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CBSE Class 12 Mathematics★ 6-8 marks · CBSE Class 12 Mathematics Chapter 11; lines, planes, and distances are high-yield for JEE

Three Dimensional Geometry

TWO coordinates (x, y) describe a point on a plane. THREE coordinates (x, y, z) describe a point in SPACE. This chapter extends geometry from 2D to 3D — covering lines, planes, angles, distances, and coplanarity. CBSE Class 12 Mathematics Chapter 11.

⏱ 20 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Find direction cosines and direction ratios of a line
2Write equations of lines in vector and Cartesian form
3Write equations of planes in various forms
4Find angles between lines, planes, and line-plane pairs
5Compute distances and test coplanarity of lines

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Why this chapter matters

Three-dimensional geometry extends coordinate geometry into space using vectors. Direction cosines, equations of lines and planes, angles, and shortest distances are essential for physics, engineering, computer graphics, and a reliable scoring topic in exams.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 12 Mathematics: Vector Algebra

Dot and cross products of vectors

Three Dimensional Geometry

'Three-dimensional geometry adds DEPTH to the flat world of 2D — a third dimension opens up a universe of possibilities.'

1. Chapter Overview

Three Dimensional Geometry extends coordinate geometry from the plane (2D) to SPACE (3D). Topics include: DIRECTION COSINES and direction ratios of a line, EQUATIONS OF A LINE IN SPACE (vector and Cartesian forms), EQUATIONS OF A PLANE, ANGLE between lines and planes, the SHORTEST DISTANCE between two lines, and COPLANARITY of lines. Vector algebra provides the most elegant framework for 3D geometry.

2. Direction Cosines and Direction Ratios

Direction Cosines (DC)

If a line makes angles α, β, γ with x, y, z axes, then: l = cos α, m = cos β, n = cos γ are the DIRECTION COSINES.
Property: l² + m² + n² = 1.

Direction Ratios (DR)

Any three numbers PROPORTIONAL to direction cosines: a, b, c.
If a, b, c are DR, then DC = (a/√(a²+b²+c²), b/√(a²+b²+c²), c/√(a²+b²+c²)).

3. Equation of a Line in Space

Vector Form

r⃗ = a⃗ + λb⃗. 'a point + direction × parameter.'
a⃗ = position vector of a point on the line. b⃗ = direction vector.

Cartesian Form

(x − x₁)/a = (y − y₁)/b = (z − z₁)/c = λ.
(x₁, y₁, z₁) is a point on the line. (a, b, c) are direction ratios.

4. Equation of a Plane

Form	Equation	Description
Normal form	r⃗ · n̂ = d	n̂ = unit normal, d = perpendicular distance from origin
General form	ax + by + cz + d = 0	a, b, c are DR of normal
Through a point	(r⃗ − a⃗) · n⃗ = 0	Plane through a point with normal n⃗
Three points	(r⃗ − a⃗) · [(b⃗ − a⃗) × (c⃗ − a⃗)] = 0	Plane through A, B, C
Intercept form	x/a + y/b + z/c = 1	a, b, c are x, y, z intercepts
Two direction vectors	r⃗ = a⃗ + λb⃗ + μc⃗	Parametric form

5. Angle Between Lines and Planes

Angle Between Two Lines

cos θ = |b⃗₁ · b⃗₂| / (|b⃗₁||b⃗₂|)
Lines are PERPENDICULAR if b⃗₁ · b⃗₂ = 0.
Lines are PARALLEL if b⃗₁ × b⃗₂ = 0⃗.

Angle Between a Line and a Plane

sin φ = |b⃗ · n⃗| / (|b⃗||n⃗|). Where b⃗ is along the line and n⃗ is normal to the plane.
'The angle between a line and a plane is the COMPLEMENT of the angle between the line and the normal.'

Angle Between Two Planes

cos θ = |n⃗₁ · n⃗₂| / (|n⃗₁||n⃗₂|)

6. Distance and Coplanarity

Distance of a Point from a Plane

d = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)

Distance Between Parallel Planes

d = |d₁ − d₂| / √(a² + b² + c²)

Shortest Distance Between Two Lines

Skew lines: d = |(b⃗₁ × b⃗₂) · (a⃗₂ − a⃗₁)| / |b⃗₁ × b⃗₂|
'The shortest distance between two skew lines is the projection of the vector connecting them onto the common perpendicular.'

Coplanarity

Two lines are COPLANAR if they lie in the same plane — the shortest distance between them is ZERO.
Condition: (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂) = 0.

7. Comparison Table: 2D vs 3D Geometry

Concept	In 2D	In 3D
Coordinates	(x, y)	(x, y, z)
Line equation	y = mx + c	Vector: r⃗ = a⃗ + λb⃗
Distance formula	√((x₂−x₁)² + (y₂−y₁)²)	√((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²)
Angle between lines	tan θ =	m₁−m₂
Perpendicular condition	m₁m₂ = −1	b⃗₁·b⃗₂ = 0

8. Common Mistakes

Confusing direction cosines with direction ratios: DC satisfy l²+m²+n²=1. DR are any proportional numbers. To convert DR to DC, divide by √(a²+b²+c²).
Wrong formula for angle between line and plane: Use sin φ, NOT cos φ. The angle between a line and a plane is NOT the same as the angle between the line and the normal.
Shortest distance formula: Make sure you use the CORRECT formula — the cross product gives a vector perpendicular to BOTH lines.
Coplanarity check: If the scalar triple product is ZERO, the lines are coplanar. Non-zero means they ARE skew.

9. CBSE Exam Focus

Direction cosines and direction ratios — finding DC from DR, verifying l²+m²+n²=1
Equation of a line in space — vector and Cartesian forms
Equation of a plane — various forms, converting between forms
Angle between two lines, line and plane, two planes
Distance of a point from a plane
Shortest distance between skew lines
Coplanarity of two lines

10. Self-Test

Q1: Find the direction cosines of a line that makes equal angles with all three axes. A1: Let l = m = n. Then l²+m²+n² = 1 ⇒ 3l² = 1 ⇒ l = ±1/√3. DC = (1/√3, 1/√3, 1/√3) or (−1/√3, −1/√3, −1/√3).

Q2: Find the vector equation of the line passing through the points (1, 2, 3) and (4, 5, 6). A2: a⃗ = î + 2ĵ + 3k̂. b⃗ = (4−1)î + (5−2)ĵ + (6−3)k̂ = 3î + 3ĵ + 3k̂. r⃗ = (î+2ĵ+3k̂) + λ(3î+3ĵ+3k̂).

Q3: Find the angle between the planes x + 2y + 2z = 9 and 2x + 3y + 6z = 8. A3: n⃗₁ = î+2ĵ+2k̂, |n⃗₁| = 3. n⃗₂ = 2î+3ĵ+6k̂, |n⃗₂| = 7. n⃗₁·n⃗₂ = 2+6+12=20. cos θ = 20/(3×7) = 20/21. θ = cos⁻¹(20/21).

Q4: Find the shortest distance between the lines r⃗ = (î+2ĵ+k̂) + λ(î−ĵ+k̂) and r⃗ = (2î−ĵ−k̂) + μ(2î+ĵ+2k̂). A4: Using the skew line formula: a⃗₂−a⃗₁ = (2−1)î+(−1−2)ĵ+(−1−1)k̂ = î−3ĵ−2k̂. b⃗₁×b⃗₂ = determinant. Compute: b⃗₁×b⃗₂ = (−2−1)î − (2−2)ĵ + (1+2)k̂ = −3î + 0ĵ + 3k̂. |b⃗₁×b⃗₂| = √(9+9) = 3√2. (a⃗₂−a⃗₁)·(b⃗₁×b⃗₂) = (1)(−3)+(−3)(0)+(−2)(3) = −3−6 = −9. d = |−9|/(3√2) = 3/√2 units.

Q5: Show that the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−4)/5 = (y−1)/2 = z are coplanar. A5: b⃗₁ = 2î+3ĵ+4k̂, b⃗₂ = 5î+2ĵ+k̂. a⃗₁ = î+2ĵ+3k̂, a⃗₂ = 4î+ĵ+0k̂. a⃗₂−a⃗₁ = 3î−ĵ−3k̂. b⃗₁×b⃗₂ = determinant: (−5)î − (−18)ĵ + (−11)k̂ = −5î+18ĵ−11k̂. (a⃗₂−a⃗₁)·(b⃗₁×b⃗₂) = 3(−5)+(−1)(18)+(−3)(−11) = −15−18+33 = 0. Scalar triple product = 0 ⇒ lines are COPLANAR.

11. Conclusion

Three-dimensional geometry transforms how we think about space:

LINES: 'A line in space needs a point and a direction — in 2D, slope was enough. In 3D, we need vectors.'
PLANES: 'A plane is defined by a normal vector and a point — the infinite set of all vectors perpendicular to the normal.'
DISTANCES: 'Shortest distance between skew lines is a uniquely 3D concept — in 2D, all lines either intersect or are parallel.'
ANGLES: 'Vectors make angle calculations elegant — just use dot products.'

'Vector algebra is the natural language of 3D space — it turns geometry into algebra.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Direction cosines

l^2 + m^2 + n^2 = 1

DCs from DRs: divide each by sqrt(a^2+b^2+c^2).

Line equations

Vector: r = a + lambda b; Cartesian: (x-x1)/a = (y-y1)/b = (z-z1)/c

A point plus a direction vector.

Angle between line and plane

sin(phi) = |b.n| / (|b||n|)

Uses sine, not cosine; phi is complement of line-normal angle.

Shortest distance (skew lines)

d = |(b1 x b2).(a2 - a1)| / |b1 x b2|

Coplanar if this scalar triple product is zero.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing direction cosines with direction ratios

✓ DCs satisfy l^2+m^2+n^2 = 1; DRs are any proportional numbers. Convert DRs to DCs by dividing by their magnitude.

WATCH OUT

✗ Using cosine for the line-plane angle

✓ The angle between a line and a plane uses sine: sin(phi) = |b.n|/(|b||n|).

WATCH OUT

✗ Mixing up the shortest-distance formula

✓ Use the cross product b1 x b2, which is perpendicular to both lines, in the numerator and denominator.

WATCH OUT

✗ Misreading the coplanarity test

✓ Lines are coplanar when the scalar triple product (a2-a1).(b1 x b2) = 0; non-zero means skew.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Direction cosines and lines

DCs, DRs, and line equations

Questions

Planes

Equations of planes and conversions

Questions

Angles, distances, coplanarity

Angles, point-plane distance, skew-line distance

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Direction Cosines

Find the direction cosines of a line making equal angles with all three axes.

▶ Show solution

If l = m = n then 3l^2 = 1, so l = +/- 1/sqrt(3). DCs are (1/sqrt3, 1/sqrt3, 1/sqrt3) or their negatives.

Q2EASY· Line Equation

Find the vector equation of the line through (1,2,3) and (4,5,6).

▶ Show solution

a = i + 2j + 3k, b = 3i + 3j + 3k. r = (i + 2j + 3k) + lambda(3i + 3j + 3k).

Q3MEDIUM· Angle

Find the angle between the planes x + 2y + 2z = 9 and 2x + 3y + 6z = 8.

▶ Show solution

n1 = (1,2,2) with |n1| = 3, n2 = (2,3,6) with |n2| = 7, n1.n2 = 20. cos(theta) = 20/21, so theta = arccos(20/21).

Q4HARD· Shortest Distance

Find the shortest distance between r = (i+2j+k) + lambda(i-j+k) and r = (2i-j-k) + mu(2i+j+2k).

▶ Show solution

a2-a1 = i - 3j - 2k. b1 x b2 = -3i + 0j + 3k with magnitude 3 sqrt(2). (a2-a1).(b1 x b2) = -9. d = 9/(3 sqrt 2) = 3/sqrt(2) units.

Q5HARD· Coplanarity

Show the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-4)/5 = (y-1)/2 = z are coplanar.

▶ Show solution

b1 = (2,3,4), b2 = (5,2,1), a2-a1 = (3,-1,-3). b1 x b2 = (-5,18,-11). (a2-a1).(b1 x b2) = -15 - 18 + 33 = 0, so the lines are coplanar.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Direction cosines satisfy l^2+m^2+n^2 = 1; DRs are proportional numbers.
•Line: r = a + lambda b (vector) or (x-x1)/a = (y-y1)/b = (z-z1)/c (Cartesian).
•Plane forms: normal, general (ax+by+cz+d=0), intercept, through three points.
•Angle between lines/planes uses cosine of dot product; line-plane uses sine.
•Point-plane distance d = |ax1+by1+cz1+d|/sqrt(a^2+b^2+c^2).
•Shortest distance between skew lines uses (b1 x b2).(a2 - a1).
•Lines are coplanar when that scalar triple product is zero.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question type	Marks each	Typical count	What it tests
Shortest distance / coplanarity	3-5	1	Skew lines and coplanarity
Lines and planes	3	1	Equations and conversions
Angles / distances	2-3	1	Angles and point-plane distance

Prep strategy

Memorise DC and DR relationships
Practise vector and Cartesian line/plane forms
Use sine for line-plane angle and cosine for line-line/plane-plane
Apply the scalar triple product for distance and coplanarity

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Computer graphics

3D geometry of lines and planes underlies rendering, collision detection, and modelling.

Engineering and architecture

Planes and distances are used in structural design and spatial layout.

Navigation and robotics

Direction vectors and shortest distances guide motion planning in three dimensions.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Convert DRs to DCs when normalisation is needed
Pick the correct plane form for the given data
Use sine for line-plane, cosine for line-line/plane-plane angles
Apply the scalar triple product for distance and coplanarity

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Find the foot of perpendicular and image of a point in a plane.
Derive the equation of a plane containing two given lines.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Mathematics examHigh

JEE Main and Advanced (3D Geometry)High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The dot product formula naturally gives the angle between the line's direction vector b and the plane's normal vector n. But the angle phi we want is between the line and the plane itself, which is the complement of the angle between the line and the normal. Since sin(phi) = cos(90 - phi), taking sin(phi) = |b.n|/(|b||n|) directly gives the line-plane angle. Using cosine here would give the angle to the normal instead.

Compare their direction vectors and use the coplanarity test. If the direction vectors are proportional, the lines are parallel. If not, compute the scalar triple product (a2 - a1).(b1 x b2): if it is zero the lines are coplanar and therefore intersect, while if it is non-zero the lines are skew (neither parallel nor intersecting). The shortest distance formula then gives zero for intersecting/coplanar lines and a positive value for skew lines.

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