In short: Three-Dimensional Geometry (Chapter 11) is a formula-driven chapter in CBSE Class 12 Maths, carrying 5-6 marks on its own and up to 14 marks when combined with Vectors (Chapter 10). It covers direction cosines, equations of a line in vector and Cartesian form, equation of a plane, angle between lines and planes, shortest distance between skew lines, and distance of a point from a plane. Expect 1-2 questions — usually a 5-marker on the equation of a line or plane with a distance calculation. Once you memorize the formulas and practice the patterns, this chapter is highly predictable.
Marks Distribution
| Topic | Type of Questions | Expected Marks |
|---|---|---|
| Direction cosines/ratios | 1-2 mark MCQ/short | 1-2 |
| Equation of line (vector + Cartesian) | 2-3 mark questions | 2-3 |
| Equation of plane | 3-5 mark questions | 3-4 |
| Angle between lines/planes | 2-3 mark questions | 2-3 |
| Shortest distance / distance from plane | 3-5 mark questions | 3-4 |
| 3D Geometry total | 5-6 | |
| Vectors + 3D Geometry combined | up to 14 |
For all Maths formulas in one place, see the CBSE Class 12 Maths formulas list.
Direction Cosines and Direction Ratios
Direction Cosines (l, m, n)
The cosines of the angles that a line makes with the positive x, y, and z axes.
l = cos α, m = cos β, n = cos γ
Fundamental relation: l² + m² + n² = 1
Direction Ratios (a, b, c)
Any set of three numbers proportional to the direction cosines.
If a, b, c are direction ratios, then:
- l = a/√(a²+b²+c²)
- m = b/√(a²+b²+c²)
- n = c/√(a²+b²+c²)
Direction Ratios of a Line Joining Two Points
If the line joins P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
- Direction ratios: (x₂-x₁), (y₂-y₁), (z₂-z₁)
Equation of a Line in 3D
Vector Form
A line passing through point A (position vector a⃗) and parallel to vector b⃗:
r⃗ = a⃗ + λb⃗
Cartesian Form
A line passing through (x₁, y₁, z₁) with direction ratios a, b, c:
(x - x₁)/a = (y - y₁)/b = (z - z₁)/c = λ
Line Through Two Points
Vector form: r⃗ = a⃗ + λ(b⃗ - a⃗)
Cartesian form: (x-x₁)/(x₂-x₁) = (y-y₁)/(y₂-y₁) = (z-z₁)/(z₂-z₁)
Equation of a Plane
Different Forms
| Form | Equation | When to Use |
|---|---|---|
| Normal form (vector) | r⃗ · n̂ = d | Normal direction and distance from origin known |
| Normal form (Cartesian) | ax + by + cz = d | Same as above |
| Point-normal form | r⃗ · n⃗ = a⃗ · n⃗ | Point on plane and normal known |
| Intercept form | x/a + y/b + z/c = 1 | Intercepts on axes known |
| Three-point form | Use determinant | Three points on the plane given |
| Through intersection | P₁ + λP₂ = 0 | Two planes and a condition given |
Plane Through a Point Perpendicular to a Vector
If the plane passes through A(x₁, y₁, z₁) and is perpendicular to n⃗ = ai + bj + ck:
a(x-x₁) + b(y-y₁) + c(z-z₁) = 0
Plane Through Three Points
For points A(x₁,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃), the plane equation is found using:
| x-x₁ y-y₁ z-z₁ | | x₂-x₁ y₂-y₁ z₂-z₁ | = 0 | x₃-x₁ y₃-y₁ z₃-z₁ |
Angle Between Two Lines
If two lines have direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂):
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁²+b₁²+c₁²) × √(a₂²+b₂²+c₂²)]
In vector form: cos θ = |b⃗₁ · b⃗₂| / (|b⃗₁| × |b⃗₂|)
Condition for perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0
Condition for parallel: a₁/a₂ = b₁/b₂ = c₁/c₂
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Angle Between Two Planes
For planes a₁x + b₁y + c₁z = d₁ and a₂x + b₂y + c₂z = d₂:
cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁²+b₁²+c₁²) × √(a₂²+b₂²+c₂²)]
Same formula as angle between lines (because normals to planes behave like direction vectors).
Angle Between a Line and a Plane
If the line has direction ratios (a, b, c) and the plane has normal (A, B, C):
sin θ = |aA + bB + cC| / [√(a²+b²+c²) × √(A²+B²+C²)]
Note: This uses sin (not cos) because we measure the angle between the line and the plane, not between the line and the normal.
Shortest Distance Between Two Skew Lines
Two lines that are neither parallel nor intersecting are called skew lines.
Vector Form
For lines r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂:
d = |(a⃗₂ - a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|
Cartesian Form
For lines (x-x₁)/a₁ = (y-y₁)/b₁ = (z-z₁)/c₁ and (x-x₂)/a₂ = (y-y₂)/b₂ = (z-z₂)/c₂:
d = |determinant| / √[(b₁c₂-b₂c₁)² + (c₁a₂-c₂a₁)² + (a₁b₂-a₂b₁)²]
Where the determinant is: | x₂-x₁ y₂-y₁ z₂-z₁ | | a₁ b₁ c₁ | | a₂ b₂ c₂ |
Distance Between Parallel Lines
For parallel lines r⃗ = a⃗₁ + λb⃗ and r⃗ = a⃗₂ + μb⃗:
d = |b⃗ × (a⃗₂ - a⃗₁)| / |b⃗|
Distance of a Point from a Plane
Distance of point (x₁, y₁, z₁) from plane ax + by + cz + d = 0:
D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
Board exam tip: This is one of the most commonly asked 3-mark questions. Remember to take the absolute value and include the constant term d.
All Formulas Quick Reference
| Formula | Expression |
|---|---|
| Direction cosines relation | l² + m² + n² = 1 |
| Equation of line (vector) | r⃗ = a⃗ + λb⃗ |
| Equation of line (Cartesian) | (x-x₁)/a = (y-y₁)/b = (z-z₁)/c |
| Equation of plane | r⃗ · n⃗ = d or ax + by + cz = d |
| Angle between lines | cos θ = |b⃗₁·b⃗₂| / (|b⃗₁||b⃗₂|) |
| Angle between line & plane | sin θ = |b⃗·n⃗| / (|b⃗||n⃗|) |
| Shortest distance (skew lines) | d = |(a⃗₂-a⃗₁)·(b⃗₁×b⃗₂)| / |b⃗₁×b⃗₂| |
| Distance of point from plane | D = |ax₁+by₁+cz₁+d| / √(a²+b²+c²) |
Solved Examples
Example 1: Equation of a Line
Q: Find the equation of a line passing through (1, 2, 3) and parallel to the vector 2i + 3j - 4k.
Solution: Vector form: r⃗ = (i + 2j + 3k) + λ(2i + 3j - 4k)
Cartesian form: (x-1)/2 = (y-2)/3 = (z-3)/(-4)
Example 2: Angle Between Two Lines
Q: Find the angle between the lines whose direction ratios are (1, 2, 3) and (2, -1, 1).
Solution: cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁²+b₁²+c₁²) × √(a₂²+b₂²+c₂²)] cos θ = |1(2) + 2(-1) + 3(1)| / [√(1+4+9) × √(4+1+1)] cos θ = |2-2+3| / [√14 × √6] cos θ = 3/√84 = 3/(2√21) θ = cos⁻¹(3/2√21)
Example 3: Distance of Point from Plane
Q: Find the distance of the point (2, 3, -5) from the plane x + 2y - 2z = 9.
Solution: Rewrite as: x + 2y - 2z - 9 = 0 D = |1(2) + 2(3) + (-2)(-5) - 9| / √(1 + 4 + 4) D = |2 + 6 + 10 - 9| / √9 D = |9| / 3 = 3 units
Example 4: Shortest Distance Between Skew Lines
Q: Find the shortest distance between the lines: r⃗ = (i + 2j + 3k) + λ(2i + 3j + 4k) and r⃗ = (2i + 4j + 5k) + μ(3i + 4j + 5k)
Solution: a⃗₁ = i + 2j + 3k, b⃗₁ = 2i + 3j + 4k a⃗₂ = 2i + 4j + 5k, b⃗₂ = 3i + 4j + 5k
a⃗₂ - a⃗₁ = i + 2j + 2k
b⃗₁ × b⃗₂ = |i j k| |2 3 4| |3 4 5| = i(15-16) - j(10-12) + k(8-9) = -i + 2j - k
|b⃗₁ × b⃗₂| = √(1 + 4 + 1) = √6
(a⃗₂ - a⃗₁) · (b⃗₁ × b⃗₂) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1
d = |1|/√6 = 1/√6 units
Example 5: Equation of a Plane
Q: Find the equation of the plane passing through (1, 1, 1) and perpendicular to the line x/2 = y/3 = z/4.
Solution: Direction ratios of the line (and hence normal to the plane): (2, 3, 4) Point on plane: (1, 1, 1)
Plane equation: 2(x-1) + 3(y-1) + 4(z-1) = 0 2x - 2 + 3y - 3 + 4z - 4 = 0 2x + 3y + 4z = 9
Board Exam Important Questions
- Find the shortest distance between the lines r⃗ = (1-t)(i) + (t-2)(j) + (3-2t)(k) and r⃗ = (s+1)(i) + (2s-1)(j) - (2s+1)(k). (5 marks) — High frequency
- Find the equation of the plane passing through (1,1,1), (1,-1,1), and (-7,3,-5). (5 marks)
- Find the distance of the point (-1, -5, -10) from the point of intersection of the line r⃗ = (2i - j + 2k) + λ(3i + 4j + 2k) and the plane r⃗ · (i - j + k) = 5. (5 marks)
- Find the angle between the line (x+1)/2 = (y-1)/3 = (z-2)/6 and the plane 3x + y - z = 7. (3 marks)
- Find the equation of a line through (1, 2, -4) and perpendicular to the lines r⃗ = (8i - 19j + 10k) + λ(3i - 16j + 7k) and r⃗ = (15i + 29j + 5k) + μ(3i + 8j - 5k). (5 marks)
- Show that the lines (x-1)/2 = (y+1)/3 = z/1 and (x+1)/5 = (y-2)/1 = z/0 are coplanar. Find the equation of the plane containing them. (5 marks)
Frequently Asked Questions
How many marks does 3D Geometry carry?
5-6 marks on its own. Combined with Vectors (Chapter 10), the entire unit carries up to 14 marks — making it the highest-weightage unit in Maths.
Which formula is most frequently tested?
The shortest distance between two skew lines and the distance of a point from a plane are the two most commonly tested formulas. Master these and you cover 60% of the chapter's board questions.
Should I use vector form or Cartesian form?
Both give the same answer. Use whichever you are more comfortable with. Vector form is usually shorter for angle calculations. Cartesian form is clearer for distance calculations. CBSE accepts both.
How do I know if two lines are skew, parallel, or intersecting?
First check if direction ratios are proportional — if yes, lines are parallel. If not, solve for the parameter values where all three coordinate equations are satisfied simultaneously. If a consistent solution exists, lines intersect; otherwise, they are skew.
Is the determinant method for shortest distance difficult?
It looks intimidating but follows a fixed pattern. Always write the determinant in the form: first row = (a⃗₂-a⃗₁), second row = b⃗₁, third row = b⃗₂. Practice 5 problems and the pattern becomes mechanical.
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