Class 12 Maths · Chapter 11
Short answer:
NCERT Class 12 Maths Chapter 11 — Three Dimensional Geometry — me total 3 exercises hain (11.1, 11.2, aur Miscellaneous), roughly 20 questions cover karte hain. Direction cosines/ratios se shuru hoke, line ka equation (vector aur Cartesian form, dono two-point aur point-direction), do lines ke beech angle, skew lines ke beech shortest distance, plane ka equation (normal form, teen points se, intercept form), do planes ke beech angle, line aur plane ke beech angle, aur point se plane ki distance — sab yahan step-by-step solve kiya gaya hai.
3D geometry me har line aur plane ko vectors ya coordinates ke through define karte hain, aur unke beech ke angles/distances nikalna is chapter ka core hai. Class 11 me humne 2D coordinate geometry padhi thi; ab teesra axis (z-axis) add hone se lines skew ho sakti hain — ye naya concept hai jo 2D me possible nahi tha. Is chapter me formula-heavy kaam hai, isliye har formula ka use-case pehchanna zaroori hai.
Chapter 11 Summary — 5 Minute Revision
1. Direction Cosines aur Direction Ratios
Agar ek line, x, y, z axes se angles α, β, γ banati hai, to l = cos α, m = cos β, n = cos γ uske direction cosines (d.c.'s) kehlate hain. Inka fundamental relation:
l² + m² + n² = 1
Direction ratios (a, b, c) kisi bhi numbers hote hain jo d.c.'s ke proportional hon (a = λl, b = λm, c = λn). Inse d.c.'s nikalne ke liye normalize karna padta hai:
l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)
Do points P(x₁,y₁,z₁) aur Q(x₂,y₂,z₂) ko joinne wali line ke direction ratios: (x₂−x₁, y₂−y₁, z₂−z₁).
2. Line ka Equation
| Form | Point + Direction diya ho | Do Points diye hon |
|---|---|---|
| Vector | r = a + λb | r = a + λ(b − a) |
| Cartesian | (x−x₁)/a = (y−y₁)/b = (z−z₁)/c | (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁) |
↔ Table ko side me swipe karein
Yaha a = position vector of a known point, b = direction vector, λ = parameter.
3. Do Lines ke beech Angle
Agar lines ke direction ratios (a₁,b₁,c₁) aur (a₂,b₂,c₂) hain:
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁²+b₁²+c₁²) · √(a₂²+b₂²+c₂²))
Parallel lines: a₁/a₂ = b₁/b₂ = c₁/c₂. Perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0.
4. Skew Lines aur Shortest Distance
Do lines skew hoti hain jab wo na parallel hon na intersect karti hon — 3D me ye possible hai kyunki lines alag planes me ho sakti hain. Unke beech shortest distance nikalne ka formula thoda alag hota hai. Do lines r = a₁ + λb₁ aur r = a₂ + μb₂ ke beech shortest distance:
d = |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|
Agar do lines parallel hain (r = a₁ + λb aur r = a₂ + μb), to distance:
d = |b × (a₂ − a₁)| / |b|
5. Plane ka Equation
| Type | Vector Form | Cartesian Form |
|---|---|---|
| Normal form (distance d se origin) | r · n̂ = d | lx + my + nz = d |
| Point + normal diya ho | (r − a) · n = 0 | a(x−x₁) + b(y−y₁) + c(z−z₁) = 0 |
| Intercept form | — | x/a + y/b + z/c = 1 |
↔ Table ko side me swipe karein
Teen points se plane: agar A, B, C plane pe hain, to normal n = AB × AC hota hai, aur plane ka equation (r − A) · n = 0 se milta hai. Cross product ka order (AB × AC ya AC × AB) sirf normal ki direction flip karta hai — plane ka equation wahi rehta hai.
6. Do Planes ke beech Angle
Planes a₁x+b₁y+c₁z=d₁ aur a₂x+b₂y+c₂z=d₂ ke beech angle (normal vectors ke beech angle):
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁²+b₁²+c₁²) · √(a₂²+b₂²+c₂²))
7. Line aur Plane ke beech Angle
Line ka direction vector b aur plane ka normal n ho, to line aur plane ke beech angle φ (normal-line angle se complementary):
sin φ = |b · n| / (|b| |n|)
Note: yaha sin use hota hai, cos nahi — kyunki φ line aur plane ke beech ka angle hai, line aur normal ke beech ka nahi.
8. Point se Plane ki Distance
Point (x₁,y₁,z₁) se plane ax+by+cz+d=0 ki distance:
Distance = |ax₁ + by₁ + cz₁ + d| / √(a²+b²+c²)
Poore Class 12 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–Q21)
Q1. Ek line x, y, z axes se equal angles banati hai. Uske direction cosines nikalo.
Angles equal hain, so l = m = n. Fundamental relation l²+m²+n²=1 use karte hain:
l² + l² + l² = 1 ⟹ 3l² = 1 ⟹ l² = 1/3 ⟹ l = ±1/√3
Isliye direction cosines = ±1/√3, ±1/√3, ±1/√3 (same sign, kyunki equal angles).
Q2. Point (2, 3, −4) aur origin ko joinne wali line ke direction cosines nikalo.
Direction ratios = (2−0, 3−0, −4−0) = (2, 3, −4).
|r| = √(2² + 3² + (−4)²) = √(4+9+16) = √29
d.c.'s = 2/√29, 3/√29, −4/√29
Q3. Points A(2,3,4) aur B(4,1,−2) ko joinne wali line ke direction cosines nikalo.
Direction ratios = (4−2, 1−3, −2−4) = (2, −2, −6).
magnitude = √(2² + (−2)² + (−6)²) = √(4+4+36) = √44 = 2√11
d.c.'s = 2/(2√11), −2/(2√11), −6/(2√11) = 1/√11, −1/√11, −3/√11
Q4. Show karo ki points (2,3,4), (−1,−2,1), (5,8,7) collinear hain.
Let A(2,3,4), B(−1,−2,1), C(5,8,7).
AB ke direction ratios = (−1−2, −2−3, 1−4) = (−3,−5,−3)
BC ke direction ratios = (5−(−1), 8−(−2), 7−1) = (6,10,6) = 2×(3,5,3) = −2×(−3,−5,−3)
Chunki BC, AB ka scalar multiple hai (parallel), aur ek common point B hai, isliye A, B, C collinear hain.
Q5. Point (1,2,3) se guzarne wali line jo vector 3î+2ĵ−2k̂ ke parallel hai, uska vector aur Cartesian equation likho.
Point a = î+2ĵ+3k̂, direction b = 3î+2ĵ−2k̂.
Vector form: r = (î+2ĵ+3k̂) + λ(3î+2ĵ−2k̂)
Cartesian form: (x−1)/3 = (y−2)/2 = (z−3)/(−2)
Q6. Points A(3,4,−7) aur B(1,−1,6) se guzarne wali line ka vector aur Cartesian equation likho.
Direction ratios = (1−3, −1−4, 6−(−7)) = (−2, −5, 13).
Vector form: r = (3î+4ĵ−7k̂) + λ(−2î−5ĵ+13k̂)
Cartesian form: (x−3)/(−2) = (y−4)/(−5) = (z+7)/13
Q7. Do lines r = î+ĵ+λ(2î+ĵ+2k̂) aur r = 2î+ĵ−k̂+μ(3î−5ĵ+2k̂) ke beech angle nikalo.
b₁ = 2î+ĵ+2k̂, b₂ = 3î−5ĵ+2k̂.
b₁·b₂ = 2(3) + 1(−5) + 2(2) = 6−5+4 = 5
|b₁| = √(4+1+4) = 3, |b₂| = √(9+25+4) = √38
cos θ = |5| / (3√38) = 5/(3√38)
θ = cos⁻¹(5/(3√38))
Q8. Lines (x−1)/2 = (y−2)/2 = (z−3)/1 aur (x+3)/(−1) = (y−5)/8 = (z−7)/4 ke beech angle nikalo.
Direction ratios: (2,2,1) aur (−1,8,4).
a₁a₂+b₁b₂+c₁c₂ = 2(−1) + 2(8) + 1(4) = −2+16+4 = 18
|d1| = √(4+4+1) = 3, |d2| = √(1+64+16) = 9
cos θ = |18|/(3×9) = 18/27 = 2/3
θ = cos⁻¹(2/3)
Q9. Check karo ki lines (x−5)/7 = (y+2)/(−5) = z/1 aur x/1 = y/2 = z/3 perpendicular hain ya nahi.
Direction ratios: (7,−5,1) aur (1,2,3).
a₁a₂+b₁b₂+c₁c₂ = 7(1) + (−5)(2) + 1(3) = 7−10+3 = 0
Dot product zero hai, isliye lines perpendicular hain.
Q10. Do lines r = î−ĵ+λ(2î+k̂) aur r = 2î−ĵ+μ(î+ĵ−k̂) ke beech shortest distance nikalo (check karo ye intersecting hain ya skew).
a₁ = î−ĵ, b₁ = 2î+k̂; a₂ = 2î−ĵ, b₂ = î+ĵ−k̂.
a₂ − a₁ = î
b₁ × b₂ = |î ĵ k̂; 2 0 1; 1 1 −1| = î(0×(−1)−1×1) − ĵ(2×(−1)−1×1) + k̂(2×1−0×1)
= î(−1) − ĵ(−3) + k̂(2) = −î+3ĵ+2k̂
(a₂−a₁)·(b₁×b₂) = (î)·(−î+3ĵ+2k̂) = −1
|b₁×b₂| = √(1+9+4) = √14
d = |−1|/√14 = 1/√14
Distance zero nahi hai, isliye lines intersecting nahi — skew hain, aur shortest distance = 1/√14 units.
Q11. Do parallel lines r = î+2ĵ+4k̂+λ(2î+3ĵ+6k̂) aur r = 3î+3ĵ−5k̂+μ(2î+3ĵ+6k̂) ke beech distance nikalo.
Same direction b = 2î+3ĵ+6k̂, so lines parallel hain. a₁ = î+2ĵ+4k̂, a₂ = 3î+3ĵ−5k̂.
a₂−a₁ = 2î+ĵ−9k̂
b × (a₂−a₁) = |î ĵ k̂; 2 3 6; 2 1 −9|
= î(3×(−9)−6×1) − ĵ(2×(−9)−6×2) + k̂(2×1−3×2)
= î(−27−6) − ĵ(−18−12) + k̂(2−6) = −33î+30ĵ−4k̂
|b×(a₂−a₁)| = √(33²+30²+4²) = √(1089+900+16) = √2005
|b| = √(4+9+36) = 7
d = √2005 / 7 units
Q12. Plane ka equation likho jo point (1,0,−2) se guzarti hai aur normal vector î+ĵ−k̂ ke perpendicular hai.
Point a=(1,0,−2), normal n = î+ĵ−k̂.
Vector form: (r − a)·n = 0 ⟹ r·(î+ĵ−k̂) = (î−2k̂)·(î+ĵ−k̂) = 1+0+2 = 3
Cartesian form: 1(x−1) + 1(y−0) − 1(z+2) = 0 ⟹ x + y − z − 3 = 0
Q13. Teen points (1,1,0), (1,2,1), (−2,2,−1) se guzarne wale plane ka equation nikalo.
A(1,1,0), B(1,2,1), C(−2,2,−1).
AB = (0,1,1), AC = (−3,1,−1)
n = AB × AC = |î ĵ k̂; 0 1 1; −3 1 −1|
= î(1×(−1)−1×1) − ĵ(0×(−1)−1×(−3)) + k̂(0×1−1×(−3))
= î(−1−1) − ĵ(0+3) + k̂(0+3) = −2î−3ĵ+3k̂
Plane: −2(x−1) − 3(y−1) + 3(z−0) = 0 ⟹ −2x−3y+3z+5 = 0 ⟹ 2x+3y−3z−5=0
Q14. Plane 2x + 3y + 4z − 12 = 0 ka intercept form likho, aur x, y, z axes pe intercepts batao.
Equation ko d=1 form me lao: 2x/12 + 3y/12 + 4z/12 = 1
x/6 + y/4 + z/3 = 1
Isliye x-intercept = 6, y-intercept = 4, z-intercept = 3.
Q15. Do planes 2x + y − 2z = 5 aur 3x − 6y − 2z = 7 ke beech angle nikalo.
Normals: n₁=(2,1,−2), n₂=(3,−6,−2).
n₁·n₂ = 2(3)+1(−6)+(−2)(−2) = 6−6+4 = 4
|n₁| = √(4+1+4) = 3, |n₂| = √(9+36+4) = 7
cos θ = |4|/(3×7) = 4/21
θ = cos⁻¹(4/21)
Q16. Check karo ki planes 7x + 5y + 6z + 30 = 0 aur 3x − y − 10z + 4 = 0 perpendicular hain.
Normals: (7,5,6) aur (3,−1,−10).
n₁·n₂ = 7(3) + 5(−1) + 6(−10) = 21−5−60 = −44
Dot product zero nahi hai, isliye planes perpendicular nahi hain.
Q17. Line (x+1)/2 = y/3 = (z−3)/6 aur plane 10x + 2y − 11z = 3 ke beech angle nikalo.
Line direction b = (2,3,6), plane normal n = (10,2,−11).
b·n = 2(10)+3(2)+6(−11) = 20+6−66 = −40
|b| = √(4+9+36) = 7, |n| = √(100+4+121) = 15
sin φ = |−40|/(7×15) = 40/105 = 8/21
φ = sin⁻¹(8/21)
Q18. Point (−1, −5, −10) ki plane r·(î−2ĵ+k̂) = 5 se distance nikalo (line r = 2î−ĵ+2k̂+λ(3î+4ĵ+2k̂) ke intersection point ki bhi).
Seedha point-to-plane distance nikalte hain. Plane Cartesian: x − 2y + z = 5, i.e. x−2y+z−5=0. Point (−1,−5,−10):
Distance = |1(−1) − 2(−5) + 1(−10) − 5| / √(1²+(−2)²+1²)
= |−1+10−10−5| / √6 = |−6|/√6 = 6/√6 = √6 units
Q19. Origin se plane 2x − 3y + 6z + 14 = 0 ki distance nikalo, aur foot of perpendicular ke direction cosines batao.
Point (0,0,0), plane 2x−3y+6z+14=0:
Distance = |2(0)−3(0)+6(0)+14| / √(4+9+36) = 14/7 = 2 units
Normal direction ratios (2,−3,6), magnitude 7, isliye d.c.'s = 2/7, −3/7, 6/7.
Q20 (Miscellaneous). Point (0,0,0) se plane 3x−4y+12z=3 pe foot of perpendicular ki distance, aur perpendicular line ka equation nikalo.
Normal direction ratios (3,−4,12), magnitude = √(9+16+144)=13.
Distance = |3(0)−4(0)+12(0)−3| / 13 = 3/13 units
Perpendicular line origin se, direction (3,−4,12) ke saath:
Cartesian: x/3 = y/(−4) = z/12
Q21 (Miscellaneous). Line r = 2î−2ĵ+3k̂+λ(î−ĵ+4k̂) aur plane r·(î+5ĵ+k̂)=5 ke beech angle nikalo, aur bataiye ye parallel hain ya nahi.
b = (1,−1,4), n = (1,5,1).
b·n = 1(1)+(−1)(5)+4(1) = 1−5+4 = 0
b·n = 0 ka matlab hai b, n ke perpendicular hai — isliye line, plane ke parallel hai (aur plane me nahi hai, kyunki point (2,−2,3) plane satisfy nahi karta: 2+(−10)+3=−5≠5).
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| D.c.'s relation | l² + m² + n² = 1 |
| D.r.'s se d.c.'s | l = a/√(a²+b²+c²), similarly m, n |
| Line — vector form | r = a + λb |
| Line — Cartesian (point+direction) | (x−x₁)/a = (y−y₁)/b = (z−z₁)/c |
| Line — Cartesian (two points) | (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁) |
| Angle between two lines | cos θ = |a₁a₂+b₁b₂+c₁c₂| / (√Σa₁² · √Σa₂²) |
| Lines parallel | a₁/a₂ = b₁/b₂ = c₁/c₂ |
| Lines perpendicular | a₁a₂+b₁b₂+c₁c₂ = 0 |
| Shortest distance — skew lines | d = |(a₂−a₁)·(b₁×b₂)| / |b₁×b₂| |
| Distance — parallel lines | d = |b×(a₂−a₁)| / |b| |
| Plane — normal form (vector) | r · n̂ = d |
| Plane — point + normal | a(x−x₁)+b(y−y₁)+c(z−z₁) = 0 |
| Plane — intercept form | x/a + y/b + z/c = 1 |
| Plane through 3 points | normal n = AB × AC, then (r−A)·n = 0 |
| Angle between two planes | cos θ = |a₁a₂+b₁b₂+c₁c₂| / (√Σa₁² · √Σa₂²) |
| Angle between line and plane | sin φ = |b·n| / (|b||n|) |
| Distance of point from plane | |ax₁+by₁+cz₁+d| / √(a²+b²+c²) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Direction ratios ko direction cosines samajh lena. D.r.'s ko normalize kiye bina l,m,n me daal dena — l²+m²+n²=1 satisfy nahi hoga. Hamesha magnitude se divide karo.
- Shortest distance formula me sign error. (a₂−a₁)·(b₁×b₂) me modulus lagana bhoolna, ya a₁, a₂ ko ulta subtract karna — final answer negative aa sakta hai, magnitude hamesha positive rakho.
- Line-plane angle me cos use karna instead of sin. Do planes ya do lines ke beech angle me cos hota hai, lekin line-plane angle me sin — kyunki ye normal-line angle se complementary hai. Formula confuse karna common mistake hai.
- Plane through 3 points me cross product order ka galat dar. AB×AC ya AC×AB — dono se plane ka same equation banta hai (bas normal ki direction flip hoti hai), lekin students isse galat samajh ke confuse ho jaate hain aur calculation dobara karte hain.
- Distance formula me absolute value bhool jaana. Point plane ke kisi bhi side pe ho sakta hai, ax₁+by₁+cz₁+d negative aa sakta hai — modulus lagana zaroori hai, warna distance negative ho jayegi jo galat hai.
- Skew lines ko parallel ya intersecting samajh lena bina check kiye. Pehle check karo direction ratios proportional hain (parallel) ya nahi; phir dekho shortest distance zero hai ya nahi (intersecting vs skew) — bina check kiye directly formula laga dena galat result de sakta hai.
Board-Style Important Questions
- 1 mark: Ek line ke direction cosines l, m, n hain. l²+m²+n² ki value kya hogi?
- 2 marks: Points (2,3,4) aur (−1,−2,1) ko joinne wali line ke direction ratios nikalo.
- 2 marks: Do lines ke direction ratios (1,2,3) aur (3,2,1) diye hain. Inke beech angle nikalo.
- 4 marks: Do skew lines diye hain (vector form me) — unke beech shortest distance nikalo.
- 4 marks: Teen diye gaye points se guzarne wale plane ka Cartesian equation nikalo.
- 4 marks: Ek point diya hai aur ek plane diya hai — point se plane ki distance nikalo, aur foot of perpendicular ke direction cosines batao.
Aksar Poochhe Jaane Wale Sawaal
Direction cosines aur direction ratios me kya farak hai?
Direction cosines (l,m,n) normalized values hain jo l²+m²+n²=1 satisfy karte hain — ye line ke axes ke saath angles ke cosine hain. Direction ratios (a,b,c) koi bhi numbers ho sakte hain jo d.c.'s ke proportional hon, unhe normalize karne se d.c.'s milte hain.
Skew lines kya hoti hain aur ye 2D me kyun possible nahi?
Skew lines wo lines hain jo na parallel hain na intersect karti hain. 2D (ek plane) me ye possible nahi kyunki do non-parallel lines hamesha intersect karengi. 3D me lines alag-alag planes me ho sakti hain, isliye wo na milen na parallel hon — tab wo skew kehlati hain.
Line aur plane ke beech angle me sin kyun use hota hai, cos nahi?
Kyunki jo formula milta hai (b·n)/(|b||n|) wo dar-asal line ke direction aur plane ke normal ke beech ka cosine deta hai. Line aur plane ke beech ka actual angle isse 90° complementary hota hai, isliye cos ki jagah sin aata hai (sin φ = cos(90°−φ)).
Plane ka equation teen points se kaise banate hain?
Do vectors banao plane ke andar (jaise AB aur AC), unka cross product lo — wo plane ka normal vector n hoga. Fir kisi ek point (jaise A) aur is normal ka use karke (r−A)·n=0 se equation milega.
Shortest distance formula me agar denominator |b₁×b₂| zero aa jaaye to kya matlab hai?
Iska matlab hai b₁ aur b₂ parallel hain (cross product zero), isliye lines parallel hain, skew nahi. Us case me skew-lines wala formula use nahi hoga — parallel-lines wala alag formula use karna padega.
Exam me distance of point from plane formula yaad rakhne ka trick kya hai?
Plane ko hamesha ax+by+cz+d=0 form me likho (right side zero). Fir point ki coordinates plug karo numerator me aur modulus lagao, denominator me normal ka magnitude √(a²+b²+c²) aata hai — bas 'plug and take modulus' yaad rakho.
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