## What's hot

Linear law
Linear law
sabariahosman

solucionario de purcell 3
solucionario de purcell 3
José Encalada

Line Plane In 3 Dimension
Line Plane In 3 Dimension
roszelan

Integration
Integration
timschmitz

Chapter 16 2
Chapter 16 2
EasyStudy3

Legendre Function
Legendre Function
Dr. Nirav Vyas

Matemaika Diskrit - 04 induksi matematik - 03
Matemaika Diskrit - 04 induksi matematik - 03
KuliahKita

Integration by Parts & by Partial Fractions
Integration by Parts & by Partial Fractions
MuhammadAliSiddique1

Pmp2
Lesson 30: The Definite Integral
Lesson 30: The Definite Integral
Matthew Leingang

Matematika teknik modul 1 b pd eksak dan linier
Matematika teknik modul 1 b pd eksak dan linier
Prayudi MT

Difference quotient algebra
Difference quotient algebra
math260

Lesson 8: Curves, Arc Length, Acceleration
Lesson 8: Curves, Arc Length, Acceleration
Matthew Leingang

Iterative methods
Iterative methods
Ketan Nayak

5.1 Defining and visualizing functions. Dynamic slides.
5.1 Defining and visualizing functions. Dynamic slides.
Jan Plaza

4.2 exponential functions and periodic compound interests pina t
4.2 exponential functions and periodic compound interests pina t
math260

Deep Learning A-Z™: Regression & Classification - Logistic Regression
Deep Learning A-Z™: Regression & Classification - Logistic Regression
Kirill Eremenko

Review himpunan
Review himpunan
andra1223

Pembuktian teorema pythagoras dengan identitas trigonometri
Pembuktian teorema pythagoras dengan identitas trigonometriRirin Skn

Triple integrals in spherical coordinates
Triple integrals in spherical coordinates
Viral Prajapati

### What's hot(20)

Linear law
Linear law

solucionario de purcell 3
solucionario de purcell 3

Line Plane In 3 Dimension
Line Plane In 3 Dimension

Integration
Integration

Chapter 16 2
Chapter 16 2

Legendre Function
Legendre Function

Matemaika Diskrit - 04 induksi matematik - 03
Matemaika Diskrit - 04 induksi matematik - 03

Integration by Parts & by Partial Fractions
Integration by Parts & by Partial Fractions

Pmp2
Pmp2

Lesson 30: The Definite Integral
Lesson 30: The Definite Integral

Matematika teknik modul 1 b pd eksak dan linier
Matematika teknik modul 1 b pd eksak dan linier

Difference quotient algebra
Difference quotient algebra

Lesson 8: Curves, Arc Length, Acceleration
Lesson 8: Curves, Arc Length, Acceleration

Iterative methods
Iterative methods

5.1 Defining and visualizing functions. Dynamic slides.
5.1 Defining and visualizing functions. Dynamic slides.

4.2 exponential functions and periodic compound interests pina t
4.2 exponential functions and periodic compound interests pina t

Deep Learning A-Z™: Regression & Classification - Logistic Regression
Deep Learning A-Z™: Regression & Classification - Logistic Regression

Review himpunan
Review himpunan

Pembuktian teorema pythagoras dengan identitas trigonometri
Pembuktian teorema pythagoras dengan identitas trigonometri

Triple integrals in spherical coordinates
Triple integrals in spherical coordinates

## Similar to Vectors2

Notes on Equation of Plane
Notes on Equation of Plane
Herbert Mujungu

Three dimensional geometry
Three dimensional geometry
nitishguptamaps

5.vector geometry Further Mathematics Zimbabwe Zimsec Cambridge
5.vector geometry Further Mathematics Zimbabwe Zimsec Cambridge
alproelearning

Impact of Linear Homogeneous Recurrent Relation Analysis
Impact of Linear Homogeneous Recurrent Relation Analysis
ijtsrd

planes and distances
planes and distances
Elias Dinsa

7.5 lines and_planes_in_space
7.5 lines and_planes_in_space
Mahbub Alwathoni

Further pure mathmatics 3 vectors
Further pure mathmatics 3 vectors
Dennis Almeida

3D Coordinate Geometry
3D Coordinate Geometry
ParasKulhari

1525 equations of lines in space
1525 equations of lines in space
Dr Fereidoun Dejahang

Sequence and Series
Solutions of AHSEC Mathematics Paper 2015
Solutions of AHSEC Mathematics Paper 2015
Nayanmani Sarma

Crack problems concerning boundaries of convex lens like forms
Crack problems concerning boundaries of convex lens like forms
ijtsrd

Class 14 3D HermiteInterpolation.pptx
Class 14 3D HermiteInterpolation.pptx
MdSiddique20

Differential geometry three dimensional space
Differential geometry three dimensional space
Solo Hermelin

Analytical Geometry in three dimension
Analytical Geometry in three dimension
SwathiSundari

Maths04
Maths04
sansharmajs

Presntation11
Presntation11
Bindiya syed

TABREZ KHAN.ppt
TABREZ KHAN.ppt
TabrezKhan733764

A Proof of Twin primes and Golbach's Conjecture
A Proof of Twin primes and Golbach's Conjecture
nikos mantzakouras

Analysis Of Algorithms Ii
Analysis Of Algorithms Ii
Sri Prasanna

### Similar to Vectors2(20)

Notes on Equation of Plane
Notes on Equation of Plane

Three dimensional geometry
Three dimensional geometry

5.vector geometry Further Mathematics Zimbabwe Zimsec Cambridge
5.vector geometry Further Mathematics Zimbabwe Zimsec Cambridge

Impact of Linear Homogeneous Recurrent Relation Analysis
Impact of Linear Homogeneous Recurrent Relation Analysis

planes and distances
planes and distances

7.5 lines and_planes_in_space
7.5 lines and_planes_in_space

Further pure mathmatics 3 vectors
Further pure mathmatics 3 vectors

3D Coordinate Geometry
3D Coordinate Geometry

1525 equations of lines in space
1525 equations of lines in space

Sequence and Series
Sequence and Series

Solutions of AHSEC Mathematics Paper 2015
Solutions of AHSEC Mathematics Paper 2015

Crack problems concerning boundaries of convex lens like forms
Crack problems concerning boundaries of convex lens like forms

Class 14 3D HermiteInterpolation.pptx
Class 14 3D HermiteInterpolation.pptx

Differential geometry three dimensional space
Differential geometry three dimensional space

Analytical Geometry in three dimension
Analytical Geometry in three dimension

Maths04
Maths04

Presntation11
Presntation11

TABREZ KHAN.ppt
TABREZ KHAN.ppt

A Proof of Twin primes and Golbach's Conjecture
A Proof of Twin primes and Golbach's Conjecture

Analysis Of Algorithms Ii
Analysis Of Algorithms Ii

## More from Math Academy Singapore

Sec 3 A Maths Notes Indices
Sec 3 A Maths Notes Indices
Math Academy Singapore

Sec 4 A Maths Notes Maxima Minima
Sec 4 A Maths Notes Maxima Minima
Math Academy Singapore

Sec 3 E Maths Notes Coordinate Geometry
Sec 3 E Maths Notes Coordinate Geometry
Math Academy Singapore

Sec 3 A Maths Notes Indices
Sec 3 A Maths Notes Indices
Math Academy Singapore

Sec 2 Maths Notes Change Subject
Sec 2 Maths Notes Change Subject
Math Academy Singapore

Sec 1 Maths Notes Equations
Sec 1 Maths Notes Equations
Math Academy Singapore

Math academy-partial-fractions-notes
Math academy-partial-fractions-notes
Math Academy Singapore

Complex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levels
Math Academy Singapore

Recurrence
Functions 1 - Math Academy - JC H2 maths A levels
Functions 1 - Math Academy - JC H2 maths A levels
Math Academy Singapore

Probability 1 - Math Academy - JC H2 maths A levels
Probability 1 - Math Academy - JC H2 maths A levels
Math Academy Singapore

### More from Math Academy Singapore(11)

Sec 3 A Maths Notes Indices
Sec 3 A Maths Notes Indices

Sec 4 A Maths Notes Maxima Minima
Sec 4 A Maths Notes Maxima Minima

Sec 3 E Maths Notes Coordinate Geometry
Sec 3 E Maths Notes Coordinate Geometry

Sec 3 A Maths Notes Indices
Sec 3 A Maths Notes Indices

Sec 2 Maths Notes Change Subject
Sec 2 Maths Notes Change Subject

Sec 1 Maths Notes Equations
Sec 1 Maths Notes Equations

Math academy-partial-fractions-notes
Math academy-partial-fractions-notes

Complex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levels

Recurrence
Recurrence

Functions 1 - Math Academy - JC H2 maths A levels
Functions 1 - Math Academy - JC H2 maths A levels

Probability 1 - Math Academy - JC H2 maths A levels
Probability 1 - Math Academy - JC H2 maths A levels

## Recently uploaded

Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
Scholarhat

Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Dr. Vinod Kumar Kanvaria

Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
TechSoup

South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)
Academy of Science of South Africa

Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Akanksha trivedi rama nursing college kanpur.

Assignment_4_ArianaBusciglio Marvel(1).docx
Assignment_4_ArianaBusciglio Marvel(1).docx
ArianaBusciglio

Main Java[All of the Base Concepts}.docx
Main Java[All of the Base Concepts}.docx
adhitya5119

A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptx
thanhdowork

How to Add Chatter in the odoo 17 ERP Module
How to Add Chatter in the odoo 17 ERP Module
Celine George

Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Mohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia

Advantages and Disadvantages of CMS from an SEO Perspective
Advantages and Disadvantages of CMS from an SEO Perspective
Krisztián Száraz

S1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptx
tarandeep35

Digital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments Unit
chanes7

Introduction to AI for Nonprofits with Tapp Network
Introduction to AI for Nonprofits with Tapp Network
TechSoup

Top five deadliest dog breeds in America
Top five deadliest dog breeds in America
Bisnar Chase Personal Injury Attorneys

A Independência da América Espanhola LAPBOOK.pdf
A Independência da América Espanhola LAPBOOK.pdf
Jean Carlos Nunes Paixão

বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
eBook.com.bd (প্রয়োজনীয় বাংলা বই)

How to Fix the Import Error in the Odoo 17
How to Fix the Import Error in the Odoo 17
Celine George

CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf
camakaiclarkmusic

PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad
AyyanKhan40

### Recently uploaded(20)

Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat

Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...

Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion

South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)

Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama University

Assignment_4_ArianaBusciglio Marvel(1).docx
Assignment_4_ArianaBusciglio Marvel(1).docx

Main Java[All of the Base Concepts}.docx
Main Java[All of the Base Concepts}.docx

A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptx

How to Add Chatter in the odoo 17 ERP Module
How to Add Chatter in the odoo 17 ERP Module

Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx

Advantages and Disadvantages of CMS from an SEO Perspective
Advantages and Disadvantages of CMS from an SEO Perspective

S1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptx

Digital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments Unit

Introduction to AI for Nonprofits with Tapp Network
Introduction to AI for Nonprofits with Tapp Network

Top five deadliest dog breeds in America
Top five deadliest dog breeds in America

A Independência da América Espanhola LAPBOOK.pdf
A Independência da América Espanhola LAPBOOK.pdf

বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf

How to Fix the Import Error in the Odoo 17
How to Fix the Import Error in the Odoo 17

CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf

PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad

### Vectors2

• 1. VECTORS II 1 Vector Product The vector product can be seen in 3 different forms, used for different types of questions. (1) Mathematical calculation of vector product: Vector (cross) product   a1 a2 a3   ×   b1 b2 b3   =   a2b3 − a3b2 −(a1b3 − a3b1) a1b2 − a2b1   (2) 3D picture of vector product: a b n Π Let Π be the plane that a and b lies. The cross product of a and b produces a vector n that is perpendicular to Π. A vector that is perpendicular to a plane is also called a normal vector to the plane. (3) Vector product in terms of sine: a × b = (|a||b| sin θ)ˆn |a × b| = |a||b| sin θ where θ is the angle between the 2 vectors. Example 1. Let a = 2i + j − 3k and b = −i + 2j + 3k, find the vectors a × b and b × a. What is the relationship between a × b and b × a? www.MathAcademy.sg 1 © 2017 Math Academy
• 2. 2 Equation of Planes m1 m2 n Π A A plane Π is defined by a point A which it passes through, and two direction vectors m1 and m2 that lies on the plane. Parametric form or Vector form r = −→ OA + λm1 + µm2, λ, µ ∈ R where m1, m2 are parallel to the plane and A is a point on the plane. Scalar-Product form Let r, a be points on the plane, n be a normal vector to the plane. Then, r · n = a · n where n = m1 × m2. Proof: For any position vector −−→ OR on the plane, since −→ AR is perpendicular to n, −→ AR · n = 0 (−−→ OR − −→ OA ) · n = 0 −−→ OR · n − −→ OA · n = 0 −−→ OR · n = −→ OA · n Cartesian Form We must first have the equation in scalar-product form. r · n = D r ·   a b c   = D   x y z   ·   a b c   = D ax + by + cz = D (Cartesian form) www.MathAcademy.sg 2 © 2017 Math Academy
• 3. Example 2. t (a) Find the equation of the following, in parametric, scalar-product and cartesian form. Plane Π which passes through the point (1, 0, 2) and is parallel to the vectors   1 0 1   and   0 1 0  . Solution: Parametric: r =   1 0 2   + λ   1 0 1   + µ   0 1 0   , λ, µ ∈ R. Scalar-Product:   1 0 1   ×   0 1 0   =   −1 0 1   r · n = a · n r ·   −1 0 1   =   1 0 2   ·   −1 0 1   r ·   −1 0 1   = −1 + 2 = 1 r ·   −1 0 1   = 1 Cartesian:   x y z   ·   −1 0 1   = 1 −x + z = 1 *Note that the following equations refer to the same plane. r ·   −1 0 1   = 1 r ·   1 0 −1   = −1 r ·   −2 0 2   = 2 www.MathAcademy.sg 3 © 2017 Math Academy
• 4. (b) The plane Π contains the point A(2, 2, −2) and the line r =   1 3 −3   + λ   −2 0 1   , λ ∈ R. Find the equation of Π1 in scalar product form. [r ·   1 3 2   = 4] (c) Find the cartesian equation of the following planes: (i) x − y plane x y z (ii) x − z plane x y z (iii) y − z plane x y z (d) Given the following equation, r ·   1 3 2   = 4, how do we create points that lies on the plane? How do we check if a point lies on a plane? www.MathAcademy.sg 4 © 2017 Math Academy
• 5. 3 Foot of perpendicular from a point to a plane P F n Π Finding foot of perpendicular from point to plane Strategy: Given equation of the plane Π : r · n = D (1) Step 1: Form the equation of the line ℓ that passes through P and is perpendicular to Π. ℓ : r = −−→ OP + λn, λ ∈ R. (2) Step 2: Intersect ℓ with Π to get the foot of the perpendicular. That is, substitute (2) into (1). ( −−→ OP + λn) · n = D Substitute λ back into (2) to get the foot of the perpendicular. Example 3. Find the foot of perpendicular from the point P(2, −1, 3) to Π : r ·   −1 1 −1   = 3. Solution: Let ℓ be the line that passes through P and perpendicular to Π. ℓ : r =   2 −1 3   + λ   −1 1 −1   , λ ∈ R. Intersect ℓ and Π,   2 − λ −1 + λ 3 − λ   ·   −1 1 −1   = 3 −2 + λ − 1 + λ − 3 + λ = 3 −6 + 3λ = 3 λ = 3 ∴ Foot of perpendicular: −−→ OF =   2 −1 3   + 3   −1 1 −1   =   −1 2 0  . www.MathAcademy.sg 5 © 2017 Math Academy
• 6. Example 4. The point P has position vector i + 2j - k and a plane Π has equation Π : x + 2y + 3z = 16. Let A be the point with position vector 9i + 2j + k. (i) Find the foot of perpendicular from P to Π. (ii) Hence, find the shortest distance between P and Π. (iii) Show that A lies on the plane Π. (iv) Let ℓ be the line that passes through A and P. Find the vector equation of the reflection of the line ℓ in the plane Π. [(i) −−→ OF =   2 4 2   (ii) √ 14] www.MathAcademy.sg 6 © 2017 Math Academy
• 7. (iii) Since   9 2 1   ·   1 2 3   = 9 + 4 + 3 = 16, A lies on the plane. (iv) Π A P F P′ ℓ −−→ OF = −−→ OP + −−→ OP′ 2 2 −−→ OF = −−→ OP + −−→ OP′ −−→ OP′ = 2 −−→ OF − −−→ OP = 2   2 4 2   −   1 2 −1   =   3 6 5   Direction vector of ℓ = −−→ OP′ − −→ OA =   3 6 5   −   9 2 1   =   6 4 4   ∴ r =   9 2 1   + µ   6 4 4   , µ ∈ R. www.MathAcademy.sg 7 © 2017 Math Academy
• 8. 4 Angles In this section, denote m as the direction vector of the line, n as the normal vector of the plane. θ ℓ1 ℓ2 m1 m2 Acute angle between two lines cos θ = m1 · m2 |m1||m2| . Π1 Π2 θ θ n2 n1 Acute angle between 2 planes cos θ = n1 · n2 |n1||n2| . ϕ θ Π ℓ m n Acute angle between line and a plane cos ϕ = m · n |m||n| . ∴ θ = 90◦ − ϕ Alternative Method sin θ = m · n |m||n| . www.MathAcademy.sg 8 © 2017 Math Academy
• 9. Example 5. Find the ACUTE angle between the following two lines: ℓ1 : r = (λ − 3)i + (2λ − 1)j + (2λ − 1)k, λ ∈ R. ℓ2 : r = (3µ − 1)i + (2µ − 1)j + (µ − 1)k, µ ∈ R. [36.7◦] Example 6. Find the acute angle between the line r = (i + 2j − k) + λ(i − j + k) and plane 2x − y + z = 4. [70.5◦] Example 7. Find the acute angle between the planes r · (i + j + k) = 3 and r · (2i + 2j − k) = 1. [54.7◦] www.MathAcademy.sg 9 © 2017 Math Academy
• 10. 5 Relationship between a line and a plane Are the line and plane parallel? YES The line lies in the plane. The line does not lie in the plane. NO The line intersects the plane at a point. m n Π : r · n = d ℓ Recall that if two vectors a and b are perpendicular, then a · b = 0. Let m be a direction vector for the line ℓ and n be a normal for the plane Π. m is parallel to the plane Π ⇔ m is perpendicular to n ⇔ m · n = 0 3 different ways to check if ℓ lies on Π Let the equations of the line ℓ and plane Π be given by ℓ : r = a + λm, λ ∈ R and Π : r · n = d. (1) Check if ℓ and Π are parallel. If they are, then m · n = 0. If we have checked that the above holds, to check if ℓ lies in the plane Π, check that (Any point on ℓ) ·n = d (2) If (a + λm) · n = d, then line lies in plane. (3) Show that 2 points on ℓ lie on Π. What about the case where ℓ does not lie on Π? www.MathAcademy.sg 10 © 2017 Math Academy
• 11. Example 8. (a) Determine the relationship between pairs of lines and planes: (i) ℓ1 : r =   1 2 2   + λ   2 −1 −1   and Π1 : r ·   1 1 1   = 5 (ii) ℓ2 : r = (i + j + k) + λ(i − 2j + k) and Π2 : r · (i + j + k) = 5 (b) Show that the line ℓ1 : r =   1 2 2   + λ   2 −1 −1   intersects the plane r ·   1 0 1   = 2 at only 1 point. [(ii) Parallel but does not lie in plane ] Solution: (i) m · n =   2 −1 −1   ·   1 1 1   = 2 − 1 − 1 = 0. Therefore, ℓ1 and Π1 are parallel. Furthermore, since   1 2 2   ·   1 1 1   = 1 + 2 + 2 = 5, ℓ1 lies in Π1. Alternative Method: Since   1 + 2λ 2 − λ 2 − λ   ·   1 1 1   = 1 + 2λ + 2 − λ + 2 − λ = 5, ∴ ℓ1 lies in Π1. www.MathAcademy.sg 11 © 2017 Math Academy
• 12. 6 Relationship between 2 planes 6.1 Parallel Planes Two planes are parallel to each other ⇐⇒ Their normals are scalar multiple of each other. Q: What is the relationship between the following 2 planes? p1 : r ·   1 −1 2   = 2, p1 : r ·   −2 2 −4   = 3. www.MathAcademy.sg 12 © 2017 Math Academy
• 13. 6.2 Line of intersection between 2 planes Any 2 non parallel, non identical planes will intersect in a line. n1 n2 ℓ Π1 Π2 Method 1: Cross the 2 normals Finding line of intersection between 2 planes The direction vector, m, of the line of intersection between Π1 and Π2 is given by m = n1 × n2, where n1, n2 are the normal vectors of Π1 and Π2 respectively. We will use a GC if there are no unknowns in n1 and n2. Method 2: Using GC Example 9. The equations of two planes p1, p2 are 2x − 4y + z = 6 x + y − 2z = 6 respectively. The planes p1 and p2 meet in a line l. Find a vector equation of l. Question: Is there an alternative method to find a common point on both planes? www.MathAcademy.sg 13 © 2017 Math Academy
• 14. 6.3 Relationship of two perpendicular planes p1 p2 n2n1 If 2 planes, p1 and p2 are perpendicular, then the following occurs: (a) n1 is parallel to p2 =⇒ p2 contains the direction vector n1, (b) n2 is parallel to p1 =⇒ p1 contains the direction vector n2. Example 10 (2012/RVHS/Prelim/I/10modified). The equations of two planes p1 and p2 are 2y − z = 0, 2x + z = 2. Find the equation of the plane p3 that passes through the point (1, 1, 1) and is perpendicular to both p1 and p2. Solution: p1 p2 p1 p2 p3 p3 is perpendicular to p1 =⇒ p3 is parallel to n1. p3 is perpendicular to p2 =⇒ p3 is parallel to n2. ∴ p3 is parallel to both n1 and n2. n3 = n1 × n2 =   0 2 −1   ×   2 0 1   =   2 −2 −4   Therefore, p3 : r ·   2 −2 −4   =   1 1 1   ·   2 −2 −4   = 2 − 2 − 4 = −4 www.MathAcademy.sg 14 © 2017 Math Academy
• 15. 6.4 Reflection of a plane in another plane. π1 π2 l 2 direction vectors of the reflected plane Suppose we want to find the reflection of plane π1 in π2. Lets call it π′ 1. We also know that π1 and π2 intersect at the line l. 1) l will also lie on the reflected plane π′ 1. Hence π′ 1 also contains direction vector of l. 2) Take a point P from π1 and reflect it in π2. This reflected point P′ will be on π′ 1. Now take a point A from l (which is also on π′ 1). π′ 1 will contain direction vector −−→ AP′. www.MathAcademy.sg 15 © 2017 Math Academy
• 16. Example 11 (2011/YJC/Prelim/I/10modified). The planes π1 and π2 have equations r · (2i − k) = 3 and r · (−i + 3j) = 2 respectively. The 2 planes intersect in a line l. (i) Find a vector equation of l. [2] (ii) Find the coordinates of the point Q which is the reflection of the point P(5, −2, 7) in π2. [4] (iii) It is given that P lies on π1. Hence or otherwise, find a vector equation of the reflection of the plane π1 in π2. [2] [(i) r =   3 2 7 6 0   + t   3 1 6   , t ∈ R (ii) (2.4, 5.8, 7) (iii) r =   3 2 7 6 0   + s   3 1 6   + t   − 9 10 −139 30 −7   , s, t ∈ R] www.MathAcademy.sg 16 © 2017 Math Academy
• 17. 7 Distance involving plane A B n Π F θ −−→ AB · ˆn −−→ AB × ˆn A B Π F ℓ A B Π2 F Π1 −−→ AB × ˆn −−→ AB × ˆn −−→ AB · ˆn −−→ AB · ˆn Proof: Dist between the point and plane | −−→ AB · ˆn| = |ˆn|| −−→ AB| cos θ = | −−→ AB| cos θ = | −→ AF| Length of projection onto plane | −−→ AB × ˆn| = |ˆn|| −−→ AB| sin θ = | −−→ AB| sin θ = | −−→ FB| 8 Distance involving line ℓ : r = a + λm, λ ∈ R A B θ | −−→ AB × ˆm| | −−→ AB · ˆm| Distance from point to line: | −−→ AB| sin θ = | −−→ AB × ˆm| Projection of −−→ AB to line: | −−→ AB| cos θ = | −−→ AB · ˆm| www.MathAcademy.sg 17 © 2017 Math Academy
• 18. Example 12. The equation of a plane π is shown below: π : r.   2 1 −2   = 6 (a) The point A has position vector 3i − j + 4k. B is another point such that −−→ OB = 3i − 6j + 2k. Find the length of the projection of AB onto the plane π. (b) Find the distance between point A and the plane. Solution: (a) A   2 1 −2   π B θ −−→ AB = −−→ OB − −→ OA =   3 −6 2   −   3 −1 4   =   0 −5 −2   | −−→ AB × ˆn| =   0 −5 −2   × 1 √ 22 + 1 + 22   2 1 −2   = 1 3   0 −5 −2   ×   2 1 −2   = 1 3   12 −4 10   = 1 3 √ 122 + 42 + 102 = 2 √ 65 3 (b) A π θ C The point C =   0 6 0   lies on the plane since   0 6 0   ·   2 1 −2   = 6. −→ AC = −−→ OC − −→ OA =   0 6 0   −   3 −1 4   =   −3 7 4   | −→ AC · ˆn| =   −3 7 4   · 1 3   2 1 −2   = 1 3 | − 6 + 7 + 8| = 3 www.MathAcademy.sg 18 © 2017 Math Academy
• 19. Example 13. Show the the line ℓ1 : r = 2i − 2j + 3k + λ(i − j + 4k) is parallel to the plane Π : r · (i + 5j + k) = 5. Find the distance between them. [ 10√ 27 units] www.MathAcademy.sg 19 © 2017 Math Academy
• 20. Example 14 (2011/VJC/Prelim/I/12). The position vectors of points A, B, C and D with respect to the origin O are −2i + 5j + 5k, 3i + j + 4k, 3i − 4j + k and i + j − 5k respectively. (i) Obtain an equation of the plane Π, containing A, B and C in the r · n = p form. (ii) Calculate the angle between Π and the line joining A and D. Hence, or otherwise, find the shortest distance of D from Π. [(a) r.   7 15 −25   = −64 (b) 39.0◦, 7.04 units ] www.MathAcademy.sg 20 © 2017 Math Academy
• 21. 9 Alternative method to find distance between two planes Distance from origin to plane If r · ˆn = d, then, Distance from origin to plane = |d| Example 15. Find the distance between the following plane and the origin: Π1 : r ·   1 5 1   = 3 Solution: Π1 : r ·   1 √ 27   1 5 1     = 3 √ 27 Therefore, distance between Π1 and origin is 3√ 27 . Π1 Π2 O d1 d2 n Distance between 2 parallel planes Π1 : r · ˆn = d1 Π2 : r · ˆn = d2 Distance between the two planes = |d1 − d2|. Note: If d1 and d2 are of the same signs, then the planes Π1 and Π2 lie on the same side of the origin. If d1 and d2 are of opposite signs, then the planes Π1 and Π2 lie on different sides of the origin. Example 16. Find the distance between the following planes and determine if they lie on the same side of the origin. Π1 : r ·   1 5 1   = 3 Π2 : r ·   1 5 1   = −2 Solution: Π1 : r ·   1 √ 27   1 5 1     = 3 √ 27 Π2 : r ·   1 √ 27   1 5 1     = −2 √ 27 Therefore, distance between Π1 and Π2 is 3√ 27 − −2√ 27 = 5√ 27 . They lie on different sides of the origin since 3√ 27 and −2√ 27 have different signs. www.MathAcademy.sg 21 © 2017 Math Academy
• 22. Distance from origin to plane If r · ˆn = d, then distance from origin to plane = |d|. Proof: (Optional) O A n Π : r · ˆn = d θ |a · ˆn| Let A be a point on Π such that its position vector is a. Then, length of projection of −→ OA onto n = |a · ˆn|. In addition, since A lies on Π, we also have a · ˆn = d. Hence, length of projection of −→ OA onto n = |a · ˆn| = |d|. www.MathAcademy.sg 22 © 2017 Math Academy
Current LanguageEnglish
Español
Portugues
Français
Deutsche
© 2024 SlideShare from Scribd