 2c − 4 = 0 ⟹ c = 2
Answer: a = −2, b = −16/3, c = 2.
Misc, Q4. O(0,0,0), A(a,0,0), B(0,b,0), C(0,0,c) — in char points se equidistant point ke coordinates nikalo.
Let P(x, y, z) equidistant ho sabhi char points se, matlab PO = PA = PB = PC.
Pehle PO² = PA² lagate hain:
x² + y² + z² = (x−a)² + y² + z²
x² = x² − 2ax + a² ⟹ 2ax = a² ⟹ x = a/2
Isi tarah PO² = PB² se y = b/2, aur PO² = PC² se z = c/2 milta hai.
Required point: (a/2, b/2, c/2).
Misc, Q5. Un points P ka set nikalo jinke liye PA² + PB² = 2k², jahan A(3,4,5) aur B(−1,3,−7) hain.
Let P(x, y, z).
PA² = (x−3)² + (y−4)² + (z−5)²
PB² = (x+1)² + (y−3)² + (z+7)²
In dono ko add karte hain aur expand karte hain:
(x−3)²+(x+1)² = 2x² − 4x + 10
(y−4)²+(y−3)² = 2y² − 14y + 25
(z−5)²+(z+7)² = 2z² + 4z + 74
PA² + PB² = 2x²+2y²+2z² − 4x − 14y + 4z + 109 = 2k²
Simplify karke required equation: 2x² + 2y² + 2z² − 4x − 14y + 4z + (109 − 2k²) = 0.
Misc, Q6. Un points P ka set nikalo jinke A(4,0,0) aur B(−4,0,0) se distances ka sum 10 hai.
Condition: PA + PB = 10, jahan P(x,y,z).
PA = 10 − PB
Dono side square karke aur simplify karke (x−4)² − (x+4)² = −16x use karte hain:
−16x = 100 − 20·PB ⟹ PB = (25 + 4x)/5
Ab PB² ko wapas expand karke set karte hain:
(x+4)² + y² + z² = (25+4x)²/25
25(x²+8x+16+y²+z²) = 625 + 200x + 16x²
25x² + 200x + 400 + 25y² + 25z² = 625 + 200x + 16x²
9x² + 25y² + 25z² = 225
Dono side 225 se divide karte hain — required equation (ek ellipsoid):
x²/25 + y²/9 + z²/9 = 1
Important Equations — Ek Nazar Me
| Concept | Formula | Note |
|---|---|---|
| Distance between two points | d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²] | 2D distance formula ka hi 3D version — bas ek extra (z₂−z₁)² term |
| Distance from origin | d = √(x² + y² + z²) | Special case jab ek point origin (0,0,0) ho |
| Section formula (internal division, ratio m:n) | ( (mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n) ) | Point R jo P(x₁,y₁,z₁) aur Q(x₂,y₂,z₂) ko m:n me andar se divide karta hai |
| Midpoint formula | ( (x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2 ) | Section formula ka special case jab m = n = 1 |
| Centroid of a triangle | ( (x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3 ) | Teeno vertices ke coordinates ka average |
| Octant sign convention | I(+,+,+) · II(−,+,+) · III(−,−,+) · IV(+,−,+) · V(+,+,−) · VI(−,+,−) · VII(−,−,−) · VIII(+,−,−) | Pehle 4 octants z > 0 wale, agle 4 z < 0 wale |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Distance formula me squares add karke square root lagana bhool jana — final answer d² likh dena, d nahi.
- Section/midpoint formula me kisi ek point ka x-coordinate doosre point ke y ya z ke saath mix kar dena — hamesha subscript se match karke hi likho.
- Octant naming me sign convention galat yaad rakhna — yeh bhool jana ki pehle 4 octants me z positive hota hai aur agle 4 me negative.
- Collinearity prove karte waqt sirf do distances equal nikal ke ruk jana — teesri (sabse badi) side ka sum-check zaroori hai, warna proof incomplete rahega.
- Isosceles/right-triangle verify karte waqt har baar square root nikal ke calculation lambi kar dena — distance² (squared form) use karo, aakhir me hi root lo agar zarurat ho.
- Centroid formula me denominator '3' likhna bhool jana (2D wale muddat purane habit se) ya sirf 2 coordinates ka average le lena teeno ki jagah.
Board-Style Important Questions
- PYQ Pattern — Octant Identification: Point (−3, 1, 2) kis octant me hai?
Solution: signs (−,+,+) hain, jo Octant II ke corresponding hain — Answer: Octant II. - PYQ Pattern — Distance Formula: P(2, 3, −4) aur Q(1, −1, 3) ke beech distance nikalo.
Solution: d = √[(1−2)²+(−1−3)²+(3+4)²] = √[1+16+49] = √66 units. - PYQ Pattern — Collinearity Proof: Dikhao ki (1, −1, 3), (2, −4, 5), (5, −13, 11) collinear hain.
Solution: AB = √[1+9+4] = √14, BC = √[9+81+36] = √126 = 3√14, AC = √[16+144+64] = √224 = 4√14. Chunki AB+BC = √14+3√14 = 4√14 = AC, points collinear hain. - PYQ Pattern — Equidistant Point Set: A(1,2,3) aur B(3,2,−1) se equidistant points ka set (equation) nikalo.
Solution: PA²=PB² set karke expand karne par x − 2z = 0 milta hai (poori working intext/exercise section me di gayi hai). - PYQ Pattern — Parallelogram Vertex + Diagonal Length: Parallelogram ke teen vertices A(3,−1,2), B(1,2,−4), C(−1,1,2) diye hain — chautha vertex D nikalo, aur diagonal AC ki length bhi batao.
Solution: diagonals bisect karte hain method se D(1,−2,8) milta hai (working Miscellaneous Q1 me di gayi hai); AC = √[(3+1)²+(−1−1)²+(2−2)²] = √[16+4+0] = √20 = 2√5 units.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 11 me kitne exercises hain aur kitne questions solve karne padte hain?
Is chapter me sirf 2 exercises (11.1 me 4 questions, 11.2 me 5 questions) aur ek Miscellaneous Exercise (6 questions) hain — total lagbhag 15 questions, jo class 11 maths ke doosre chapters se kaafi kam hai. Isliye yeh chapter jaldi complete ho jaata hai agar concept clear ho.
Kya Class 11 Chapter 11 me direction cosines/ratios bhi aate hain?
Nahi. Direction cosines, direction ratios, line aur plane ki equations — yeh sab Class 12 ke Three Dimensional Geometry chapter me aata hai. Class 11 ka yeh chapter sirf coordinate system introduce karta hai — axes, octants, distance formula aur section formula tak hi limited hai.
Section formula 2D wale Straight Lines chapter se kaise alag hai?
Formula ka logic same hai — bas ek extra z-coordinate add ho jaata hai. Agar tumne class 11 maths straight lines ncert solutions me section formula practice ki hai, to yahan bas (mz₂+nz₁)/(m+n) wala extra term yaad rakhna hai, baaki sab same hai.
Kya is chapter se board exam me marks-heavy questions aate hain?
Is chapter se zyadatar direct-formula type questions (distance, octant, section formula) aate hain, jabki Miscellaneous Exercise wale locus-type questions thode application-based aur tricky hote hain. Exact marks-weightage CBSE ke official curriculum/sample-paper document se hi confirm karna sahi rahega — yahan koi specific marks value claim nahi kiya ja raha.
Class 11 maths ke sabhi formulas ek jagah kaise revise karein — including yeh chapter?
Har chapter ke solution page ke saath ek formula table milta hai (jaisa upar diya hai) — Sets (chapter 1), Trigonometric Functions (chapter 3), Sequences and Series, Straight Lines, Conic Sections, Limits and Derivatives sab ke formulas ekjagah collect karke apna khud ka class 11 maths formulas pdf download-ready revision sheet bana sakte ho.
Chapter 11 padhne se pehle kaunse chapters revise kar lene chahiye?
Coordinate geometry ka basic logic Straight Lines chapter (2D) se aata hai — us chapter ke distance/section formula clear hone chahiye. Agar wahan doubt hai to pehle class 11 maths straight lines ncert solutions revise kar lo, phir yeh 3D chapter bahut smooth lagega.
Class 11 Maths — Saare Chapters
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Class 11 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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