Three Dimensional Geometry — Mathematics Class 12 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Three Dimensional Geometry (Class 12) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Three Dimensional Geometry (CBSE & HBSE)
Three Dimensional Geometry carries 10-12 marks in CBSE. Questions cover direction cosines/ratios, equations of lines and planes (vector and Cartesian), angle between lines/planes, and distance from point to plane.
Direction Cosines, Ratios and Equations of Lines
Direction Cosines and Direction Ratios
Direction Cosines (DCs) l, m, n: cosines of the angles α, β, γ that a line makes with positive x, y, z axes. l = cosα, m = cosβ, n = cosγ
Key property: l² + m² + n² = 1 (always!)
Direction Ratios (DRs) a, b, c: any numbers proportional to l, m, n. DCs from DRs: l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)
DC of axes:
- x-axis: (1,0,0)
- y-axis: (0,1,0)
- z-axis: (0,0,1)
Equation of a Line in Space
Passing through point A(x₁,y₁,z₁) with direction ratios a,b,c:
Vector form: r⃗ = a⃗ + λb⃗ where a⃗ = x₁î+y₁ĵ+z₁k̂, b⃗ = aî+bĵ+ck̂
Cartesian form: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
Passing through two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂): (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)
Angle Between Two Lines
If b⃗₁ and b⃗₂ are direction vectors (or a₁,b₁,c₁ and a₂,b₂,c₂ are DRs): cosθ = |b⃗₁·b⃗₂|/(|b⃗₁||b⃗₂|) = |a₁a₂+b₁b₂+c₁c₂|/[√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)]
- Parallel lines: a₁/a₂ = b₁/b₂ = c₁/c₂
- Perpendicular lines: a₁a₂+b₁b₂+c₁c₂ = 0
Shortest Distance Between Skew Lines
For lines r⃗ = a⃗₁+λb⃗₁ and r⃗ = a⃗₂+μb⃗₂: SD = |(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)|/|b⃗₁×b⃗₂|
For parallel lines: SD = |(a⃗₂−a⃗₁)×b⃗|/|b⃗|
CBSE Tip: For skew lines, the shortest distance formula uses scalar triple product in numerator. Memorize the denominator as |b⃗₁×b⃗₂|. For the angle, always take the acute angle (use absolute value).
Equations of Planes
Equation of a Plane
Vector form: r⃗·n⃗ = d (where n⃗ is the normal vector)
Cartesian form: ax + by + cz = d (where (a,b,c) are DRs of normal)
Plane through point (x₁,y₁,z₁) with normal (a,b,c): a(x−x₁) + b(y−y₁) + c(z−z₁) = 0
Intercept form: x/p + y/q + z/r = 1 (intercepts p,q,r)
Plane Through Three Points
Three points determine a unique plane. Set up: a(x−x₁)+b(y−y₁)+c(z−z₁)=0 and substitute the other two points to find a,b,c.
Alternatively, use the determinant:
| x−x₁ y−y₁ z−z₁ |
|---|
|x₂−x₁ y₂−y₁ z₂−z₁| = 0
| x₃−x₁ y₃−y₁ z₃−z₁ |
|---|
Angle Between Two Planes
Between a₁x+b₁y+c₁z=d₁ and a₂x+b₂y+c₂z=d₂: cosθ = |a₁a₂+b₁b₂+c₁c₂|/[√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)]
- Parallel planes: a₁/a₂ = b₁/b₂ = c₁/c₂
- Perpendicular planes: a₁a₂+b₁b₂+c₁c₂ = 0
Distance from Point to Plane
Distance from P(x₁,y₁,z₁) to plane ax+by+cz=d: d = |ax₁+by₁+cz₁−d|/√(a²+b²+c²)
Angle Between Line and Plane
For line with direction (a,b,c) and plane Ax+By+Cz=D: sinθ = |aA+bB+cC|/[√(a²+b²+c²)·√(A²+B²+C²)]
CBSE Tip: The distance formula from point to plane is the most frequently tested result in this chapter. Also know how to find the foot of perpendicular from a point to a plane and the image of a point in a plane.
Intersection of Lines and Planes
Coplanarity of Two Lines
Lines r⃗ = a⃗₁+λb⃗₁ and r⃗ = a⃗₂+μb⃗₂ are coplanar iff: (a⃗₂−a⃗₁)·(b⃗₁×b⃗₂) = 0 In Cartesian: |x₂−x₁ y₂−y₁ z₂−z₁; a₁ b₁ c₁; a₂ b₂ c₂| = 0
Plane Containing Two Lines
If coplanar, the equation of the plane containing them: Use the normal n⃗ = b⃗₁×b⃗₂ and any point on either line.
Image of a Point in a Plane
To find image Q of P(x₁,y₁,z₁) in plane ax+by+cz=d:
- Line PQ is perpendicular to plane: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c = k
- Foot M of perpendicular: substitute x=x₁+ak, y=y₁+bk, z=z₁+ck in plane
- Find k; foot M = (x₁+ak, y₁+bk, z₁+ck)
- Image Q: xQ = 2xM−x₁, yQ = 2yM−y₁, zQ = 2zM−z₁
Summary: Line and Plane Key Formulas
| Quantity | Formula | ||||
|---|---|---|---|---|---|
| Line through (x₁,y₁,z₁), DRs (a,b,c) | (x−x₁)/a=(y−y₁)/b=(z−z₁)/c | ||||
| Plane through (x₀,y₀,z₀) ⊥ (a,b,c) | a(x−x₀)+b(y−y₀)+c(z−z₀)=0 | ||||
| Distance point to plane | ax₁+by₁+cz₁−d | /√(a²+b²+c²) | |||
| SD between skew lines | (a⃗₂−a⃗₁)·(b⃗₁×b⃗₂) | / | b⃗₁×b⃗₂ |
CBSE Tip: For a 5-mark question, CBSE may ask to find the plane through a point containing a given line, or the image of a point in a plane, or the foot of the perpendicular from a point to a line. Master the perpendicular foot technique.
Frequently asked questions
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Yes. The Three Dimensional Geometry notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Three Dimensional Geometry chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Three Dimensional Geometry.