Curriculum

Mathematics Grade XII

0/13

Text lesson

Chapter 11: Three Dimensional Geometry

Class 12 • Mathematics • Chapter 11

Three Dimensional Geometry

Describing points, lines and planes in space using direction cosines, vectors and equations.

CBSE / NCERT Aligned

School Revise

Chapter Roadmap

Direction Cosines and Ratios • Equation of a Line • Angle Between Two Lines • Equation of a Plane • Angle Between Planes • Distance of a Point from a Plane

Geometry in All Three Dimensions

On a flat page we locate a point with two numbers. In the real world we need three, because space has length, width and height. Three dimensional geometry extends coordinate geometry into space, giving exact descriptions of points, the lines joining them and the planes they lie in.

The key new idea is direction. A line in space is fixed by a point on it and the direction it heads, captured by direction cosines or direction ratios. Planes are fixed by a point and a normal direction. With these, we measure angles between lines and planes and find the distance from a point to a plane, the everyday questions of architecture, engineering and graphics.

Foundation

If a line makes angles α, β, γ with the axes, its direction cosines are l = cosα, m = cosβ, n = cosγ and satisfy l² + m² + n² = 1. Any proportional triple (a, b, c) gives the direction ratios of the same line.

Key Terms You Must Know

Term

Meaning

Example

Direction cosines

Cosines of the angles a line makes with the axes; l, m, n.

l² + m² + n² = 1

Direction ratios

Any numbers proportional to the direction cosines.

(2, 3, 6)

Equation of a line

Point plus direction, in vector or cartesian form.

(x−1)/2 = (y−2)/3 = z/1

Normal to a plane

A direction perpendicular to the whole plane.

coefficients in ax+by+cz=d

Angle between planes

The angle between their normals.

cosθ from normals

Distance from a plane

Perpendicular distance of a point to a plane.

|ax₀+by₀+cz₀−d|/√(a²+b²+c²)

Core Concepts, Step by Step

1. Direction Cosines and Direction Ratios

The direction of a line is recorded by the cosines of the angles it makes with the x, y and z axes, written l, m and n. These always satisfy l² + m² + n² = 1. Often it is easier to use any proportional triple of direction ratios (a, b, c); dividing by √(a² + b² + c²) converts them back to direction cosines.

A point, a line and a plane sketched against the three coordinate axes

Three dimensional coordinate axes x, y and z with a line and plane illustrating spatial geometry

2. Equation of a Line in Space

A line is fixed by one point on it and its direction. In vector form it is r = a + λb, where a is the position vector of a point and b the direction. In cartesian form it becomes (x − x₁)/a = (y − y₁)/b = (z − z₁)/c, with (a, b, c) the direction ratios.

3. Angle Between Two Lines

The angle between two lines depends only on their directions. Using direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂), the cosine of the angle is the dot product over the product of magnitudes: |a₁a₂ + b₁b₂ + c₁c₂| divided by the two magnitudes. The lines are perpendicular when this dot product is zero and parallel when the ratios are proportional.

4. Equation of a Plane

A plane is fixed by a point and a normal direction perpendicular to it. Its general cartesian equation is ax + by + cz = d, where (a, b, c) is the normal. The intercept form x/p + y/q + z/r = 1 reads off where the plane crosses each axis, which is handy for sketching.

5. Angle Between Planes and a Line

Two planes meet at an angle equal to the angle between their normals, found by the dot-product formula. For a line and a plane, the relevant angle is between the line’s direction and the plane’s normal; the angle the line makes with the plane itself is its complement, so we use sine rather than cosine.

6. Distance of a Point from a Plane

The perpendicular distance from a point (x₀, y₀, z₀) to the plane ax + by + cz = d is |ax₀ + by₀ + cz₀ − d| divided by √(a² + b² + c²). The absolute value keeps the distance positive, and the denominator scales by the length of the normal. This single formula answers a huge range of spatial problems.

Key Results with Proofs

Result 1: Direction Cosines Sum to One

Statement. For any line, l² + m² + n² = 1.

Proof

Direction cosines are a unit vector’s components, so their squares sum to one.

Take a direction with ratios (a, b, c) and length r = √(a² + b² + c²). – start from direction ratios

The direction cosines are l = a/r, m = b/r, n = c/r. – normalise

Then l² + m² + n² = (a² + b² + c²)/r². – square and add

Since r² = a² + b² + c², this equals 1. – simplify

This is why only two of the three direction cosines are independent.

Result 2: Angle Between Two Lines

Statement. cosθ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁²+b₁²+c₁²) √(a₂²+b₂²+c₂²)).

Proof

The angle between lines is just the angle between their direction vectors.

Each line’s direction is the vector of its direction ratios. – directions as vectors

The angle between two vectors uses the dot product over the magnitudes. – apply the dot-product angle

a₁a₂ + b₁b₂ + c₁c₂ is the dot product of the two direction vectors. – compute the dot product

Dividing by the magnitudes gives cosθ; the modulus takes the acute angle. – the formula

Perpendicular lines have a₁a₂ + b₁b₂ + c₁c₂ = 0.

Result 3: Distance from a Point to a Plane

Statement. The distance from (x₀, y₀, z₀) to ax + by + cz = d is |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²).

Proof

The distance is the projection of the point-to-plane gap onto the unit normal.

The normal direction to the plane is (a, b, c). – identify the normal

Drop a perpendicular from the point to the plane along this normal. – perpendicular foot

Projecting the gap onto the unit normal gives (ax₀ + by₀ + cz₀ − d)/√(a²+b²+c²). – project onto the normal

Taking the absolute value yields the positive distance. – make it positive

If the point lies on the plane the numerator is 0, as expected.

Worked Examples

Example 1

Question: Find the direction ratios and direction cosines of the line through (1, 2, 3) and (4, 6, 3).

▶ Show full working

Subtract coordinates, then normalise.

Direction ratios = (4 − 1, 6 − 2, 3 − 3) = (3, 4, 0). – subtract

Magnitude = √(9 + 16 + 0) = 5. – length

Direction cosines = (3/5, 4/5, 0). – divide by 5

Answer: Ratios (3, 4, 0); cosines (3/5, 4/5, 0).

Example 2

Question: Find the direction cosines of the x-axis.

▶ Show full working

It points purely along x.

The x-axis makes 0° with x and 90° with y and z. – angles

So the cosines are (cos0, cos90, cos90) = (1, 0, 0). – cosines

Answer: (1, 0, 0).

Example 3

Question: Verify that (1/√3, 1/√3, 1/√3) are valid direction cosines.

▶ Show full working

Check the sum of squares.

Sum of squares = 1/3 + 1/3 + 1/3. – square and add

= 1, so they are valid direction cosines. – check equals 1

Answer: Yes, the squares sum to 1.

Example 4

Question: Find the angle between lines with direction ratios (1, 1, 0) and (1, 0, 1).

▶ Show full working

Use the dot-product formula.

Dot product = 1(1) + 1(0) + 0(1) = 1. – dot product

Each magnitude is √2, so cosθ = 1/(√2 × √2) = 1/2. – cosine

θ = 60°. – angle

Answer: 60°.

Example 5

Question: Find the distance from (1, 2, 3) to the plane x + 2y + 2z = 9.

▶ Show full working

Use the distance formula.

Numerator = |1 + 4 + 6 − 9| = |2| = 2. – substitute the point

Denominator = √(1 + 4 + 4) = 3. – length of the normal

Distance = 2/3. – divide

Answer: 2/3.

Example 6

Question: Find the equation of the plane through (1, 0, 0), (0, 1, 0) and (0, 0, 1).

▶ Show full working

Use the intercept form.

The intercepts on the axes are 1, 1 and 1. – read intercepts

Intercept form x/1 + y/1 + z/1 = 1. – write it

So x + y + z = 1. – clear

Answer: x + y + z = 1.

Example 7

Question: Find the angle between the planes x + y + z = 1 and x − y + z = 2.

▶ Show full working

Use the normals.

Normals are (1, 1, 1) and (1, −1, 1). – read the normals

Dot product = 1 − 1 + 1 = 1; each magnitude is √3. – compute

cosθ = 1/3. – the cosine

Answer: cosθ = 1/3.

Example 8

Question: Are the lines with direction ratios (2, −1, 2) and (1, 2, 0) perpendicular?

▶ Show full working

Check the dot product.

2(1) + (−1)(2) + 2(0) = 2 − 2 + 0 = 0. – dot product

A zero dot product means the lines are perpendicular. – interpret

Answer: Yes, they are perpendicular.

Example 9

Question: Find λ so the planes 2x + λy + 3z = 1 and x − 2y + z = 4 are perpendicular.

▶ Show full working

Perpendicular planes have perpendicular normals.

Normals (2, λ, 3) and (1, −2, 1); set their dot product to 0. – condition

2 − 2λ + 3 = 0, so 5 − 2λ = 0. – solve

λ = 5/2. – value

Answer: λ = 5/2.

Example 10

Question: Write the cartesian equation of the line through (2, −1, 3) with direction ratios (1, 4, 2).

▶ Show full working

Point plus direction.

Use (x − x₁)/a = (y − y₁)/b = (z − z₁)/c. – template

(x − 2)/1 = (y + 1)/4 = (z − 3)/2. – substitute

Answer: (x − 2)/1 = (y + 1)/4 = (z − 3)/2.

Example 11

Question: Find the distance from the origin to the plane 2x + 3y + 6z = 14.

▶ Show full working

Use the distance formula at (0, 0, 0).

Numerator = |0 + 0 + 0 − 14| = 14. – substitute origin

Denominator = √(4 + 9 + 36) = 7. – length of normal

Distance = 14/7 = 2. – divide

Answer: 2.

Example 12

Question: Find the angle between the line with direction ratios (2, 2, 1) and the plane with normal (1, 2, 2).

▶ Show full working

Use sine with the normal.

sinθ = |2(1) + 2(2) + 1(2)| / (√(4+4+1) × √(1+4+4)). – line-plane angle

= |2 + 4 + 2| / (3 × 3) = 8/9. – compute

So θ = sin⁻¹(8/9). – angle

Answer: θ = sin⁻¹(8/9).

Where You Meet This in Real Life

Architecture

Roof planes, ramps and supports are defined by their normals and the angles between them.

Computer graphics

Every 3D scene uses planes and normals to decide visibility and lighting of surfaces.

Aviation

An aircraft’s heading in space is a direction in three dimensions, described by direction cosines.

Robotics

Positioning a robot arm requires lines and planes in space and the angles between them.

Geology and mining

The orientation of rock layers is recorded as a plane with a measured normal direction.

Practice Sets A–D

Practice Set A – Basics

A1. Find the direction ratios of the line through (0, 0, 0) and (2, 3, 6).

▶ Reveal full working

Subtract.

(2, 3, 6).

Answer: (2, 3, 6).

A2. Find the direction cosines of the y-axis.

▶ Reveal full working

Pure y direction.

(0, 1, 0).

Answer: (0, 1, 0).

A3. State the normal to the plane 3x − y + 2z = 5.

▶ Reveal full working

Read coefficients.

(3, −1, 2).

Answer: (3, −1, 2).

A4. Do (1, 0, 0) satisfy l² + m² + n² = 1?

▶ Reveal full working

Check.

1 + 0 + 0 = 1, yes.

Answer: Yes.

Practice Set B – Conceptual

B1. Why do direction cosines satisfy l² + m² + n² = 1?

▶ Reveal full working

They form a unit vector.

They are the components of a unit vector along the line. – unit vector

The squared length of a unit vector is 1. – squares sum to 1

Answer: Because they are the components of a unit direction vector.

B2. How do you find the angle between two planes?

▶ Reveal full working

Use normals.

Find the angle between their normal vectors with the dot-product formula.

Answer: From the angle between their normals.

B3. When are two lines parallel in space?

▶ Reveal full working

Proportional ratios.

When their direction ratios are proportional.

Answer: When their direction ratios are proportional.

B4. Why is the line-plane angle found using sine?

▶ Reveal full working

Complement of the normal angle.

The line makes an angle with the normal; the angle with the plane is its complement, so cosine becomes sine.

Answer: Because it is the complement of the angle with the normal.

Practice Set C – Application / Numerical

C1. Find the distance from (2, 1, 0) to the plane 2x + y + 2z = 3.

▶ Reveal full working

Distance formula.

Numerator |4 + 1 + 0 − 3| = 2; denominator √(4 + 1 + 4) = 3. – compute

Distance = 2/3. – divide

Answer: 2/3.

C2. Find the angle between lines (1, 2, 2) and (2, 2, 1).

▶ Reveal full working

Dot-product angle.

Dot = 2 + 4 + 2 = 8; magnitudes 3 and 3. – compute

cosθ = 8/9. – cosine

Answer: cosθ = 8/9.

C3. Find where the plane x + y + z = 6 meets the x-axis.

▶ Reveal full working

Set y = z = 0.

x = 6, so the point is (6, 0, 0). – substitute

Answer: (6, 0, 0).

C4. Find the direction cosines of (6, 8, 0).

▶ Reveal full working

Normalise.

Magnitude √(36 + 64) = 10; cosines (6/10, 8/10, 0) = (3/5, 4/5, 0). – divide

Answer: (3/5, 4/5, 0).

Practice Set D – HOTS / Multi-step

D1. Find the distance between the parallel planes x + 2y + 2z = 6 and x + 2y + 2z = 12.

▶ Reveal full working

Take a point on one and use the distance to the other.

On the first plane take (6, 0, 0). – point on plane 1

Distance to second: |6 + 0 + 0 − 12|/√(1 + 4 + 4) = 6/3. – apply formula

= 2. – value

Answer: 2.

D2. Show that the line through (1, 2, 3) with ratios (1, 1, 1) is parallel to the plane x + y − 2z = 5.

▶ Reveal full working

Line is parallel to a plane when its direction is perpendicular to the normal.

Normal of the plane is (1, 1, −2); line direction is (1, 1, 1). – identify

Dot product = 1 + 1 − 2 = 0, so direction is perpendicular to the normal. – check

A direction perpendicular to the normal lies along the plane, so the line is parallel. – conclude

Answer: Yes, the direction is perpendicular to the normal, so the line is parallel to the plane.

D3. Find the angle between the planes 2x − y + 2z = 3 and x + 2y − 2z = 1.

▶ Reveal full working

Use the normals.

Normals (2, −1, 2) and (1, 2, −2); dot = 2 − 2 − 4 = −4. – dot product

Each magnitude is 3, so cosθ = |−4|/9 = 4/9. – cosine

Answer: cosθ = 4/9.

D4. Find the foot of the perpendicular distance check: is (1, 1, 1) closer to plane x + y + z = 3 or x + y + z = 6?

▶ Reveal full working

Compute both distances.

To x + y + z = 3: |3 − 3|/√3 = 0, the point is on it. – first plane

To x + y + z = 6: |3 − 6|/√3 = 3/√3 = √3. – second plane

So (1, 1, 1) lies on the first plane and is √3 from the second. – conclude

Answer: It lies on x + y + z = 3 and is √3 from x + y + z = 6.

Chapter Summary

Everything in One Glance

Direction Cosines

l, m, n are cosines of the axis angles with l² + m² + n² = 1; ratios are any proportional triple.

Line in Space

Fixed by a point and a direction: r = a + λb, or the symmetric cartesian form.

Angle Between Lines

cosθ from the dot product of direction ratios over the magnitudes.

Plane

ax + by + cz = d with normal (a, b, c); intercept form aids sketching.

Angles with Planes

Plane-plane angle from normals (cosine); line-plane angle from the complement (sine).

Distance

|ax₀ + by₀ + cz₀ − d|/√(a² + b² + c²) from a point to a plane.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find the direction ratios of the line through (1, 1, 1) and (3, 4, 1).

State the direction cosines of the z-axis.

Give the normal to the plane x − 2y + 3z = 7.

Find the angle between lines (1, 0, 0) and (0, 1, 0).

Find the distance from the origin to x + 2y + 2z = 9.

Are lines (1, 2, 2) and (2, −2, 1) perpendicular?

Where does x + y + z = 4 meet the y-axis?

Find the direction cosines of (3, 0, 4).

School Revise Virtual Lab

Practice the concepts in this chapter with interactive simulations and visual tools.

Open the Virtual Lab →

Class 12 Mathematics Chapter 11: Three Dimensional Geometry, Complete Notes and Practice

These free Class 12 Maths Three Dimensional Geometry notes follow the NCERT 2026 to 27 syllabus and cover direction cosines and ratios, the equation of a line, the angle between two lines, the equation of a plane, angles between planes and a line, and the distance of a point from a plane, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.