Three Dimensional Geometry – Complete Guide For Class 12 Math Chapter 11

Welcome to iPrep, your Learning Super App. Our learning resources for the chapter, Three Dimensional Geometry in Mathematics for Class 12th are designed to ensure you grasp this concept with clarity and perfection. Whether you’re studying for an upcoming exam or strengthening your concepts, our engaging animated videos, practice questions and notes offer you the best of integrated learning with interesting explanations and examples. 

The chapter Three Dimensional Geometry explores the spatial positioning of points, lines, and planes using a three-coordinate system. This chapter begins by introducing the concepts of three mutually perpendicular planes, forming the x, y, and z axes, essential for locating points in space. It then delves into direction cosines and direction ratios, which are critical for understanding the orientation of lines. The chapter covers equations of lines and planes in both vector and Cartesian forms, providing methods to calculate the angles between them.

Additionally, it discusses the conditions for lines to be parallel or perpendicular and introduces skew lines, which are non-intersecting and non-parallel. The shortest distance between skew lines and parallel lines is also covered, along with practical examples to illustrate these concepts. This chapter lays a robust foundation for analyzing spatial relationships and solving complex geometric problems in three dimensions.

Three Dimensional Geometry

In the realm of geometry, understanding the position of objects in Three Dimensional space is crucial. Whether it’s tracking the flight of a passenger jet or the trajectory of a ball, Three Dimensional coordinate geometry provides the tools to pinpoint locations and describe movements precisely.

Three Mutually Perpendicular Planes

The chapter Three Dimensional Geometry gives a thorough understanding of 3 mutually perpendicular planes. It includes-

• These planes are called coordinate planes.
• In three dimensional coordinate geometry, the position of a point is determined by these three planes.

Direction Angles of a Line

• Directed line L passing through the origin makes angles with x, y, and z-axes, respectively.
• , and are called direction angles.

Direction Cosines of a Line

The chapter three dimensional geometry also thoroughly introduces the concept of direction cosines of a line. This involves-

• Cosine of direction angles , and ,  and  ,i.e., cos , cos, and cos respectively, are called direction cosines of the line L. 
• cos   = l, cos = m and cos = n

Example:

If a line makes angles of 0°, 45° and 60° with the positive direction of x, y and z-axis, then

l = cos 0° = 1, m = cos 45° = 1/√2 and n = cos 60° = 1/2

Direction Ratios of a Line

Choose any three numbers a, b, c such that a ∝ l, b ∝ m, and c ∝ n, where l, m, and n are direction cosines of the line. Then a, b, and c are called the direction ratios of that line.

Relation Between the Direction Cosines of a Line

we find x = lr, y = mr and z = nr.

Hence, l² + m² + n² = 1 (Therefore,  x² + y² + z² = r² )

Direction Cosines of a Line Passing Through Two Points

The direction cosines of the line are

 x₂ – x₁/AB , y₂ – y₁/AB , z₂ – z₁/AB

where AB = √(x₂ – x₁)² +  (y₂ – y₁)² + (z₂ – z₁)²

Direction Ratios of a Line Segment joining Two Points

Given points : A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

The direction cosines of the line AB are  x₂ – x₁/AB , y₂ – y₁/AB , z₂ – z₁/AB

Relation between Direction Cosines and Direction Ratios

Let a, b, and c be the direction ratios of a line and l, m, and n be the direction cosines of that line. Then

l = ± a /√a² + b² + c², m = ± b /√a² + b² + c²,  n = ± c /√a² + b² + c²

Equation of a Line Through a Given Point and Parallel to a Given Vector

Vector Form

AP = λb, where λ is some real number.

⇒ r = a  + λb, where λ ∈ R.

Cartesian Form

b = ai + bj  + ck 

Example:

Find the vector equation of a line that passes through the point with position vector  2i – j + 4k and is in the direction of i + j – 2k. Also, reduce it to cartesian form.

Solution: Here, a = 2i – j + 4k and b =  i + j – 2k. So, the vector equation of the required line is

r = (2i – j + 4k)  + λ (i + j – 2k)             …(i)

where λ is a parameter.

Cartesian Form: Putting r = xi + yj + zk in (i), we get 

xi + yj + zk = (2i – j + 4k)  + λ (i + j – 2k) 

On comparing, we get

x = 2 + λ, y = -1 + λ and z = 4 – 2λ

Eliminating λ, we get the cartesian form

x -2 / 1 = y +1 = z – 4 /-2

Angle Between Two Lines

When given lines in vector Form:

Lines: r = a₁ + λb₁ and r = a₂ + b₂, where λ, ∈ R

If θ is the acute angle between the lines, then

cos θ = b₁.b₂ /|b₁||b₂|

Lines will be perpendicular if b₁.b₂ = 0 and parallel if b₁ = λb₂ for some scalar λ.

• When direction cosines or direction ratios are given 

Condition for Perpendicular Lines

Condition for Parallel Lines

Skew Lines

• These lines are neither intersecting nor parallel.
• Such a pair of lines is noncoplanar.

These lines are called skew lines.

Shortest Distance between Two Skew Lines

Vector Form

Lines: r = a₁ + λb₁ and r = a₂ + b₂, where λ, ∈ R

 Shortest distance between the lines = | (b₁ x b₂). (a₂ – a₁) / |b₁ x b₂||

Cartesian Form

Lines : l₁ : x – x₁/a₁ = y – y₁/b₁ = z – z₁/z₁ and l₂ : x – x₂/a₂ = y – y₂/b₂ = z – z₂/z₂

Shortest Distance –

Distance Between Two Parallel Lines

 r = a₁ + λb₁, where λ ∈ R

 r = a₂ + b₂,  where ∈ R

d =| b x (a₂ – a₁) / |b||

Equation of a Plane in Normal Form

Vector Form

ON is the normal vector from the origin to the plane

and d is the perpendicular distance of the plane from the origin.

Here, n is the unit vector along the normal vector.

Cartesian Form

n: Unit vector along the normal(ON) to the plane.

l, m, n. : Direction Cosines of n 

Plane Perpendicular to a Given Vector and Passing Through a Given Point 

Vector Form

Cartesian Form

Equation of a Plane Passing Through Three Non-Collinear Points

Vector Form

The vector (RS × RQ) is perpendicular

to the plane containing points R, S, and Q.

The equation of the plane (in vector form) is:

Cartesian Form

Non collinear points : R(x₁, y₁, z₁), S(x₂, y₂, z₂ ) and Q(x₃, y₃, z₃)

Let P(x, y, z) be any point P on the plane with position vector r.

RP = r – a  = (x – x₁) i  + (y – y₁)j + (z – z₁) k

RS = b – a  = (x₂ – x₁) i + (y₂ – y₁)j+ (z₂ – z₁) k

RQ = c – a  = (x₃ – x₁) i + (y₃ – y₁)j+ (z₃ – z₁) k

Substituting these values in the equation of a plane in vector form i.e., 

(r – a ). [(b – a) x (c – a)] = 0 

We obtain, 

= 0

which is the equation of the plane in Cartesian form passing through three noncollinear points (x₁, y₁,z₁),  (x₂, y₂, z₂) and (x₃, y₃, z₃)

Intercept Form of the Equation of a Plane

Within the chapter three dimensional geometry, there is also the mention of the Intercept Form of the Equation of a Plane.

Plane Passing Through the Intersection of Two Given Planes Vector Form

Let us observe two coplanar lines.

Vector Form

Lines:   r = a₁ + λb₁ and r = a₂ + b₂

Lines will be coplaner if (a₂ – a₁). (b₁ x b₂) = 0

Cartesian Form

Lines : x – x₁/a₁ = y – y₁/b₁ = z – z₁/z₁ and  x – x₂/a₂ = y – y₂/b₂ = z – z₂/c₂

Lines will be coplanar 

= 0

Angle Between Two Planes

Distance of a Point from a Plane

Example:

Find the distance of the point 2 i + j  + k  from the plane r. (i – 2j + 4 k )  = 9.

Solution:

We know that the perpendicular distance of a point P with position vector a from the plane r. n = d is given by (a. n)- d / |n|

Here, a = 2 i + j  + k,   n = i – 2j + 4 k and d = 9.

On substituting these values in (a. n)- d / |n|, we get

(a. n)- d / |n| = |(2 I + j  + k). (i – 2j + 4 k ) – 9| /|i – 2j + 4 k |

                         = | 4-9| / |√21| = 5/√21

Angle Between a Line and a Plane

Let’s Conclude

In conclusion, Three Dimensional Geometry forms a vital chapter in CBSE Class 12 Mathematics, offering essential tools to navigate spatial concepts. This chapter, Three Dimensional Geometry – Complete Guide for Class 12 Math Chapter 11, covers everything from understanding direction cosines and direction ratios to calculating distances between skew lines and parallel planes. Mastering these concepts will allow students to analyze and solve complex geometric problems with ease.

At iPrep, we provide comprehensive resources for mastering Three Dimensional Geometry, ensuring students gain clarity and confidence in their understanding. With our interactive learning materials, students can strengthen their foundation in this crucial chapter and excel in their exams.