Vectors and Moments

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Mr . Sherif Yehia Al Maraghy https://twitter.com/Mr_Sherif_yehia

01009988836 – 01009988826 Email : [email protected]

2

3 _-1

-4

 

0 ,1 The unit vector: A vector whose norm is 1

 

   

 

   

2 2 2 2 2 2 2 2

it is denoted by " i " vector and " j " vector where :

i 1,0 i x y 1 0 1 j 0 ,1 j x y 0 1 1               j

 

i  1,0 O y' y x x'











 



1 1 2 2 2 2 1 1 2 1 2 1 2 1 2 1

Very important note: If A x , y And B x , y

AB B A x , y x , y x x , y y x x i y y j

 

          

Vectors and Moments

First: Vectors

As we said before, Vectors is a physical quantity which has both magnitude and direction

Shapes of vectors

1st shape: We can write vector as

 

a ,b 2nd shape: we can write vector as a aiˆ b jˆ So for example: F 

 

2 ,3 means F 2iˆ 3 j Or Hˆ 



-4 ,-1 means H



-4iˆ ˆ j

This is called the position vector for any point

---

The length of the vectors “ Norm of the vector”













   

2 2

If A x , y Then A x y is the length of vector A and it is denoted by Norm A Example : If A 3 ,-4 Then the norm length of A 3 -4 5

  

   

---

The unit vectors

" i " and " j "

(2)

rd _-₅₅

-o

Let A and B be two non Zero vectors , Let us represent them by OC and OD ,Then the angle COD where 180 is called the smallest angle

Between these two vectors while the angle COD of measu      o

re more than 180 is called the biggest angle between them.

 O

_{ }

fig 1 a b D C  O

_{ }

fig 2 a b D C Small angle

The scalar product of two vectors is a scalar quantity which is equal to the norm of the first vector multiplied by Cosine the angle between them .

A

B



A

B Cos



where 0 



180o





Scalar Algebraic Product

Vector Product



A

B

o

30

In this chapter, we will discuss the two kinds of vector multiplication which are:

First kind: Algebraic Product

It is denoted by

Some definitions

(1) The angle between the two vectors:

Very important note:

But we prefer to use the small angle

(2) The scalar product of two vectors :

--- Example (1)

Answer

 

We can get the small angle between the two vectors A and B by Representing them by drawing

Or the same point " see figures 1 and 2 " If is the measur

the angle between two vectors :

Outwards Inwards

o o

e of the angle the two vectors 0 180 the measure of the angle between them 360

       small So big o

Find A B where A is a vector of magnitude 5 and direction 30 north of west and B is a vector of magnitude 8 towards south

o

A 5 And B 8 , and the measure of the small angle between A and B is 120 So A B A B Cos A B 5 8 Cos120 -20        

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01009988836 – 01009988826 Email : [email protected] B 40 A 65  Example (2) Answer --- Example (3) Answer --- Example (4) Answer

If A and B are two vectors such that A 15 , B 12 and A B -135 , Then find The measure of the angle between A and B

  

o A B  A B Cos



  -13515 12 Cos



 Cos



-0.75   



138 36'

The vector A is of magnitude 65 towards west , B is a vector of magnitude 40 and direction 5

east of south and with angle where Tan , Then find A B 12







o







A B A B Cos 90 A B A B - Sin -5 A B 65 40 -1000 13            5 12



13

 

1 2 3 1 2

ABC is a right angled triangle at B where AB 6 cm , BC 8 cm , The vectors F , F and F of magnitude 150 , 200 , 250 gm.wt act along BA , BC and CA , Find :

1 F F  

 

2 3

 

1 3 2 F F 3 F F

 







 

o o 1 2 1 2 1 2 2 3 1 ABC 90 : F F 150 200 Cos 90 0 Note When F F F F 0 directly

2 The angle between F and F must be inward Or outward So to get the measure of the smaller angle

       









 

2 3 2 3 o 2 3 2 3 1 3 1 3

Between F and F , we have to draw the ray BC ACD is the angle between F and F

-8 F F F F Cos 180 200 250 -Cos 50000 -40000 10 3 F F F F             _ _    Cos 150 250 6 22500 10     



C D B A E



3 F 1 F 2 F 8 cm 10 cm 6 cm

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rd _-₅₅

-The Algebraic Projection of a vector in the direction of another vector

Definition

Conclusion: The algebraic Projection of a vector in the direction of another vector:

--- Example (5)

Answer

The algebraic projection of vector B in the direction of the vector A is defined to be the scalar Quantity B Cos , where B  B and is the measure of the smaller angle between A and B

B B A A   D D C C O O

The projection of B in the

Direction of A is  B Cos  B A O _A B D A B C D C o

The projection of B in the direction

of A is  B Cos 90 0 o

The projection of B in the direction of A is  B Cos 180 - B

The magnitude of the first vector × Cos the angle between the two vectors









So A B ABCos

A B A BCos A The algebraic projection of B in the direction of A

 



  

o

A is a vector where A 30 and B is a vectors B 20 and the angle between A and B is 75 , then find the algebraic projection of A in the direction of B and also the algebraic Projec

 

tion of B in direction of A .

o o

The algebraic projection of A in the direction of B A Cos 30 Cos75 7.768 The algebraic projection of B in the direction of A B Cos 20 Cos75 5.176

 

  

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01009988836 – 01009988826 Email : [email protected] B 12 A 18 o 30 o 60 Example (6) Answer --- Example (7) Answer o o

A is a vector of magnitude 18 and direction 60 north of east , B is another vector of magnitude 12 and direction 30 south of west , Find the algebraic projection of the two vectors A and B in the direction of the other .

o o o o

o

The smaller angle between the two vectors 30 90 30 150 The algebraic projection of A in the direction of B :

A Cos 18 Cos150 -9 3 The algebraic projection of B in the di

         o rection of A : B Cos 12 Cos150 -6 3 A B C D  10 7.5 A B C D 10 7.5 12.5 

 

ABCD is a rectangle in which AB 7.5 cm , BC 10 cm , Find : a The algebraic projection of the vector CB in the direction of AC b The algebraic projection of the vector BD in the direction of

 

BA

 

a Don' t forget that the direction of the two vectors must be either inwards Or outwards , so we must extend AC and get the angle , so the algebraic projection of CB in the direct





   

 

o 2 2 ion of AC : CB Cos CB Cos 180 - CB Cos

10 4 where AC 7.5 10 12.5 cm And Cos

12.5 5 - 4

So - CB Cos 10 -8 5

b The algebraic projection of BD in the d                 _ _   7.5 irection of BA : BD Cos 12.5 7.5 12.5    

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rd _-₅₆

-The scalar product of two Non Zero vectors is ve Or -ve quantity Or zero according to the measure of the between the two vectors



 

 

a If is an acute angle Then A B is ve

b If is an obtuse angle Then A B is ve

c If is a right angle Then A B is Zero and if : A B



     o o o 0 < θ < 90 90 < θ < 180 π θ = 2 0 , Then either A 0 Or B 0 Or A B    

Properties of the Scalar Product between two vectors

Rule (1)

Note (1): The units for measuring the scalar quantity A B is equal to the unit of : A multiplied by the unit of B

For example:

If A is a unit vector , whose norm is the and B is a displacement whose norm is Then the unit of the calr Product A B is

Newton meter

Newton meter Note (2):

---

Rule (2)

The scalar product of any vector into itself is equal to the square of its magnitude

2

So for any vector a a a where a a

---

Rule (3)





Example: 3 a 2b 6 a b  a 6 b 6 a b ---

Rule (4)

--- A B  A B Cos  B A Cos B A

 

  



For any two vectors a and b and for any scalar m: m a b  a m b m a b







 



For any three vectors a , b and c , The distributive law holds: a  b c  a c  b c

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

 



a  a Cos



i a Sin



j a Sin



a Cos



a o o o

if the magnitude of the vector a is 3 and 110 , find the Component form. 3 Cos 110 3Sin 110 -1.026 2.82





   

Example :

a i j i j

F  4i 2 j in the direction of the vector

Rule (5)

---

Rule (6)

---

Rule (7)

Perpendicular resolution of vector “ Geometric Component”

---

Rule (8)

Example: Find the algebraic component of the vector AB where A



-2 ,3 And B







1,-1



Answer



 



   



 



2 2 AB The algebraic Component of F in the direction of AB is F

AB And AB B A i j - 2i 3 j 3i 4 j And AB 3 - 4 5 3i 4 j AB 12 8 F 4 i 2 j 5 AB             _     0.8 5  i i j j1 , i j j i0 1ˆ 2 ˆ 1ˆ 2ˆ 1 1 2 2 If a a i a j and b b i b j Then a b a b a b       a The algebraic Component of F in the direction of a is equal to F

a



 





















 



1 2 1 2 1 1 1 2 2 1 2 2 1 1 1 2 2 1 2 2 1 1 2 2 st st nd nd 1 2 1 2 Proof : ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b a i a j b i b j a i b i a i b j a j b i a j b j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b i i a b i j a b j i a b j j a b a b ˆ ˆ ˆ ˆ a b a i a j b i b j 1 1 2 2                     Conclusion : Don' t forget : i i j j 1 , i j j i 0        

(8)

rd _-₅₆

-If A 3i 4 j and B 5i 12 j , Then find A B and the measure of the angle included between the two vectors .

   

If A 3 i j and B 4 j , Then find the algebraic projection of each of the two vectors In the direction of each other .

  

If A  3i j and B  2i k j , Then find the value of k so that A B

Mixed examples on the whole lesson



The analysis being referred to two perpendicular directions and i and j are the unit vectors in these two directions



Example (1) Answer --- Example (2) Answer --- Example (3) Answer



 



   

o o ˆ ˆ ˆ ˆ A B 3i j 2i k j 3 2 1 k 6 k And A B 90 Cos 90 0 A B A B Cos A B 0 6 k 0 k 6



                       



 



   

2 2 2 2 2 2 -1 o ˆ ˆ ˆ ˆ

There is no angle given, then : A B 3i 4 j 5i 12 j 15i 48 j 15 48 63 A 3 4 5 And B 5 12 13

A B 63 63 63

A B A B Cos Cos Cos 14 15'

13 5 65 65 A B                           









   

 

2

 

₂

   

₂ ₂

The algebraic projection of A in the direction of B : A Cos So we have to get Cos ˆ ˆ ˆ ˆ A B 3 i j 0 i 4 j 3 0 1 4 4 A 3 1 2 And B 0 4 4 A B A B A B Cos Cos A B                      4 1 -11 o Cos 60 2 4 2 2 1 The algebraic projection of A in the direction of B : A Cos 2 1

2 1 The algebraic projection of B in the direction of A : B Cos 4 2

2                 

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If A  2i 5 j and B  i k j , Then find the value of k so that A // B

o

If A k i j and B 12i 5 j , Then find the value of k that makes the measure of the Angle between A and B equals 45

    Example (4) Answer --- Example (5) Answer ---



 



   









o 2 2 2 2 ₂ 2 o 2 2 ₂ ₂ ₂ A // B 0 And A B A B Cos Where A B 2i 5 j i k j 2 5k And A 2 5 29 And B 1 k k 1 2 5k 29 k 1 Cos0 2 5k 29 k 1 So by squaring both: 2 5k 29 k 1 4 20k 25k 29k 2                                     





2 2 9 4k 20k 25 0 2k 5 0 2k 5 k 2.5            

 



 



   









o 2 2 ₂ 2 2 2 2 2 2 2 A B A B Cos 45 1 Where A B k i j 12i 5 j 12 k 5 And A k -1 k 1 And B 12 -5 13 1 169 12 k 5 13 k 1 So by squaring both: 12 k 5 k 1 2 2 169k 169 144 120k 25 288 240k 50 169k 2                                  







2 2 169 0 119k 240k 119 0 17k 7 7k 17 0 7 -17

k "agreed" And k "refused"

17 7

b b 4ac Note : We can use formula to find k : k

2a                  

(10)

rd _-₅₆

- 

 

ABCD is a trapezium in which AD // BC , m A m B , AD 16 cm , BC 21 cm

2 And AB 12 cm , Find :

1 The algebraic projection of AD in the direction of CB 2 The algebraic projection of BD in the di



     



 

rection of CB 3 The algebraic projection of CD in the direction of AB

Example (6) Answer C D 21 cm 16 cm 12 cm A B C D 16 cm 16 cm 12 cm A B C D 16 cm 12 cm A B 5 cm 12 cm 5 cm 12 cm 13 cm   16 cm

 

o

1 The measure of the angle between AD and CB is equal 180 Thus the

of AD in the direction of CB is

AD Cos The angle between AD and CB

algebraic projection





 

o 16Cos180 -16

2 Don' t forget that the direction of the two vectors must be either inwards Or outwards , so we must extend CB  





   

o 2 2

and get the angle , so the algebraic projection of BD in the direction of CB :

BD Cos BD Cos 180 - BD Cos

16 4 where BD 16 12 20 cm And Cos

20 5 So -            

 

- 4 BD Cos 20 -16 5

3 We must extend CD , so that angle EDF is the angle between CD And AB

The algebraic projection of CD in the direction of AB : CD Cos 180  _ _    



o



12 - CD Cos -13 -12 cm 13        _ _  

(11)

 





 





 



  

ABCD is a square , the length of whose side is 8 cm , find : 1 a AB CD b 2 BC AD c BC -2 DB d AC BD 4      

 





o a AB CD AB CD Cos 8 8 Cos180 64 -1 -64 1 1 1 b 2 BC AD 2 BC AD BC AD Cos 4 4 2 1 2



        _ _  _         

 





 



  

o o o 8 8 Cos0 32 c BC -2 DB -2 BC DB Cos -2 8 8 2 Cos135 128 d AC BD AC BD Cos 8 8 Cos 90 0



             Example (7) Answer --- C D A B 8 cm C D A B 8 cm C D A B 8 cm  8 2

(12)

rd _-₅₅

- 





 



  

ABCD is a rectangle which AB 6 cm , BC 8 cm , find : 5 a -2 AB AC c BC DB d AC BD 2        

   

 





 

2 2 AC 8 6 10 cm a -2 AB AC -2 AB AC Cos 6 -2 6 10 -72 10 5 5 b BC DB BC DB 2 2 5 BC BD C 2



            _         





 

   





o 2 2 os 180 5 - BC BD Cos 2 5 - 8 8 10 -160 2 10 c In AMB: AC 6 8 10 cm 1

AM AC 5 cm Median from right angle 2 Also BM AM 5 cm Prop



               





     

  



  

2 2 2 erties of a rectangle 5 5 6 7 So in AMB : Cos 2 5 5 25 7 AC BD AC BD Cos 10 10 28 25



           Example (8) Answer --- C D A B 8 cm C D A B C D A B  10 cm  10 cm  8 cm 8 cm 6 cm  8 cm 6 cm 6 cm M

(13)





 





 



  

o

ABCD is a parallelogram in which m BAC 90 , BC 2 AB 8 cm , find:

1 1 a CA CB b AD DC c AB 3 AD d 7 BD AC 3 2                

   

 





2 2 o AC 8 4 4 3 cm 4 3 a CA CB CA CB Cos 4 3 8 48 8 1 1 1 b AD DC AD DC 3 2 6 1 AD DC Cos 180 6



             _               _{ }   

 













o 1 - AD DC Cos 6 1 4 8 - 8 4 -6 8 3 c AB 3 AD 3 AB AD Cos 90 3 4 8 - Sin 4 -32 3 -1 8



              

 



  

2 ₂ 6 3 1 d In AMB: AM AC 2 3 cm 2 MB 2 3 4 2 7 cm BD 4 7 7 BD AC 7 BD AC Cos 2 3 7 4 7 4 3 336 2 7



                Example (9) Answer --- C D A B 8 cm  8 cm 4 cm 4 3 C D A B 8 cm  8 cm 4 cm 4 3 C D A B 8 cm 8 cm 4 cm 4 3 C D A B 8 cm  8 cm 4 cm 2 3 

(14)

rd _-₅₅

-



If ABCD is a square , then prove that : AB  BC BD Zero





o o AB BC BD AB BD BC BD AB BD Cos135 BC BD Cos 45 -1 1 8 8 2 8 8 2 Zero 2 2            _ _  _ _    



 



2 2

Prove that for any two vectors a and b : a b a b a b where a a

and b b      

   

Let a 3i 2 j , b -i 4 j And c -2 j 1

Then find with respect to i and j the vector: - c 2 b -3 a c 5           Example (10) Answer --- Example (11) Answer



a b

 

a b



a a  a b  b a b b a2 b2 --- Example (12) Answer

   

    



   

 

Let a 3i 2 j , b -i 4 j And c -2 j 1

Then find with respect to i and j the vector: - c 2 b -3 a c 5 2 b -3 a -6 b a -6 3 -1 2 4 -30 1 1 - c 2 b -3 a c - c -30 c - c 6 c 5 5             _  _     _ _     -7 c -7 -2j

 

14 j --- C D A B 8 cm

(15)











   



   



   



   



 



2 2 2 2 AB AC AB AC Cos BAC AB AC Cos BAC AB AC AB B A - 4 ,3 3 ,2 -7 ,1 -7 i j AB -7 1 50 AC C A - 3 ,-6 3 ,2 -6 ,-8 -6 i 8 j AC -6 -8 10 AB AC -7 i j -6 i 8 j                              





o 42 8 34 34 Cos BAC 61 16' 10 50



         

     

   



   



   



 



2 2 2 2 o AB B A 2 ,3 1,1 1,2 i 2 j And AB 1 2 5 And AC C A 5 ,-1 1,1 4 ,-2 4i 2 j And AC 4 2 2 5 AB AC i 2 j 4i 2 j 4 4 0 90 AB AC

The algebraic component of F in th



                               



 





 



e direction of AB is equal to : i 2 j AB F 7 i 4 j 3 5 Newton 5 AB

And The algebraic component of F in the direction of AC is equal to : 4 i 2 j AC F 7 i 4 j 2 5 2 5 AC           Newton Example (13)

 













ABC is a triangle in which A 3,2 , B -4 ,3 and C -3,-6 , Then find m BAC Answer

--- Example (14)

 





If A 1,1 , B 2 ,3 and C 5 ,-1 , Then prove that AB AC , Then find the algebraic Component of F 7i 4 j Newton in the two directions AB and AC .

      Answer --- C A B

(16)

01009988836 – 01009988826 Email : [email protected] rd _-₅₆

-  -  - 

   



   



   

  



2 2 2 2 AB BC AB BC Cos where : AB B A 7 ,5 4 ,1 3 ,4 And AB 3 4 5 And BC C B -5 ,-4 7 ,5 -12 ,-9 And BC -12 -9 15 3 ,4 -12 ,-9 AB BC -36 36 Cos 5 15 75 AB BC



                       

  



   

-24 25

The algebraic component of AB in the direction of BC is equal to : -12 ,-9 3 -12 4 -9

BC

AB 3 ,4 - 4.8

15 15

BC

The algebraic component of BC in the direction of AB is equ

    



  

   

al to : 3 ,4 3 -12 4 -9 AB BC -12 ,-9 -14.4 5 5 AB      Example (15)

 





If A 4 ,1 , B 7 ,5 and C -5 ,-4 , Find Cos where is the angle between the two Vectors AB and BC , Then determine the algebraic Component of each of the two vectors

in the direction of the other .



  

Answer

(17)

Second kind: Vector Product

It is denoted by







ˆ

A

B

A

B Sin



k where :







Rule

ˆ A A And B B and is the measure of the small angle between A and B and k is the unit vector perpendicular to the plane containing the two vectors A and B .

The direction of the un



 

ˆ

it vector k is determined according to the right hand rule which states that:

If the curved fingers of the right hand indicates the rotation of the vector A towards vector B Throught the smaller angle , then the thumb will indicate the direction of k as shown in figuresˆ





the vector product A B is a And not and its direction is Perpendicular to the plane containing the vectors A and B in the direction which is determined according to



Note that : vector scalar

the right hand rule and the unit vector in the direction of A B

A B

ˆ ˆ

is the vector k which is equal to k

AB Sin



 

(18)

rd _-₅₆

-Properties of the Vector Product between two vectors

Rule (1)

Note :









For the two vectors A B and B A have the same magnitude But have different direction ---

Rule (2)

o o o

If A // B , Then A B 0

for , In this case is equal to Zero or 180 and hence Sin 0 Sin 180 0 If A B 0 , Then either A 0 Or B 0 Or



        Note : A // B ---

Rule (3)

---

Rule (4)









ˆ A B - B A  A B Sin k





o o

The vector product of any vector into itself is equal zero as Sin0 0 So for any vector : a a zero

ˆ

Proof: a a a a Sin 0 k Zero



 

  

The Right hand system





o o

In the given figure:

OX and OY are two perpendicular directions

ˆ ˆ

i and j are two unit vectors in these directions respectively:

ˆ ˆ

ˆ ˆ ˆ ˆ

So i j i j Sin 90 1 1 Sin 90 k k

ˆ

Where k is the unit vector

     

ˆ ˆ

perpendicular to plane OXY containing i and j And in the direction of the right hand rule.

ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

From this: we can say : i j k k i j j k i And when they are in th

     

e anticlock wise :

ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

From this: we can say : j i -k i k - j k j -i ˆ ˆ

ˆ ˆ ˆ ˆ

But i i j j k k 0

ض ع ب ه ب ش ة جا ح ىا

     

(19)

Rule (5)









Example: 3 a  2b 6 a  b  a 6 b 6 a  b ---

Rule (6)

---

Rule (7)

---

Important Remark

---









1 2 1 2 1 2 2 1 st nd nd st ˆ ˆ ˆ ˆ ˆ If a a i a j and b b i b j Then a b a b a b k ˆ Or a b 1 2 2 1 k            



 



 





 





1 2 1 2 1 1 1 2 2 1 2 2 1 1 1 2 2 1 2 2 1 2 2 1 1 2 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Proof : a b a i a j b i b j a i b i a i b j a j b i a j b j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b i i a b i j a b j i a b j j ˆ ˆ ˆ 0 a b k a b - k 0 a b a b k Don' t forget : i i j                             





ˆ ˆ j 1 , i j k & j i - k ˆ ˆ ˆ

Where i , j , k are the right system of unit vectors

       

 

 

  



For any two vectors a and b and for any scalar m: m a  b  a  m b m a  b







 



For any three vectors a , b and c , The distributive law holds: a  b  c  a  c  b  c  O D N C A B We know that A B AB Sin

If OC represents the vector A And if OD represents the vector B

And AB Sin OC OD Sin The surface area of the parallelogram OCND



 

 

 

(20)

rd _-₅₆

(21)

01009988836 – 01009988826 Email : [email protected] Example (1) Answer --- Example (2) Answer --- Example (3) Answer

 





 





 





 





ˆ ˆ ˆ ˆ ˆ ˆ Let A 6 i 15 j , B 5i 3 j , C -i j , find : a A B C b A B C c B A C d C B A             





 







 









 







 







 







st nd nd st ˆ ˆ ˆ

Let i , j , k be a right system of unit vectors

ˆ ˆ ˆ ˆ ˆ ˆ a A B 6 i 15 j 5i 3 j 11i 12 j ˆ ˆ ˆ ˆ ˆ ˆ ˆ A B C 11i 12 j -i j 1 2 2 1 k -11 12 k k ˆ ˆ ˆ ˆ ˆ ˆ b A B 6 i 15 j 5i 3 j -18 75 k -93 k                             









_

_{ }

_

 







 







o o ˆ ˆ ˆ ˆ ˆ ˆ ˆ A B C -93 k -i j -93 k -i -93 k j ˆ ˆ ˆ ˆ 93 k i Cos 90 93 k j Cos 90 0 0 0 ˆ ˆ ˆ ˆ ˆ ˆ c A C 6 i 15 j -i j -6 15 k 9k B A                    











   

 



 











 

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ C 5i 3 j 9k 45 i k 27 j k 45 - j 27 i -27i 45 j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ d C B -i j 5i 3 j 6 i 15 j 3 5 k 6 i 15 j 48 k i 120 k j               _    _          ˆ ˆ 48 j120 i ˆ ˆ ˆ ˆ ˆ

If A  i 3 j , B mi 2 j , A B -14k , Then find the value of m .



 







ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ A B -14 k i 3 j mi 2 j -14 k -2 3m k -14 k -2 3m -14 m 4                 



 







 



 



 

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

Let C mi n j And A C -19 k -2i 5 j mi n j k -14 k

ˆ ˆ -2n 5m k -14 k -2n 5m -14 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ And C B 6 k mi n j 3i 4 j k 6 k 4m 3n 6 2        _    _                 _    _         ˆ ˆ ˆ ˆ

If A -2i 5 j and B 3i 4 j , find the vector C in the plane A and B such that:

ˆ ˆ

A C -14k And C B 6k

   

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01009988836 – 01009988826 Email : [email protected] rd _-₅₅

-

   





   





 



ˆ ˆ ˆ ˆ AB B A -1,5 2 ,1 -3 ,4 -3i 4 j BA - AB 3i 4 j ˆ ˆ ˆ ˆ AC C A - 4 ,-1 2 ,1 -6 ,-2 -6 i 2 j CA - AC 6 i 2 j ˆ ˆ ˆ ˆ BC C B - 4 ,-1 -1,5 -3 ,-6 -3i 6 j CB - BC 3i 6 j AB AC                                  



 



























ˆ ˆ ˆ ˆ ˆ ˆ -3i 4 j -6 i 2 j 6 24 k 30k ˆ ˆ ˆ ˆ ˆ BA CB 3i 4 j 3i 6 j 18 12 30k ˆ ˆ ˆ ˆ ˆ BC CA -3i 6 j 6 i 2 j -6 36 30k                     

 





 

ˆ ˆ ˆ ˆ ˆ ˆ ˆ A 3i 2 j , B -i j , C i 5 j And D 6i ˆ ˆ ˆ ˆ ˆ ˆ a AB B A - 4 i 3 j And CD D C 5i 5 j AB CD 20 15 k 5k ˆ ˆ b AC C A -2i 7 j & BA A B                           







 







 







ˆ ˆ ˆ ˆ 4 i 3 j & DC C D -5i 5 j ˆ ˆ ˆ ˆ ˆ ˆ ˆ BA DC -i 2 j AC BA DC -2i 7 j -i 2 j - 4 7 3k ˆ ˆ ˆ ˆ ˆ ˆ c BC C B 2i 4 j And AB AC - 4 i 3 j -2i 7 j 8 21 29 BC AB                             























ˆ ˆ ˆ ˆ AC 29 2i 4 j 58 i 116 j ˆ ˆ ˆ ˆ ˆ ˆ BC AB AC BD 58 i 116 j 7 i j -58 812 k -870 k            Example (4) Answer --- Example (5) Answer --- Example (6) Answer ---



 



 



 



 

ˆ ˆ ˆ ˆ ˆ ˆ

Let C mi n j And A C 21 -2i 5 j mi n j 21 -2m 5n 21 1

ˆ ˆ ˆ ˆ ˆ ˆ ˆ

And C B -8 k mi n j i j k -8 k - m n -8 2 ˆ ˆ From 1 and 2 and by simultanous : m 3 And n 5 C 3i 5 j

                  _    _              

 





ˆ

If A  2 ,3 and B  1,-1 , Then find the vector C such that: A C 21 & C B -8 k

 









If A  2 ,1 , B  -1,5 and C  -4 ,-1 Then prove that: AB AC BA CB BC CA









 





 





If A 3 ,-2 , B -1,1 , C 1,5 and D 6 ,0 , Then find:

a AB CD b AC BA DC c BC AB AC BD

   

(23)

 



   













 



   

 



  

















 



 



  





  





ˆ ˆ a a c i j 4 j 4 0 k 4k ˆ ˆ ˆ ˆ a c b 4k -i 2 j -4 k i 8 k j -4 0 8 0 0 ˆ ˆ ˆ b b c -i 2 j 4 j -i 2 j a b c i j -i 2 j - 2 1 k -3k c a b i j -i 2 j -1 2 -3 b -3c -3 b c -3 -i 2 j 4 j -3 - 4 0                                                

 



  



 

    

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ˆ ˆ k 12k d b c -i 2 j 4 j 0 8 8 a b -3 ˆ ˆ b c a a b c 8 a -3 c -24 a c -24 4k -96 k                 

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Let C a i b j , So C // A C A 0 ˆ ˆ a i b j 3i 4 j -4 a 3b k - k 4 a 3b 0 4 a 3b 0 1 ˆ ˆ ˆ ˆ C B 62k a i b j 4i 5 j 62k 5a 4b k 62k 5a 4b 62 2 , Then from 1 and 2 : b -8 & c 6 C

                                          6 i8 j

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o 2 2 2 2 2 2 2 2 ˆ ˆ ˆ a 2i 3 j mi 2 j -5k 4 3m k -5k 4 3m -5 m 3 b A B 2i 3 j mi 2 j 0 2m 6 0 m -3 2i 3 j mi 2 j A B 1 2m 6 c Cos Cos 45 2 A B ₂ ₃ _m ₂ 13 m 4

By squaring both sides : 13 m 4 2 2m 6 13m 52 2                                             

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4m 24m36 Example (7) Answer --- Example (8) Answer --- Example (9) Answer

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         

Let A i j , B -i 2 j and C 4 j , Then find:

a A C B b A B C c B A B C d B C A A B C

ˆ If A  3i 4 j , B  4i 5 j , Then find the vector C where C // A , C B 62k

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If A 2 ,3 And B m ,2 , find the value of m in each of the following cases: ˆ

a A B -5 k b A B

(24)

01009988836 – 01009988826 Email : [email protected] rd _-₅₅

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ˆ ˆ ˆ ˆ ˆ ˆ ˆ A -2 j , B 6 i 4 j , C 7i 4 j And D -2i 4 j ˆ ˆ ˆ ˆ AC C A 7 i 6 j And BD D B - 8 i 8 j ˆ ˆ ˆ ˆ ˆ ˆ AC BD 7 i 6 j - 8 i 8 j 56 48 k 104 k

Area of ABCD Area of ABC Area of ACD W                            

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1

here Area of ABC AB AC Sin

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But we know that : AB AC Sin AB BC

1 Area of ABC AB BC 2 ˆ ˆ ˆ ˆ AB B A 6 i 2 j And BC C B i 8 j ˆ ˆ ˆ ˆ ˆ ˆ AB BC 6 i 2 j i 8 j 48 2 k 50 k

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1 1 Area of ABC AB BC 50 25 1 2 2 1 1

Also Area of ACD AC AD Sin AC AD

2 2 ˆ ˆ ˆ ˆ AC C A 7 i 6 j And AD D A -3i 6 j ˆ ˆ ˆ ˆ ˆ ˆ AC AD 7 i 6 j -3i 6 j 42 18 k 60 k Area of

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                                   ACD 1 AC AD 1 60 30

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Area of ABCD Area of ABC Area of ACD 25 30 55

                Example (10) Answer ---

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If A 0 ,-2 , B 6 ,-4 , C 7 ,4 and D -2 ,4 are vertices of a quadrilateral , then find AC BD and calculate the area of ABCD

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