Three Dimensional Geometry extends your understanding of lines and planes from 2D to 3D space. This chapter is very important for CBSE boards (6–8 marks) and frequently tested in competitive exams. It uses concepts from Vector Algebra, so make sure you’re comfortable with dot and cross products.
Key Concepts
1. Direction Cosines and Direction Ratios
Direction Cosines: l = cos α, m = cos β, n = cos γ
Relation: l² + m² + n² = 1
Direction Ratios: a, b, c (proportional to l, m, n)
l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)
2. Equation of a Line in 3D
Vector Form
r→ = a→ + λb→
Line through two points A(a→) and B(b→):
r→ = a→ + λ(b→ − a→)
Cartesian Form
(x − x₁)/a = (y − y₁)/b = (z − z₁)/c
Through two points (x₁, y₁, z₁) and (x₂, y₂, z₂):
(x − x₁)/(x₂ − x₁) = (y − y₁)/(y₂ − y₁) = (z − z₁)/(z₂ − z₁)
3. Angle Between Two Lines
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁²+b₁²+c₁²) × √(a₂²+b₂²+c₂²))
Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂
Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0
4. Shortest Distance Between Two Lines
Skew Lines (non-parallel, non-intersecting)
Shortest Distance = |(a₂→ − a₁→) · (b₁→ × b₂→)| / |b₁→ × b₂→|
If shortest distance = 0, the lines intersect.
Parallel Lines
Distance = |b→ × (a₂→ − a₁→)| / |b→|
5. Equation of a Plane
Different Forms
| Form | Equation | When to Use |
|---|---|---|
| Normal form (vector) | r→ · n̂ = d | Given normal and distance from origin |
| General form | ax + by + cz + d = 0 | Standard form; (a,b,c) is normal |
| Point-normal (vector) | (r→ − a→) · n→ = 0 | Given a point and normal |
| Point-normal (Cartesian) | a(x−x₁) + b(y−y₁) + c(z−z₁) = 0 | Given point (x₁,y₁,z₁) and normal (a,b,c) |
| Intercept form | x/a + y/b + z/c = 1 | Given x, y, z intercepts |
| Three-point form | Use determinant | Given 3 non-collinear points |
6. Angle Between Two Planes
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁²+b₁²+c₁²) × √(a₂²+b₂²+c₂²))
Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂
Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0
7. Distance of a Point from a Plane
D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
8. Angle Between a Line and a Plane
sin φ = |b→ · n→| / (|b→| × |n→|)
Important Definitions
| Term | Definition |
|---|---|
| Direction Cosines | Cosines of angles a line makes with coordinate axes (l, m, n) |
| Skew Lines | Lines that are neither parallel nor intersecting |
| Normal to a Plane | Vector perpendicular to every line lying in the plane |
| Coplanar Lines | Lines that lie in the same plane (shortest distance = 0) |
Solved Examples — NCERT-Based
Example 1: Find the equation of the line through (1, 2, 3) with direction ratios 4, 5, 6
Solution:
Cartesian form: (x−1)/4 = (y−2)/5 = (z−3)/6
Vector form: r→ = (î + 2ĵ + 3k̂) + λ(4î + 5ĵ + 6k̂)
Example 2: Find the angle between planes 2x + y − 2z = 5 and 3x − 6y − 2z = 7
Solution:
n₁→ = (2, 1, −2), n₂→ = (3, −6, −2)
cos θ = |6 − 6 + 4| / (√9 × √49) = 4/21
θ = cos⁻¹(4/21)
Example 3: Find the distance of point (2, 5, −3) from the plane 6x − 3y + 2z − 4 = 0
Solution:
D = |6(2) − 3(5) + 2(−3) − 4| / √(36 + 9 + 4)
= |12 − 15 − 6 − 4| / √49 = |−13|/7 = 13/7 units
Example 4: Find the shortest distance between the lines:
r→ = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂) and r→ = (4î + 5ĵ + 6k̂) + μ(2î + 3ĵ + k̂)
Solution:
a₁→ = (1,2,3), b₁→ = (1,−3,2), a₂→ = (4,5,6), b₂→ = (2,3,1)
a₂→ − a₁→ = (3, 3, 3)
b₁→ × b₂→ = (−3−6)î − (1−4)ĵ + (3+6)k̂ = −9î + 3ĵ + 9k̂
|b₁→ × b₂→| = √(81+9+81) = √171 = 3√19
(a₂→ − a₁→) · (b₁→ × b₂→) = −27 + 9 + 27 = 9
SD = |9|/(3√19) = 3/√19 units
Important Questions for Board Exams
1 Mark Questions
- Find the direction cosines of a line equally inclined to all three axes.
- Write the equation of the plane with intercepts 2, 3, 4 on the axes.
2 Mark Questions
- Find the distance of the point (3, −2, 1) from the plane 2x − y + 2z + 3 = 0.
- Find the angle between the lines with direction ratios (1, 1, 2) and (√3−1, −√3−1, 4).
3 Mark Questions
- Find the equation of the plane passing through (1, 1, −1), (6, 4, −5), (−4, −2, 3).
- Find the shortest distance between the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−2)/3 = (y−4)/4 = (z−5)/5.
5 Mark Questions
- Find the equation of the plane passing through the intersection of planes x + 3y + 6 = 0 and 3x − y − 4z = 0 whose perpendicular distance from origin is unity.
- Find the foot of perpendicular and the perpendicular distance of the point (1, 2, 3) from the line (x−6)/3 = (y−7)/2 = (z−7)/(−2). Also find the image of the point.
Quick Revision Points
- Direction cosines: l² + m² + n² = 1 (always!)
- Line equation: r→ = a→ + λb→ or (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
- Plane equation: r→ · n→ = d or ax + by + cz + d = 0
- Parallel lines ⟹ proportional direction ratios
- Perpendicular lines ⟹ a₁a₂ + b₁b₂ + c₁c₂ = 0
- Distance from plane: |ax₁+by₁+cz₁+d| / √(a²+b²+c²)
- Skew lines: use cross product formula for shortest distance
- Line-plane angle uses sin, plane-plane angle uses cos
Chapter Navigation
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Related Chapters in Class 12 Maths
Practice What You Learned
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