Showing posts with label class 12. Show all posts
Showing posts with label class 12. Show all posts

CBSE Date Sheet 2025 PDF: Steps to Download Time Table Online @cbse.gov.in

CBSE Date Sheet 2025 Class 10th



Check out the schedule for all subjects below.

Date and Time

Subject Code

Subject Name

Saturday, 15th February, 2025

10:30 AM - 01:30 PM

101

English (Communicative)

184

English (Language and Literature)

Monday, 17th February, 2025

10:30 AM - 12:30 PM

036

Hindustani Music (Per Ins)

131

Rai

132

Gurung

133

Tamang

134

Sherpa

254

Elements of Book Keeping & Accountancy

418

Physical Activity Trainer

Tuesday, 18th February, 2025

  403

Security

404

Automotive

405

Introduction to Financial Markets

406

Introduction to Tourism

407

Beauty & Wellness

408

Agriculture

409

Food Production

410

Front Office Operations

411

Banking & Insurance

412

Marketing & Sales

414

Apparel

415

Multi-Media

416

Multi Skill Foundation Course

419

Data Science

420

Electronics & Hardware

421

Foundation Skills for Sciences

422

Design Thinking & Innovation

Thursday, 20th February, 2025

10:30 AM - 01:30 PM

086

Science

Saturday, 22nd February, 2025

10:30 AM - 01:30 PM

018

French

119

Sanskrit (Communicative)

122

Sanskrit

Tuesday, 25th February, 2025

10:30 AM - 01:30 PM

087

Social Science

Thursday, 27th February, 2025

10:30 AM - 01:30 PM

003

Urdu Course-A

005

Bengali

006

Tamil

009

Marathi

010

Gujarati

011

Manipuri

308

Urdu Course-B

Friday, 28th February, 2025

10:30 AM - 01:30 PM

002

Hindi Course-A

008

Hindi Course-B

Saturday, 1st March, 2025

10:30 AM - 12:30 PM

049

Painting

Monday, 3rd March, 2025

10:30 AM - 12:30 PM

413

Health Care

Wednesday, 5th March, 2025

10:30 AM - 01:30 PM

154

Elements of Business

10:30 AM - 12:30 PM

401

Retail

Thursday, 6th March, 2025

10:30 AM - 01:30 PM

017

Tibetan

020

German

076

National Cadet Corps

088

Bhoti

089

Telugu - Telangana

092

Bodo

093

Tangkhul

094

Japanese

095

Bhutia

096

Spanish

097

Kashmiri

098

Mizo

099

Bahasa Melayu

Monday, 10th March, 2025

10:30 AM - 01:30 PM

041

Mathematics Standard

241

Mathematics Basic

Wednesday, 12th March, 2025

10:30 AM - 01:30 PM

007

Telugu

016

Arabic

021

Russian

023

Persian

024

Nepali

025

Limboo

026

Lepcha

031

Carnatic Music (Vocal)

032

Carnatic Music (Melodic Instruments)

10:30 AM - 12:30 PM

033

Carnatic Music (Percussion Instruments)

034

Hindustani Music (Vocal)

035

Hindustani Music (Melodic Instruments)

136

Thai

Thursday, 13th March, 2025

10:30 AM - 01:30 PM

064

Home Science

Monday, 17th March, 2025

10:30 AM - 01:30 PM

012

Punjabi

013

Sindhi

017

Malayalam

018

Odia

014

Assamese

015

Kannada

091

Kokborok

Tuesday, 18th March, 2025

10:30 AM - 12:30 PM

165

Computer Applications

402

Information Technology

417

Artificial Intelligence

Also Read: CBSE Date Sheet 2025: Class XII Time Table PDF Out at cbse.gov.in

CBSE Class 12 Date Sheet 2025

Have a look at the following table for the schedule of all subjects:

Date and Time

Subject Code

Subject Name

Saturday, 15th February, 2025

10:30 AM - 01:30 PM

66

Entrepreneurship

Monday, 17th February, 2025

10:30 AM - 01:30 PM

48

Physical Education

Tuesday, 18th February, 2025

10:30 AM - 12:30 PM

35

Hindustani Music (Melodic Instruments)

36

Hindustani Music (Percussion Instruments)

821

Multi-Media

804

Automotive

813

Health Care

844

Data Science

847

Electronics & Hardware

Wednesday, 19th February, 2025

10:30 AM - 01:30 PM

809

Food Production

824

Office Procedures & Practices

830

Design

10:30 AM - 12:30 PM

342

Early Childhood Care & Education

Thursday, 20th February, 2025

10:30 AM - 01:30 PM

817

Typography & Computer Applications

Friday, 21st February, 2025

10:30 AM - 01:30 PM

42

Physics

Saturday, 22nd February, 2025

10:30 AM - 01:30 PM

54

Business Studies

833

Business Administration

Monday, 24th February, 2025

10:30 AM - 01:30 PM

29

Geography

Tuesday, 25th February, 2025

10:30 AM - 01:30 PM

118

French

822

Taxation

829

Textile Design

10:30 AM - 12:30 PM

843

Artificial Intelligence

Thursday, 27th February, 2025

10:30 AM - 01:30 PM

43

Chemistry

Friday, 28th February, 2025

10:30 AM - 01:30 PM

805

Financial Markets Management

807

Beauty & Wellness

828

Medical Diagnostics

Saturday, 1st March, 2025

10:30 AM - 01:30 PM

46

Engineering Graphics

10:30 AM - 12:30 PM

57

Bharatanatyam - Dance

58

Kuchipudi - Dance

59

Odissi - Dance

60

Manipuri - Dance

61

Kathakali - Dance

10:30 AM - 01:30 PM

816

Horticulture

823

Cost Accounting

836

Library & Information Science

Monday, 3rd March, 2025

10:30 AM - 01:30 PM

74, 801

Legal Studies; Retail

Tuesday, 4th March, 2025

10:30 AM - 01:30 PM

76, 811

National Cadet Corps; Banking

Wednesday, 5th March, 2025

10:30 AM - 01:30 PM

808, 812

Agriculture; Marketing

Thursday, 6th March, 2025

10:30 AM - 01:30 PM

837

Fashion Studies

Friday, 7th March, 2025

10:30 AM - 01:30 PM

835

Mass Media Studies

848

Design Thinking & Innovation

Saturday, 8th March, 2025

10:30 AM - 01:30 PM

41

Mathematics

241

Applied Mathematics

Monday, 10th March, 2025

10:30 AM - 12:30 PM

806

Tourism

10:30 AM - 01:30 PM

827

Air-Conditioning & Refrigeration

831

Salesmanship

Tuesday, 11th March, 2025

10:30 AM - 01:30 PM

1

English Elective

301

English Core

Wednesday, 12th March, 2025

10:30 AM - 01:30 PM

841

Yoga

845

Physical Activity Trainer

Thursday, 13th March, 2025

10:30 AM - 01:30 PM

803

Web Application

Saturday, 15th March, 2025

10:30 AM - 01:30 PM

002

Hindi Elective

302

Hindi Core

Monday, 17th March, 2025

10:30 AM - 01:30 PM

003

Urdu Elective

022

Sanskrit Elective

10:30 AM - 12:30 PM

031

Carnatic Music Vocal

032

Carnatic Music Melodic Instruments

033

Carnatic Music Percussion Instruments

056

Kathak - Dance

10:30 AM - 01:30 PM

303

Urdu Core

810

Front Office Operations

814

Insurance

818

Geospatial Technology

819

Electrical Technology

Tuesday, 18th March, 2025

10:30 AM - 12:30 PM

049

Painting

050

Graphics

051

Sculpture

052

Applied Art (Commercial Art)

Wednesday, 19th March, 2025

10:30 AM - 01:30 PM

30

Economics

Thursday, 20th March, 2025

10:30 AM - 12:30 PM

034

Hindustani Music (Vocal)

Friday, 21st March, 2025

10:30 AM - 01:30 PM

045

Biotechnology

073

Knowledge Tradition & Practices of India

188

Bhoti

191

Kokborok

820

Electronic Technology

825

Shorthand (English)

826

Shorthand (Hindi)

834

Food Nutrition & Dietetics

Saturday, 22nd March, 2025

10:30 AM - 01:30 PM

028

Political Science

Monday, 24th March, 2025

10:30 AM - 01:30 PM

322

Sanskrit Core

Tuesday, 25th March, 2025

10:30 AM - 01:30 PM

44

Biology

Wednesday, 26th March, 2025

10:30 AM - 01:30 PM

55

Accountancy

Thursday, 27th March, 2025

10:30 AM - 01:30 PM

39

Sociology

Saturday, 29th March, 2025

10:30 AM - 01:30 PM

65

Informatics Practices

83

Computer Science

802

Information Technology

Tuesday, 1st April, 2025

10:30 AM - 01:30 PM

27

History

Wednesday, 2nd April, 2025

10:30 AM - 01:30 PM

104

Punjabi

105

Bengali

106

Tamil

107

Telugu

108

Sindhi

109

Marathi

110

Gujarati

111

Manipuri

112

Malayalam

113

Odia

114

Assamese

115

Kannada

116

Arabic

117

Tibetan

120

German

121

Russian

123

Persian

124

Nepali

125

Limboo

126

Lepcha

189

Telugu Telangana

192

Bodo

193

Tangkhul

194

Japanese

195

Bhutia

196

Spanish

197

Kashmiri

198

Mizo

Thursday, 3rd April, 2025

10:30 AM - 01:30 PM

064

Home Science

Friday, 4th April, 2025

10:30 AM - 01:30 PM

037

Psychology

Keep checking this page for a detailed subject-wise schedule!

Share:

Class 12 exams postponed, cancels Class 10 exams

 


CBSE Class 12 exams postponed, cancels Class 10 exams

The Ministry of Education has decided to postpone the class 12 exams and cancel the class 10 board exams. For CBSE class 12 boards, the situation will be reviewed on June 1 and revised dates will be announced thereafter.

For CBSE class 10 students, result will be prepared on the basis of an “objective criterion” to be developed by the Board. Any candidate who is not satisfied with the marks allocated to him/her on this basis will be given an opportunity to sit in an exam as and when the conditions are conducive to hold the exams. Last year too CBSE had devised a new scheme for class 10 students.

 



 

Share:

New CBSE Syllabus 2021-22 for 9th, 10th, 11th, 12th Released: Applicable for CBSE Academic Session 2021-22







Subject-wise New CBSE Syllabus 2021-22 (PDF) for 9th, 10th, 11th & 12th has been officially released online at cbseacademic.nic.in. The new CBSE Syllabus is applicable for CBSE Academic Session 2021-22. The board has already instructed CBSE Schools to start new CBSE Academic Session 2021-22 from April and the board has released the new syllabus before April 2021. 

Also Check: CBSE Date Sheet 2021 (Revised) - 10th & 12th Board Exams 2021: Download PDF Now!

CBSE Syllabus 2021-22 for Class 9 & CBSE Syllabus 2021-22 for Class 10 

⇒ Download CBSE Syllabus 2021-22 for Maths: Class 9 and Class 10

 

⇒ Download CBSE Syllabus 2021-22 for Science: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Social Science: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Computer Applications: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Hindi Course-B: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Hindi Course-A: Class 9 and Class 10

With the links given above, students of CBSE Class 9th & Class 10th can check & download the latest CBSE Syllabus 2021-22 (PDF). 

CBSE Syllabus 2021-22 for Class 11 & CBSE Syllabus 2021-22 for Class 12

⇒ Download CBSE Syllabus 2021-22 for Physical Education: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for (English Core): Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for (English Elective): Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Physics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Chemistry: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Biology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Mathematics: Class 11 and Class 12

 Download CBSE Syllabus 2021-22 for Applied Mathematics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Biotechnology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Physics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Information Prac: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Computer Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Accountancy: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Business Studies: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Psychology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for History: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Political Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Hindi Core: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Hindi Elective: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Geography: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Home Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Sociology: Class 11 and Class 12

With the links given above, students of CBSE Class 9th & Class 10th can check & download the latest CBSE Syllabus 2021-22 (PDF). 

CBSE Syllabus 2021-22 is the most important resource for the students & teachers as well. New CBSE Syllabus provides a roadmap for the academic year. It also provides details about chapters, topics, practical to be studied in the entire academic year. It also provides information about the format of the question paper & important details about practical exams.

Students are also advised to check the YouTube channel of Jagran Josh where students can also access important videos with tips & strategies.

Students preparing for CBSE 10th & 12th board exams 2021 can also check important questions, previous year papers, tips & strategies for exam preparation.

Share:

Class 12 Chapter 11 (Three Diamensional Geometry) Class Notes Part II


  Three Dimensional Geometry 


1. CENTRAL IDEA OF 3D  
There are infinite number of points in space. We want to identify each and every point of space with the help of three mutually perpendicular coordinate axes OX, OY and OZ. 
2. AXES 
Three mutually perpendicular lines OX, OY, OZ are considered as three axes. 
3. COORDINATE PLANES 
Planes formed with the help of x and y axes is known as x-y plane similarly y and z axes y – z plane and with z and x axis z - x plane. 
4. COORDINATE OF A POINT 
Consider any point P on the space drop a perpendicular form that point to x - y plane then the algebraic length of this perpendicular is considered as z-coordinate and from foot of the perpendicular drop perpendiculars to x and y axes these algebraic length of perpendiculars are considered as y and x coordinates respectively. 
5. VECTOR REPRESENTATION OF A POINT IN SPACE 
If coordinate of a point P in space is (x, y, z) then the position vector of the point P with respect to the same origin is \(x\hat i + y\hat j + z\hat k\).
6. DISTANCE FORMULA
Distance between any two points (x1, y1, z1) and (x2, y2, z2) is given as \(\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \)
Vector Method
 We know that if position vector of two points A and B are given as \(\overrightarrow {OA} \) and \(\overrightarrow {OB} \) then
\[ \Rightarrow |\overrightarrow {AB} | = |\overrightarrow {AB}  - \overrightarrow {OA} |\] \[ \Rightarrow |\overrightarrow {AB} | = |({x_2}\hat i + {y_2}\hat j + {z_2}\hat k) - ({x_1}\hat i + {y_1}\hat j + {z_1}\hat k)|\] \[ \Rightarrow |\overrightarrow {AB} | = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]

7. DISTANCE OF A POINT P FROM COORDINATE AXES

Let PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZ respectively, then \(PA = \sqrt {{y^2} + {z^2}} ,\;\;PB = \sqrt {{z^2} + {x^2}} ,\;\;PC = \sqrt {{x^2} + {z^2}} \)

8. SECTION FORMULA

(i) Internal Division : 

If point P divides the distance between the points A (x₁, y₁, z₁,) and B (x₂, y₂, Z₂,) in the ratio of m: n (internally). The coordinate of P is given as

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths


Note:

All these formulae are very much similar to two dimension coordinate geometry.

9. CENTROID OF A TRIANGLE

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

10. INCENTRE OF TRIANGLE ABC

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

11. CENTROID OF A TETRAHEDRON

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

12. RELATION BETWEEN TWO LINES 

Two lines in the space may be coplanar and may be none coplanar. Non coplanar lines are called skew lines if they never intersect each other. Two parallel lines are also non intersecting lines but they are coplanar. Two lines whether intersecting or non intersecting, the angle between them can be obtained.

13. DIRECTION COSINES AND DIRECTION RATIOS 

Direction cosines : Let a, ß, be the angles which directed line makes with the positive directions of the  axes of x, y and z respectively, the \(\cos \alpha ,\cos \beta ,\cos \gamma \) are called the direction cosines of the line. The direction cosine denoted (l, m, n).

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths


(ii) If l, m, n, be the direction cosines of a lines, then + m² + n² = 1

(iii) Direction ratios: Let a, b, c be proportional to the direction cosines, l, m, n, then a, b, c are called the direction ratios.

If a, b, c are the direction ratio of any line L the \(a\hat i + b\hat j + c\hat k\) will be a vector parallel to the line L.

If l, m, n are direction cosine of line L then \(l\hat i + m\hat j + n\hat k\) is a unit vector parallel to the line L.

(iv) If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

(v) If OP = r, when O is the origin and the direction cosines of OP are l, m, n then the coordinates of P are (lr, mr, nr).

If direction cosine of the line AB are l, m, n, | AB |= r, and the coordinate of A is (x₁, y₁, z₁,) then the coordinate of B is given as (x₁, + rl, y₁, + rm, z₁ + rn)

(vi) If the coordinates P and Q are (x₁, y₁, z₁) and (x₂, y₂, z₂) then the direction ratios of line PQ are, a = x₂ − x₁, b = y₂ - y₁ and c = z₂ - z₁, and the direction cosines of line PQ are l = \(\dfrac{{{x_2} - {x_1}}}{{|\overrightarrow {PQ} |}}\), m = \(\dfrac{{{y_2} - {y_1}}}{{|\overrightarrow {PQ} |}}\) and n = \(\dfrac{{{z_2} - {z_1}}}{{|\overrightarrow {PQ} |}}\)


(vii) Direction cosines of axes : Since the positive x-axis makes angles 0°, 90°, 90° with axes of x, y and z respectively. Therefore

Direction cosines of x-axis are (1,0, 0)
Direction cosines of y-axis are (0, 1, 0)
Direction cosines of z-axis are (0, 0, 1)

 
14. ANGLE BETWEEN TWO LINE SEGMENTS

If two lines having direction ratios a₁, b₁, c₁, and a₂, b₂, c₂, respectively then we can consider two vector parallel to the lines as \({a_1}\hat i + {b_1}\hat j + {c_1}\hat k\) and \({a_2}\hat i + {b_2}\hat j + {c_2}\hat k\) and angle between them can be given as: \[\cos \theta  = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2}  \cdot \sqrt {a_2^2 + b_2^2 + c_2^2} }}\]
(i) The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0
(ii) The lines will be parallel, if \(\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\).
(iii) Two parallel lines have same direction cosines i.e. l1 = l2, m1 = m2 and n1 = n2.



15. PROJECTION OF A LINE SEGMENT ON A LINE 

(i) If the coordinates P and Q are (x₁, Y₁, Z₁,) and (X₂, Y₂, Z₂) then the projection of the line segments PQ on a line having direction cosines l, m, n is |l(x₂ - x₁) + m(y₂ - y₁) + n(z₂ - z₁)|
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

 

A PLANE 

If line joining any two points on a surface lies completely on it then the surface is a plane.
OR
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this surface is called a plane. This fixed line is called the normal to the plane.

16. EQUATION OF A PLANE 

(i) Normal form of the equation of a plane is lx + my + nz = p, where, l, m, n are the direction cosines of the normal to the plane and p is the distance of the plane from the origin.

(ii) General form : ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the direction ratios of the normal to the plane.

(iii) The equation of a plane passing through the point (x₁, y₁, z₁) is given by \[a({x_1} - {x_2}) + b({y_1} - {y_2}) + c({z_1} - {z_2}) = 0\] where a, b, c are the direction ratios of the normal to the plane.

(iv) Plane through three points : The equation of the plane through three non-collinear points (x1, Y1, z1), (x2, y2, z2) and (x3, y3, z3) is:   \[\left| {\begin{array}{*{20}{c}}x&y&z&1\\{{x_1}}&{{y_1}}&{{z_l}}&1\\{{x_2}}&{{y_2}}&{{z_2}}&1\\{{x_3}}&{{y_3}}&{{z_3}}&1\end{array}} \right| = 0\]
 
(v) Intercept Form: The equation of a plane cutting intercept a, b, c on the axes is \(\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\)

(vi) Vector Form : The equation of a plane passing through a point having position vector \({\vec a}\) and normal to vector \({\vec n}\) is \((\vec r - \vec a) \cdot \vec n\) or \(\vec r \cdot \vec n = \vec a \cdot \vec n\).

Note: 
(a) Vector equation of a plane normal to unit vector \({\hat n}\) and at a distance d from the origin is: \(\vec r \cdot \hat n = d\)

(b) Planes parallel to the coordinate planes 

(i) Equation of yz - plane is x = 0
(ii) Equation of xz - plane is y = 0
(iii) Equation of xy - plane is z = 0

(c) Planes parallel to the axes : 

If a = 0, the plane is parallel to x-axis i.e. equation of the plane parallel to the x-axis is by + cz + d = 0. Similarly, equation of planes parallel to y-axis and parallel to z-axis are ax + cz + d = 0 and ax + by + d = 0 respectively.

(d) Plane through origin : Equation of plane passing through origin is ax + by + cz = 0.

(e) Transformation of the equation of a plane to the normal form: To reduce any equation ax + by + cz - d = 0 to the normal form, first write the constant term on the right hand side and make it positive, then divided each term by \(\sqrt {{a^2} + {b^2} + {c^2}} \), where a, b, c are coefficients of x, y and z respectively \[\dfrac{{ax}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} + \dfrac{{by}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} + \dfrac{{cz}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} = \dfrac{d}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }}\]
Where (+) sign is to be taken if d > 0 an (-) sign is to be taken if d < 0.

(f) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0 distance between two parallel planes ax + by + cz + d1 = 0 and ax + dy + xz + d2 = 0 is given as: \[\frac{{|{d_1} - {d_2}|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
 
(g) Equation of a plane passing through a given point and parallel to the given vectors: The equation of a plane passing through a point and having position vector \({\vec a}\) and parallel to \({\vec b}\) and \({\vec c}\) is \[\vec r = \vec a + \lambda \vec b + \mu \vec c\] parametric form (where \(\lambda \) and \(\mu \) are scalers).
OR
\(\vec r \cdot (\vec b \times \vec c) = \vec a \cdot (\vec b \times \vec c)\) (non parametric form)

(h) A plane ax + by + cz + d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio: \(\left( { - \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{a{x_2} + b{y_2} + c{z_2} + d}}} \right)\)
 
(i) The xy-plane divides the line segment joining the point (x1, y1, z1) and (x2, y2, z2) in the ratio \( - \dfrac{{{z_1}}}{{{z_2}}}\). Similarly yz - plane divides in \( - \dfrac{{{x_1}}}{{{x_2}}}\) and zx - plane divides in \( - \dfrac{{{y_1}}}{{{y_2}}}\).


17. ANGLE BETWEEN TWO PLANES

(i) Consider two planes ax + by + cz + d = 0 and a'x + b'y + c'z+ d' = 0. Angle between these planes is the angle between their normal. Since direction ratios of their normal are (a, b, c) and (a', b', c') respectively, hence \(\theta \) the angle between them is given by: \[\cos \theta  = \frac{{aa' + bb' + cc'}}{{\sqrt {{a^2} + {b^2} + {c^2}}  \cdot \sqrt {a{'^2} + b{'^2} + c{'^2}} }}\]
Planes are perpendicular if aa' + bb' + c'' =0 and planes are parallel, if \(\dfrac{a}{{a'}} = \dfrac{b}{{b'}} = \dfrac{c}{{c'}}\).

(ii) The angle \(\theta \) between the plane \(\vec r \cdot {{\vec n}_1} = {d_1}\) and \(\vec r \cdot {{\vec n}_1} = {d_1}\) is given by \[\cos \theta  = \dfrac{{{{\vec n}_1} \cdot {{\vec n}_2}}}{{|{{\vec n}_1}||{{\vec n}_2}|}}\]
Planes are perpendicular if \({{{\vec n}_1} \cdot {{\vec n}_2} = 0}\) and planes are parallel if \({{{\vec n}_1} = \lambda {{\vec n}_2}}\) .


18. A PLANE AND A POINT 

(i) Distance of the point (x', y', z') from the plane ax + by + cz + d = 0 is given by \[\dfrac{{ax' + by' + cz' + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
(ii) The length of the perpendicular from a point having position vector \({\vec n}\) to the plane \(\vec r \cdot {\vec n} = {d}\) is given by \[p = \dfrac{{|\vec a \cdot \vec n - d|}}{{|\vec n|}}\]


19. ANGLE BISECTORS

(i) The equations of the planes bisecting the angle between two given planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are\[\dfrac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} =  \pm \dfrac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }}\]
 
(ii) Equation of bisector of the angle containing origin: First make both the constant terms positive. then the positive sign in \(\dfrac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} =  \pm \dfrac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }}\) gives the bisector of the angle which contains the origin.

(iii) Bisector of acute/obtuse angle : First make both the constant terms positive. Then
     a1a2 + b1b2 + c1c2 > 0 ⇒ origin lies obtuse angle
     a1a2 + b1b2 + c1c2 < 0 ⇒ origin lies in acute angle


20. FAMILY OF PLANES

(i) Any plane passing through the line of intersection of non- parallel planes OR equation of the plane through the given line in non symmetrical form. a1x + b1y + c1z + d1 =0 and a2x + b2y + c2z + d2 = 0 is  a1x + b1y + c1z+ d1 + λ(a2x + b2y + c2z + d2) = 0
 
(ii) The equation of plane passing through the intersection of the planes \(\vec r \cdot {{\vec n}_1} = {d_1}\) and \(\vec r \cdot {{\vec n}_2} = {d_2}\) is \(\vec r \cdot ({{\vec n}_1} + \lambda {{\vec n}_2}) = {d_1} + \lambda {d_2}\).

(iii) Plane through a given line : Equation of any plane through the line in symmetrical form \(\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}\) is A(x - x1) + B(y - y1) + c(z - z1) = 0, where Al + Bm + Cn = 0.


21. AREA OF A TRIANGLE
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle then Area of triangle is \(\Delta  = \sqrt {\Delta _x^2 + \Delta _y^2 + \Delta _z^2} \), where \({\Delta _x} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{y_1}}&{{z_1}}&1\\{{y_2}}&{{z_2}}&1\\{{y_3}}&{{z_3}}&1\end{array}} \right|\), \({\Delta _y} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{z_1}}&{{x_1}}&1\\{{z_2}}&{{x_2}}&1\\{{z_3}}&{{x_3}}&1\end{array}} \right|\) and \({\Delta _z} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|\)
 
Vector Method: From Two vectors \(\overrightarrow {AB} \)  and \(\overrightarrow {AC} \). then area is given by \[\dfrac{1}{2}|\overrightarrow {AB}  \times \overrightarrow {AC} | = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right|\]
 


22. VOLUME OF A TETRAHEDRON 

Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, y3) and D(x4, y4, y4,) is given by: \[V = \frac{1}{6}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&{{z_1}}&1\\{{x_2}}&{{y_2}}&{{z_2}}&1\\{{x_3}}&{{y_3}}&{{z_3}}&1\\{{x_4}}&{{y_4}}&{{z_4}}&1\end{array}} \right|\]
 
 
23. EQUATION OF A LINE 

(i) A straight line in space is characterized by the intersection of two planes which are not parallel and therefore, the equation of a straight line is a solution of the system constituted by the equations of the two planes, a1x + b1y + c1z +d1 = 0 and a2x + b2y + c2z + d2 = 0. This form is also known as non-symmetrical form.
(ii) The equation of a line passing through the point (x1, y1, z1,) and having direction ratios a, b, c is \[\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c} = r\] This is called the symmetric form. A general point on the line is given by (x1 + ar, y1 + br, z1 + cr).
(iii) Vector Equation: Vector equation of a straight line passing through a fixed point with position vector \({\vec a}\) and parallel to a given vector \({\vec b}\) is \(\vec r = \vec a + \lambda \vec b\) where \(\lambda \) is scalar.
 
(iv) The equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is: \[\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}\]
(v) Vector equation of a straight line passing through two points with position vectors \({\vec a}\) and \({\vec b}\) is \(\vec r = \vec a + \lambda (\vec b - \vec a)\).
(vi) Reduction of cartesion form of equation of a line to vector form and vice versa \[\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c} \Leftrightarrow \vec r = ({x_1}\hat i + {y_1}\hat j + {z_1}\hat k) + \lambda (a\hat i + b\hat j + c\hat k)\]
 

NOTE:
Straight lines parallel to coordinate axes: 
 Straight Lines
Equation
(i)Through Origin
y = mx, z = nx
(ii) x - axis
y = 0, z = 0
(iii) y - axis
x = 0, z = 0
(iv) z - axis
x = 0, y = 0
(v) Parallel to x - axis
y = q, z = r
(vi) Parallel to y - axis
x = p, z = r
(vii) Parallel to z - axis
x = p, y = q

 
24. ANGLE BETWEEN A PLANE AND A LINE

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

 
25. CONDITION FOR A LINE TO LIE IN A PLANE
 
(i) Cartesian form: Line \(\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}\) would lie in a plane ax + by + cz + d = 0, if ax1 + by1 + cz1 + d = 0 and al + bm + cn = 0.

(ii) Vector form: Line \(\vec r = \vec a + \lambda \vec b\) would lie in the plane \(\vec r \cdot \vec n = d\) if \(\vec b \cdot \vec n = 0\) and \(\vec a \cdot \vec n = d\).
 
 
26. COPLANER LINES

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
27. SKEW LINES 

(i) The straight lines which are not parallel and non-coplanar i.e. non-intersecting are called skew lines.
If \[\Delta  = \left| {\begin{array}{*{20}{c}}{\alpha ' - \alpha }&{\beta ' - \beta }&{\gamma ' - \gamma }\\l&m&n\\{l'}&{m'}&{n'}\end{array}} \right| \ne 0\]
then the lines are skew.
 
(ii) Vector Form: For lines \({{\vec a}_1} + \lambda {{\vec b}_1}\) and \({{\vec a}_2} + \lambda {{\vec b}_2}\) to be skew, when \(({{\vec b}_1} \times {{\vec b}_2})({{\vec a}_2} - {{\vec a}_1}) \ne 0\) or \([{{\vec b}_1}{{\vec b}_2}({{\vec a}_2} - {{\vec a}_1})] \ne 0\) 
 
(iii) Shortest distance between the two parallel lines \(\vec r = {{\vec a}_1} + \lambda \vec b\) and \(\vec r = {{\vec a}_2} + \mu \vec b\) is \[d = \left| {\frac{{({{\vec a}_2} - {{\vec a}_1}) \times \vec b}}{{|\vec b|}}} \right|\]


28. COPLANARITY OF FOUR POINTS 

The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are coplaner, then \[\begin{array}{l}\left| {\begin{array}{*{20}{c}}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\\{{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}}\end{array}} \right| = 0\\\end{array}\]


29. SIDES OF. PLANE 

A plane divides the three dimensional space two equal parts. Two points A (x1,y1, z1,) and B (x2, y2, z2) are on the same side of the plane ax + by + cz + d = 0 if ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d and both positive or both negative and are opposite side of plane if both of these values are in opposite sign.

30. LINE PASSING THROUGH THE GIVEN POINT (x1 y1 z1) AND INTERSECTING BOTH THE LINES    (P1 = 0, P2 = 0)    AND     (P3 = 0, P4 = 0) 
Get a plane through (x1, y1, z1) and containing the line (P1 = 0, P2 = 0) as P5 = 0
Also get a plane through (x1, y1, z1) and containing the line P3 = 0, P4 = 0 as P6 = 0
Equation of the required line is (P5 = 0, P6 = 0)

31. TO FIND IMAGE OF A POINT W.R.T. A LINE

Let is a given line. Let (x', y', z') is the image of the point P(x1, y1, z1) with respect to the line L.
then,
(i) a(x1 - x') + b(y1 - y') +c(z1 - z') = 0
(ii) \(\dfrac{{\dfrac{{x + {x_1}}}{2} - {x_2}}}{a} = \dfrac{{\dfrac{{{y_1} - y'}}{2} - {y_2}}}{b} = \dfrac{{\dfrac{{{z_1} - z'}}{2} - {z_2}}}{c} = \lambda \)
from (ii) get the value of x', y', z' in terms of λ as x' = 2aλ + 2x2 - x1, y' = 2bλ + 2y2 - y1 and z' = 2cλ + 2z2 - z1. Now, put the values of x', y', z' in (i) get λ and resubstitute the value of λ to get (x', y', z').

 

32. TO FIND IMAGE OF A POINT W.R.T. A PLANE

Let P(x1, y1, z1) is a given point and ax + by + cz + d = 0 is given plane. Let (x', y', z') is the image point, then
  1. x' - x1 = λa, y' - y1 = λb, z' - z1 = λc, \( \Rightarrow \) x' = λa + x1, y' = λb + y1, z' = λc + z1,
  2.  \(a\left( {\dfrac{{x' + {x_1}}}{2}} \right) + b\left( {\dfrac{{y' + {y_1}}}{2}} \right) + c\left( {\dfrac{{z' + {z_1}}}{2}} \right) + d = 0\)

 from (i) put the values of x', y', z' in (ii) and get the values of λ and re-substitute in (i) to get (x', y', z').



Share:

Class 12 Chapter 10 (Vector Algebra) Class Notes Part II

 

Product of Two Vectors

Considering examples:-

Suppose we have a wooden stick and in order to move the stick we will apply some force onto it.

The stick will move this shows some work is done in moving the stick.

Work done = Force × Displacement

Where Force and Displacement both are vector quantities and the work done is a scalar quantity.

 

Consider the case when the wooden stick is fixed on the wall with the help of a nail.

Again force is applied the wooden stick will start rotating.

Torque = Force × Distance 

Where Force and Distance both are vector quantities and the torque is also a vector quantity.

 Whenever cross product is considered then the resultant will be a vector quantity.

 Whenever dot product is considered then the resultant is a scalar quantity.


Scalar (or Dot Product) of two vectors

The scalar  product of  two nonzero vectors \({\overrightarrow a }\) and \({\overrightarrow b }\) is denoted by \(\overrightarrow a  \cdot \overrightarrow b \). It is defined as:  \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \), Where θ = angle between \(\overrightarrow a \) and \(\overrightarrow b \), \(|\overrightarrow a |\) = magnitude of \(\overrightarrow a \) and \(|\overrightarrow a |\) = magnitude of \(\overrightarrow b \).

 

Derivation with the help of example:-

Consider a log of wood and if a  Force F  is  applied  on it, the log  of wood moves. From the  figure  we  can  see  that the θ  is the angle  between displacement and force.

Force can  be  written into 2 components i.e. Fcosθ and Fsinθ.

Work done = (Fx Displacement + Fy Displacement), Where FX = Force in x direction and FY = Force in y direction.

Work done = \(F\cos \theta \overrightarrow D  + F\sin \theta \), Where \(\overrightarrow D \) = displacement in x direction and displacement in y direction is 0.

Therefore Work done = \(F\cos \theta \overrightarrow D  \Rightarrow W = |\overrightarrow F ||\overrightarrow D |\cos \theta \) 

 

Observations:-

\(\overrightarrow a  \cdot \overrightarrow b \) is a real number.

Let \(\overrightarrow a \) and \(\overrightarrow b \) be two  nonzero  vectors, then \(\overrightarrow a  \cdot \overrightarrow b  = 0\), if and only if  \(\overrightarrow a \) and  \(\overrightarrow b \) are perpendicular to each other i.e. \(\overrightarrow a  \cdot \overrightarrow b  = 0\) iff \(\overrightarrow a \) ⊥ \(\overrightarrow b \).

( Î¸ = 900 then cos900 = 0 therefore  \(\overrightarrow a  \cdot \overrightarrow b  = 0\)).

If θ = 0, then \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\)

 

If θ = Ï€, then \(\overrightarrow a  \cdot \overrightarrow b  =  - |\overrightarrow a ||\overrightarrow b |\)

 

Therefore \(\left( {\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1} \right)\) as  they  are in same  direction which shows they are  parallel to each other and \(\left( {\widehat i \cdot \widehat k = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0} \right)\) as they are perpendicular to each other. The scalar product is commutative i.e. \(\overrightarrow a  \cdot \overrightarrow b  = \overrightarrow b  \cdot \overrightarrow a \).

 

Angle between Vectors: Scalar (or Dot product)

To Prove: \(\cos \theta  = \dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}\) or \(\theta  = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}} \right)\).

Consider \({\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta }\)   (equation 1)

Dividing both the sides of equation(1)   by \({|\overrightarrow a ||\overrightarrow b |}\) we get,

\(\cos \theta  = \dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}\) or \(\theta  = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}} \right)\).


Properties of scalar product

Let \(\overrightarrow a \) and \(\overrightarrow b \) be any two vectors and λ be any scalar. Then \[(\lambda \overrightarrow a ) \cdot \overrightarrow b  = \lambda (\overrightarrow a  \cdot \overrightarrow b ) = \overrightarrow a  \cdot (\lambda \overrightarrow b )\]

Derivation of scalar product:-

To derive :- (\overrightarrow a  \cdot \overrightarrow b ) = (a1 b1) +( a2 b2) + (a3 b3)

Consider 2 vectors such that \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)  and \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\).

\(\overrightarrow a  \cdot \overrightarrow b  = ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k) \cdot ({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k)\)

\( = ({a_1}.{b_1})(\widehat i.\widehat i) + ({a_1}.{b_2})(\widehat i.\widehat j) + ({a_1}.{b_3})(\widehat i.\widehat k) + ({a_2}.{b_1})(\widehat j.\widehat i) + ({a_2}.{b_2})(\widehat j.\widehat j) + ({a_2}.{b_3})(\widehat j.\widehat k) + ({a_3}.{b_1})(\widehat k.\widehat i) + ({a_3}.{b_2})(\widehat k.\widehat j) + ({a_3}.{b_3})(\widehat k.\widehat k)\) equation (1)

Using \(\left( {\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1} \right)\)  and \(\left( {\widehat i \cdot \widehat k = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0} \right)\)  equation(1) becomes

Therefore \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)  and \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\).  Hence proved.


Projection of a vector on a line

Suppose a vector \(\overrightarrow {AB} \) makes  an angle θ with a given directed line l (say). Then the projection of \(\overrightarrow {AB} \) on l is a  vector \(\overrightarrow {p} \) (say) with magnitude | AB| cosθ, and the direction of \(\overrightarrow {p} \) being the same (or opposite) to that of the line l, depending upon whether cos θ is positive or negative.

The vector \(\overrightarrow {p} \) is called the projection vector and its magnitude |\(\overrightarrow {p} \)|  is called as the projection of the vector \(\overrightarrow {AB} \) on line l

Problem:-

Find the angle between two vectors \(\overrightarrow {a} \) and \(\overrightarrow {b} \) with magnitudes \(\sqrt 3 \) and 2, respectively having  \(\overrightarrow a  \cdot \overrightarrow b  = \sqrt 6 \).

Answer: It is given that, \({|\overrightarrow a | = \sqrt 3 ,|\overrightarrow b | = 2}\) and \({\overrightarrow a  \cdot \overrightarrow b  = \sqrt 6 }\).

\(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \)

Therefore, \(\sqrt 6  = \sqrt 3  \times 2 \times \cos \theta \)

\( \Rightarrow \cos \theta  = \dfrac{1}{{\sqrt 2 }}\)

\( \Rightarrow \theta  = \dfrac{\pi }{4}\)

Hence, the angle between the given vectors \(\overrightarrow {a} \) and  \(\overrightarrow {b} \) is \(\dfrac{\pi }{4}\).

 

Problem:  Find the angle between the vectors \(\widehat i - 2\widehat j + 3\widehat k\) and \(3\widehat i - 2\widehat j + \widehat k\)?

Answer:  The given vectors are \(\overrightarrow a  = \widehat i - 2\widehat j + 3\widehat k\) and \(\overrightarrow b  = 3\widehat i - 2\widehat j + \widehat k\).

\(|\overrightarrow a | = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}}  = \sqrt {14} \) and \(|\overrightarrow b | = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}}  = \sqrt {14} \).

Now \(\overrightarrow a  \cdot \overrightarrow b  = (\widehat i - 2\widehat j + 3\widehat k) \cdot (3\widehat i - 2\widehat j + \widehat k)\) = 1.3 + (-2) (-2) + 3.1 = (3 + 4 + 3) = 10.

Also, we know that \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \)

Therefore, \(10 = \sqrt {14} \sqrt {14} \cos \theta  \Rightarrow \cos \theta  = \left( {\dfrac{5}{7}} \right)\)

\( \Rightarrow \theta  = {\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)\)

 

Problem:  Find the projection of the vector î – ĵ  on the vector î + Äµ.

Answer:  Let \(\overrightarrow a \) = ( î – ĵ) and \(\overrightarrow b \) = ( î + Äµ)

Now, the projection of vector \(\overrightarrow {a} \) on \(\overrightarrow {b} \) is given by,

\({\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow b |}} = \dfrac{{1 - 1}}{{\sqrt {{1^2} + {1^2}} }} = 0}\)

Hence, the projection of vector \({\overrightarrow a }\) on \({\overrightarrow b }\) is  0.

Problem:-

Find the projection of the vector (î +3 Äµ +7 k̂) on the vector (7î - Äµ +8 k̂).

Answer:-

Let \(\overrightarrow a  = \widehat i + 3\widehat j + 7\widehat k\) and \(\overrightarrow b  = 7\widehat i - \widehat j + 8\widehat k\)

 Now, projection of vector \(\overrightarrow a \) on \(\overrightarrow b \) is given by,

\(\dfrac{{\vec a \cdot \vec b}}{{|\vec b|}} = \dfrac{{7 - 3 + 36}}{{\sqrt {{1^2} + {{( - 1)}^2} + {8^2}} }} = \dfrac{{60}}{{\sqrt {114} }}\) 
Problem:-

Show that each of the given three vectors is a unit vector: \(\dfrac{{2\hat i + 3\hat j + 6\hat k}}{7},\;\dfrac{{3\hat i - 6\hat j + 2\hat k}}{7},\;\dfrac{{6\hat i + 2\hat j - 3\hat k}}{7}\). Also, show that they are mutually perpendicular to each other.

Answer:-

Let \(\vec a = \dfrac{{2\hat i + 3\hat j + 6\hat k}}{7} = \dfrac{2}{7}\hat i + \dfrac{3}{7}\hat j + \dfrac{6}{7}\hat j,\),   \(\vec b = \;\dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat j,\),  \(\vec c = \;\dfrac{{6\hat i + 2\hat j - 3\hat k}}{7} = \dfrac{6}{7}\hat i + \dfrac{2}{7}\hat j - \dfrac{3}{7}\hat j\)

\(|\vec a| = \sqrt {{{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{6}{7}} \right)}^2}}  = \sqrt {\dfrac{{4 + 9 + 36}}{{49}}}  = 1\),  \(|\vec b| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{{ - 6}}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2}}  = \sqrt {\dfrac{{9 + 36 + 4}}{{49}}}  = 1\),  \(|\vec c| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{{ - 3}}{7}} \right)}^2}}  = \sqrt {\dfrac{{36 + 4 + 9}}{{49}}}  = 1\).

Thus, each of the given three vectors is a unit vector.

\(\vec a \cdot \vec b = \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{3}{7} \times \dfrac{{ - 6}}{7} + \dfrac{6}{7} \times \dfrac{2}{7} = \dfrac{6}{{49}} - \dfrac{{18}}{{49}} + \dfrac{{12}}{{49}} = 0\).

\(\vec b \cdot \vec c = \dfrac{3}{7} \times \dfrac{6}{7} + \dfrac{{ - 6}}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{{ - 3}}{7} = \dfrac{{18}}{{49}} - \dfrac{{12}}{{49}} - \dfrac{6}{{49}} = 0\).

\(\vec c \cdot \vec a = \dfrac{6}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{{ - 3}}{7} \times \dfrac{6}{7} = \dfrac{{12}}{{49}} + \dfrac{6}{{49}} - \dfrac{{18}}{{49}} = 0\).

Hence, the given three vectors are mutually perpendicular to each other.

 

Problem:-

 Find \(|\vec a|\)  and \(|\vec b|\), if \((\vec a + \vec b) \cdot (\vec a - \vec b) = 8\) and \(|\vec a|\) = 8 \(|\vec b|\).

Answer:-

\( \Rightarrow (\vec a \cdot \vec a - \vec a \cdot \vec b + \vec b \cdot \vec a - \vec b \cdot \vec b) = 8\)

\( \Rightarrow |\vec a{|^2} - |\vec b{|^2} = 8\) \( \Rightarrow {(8|\vec b|)^2} - |\vec b{|^2} = 8\)  using \(|\vec a|\) = 8 \(|\vec b|\).

\( \Rightarrow 64|\vec b{|^2} - |\vec b{|^2} = 8\) \( \Rightarrow 63|\vec b{|^2} = 8\)

\( \Rightarrow |\vec b| = \sqrt {\dfrac{8}{{63}}}  = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}\)

\( \Rightarrow |\vec a| = 8|\vec b| = 8 \times \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }} = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}\)

 

Problem:-

Evaluate the product (3 a ⃗ - 5 b ⃗). (2 a ⃗ + 7 b ⃗).

Answer:-

(3 a ⃗ - 5 b ⃗). (2 a ⃗ + 7 b ⃗)

= (3 a ⃗. 2 a ⃗ + 3 a ⃗. 7 b ⃗ - 5 b ⃗.2 a ⃗ - 5 b ⃗.7 b ⃗)

= (6 a ⃗. a ⃗+21 a ⃗. b ⃗-10 a ⃗ b ⃗-35 b ⃗. b ⃗)

=6 I a ⃗I2 + 11 a ⃗. b ⃗ - 35 I b ⃗I2

Problem:-

Find the magnitude of two vectors a ⃗ and b ⃗, having the

same magnitude and such  that  the angle between them is 60°  and

their scalar product is (1/2).

Answer:-

Let θ be the angle between the vectors a ⃗ and b ⃗.

It is given that I a ⃗I = I b ⃗ I, (a ⃗. b ⃗) =(1/2), and θ  = 60°.       (1)

We know that a ⃗. b ⃗ = I a ⃗I I b ⃗ I cos θ.

Therefor (1/2) = I a ⃗I I a ⃗ I cos 60° Using (1)

=> (1/2) = I a ⃗I2 x (1/2)

=> I a ⃗I2 = 1

=> I a ⃗I = I b ⃗I = 1

Problem:-

Find I x ⃗I,  if for a unit vector a ⃗, (x ⃗- a ⃗) (x ⃗+ a ⃗) =12?

Answer:-

(x ⃗- a ⃗) (x ⃗+ a ⃗) = 12

=> (x ⃗. x ⃗ + x ⃗. a ⃗ - a ⃗. x ⃗ - a ⃗.a ⃗)  = 12

=> I x ⃗I2 - I a ⃗I2  = 12

=> I x ⃗I2 – 1 = 12   (I a ⃗I = 1 as a ⃗ is a unit vector).

=> I x ⃗I2  = 13

Therefore I x ⃗I =  √ (13).

 

Problem:-

If a ⃗ = 2î +2 Äµ +3 k̂, b ⃗=-î +2 Äµ +k̂ and c ⃗ = 3î +ĵ  are such that a ⃗+ λ b ⃗ 

Is  perpendicular to c ⃗, then find the value of λ.

Answer:-

The given vectors are a ⃗ = 2î +2 Äµ +3 k̂,  b ⃗=-î +2 Äµ +k̂ and c ⃗ = 3î + ĵ

Now, a ⃗+ λ b ⃗ = (2î +2 Äµ +3 k̂) + λ (-î +2 Äµ +k̂)

= (2- λ) Ã® + (2+2 λ) Äµ + (3+ λ) k̂

If (a ⃗+ λ b ⃗) is perpendicular to c ⃗, then

(a ⃗+ λ b ⃗). c ⃗=0.

[(2- Î») î +(2+2 Î») Äµ+(3+ Î») k̂].(3 Ã®+ ĵ)=0

(2- Î») (3) + (2+2 Î») (1) + (3+ Î») (0) =0

=> 6- 3 λ + 2+ 2 λ =0

=>- λ+8=0

=> λ=8.

Hence, the required value of λ is 8.

 

Problem:-

Show that: (I a ⃗ I b ⃗) +( I b ⃗ I a ⃗) is perpendicular to

 (I a ⃗ I b ⃗) -( I b ⃗ I a ⃗), for any two nonzero vectors a ⃗ and b ⃗.

Answer:-

((I a ⃗ I b ⃗) + (I b ⃗ I a ⃗)). ((I a ⃗ I b ⃗) - (I b ⃗ I a ⃗))

= (I a ⃗I2 b ⃗. b ⃗) - (I a ⃗ I I b ⃗ I) (b ⃗. a ⃗) + (( I b ⃗ I I a ⃗ I a ⃗. b ⃗) – (I b ⃗I2 a ⃗. a ⃗))

= I a ⃗I2 I b ⃗I2 - I b ⃗I2 I a ⃗I2

=0

Hence (I a ⃗ I b ⃗) +( I b ⃗ I a ⃗)  and (I a ⃗ I b ⃗) -( I b ⃗ I a ⃗),

are perpendicular to each other.

 

Problem:-

If (a ⃗. a ⃗=0) and (a ⃗. b ⃗=0),  then what can be concluded about  the vector b ⃗?

Answer:-

It is given that (a ⃗. a ⃗=0) and (a ⃗. b ⃗=0).

Now,

(a ⃗. a ⃗)=0 => I a ⃗ I2 =0 => I a ⃗ I =0

Therefore a ⃗  is  a  zero vector.

Hence, vector b ⃗ satisfying  (a ⃗. b ⃗) can  be  any  vector.

 

 

Problem:-

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ ABC. [∠ ABC is the angle between the vectors BA ⃗ and BC ⃗]

Answer:-

The vertices of ∆ABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2).

Also, it is given that ∠ ABC is the angle between the vectors BA ⃗ and BC ⃗.

BA ⃗= {1-(-1)} Ã® + (2-0) ĵ + (3-0) k̂ = 2î +2 Äµ +3 k̂

BC ⃗ = {0-(-1)} Ã® + (1-0) ĵ + (2-0) k̂ =î + Äµ +2k̂

Therefore (BA ⃗. BC ⃗) = (2î +2 Äµ +3 k̂). (î + Äµ +2k̂)

=2x1+2x1+3x2 = 2+2+6 =10

| BA ⃗|=√ (2)2+ (2)2+ (3)2 = √4+4+9=√17

| BC ⃗|=√1+1+ (2)2 = √6.

Now, it is known that:

(BA ⃗. BC ⃗) = | BA ⃗|| BC ⃗| cos [∠ ABC]

Therefore, 10 =√17 x√6 cos [∠ ABC]

=> cos [∠ ABC] = (10)/ (√17 x√6)

=> ∠ ABC = cos-1(10/√102)

Problem:-

Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

Answer:-

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

Therefore    AB ⃗ = (2-1) Ã® + (6-2) Äµ + (3-7) k̂ = î+ 4 ĵ -4 k̂

 BC ⃗ = (3-2) Ã® + (10-6) Äµ + (-1-3) k̂ = î+ 4 ĵ -4 k̂

AC ⃗ = (3-1) Ã® + (10-2) Äµ + (-1-7) k̂= 2î+ 8 ĵ -8 k̂

I AB ⃗I=√ ((1)2 + (4)2+ (-4)2) = √1+16+16= √33

I BC ⃗I=√ ((1)2 + (4)2+ (-4)2) = √1+16+16= √33

I AC ⃗I=√ ((2)2 + (8)2+ (8)2) =√4+64+64 =√132=2√33

Therefore I AC ⃗I = I AB ⃗I + I BC ⃗I

Hence, the given points A, B, and C are collinear.


Vector (Cross) Product

The vector product of two nonzero vectors  a-> and  b-> is  denoted  by a->  × b->.

Vector product is given as:- (a× b) =| a || b | sin θ nˆ

Where n ˆ = unit vector perpendicular to both  a-> and b->.  The value of n ˆ is  given by the right hand thumb rule.

 

Consider a wooden stick which is fixed in the wall with the help of a nail.

If we apply force as shown in the figure the stick will topple.

Torque produced = F ⃗x  r ⃗

θ = angle between  the  r ⃗ and  the  force 

The 2 components of force will be F cos θ (vertical) and F sin θ (horizontal).

Form the figure we can see as F cos θ  and  r ⃗  are both in  same    directions  so  Fcos θ  won’t produce any effect.

Therefore Torque =| F ⃗ || r ⃗ | sin θ.

Conclusion:-

Result of cross product of any 2 vectors is a vector.

Direction is perpendicular to both inputs (F and r).

 

Vector Product: Right handed rectangular coordinate systems

In order to  find  the  vector product of 2  vectors    a->  and    b->, then  the direction of   a->  × b-> will  be  given  by the thumb.

 

Observations:-

a->  × b-> is a vector.

Let a⃗and b⃗  be two nonzero vectors. Then a⃗ × b⃗ = 0  if and  only  if a⃗ and b⃗ are parallel (or collinear) to  each  other, i.e., a⃗ × b⃗ = 0 ⃗ ⇔ a⃗ || b ⃗.

If θ = (Ï€/2) then ( a-> x  b->) =|  a->  ||  b-> | nˆ.

îx î =  Äµ x ĵ =   k̂ x k̂ =0  and  î x Äµ = k̂ ,  ĵ x k̂ = Ã®  and  k̂ x Ã®  = ĵ

 

sin θ =(| a-> x  b->|)/(|  a->|| b->|)

It is always true that the vector product is not commutative, as

( a-> x  b->) = - ( b-> x  a->). The magnitudes are same but the directions are different.

 

Cross  Product  as  Area of triangle

If  \({\overrightarrow a }\) and  \({\overrightarrow b }\)  represent the adjacent sides  of a  triangle then its area is given as : \(\dfrac{1}{2}\left| {\overrightarrow a  \times \overrightarrow b } \right|\)

 

By the definition of area of triangle, from the figure:

Area of triangle ABC = \(\dfrac{1}{2}AB \times CD\)

But \(AB = |\vec b|\) (given) and \(CD = |\vec a|\sin \theta \).

Thus area of triangle ABC = \(\frac{1}{2}|\vec b||\vec a|\sin \theta  = \frac{1}{2}|\vec a \times \vec b|\).

Cross Product as Area of Parallelogram

If  a-> and  b-> represent the sides of a parallelogram, then area is given as \(|\vec a \times \vec b|\).

 

Area of parallelogram ABCD =  DE

But AB=| b->| (given)  and  DE = |  a->| sin θ.

Therefore Area of parallelogram ABCD = | b->||  a->| sin θ = |  a-> x  b->|.

Vector (Cross) Product: Distributive Property

If  a-> ,  b-> and c ⃗ are  any three vectors and λ be a scalar, then:- 

a× (b + c ) = a × b + a × c .

λ(a× b) = (λ a) x (b) = a x (λ b)

Vector(Cross) Product: General Vector:-

Let a ⃗and b⃗ be two vectors and in component form they are given as:

(a1î +a2 Äµ +a3 k̂) and (b1 Ã® +b2 Äµ +b3 k̂) respectively.

To Prove:-

 

Using ( a-> x b->) = (a1 Ã® +a2 Äµ +a3 k̂) x (b1 Ã® +b2 Äµ +b3 k̂)

= a1 b1 (î x î) + a1 b2 (î x ĵ) + a1 b3 (î x k̂) + a2 b1 (ĵ x î) + a2b2 (ĵ x ĵ)

+ a2 b3 (ĵ x k̂) + a3 b1 (k̂ x î) + a3 b2 (k̂ x ĵ) + a3 b3 (k̂ x k̂)

= a1 b2 (î x ĵ) - a1 b3 (k̂ x î) - a2 b1 (î x ĵ) + a2 b3 (ĵ x k̂) + a3 b1 (k̂ x î)

- a3 b2 (ĵ x k̂)

(Using  î x î = ĵ x ĵ  = k̂ x k̂ = 0  and  î x k̂ = - k̂ x î,   ĵ x î  = - î x ĵ 

and  k̂ x ĵ  = - ĵ x k̂)

= a1 b2 k̂ - a1 b3 Äµ - a2 b1 k̂ + a2 b3 Ã® + a3 b1 Äµ - a3 b2 Ã®

Problem: -

Find a unit vector perpendicular to each ( a-> +  b->) and ( a-> -  b->), where  a->  = î +ĵ + k̂ and  b->= î +2 Äµ +3 k̂?

Now | c ⃗| = √ (4+16+4) = √24 = 2√6.

Therefore, the required unit vector is (c ⃗)/ (| c ⃗|)

= (-1/√6) Ã® + (2/ √6) Äµ – (1/√6) k̂

 

Problem:-

If a unit vector a⃗ makes  an  angle (Ï€/3)  with î,  (Ï€/4)  with ĵ and an acute angle  θ  with k̂, then find θ and hence,  the components of a⃗.

Answer:-

Let unit vector have (a1, a2, a3) components. a⃗ = a1î + a2ĵ + a3k̂.

Since a⃗ is a unit vector, | a⃗|=1.

Also it is given that a⃗ makes angles (Ï€/3) with î and (Ï€/4) ĵ with, an

acute angle θ with k̂.

Then we have:

cos (Ï€/3) = (a1/|a|)

=> (1/2) = a1 (because |a|= 1)

cos (Ï€/4) = (a2/|a|)

=> (1/√2) = a2   (because |a|= 1)

Also, cos θ = (a3/|a|)

=> a3 = cos θ

Now, |a|= 1

=> √ ((a1)2+ (a2)2+ (a3)2) =1

=> (1/2)2 + (1/√2)2 + cos 2 θ=1

=> (1/4) + (1/2) + cos 2 θ=1

=> (3/4) cos 2 θ=1

=> cos 2 θ = 1 – (3/4) (1/4)

=>cos θ = (1/2) => θ = (π/3)

Therefore a3 = cos (Ï€/3) = (1/2)

Hence, θ = (Ï€/3) and the components of a⃗ = ((1/2), (1/√2), (1/2))

Problem:  Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Answer:  The vertices of triangle ABC  are given as A (1, 1, 2), B (2, 3, 5)  and C (1, 5, 5).

The adjacent sides  \(\overrightarrow {AB} \) and \(\overrightarrow {BC} \) of ∆ABC  are given as:

\(\overrightarrow {AB}  = (2 - 1)\widehat i + (3 - 1)\widehat j + (5 - 2)\widehat k =   \widehat i + 2\widehat j + 3\widehat k \) and \(\overrightarrow {BC}  = (1 - 2)\widehat i + (5 - 3)\widehat j + (5 - 5)\widehat k =  - \widehat i + 2\widehat j\)

Area of ∆ABC = \(\dfrac{1}{2}\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right|\).

\(\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right| = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&2&3\\{ - 1}&2&0\end{array}} \right| = | - 6\widehat i - 3\widehat j + 4\widehat k|\)

Therefore \(\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right| = \sqrt {{{( - 6)}^2} + {{( - 3)}^2} + {4^2}}  = \sqrt {36 + 9 + 16}  = \sqrt {61} \) 

Hence, the area of triangle ABC is \(\dfrac{{\sqrt {61} }}{2}\) square units.


Triple product

Vector product of two vectors can be made to undergo dot or cross product with any third vector.
(a) Scalar tripple product:-
  • For three vectors A, B, and C, their scalar triple product is defined as A . (B × C) = B . (C × A) = C . (A × B) obtained in cyclic permutation.
  • Switching the two vectors in the cross product negates the triple product, i.e.:   C . (A x B) = - C . (B × A)
  • If A = (Ax, Ay, Az) , B = (Bx, By, Bz) , and C = (Cx, Cy, Cz) then, A . (B × C) is the volume of a parallelepiped having A, B, and C as edges and can easily obtained by finding the determinant of the 3 x 3 matrix formed by AB, and C. \[A.(B \times C) = \left| {\begin{array}{*{20}{c}}{{A_x}}&{{A_y}}&{{A_z}}\\{{B_x}}&{{B_y}}&{{B_z}}\\{{C_x}}&{{C_y}}&{{C_z}}\end{array}} \right|\]
  • If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the "parallelepiped" defined by them would be flat and have no volume.
(b) Vector Triple Product:-
  • For vectors A, B, and C, we define the vector tiple product as A × (B × C) = B(A . C) - C(A - B)
  • Note that (A × B) ×C ≠ A× (B × C)



Share:

Keep Learning

"The beautiful thing about learning is that no one can take it away from you”

Check Out Some Popular Posts from Our Site

Tags

Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's.

Facebook

Labels

Recent Posts

Unordered List

  • Coming soon.
  • Coming Soon.
  • Coming Soon.