Math Class 12 JEE NEET Study Material pdf

Shortest Distance Between Two Lines in 3D Space, Solved Examples | Class 12 Math Notes Study Material Download Free PDF

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The shortest distance between two lines in three-dimensional space is the length of the perpendicular segment drawn from a point on one line to the other line. This distance can be found using vector calculus or analytical geometry techniques, such as finding the vector equation of each line and calculating the distance between them. The formula involves determining the dot product and cross product of vectors associated with the lines. This concept is essential in various fields like physics, engineering, and computer graphics for determining proximity and collision detection.

In 3D space, two lines can either intersect each other at some point, parallel to each other or they can neither be intersecting nor parallel to each other also known as skew lines.

  • In the case of intersecting lines the shortest distance between them is 0.
  • For parallel lines, the length of the line joining the two parallel lines or the length of the line perpendicular to both the parallel lines has the shortest distance.
  • In the case of skew lines, the shortest distance is the line perpendicular to both of the given lines.

Note: The alphabets written in bold represent vector. ‘x’ denotes cross product (vector product).

Shortest Distance Between Two Parallel Lines 

Considering 2 lines in vector form as:

v1 = a1 + c × b

v2 = a2 + d × b

Where,

  • c and d are the constants.
  • b = parallel vector to both the vectors v1 and v2
  • a1, a2  are the position vector of some point on v1 and v2 respectively

Shortest distance = |b × (a2 – a1)| / |b|

Examples on Shortest Distance Between Two Parallel Lines

Example 1: For the following lines in 3D space.

v1 = i – 2j + (i – j + k)

v2 = i – 3j + k +(i – j + k)

Find the shortest distance between these lines?

Solution:

v1: i – 2j + (i – j + k)

v2: i – 3j + k + (i – j + k)

Here, b = i – j + k

a1 = i -2j

a2 = i – 3j + k

⇒ a2 – a1 = -j + k

Thus, |b| = √3 = 1.73

⇒ |b × (a2 – a1)| = √2 = 1.41

Shortest distance = |b × (a2 – a1)|/|b| = 1.41/1.73 = 0.815

Example 2: For the following lines in 3D space.

v1 = i – j – k + (2i – 3j + 4k)

v2 = 2i – 3j + k + 3(2i – 3j + 4k)

Find the shortest distance between these lines?

Solution: 

The vector can be written in form as:

v1 = i – j – k + (2i – 3j + 4k)

v2 = 2i – 3j + k + 3(2i – 3j + 4k)

Here, b = 2i – 3j + 4k

|b| = √(2)2 + (-3)2+ (4)2 = 5.385

a1 = i – j – k

a2 = 2i – 3j + k

a2 – a1 = i – 2j + 2k

b × (a2 – a1) = 2i – k

|b × (a2 – a1)| = √(2)2 + (1)2 = 2.236

Now applying the shortest distance formula for parallel lines = |b × (a2 – a1)|/|b| = 2.236/5.385 = 0.415

Example 3: Given two lines in the cartesian format as:

V1: (x – 2)/2 = (y – 1)/3 = (z)/4

V2: (x – 3)/4 = (y – 2)/6 = (z – 5)/8

Find the shortest distance between these lines.

Solution:

The displacement vector of V1 is 2i + 3j + 4k, for V2 is 4i + 6j + 8k

The displacement vector V2 is a multiple of V1 as, 

4i + 6j + 8k = 2 (2i + 3j + 4k)

So the two given lines are parallel to each other.

a1 = 2i + j + 0k

a2 = 3i + 2j + 5k

a2 – a1 = i + j +5k

b = 2i + 3j + 4k

|b| = √(2)2 + (3)2 + (4)2 = 5.385

b × (a2 – a1) = 11i – 6j – k 

|b × (a2 – a1)| = 12.569

shortest distance = |b × (a2 – a1)|/|b| = 12.569/5.385 = 2.334

Shortest Distance Between Skew Lines

Considering 2 lines in vector form as:

v1 = a1 + c × b1

v2 = a2 + d × b2

Here, c and d are the constants.

The shortest distance 2 skew lines = |(b1 × b2)(a2 – a1)|/|(b1 × b2)|

Note: If two lines are intersecting then’s the shortest distance considering the two lines skew will automatically come out to be zero.

Examples on Shortest Distance Between Skew Lines

Example 1: Given two lines in vector form as:

  • V1: i – j + 2i + j + k
  • V2: i + j + 3i – j – k

Find the shortest distance between these lines.

Solution:

The given lines are skew lines. 

b1 = 2i + j + k

b2 = 3i – j – k

a1 = i – j

a2 = i + j 

⇒ a2 – a1 = 2j

Thus, (b1 × b2) = 5j – 5k

Shortest distance = |(b1 × b2)(a2 – a1)|/|(b1 × b2)| = 10/7.07 = 1.41

Example 2: Given two lines in vector form as:

  • v1: 2i – j + 5(3i – j + 2k)
  • v2: i – j + 2k + 2(i + 3j + 4k)

Find the shortest distance between these lines.

Solution:

The given lines are skew lines. 

Shortest distance = |(b1 × b2)(a2 – a1)|/|(b1 × b2)|

b1 = 3i – j + 2k

b2 = i + 3j + 4k

a1 = 2i – j 

a2 = i – j + 2k

⇒ a2 – a1 = -i + 2k

Thus, (b1 × b2) = -10i – 10j + 10k

⇒ |b1 × b2| = 17.320

Thus, |(b1 × b2)(a2 – a1)| = 40 

Shortest distance = 40/17.320 = 2.309

Example 3: Given 2 lines in the cartesian form, find the shortest distance between them.

  • V1: (x – 1)/2 = (y – 1)/3 = (z)/4
  • V2: (x)/1 = (y – 2)/2 = (z – 1)/3 

Solution:

a1 = i + j 

a2 = -2j + k

b1 = 2i + 3j + 4k

b2 = i + 2j + 3k

⇒ a2 – a1 = -3i – j + k

Thus, (b1 × b2) = i – 2j + k

⇒ |b1 × b2| = 2.44

Shortest distance =|(i – 2j + k)( -3i – j + k)|/2.44 = 0

Since, shortest distance is zero it means these two lines are intersecting lines.

FAQs on Shortest Distance Between Two Lines

What is the Equation of Line in 3D?

A vector equation of a line in 3D space is given by r = a + tv.

What is the shortest path between two lines?

The shortest distance between two lines in space can be found by determining the distance between the parallel planes that contain these lines.

What is the shortest distance between a point and a line?

Shortest distance between point P with coordinates (x1​, y1​, z1​), and line represented by a vector equation r=a+tv, is given by d = ∣(P−a)⋅n∣​/∣v∣.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

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CBSE Class 12 Maths Syllabus 2025-26 with Marks Distribution

The table below shows the marks weightage along with the number of periods required for teaching. The Maths theory paper is of 80 marks, and the internal assessment is of 20 marks which totally comes out to be 100 marks.

CBSE Class 12 Maths Syllabus And Marks Distribution 2023-24

Max Marks: 80

No.UnitsMarks
I.Relations and Functions08
II.Algebra10
III.Calculus35
IV.Vectors and Three – Dimensional Geometry14
V.Linear Programming05
VI.Probability08
Total Theory80
Internal Assessment20
Grand Total100

Unit-I: Relations and Functions

1. Relations and Functions

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operations on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability

Continuity and differentiability, derivative of composite functions, chain rule, derivative of inverse trigonometric functions like sin-1 x, cos-1 x and tan-1 x, derivative of implicit functions. Concept of exponential and logarithmic functions.
Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives

Applications of derivatives: rate of change of quantities, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).

3. Integrals 

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

dy/dx + py = q, where p and q are functions of x or constants.

dx/dy + px = q, where p and q are functions of y or constants.

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

Students can go through the CBSE Class 12 Syllabus to get the detailed syllabus of all subjects. Get access to interactive lessons and videos related to Maths and Science with ANAND CLASSES’S App/ Tablet.

Frequently Asked Questions on CBSE Class 12 Maths Syllabus 2025-26

Q1

Is Calculus an important chapter in the CBSE Class 12 Maths Syllabus?

Yes, Calculus is an important chapter in the CBSE Class 12 Maths Syllabus. It is for 35 marks which means that if a student is thorough with this chapter will be able to pass the final exam.

Q2

How many units are discussed in the CBSE Class 12 Maths Syllabus?

In the CBSE Class 12 Maths Syllabus, about 6 units are discussed, which contains a total of 13 chapters.

Q3

How many marks are allotted for internals in the CBSE Class 12 Maths syllabus?

About 20 marks are allotted for internals in the CBSE Class 12 Maths Syllabus. Students can score it with ease through constant practice.

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