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Te tree representations o% te 'osition vector o% point A

Vectors Introduction- Lesson1 To solve vector problems at te C!"C level# $ou will need to understand te %ollowing terms • 'osition vector# • Displacement vector# • Collinear vectors# • Equal vectors# • Parallel vectors and resultant vectors.

are

2

a#

/

or

⃗ O  A

Displacement Vector

(n example will be used to illustrate eac point Vector Definition:

( vector is a )uantit$ tat as a magnitude *si+e, and a direction. * sown b$ an arrow, In te diagram above O A and O  B are called position vectors  because teir starting points are taen relative to te origin *,. ⃗

⃗

Te starting starting point o% vector  A  B is not te origin# so te term displacement vector is used to di%%erentiate between between tis vector and te position vector. ⃗

Vector Representation Vector ( is te vector going %rom ( to . Tree di%%erent representations o% te vector ( are

1. 2. /.

A  B ⃗

 /

)

 - as two letters letters wit an overead overead arrow  - as a column column matrix matrix *column vector,

Example 1

O   P 

'oints '*/# 2, and and 3*-1# -/, ave position vectors vectors and O   relative to te origin . Q

O  Q as column vectors

1.

"xpress

O   P   and

2.

"xpress

 P   Q  as a column vector 

/.

4ind 4ind te te len lengt gt  o% o%

 P   Q

m - as a lowercase letter 

Te size * modulus# modulus# magnitude# magnitude# lengt , o% vector ( is %ound using '$tagoras   2 / 2 = 0

Solution: Te position vectors can be %ound directl$ %rom te coordinates o% ' and 3

1,

Position vector

O⃗  P =

/ 2

 

 = −1 OQ −/

a

'oint (*2#/, (*2#/, can be viewed as being displaced %rom te origin # b$ a vector called te position vector were 5 cxcDirect Institute  - 678 89-2770 mail admin:cxcDirect.org website www.cxcdirect.org

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⃗ indin!  P Q Te displacement vector  PQ  PQ is te vector going %rom ' to 3 were  so ow do we %ind tis vector> vector>

⃗

ind

Vector equation

⃗ @  AC 

⃗ ⃗ OD OD  BO  ⃗ BO

bAc

 AD  AD @

@ @

 BM  @

⃗

 MA  MA @

Imagine tat $our starting position is point ' and $ou wis to get to point 3. ?ote tat te onl$ now pat pat or course is to travel %irst %rom ' to # and ten %rom  to 3. Tis means tat we can get %rom ' to 3 using te two vectors tat we alread$ now. i.e

⃗  A  PO

⃗

 PQ  PQ  @

Parallel Vectors and Equal Vectors

⃗ OQ

⃗ # but wat  ?ote care%ull$ tat we ave te vector O  P  we need is te vector  P  ⃗ O . Tis is owever easil$ %ound owever# since te vector ⃗ vector O P   P ⃗ O is simpl$ te reverse o% te vector i.e

⃗

2, so 

 PQ  PQ @

 PO  PO

⃗

 @

⃗ @ −OP 

−

/ 2

)

−

/ 2

) )

Vector c is parallel to vector " i%  c # $"%  and  is a constant *scalar ,.

−1 −/

A

− −0

@

)

so i% a vector is a constant *scalar, multiple o% anoter vector# ten te$ are parallel. "xample

/. Lengt *magnitude, o%

 =  −  P Q −   −0 2

2

@ 8.

iven a b =



c=

12 6

)

Solution:

D f 

 ?ow

M d e

A

)

'rove tat te vectors are parallel.

Vector Equation Exercise:

a

/ 2

 so

C

O

)

can be written as



c =  ×b

/ 2

)

*tae  as a %actor,

*  @  a constant,

⇒  c @ b * so b and c are parallel,

c

b

12 6

B

In te diagram above# te points are (# # C # ; # and # and te vectors are a#b#c#d#e# and %. ⃗ =d  !o %or example  M  C 

Equal Vectors

Vector e is e)ual to vector & i% te$ bot ave te same magnitude and direction. direction. * so e @ , , It %ollows tat e)ual vectors are also parallel vectors

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)esson * "xample1 -Vectors

Collinear vectors ' on a strai!&t line( C

Two points ( and  ave position vectors  −2   O  # were  is te  B  @ O   A  @ 0 2 origin * =#=,. Te point  lies on te line ( suc tat 1  x  AG = AB . "xpress in te %orm # /  y     A B  A   G  position vector O  G

 

B





A collinear vectors

Two vectors are collinear i% one vector is a scalar ⃗  and  B C  ⃗    are multiple o% te oter vector. I%  A B C ⃗ = k.A B ⃗ E collinear *on a straigt line,# ten  B C  were  is a scalar *constant, *?ote tat tis is te same condition %or parallel vectors,

A(- 2, 5)

G

B(4,2)

Example: ive iven n 'oin 'oints ts (*-2 (*-2#1 #1, , *2# *2#/, /, and and C*6# C*6#8, 8, . Bse a vector metod to prove tat te points are collinear

0 GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

 1( indin!

Solution:

⃗  A  B

⃗  is te vector going %rom ( to   ?ow  A  B Tat is 4irst go %rom ( to  and ten ten %rom  to  C

⃗  A ⃗  @  A O so  A B B

 but

⃗  A O

so

 @  A B

−O⃗  A  @

 @

A

−  − 2 0

o

now

 A⃗ G

1

@

/

 so we need to %ind te two displacement vectors  A B and  B   C  C   and ten establis te relationsip between tem.

⃗ +O B ⃗  @ −  A O

⃗  A B

 @

⃗   B C  C

⃗ + O C  ⃗  @  @  B O

 but

 8 /

@

1.0

 2

)

)   ) ) 

−

A

1

2 /

2 /

A

6

8

−  − 2 0

)

)    A

 2

8

@

−/

⃗ *( indin!  A G

Fe need to prove tat : ⃗ = k.A B ⃗ E. * te condition %or collinearit $,  B C 

 − 2

⃗ O B

@

@

 2

8 /

+( indin!

 ?ow

⃗  @ 1
  8

−/

 @

  2

−1

O  G

O  G

@ @

 A  A G  O A 2 −2 A −1 0

     @

= 

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

 so

⃗  B⃗ C = k.A  B

  @ 1.0

so te tree points are on a straigt line *collinear.,

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Class (ctivit$ 1 Te position vectors o% points (#  and C are

⃗ = O  A

8 2

#

⃗ = O B a

"xpress in te %orm

/ 

⃗  B C 

⃗ #  B A

2, !tate one geometrical relationsip between ( and C /, I% 'oint ; is te mid point o% ( 4ind 4ind te coordinates o% ;. Example *:

/a

@ @

⃗ @  D C 

⇒

# and O C  ⃗ = 12 −2

# vectors

b

⃗ /D A ⃗ D  A

giving so

a

a A  * b J /a, a A  b - 1.0 a b-a  * b J a, ⃗    D O +O⃗ X    +O⃗ X    −O⃗ D − 2a + 2b @ 2 (b − a )

@ @ @   @ indin!  D X  @ @

Two Geometrical Relationsi!s" ii#  DX and DC are !arallel$ !arallel$ %&# DX ' ( DC  iii# D$ C and X are are collin collinear ear % on a strai) strai)tt line#

A

Class Activit, * Te 'osition vectors o% K and  are 

D C

O   R = O

B

  −2

 and

/

O   *  =

  1

−1

X

 in te %orm  R * 

1. "xpress

In te diagram above C is te mid point o% ( and  is te mid point o% H# and D is suc tat D @ 2D(. ⃗ = /a  and Te vectors a and " are suc tat O  A ⃗ =b O B "xpress te %ollowing in terms o% a  and    A    A B  X  C  C   D  C  C  and  D  

* 4ind te lengt

  R  * 

 a b

/. iven tat anoter point is suc tat

 =  R T  T

 6

2

4ind te coordinates o% T

b GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

!tate two geometrical relationsip between DH and DC !tate one geometrical relationsip between te points D# C and H GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

Mint Te coordinates coordinates o% T comes comes %rom te position vector vector O  T  T  .........................................................

Class Activit, + (CD is a )uadrilateral suc tat  indin!  A B  @ now  A B @ @ indin!

O⃗  A =

⃗ + O  B ⃗  A O

−O⃗  A +O⃗  B @

−6 

)



O⃗  B =

−0 7

)



O⃗ C =

1 

)

−/a + b

b −/a

; is a point on ( suc tat te ratio ;( @ 1 'rove tat (C; is a parallelogram

 A   C 

now C is mid point (    @  * b J /a, ⇒ A  C  C  @   A  B

Solution: GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

indin! now

 D  C      #  D C  C = D   A  A C  C

were D @ 2D(

/int: Te opposite sides o% a parallelogram are e)ual.# so tis )uestion is testing tat $ou now ow to prove tat two vectors are e)ual. GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

and so

⃗ + D A ⃗ O D ⃗ + D  A ⃗ 2 D  A

@ /a @ /a

* given,

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Class Avtivit, 0

Te position vectors ( and  relative to te origin are a and " respectivel$.

Te point ' is on ( suc tat ' @ 2'( Te point ; is on ( suc tat ; @ ;( A

B

b a

O

 is produced to ? suc tat  @ ? Express in terms o% a and b# te vectors 1.

+  A B

+  P  A

  P  M 

2.

use a vector metod to 'rove tat 'oints '# ; and ? are collinear 

/.

Calculate  te lengt (? given tat 

a@

 8 2

and b @

 1 2

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

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Vector lesson - !olutions to activit$ )uestions

(ctivit$ 2 T

Activit, 1  R T  ⃗

R(-2,3)

B(3,4)

O T  ⃗

A(6,2

O J(1, J(1, -1)

O C(12, -2)

⃗  R *  ⃗ @  B A ⃗ @  B C   so

⃗   B C  C

⃗ + O  A ⃗  B O

@

⃗ + O⃗ C  @ BO

−

/

A



/ − 

)

A

 8 2

12 −2

−2 @

⃗  * geometrical relationsip, relationsip, / B A

 @

@

/

@

)

⃗ + O *  ⃗ @  R O

9 −8

)

−

 − 2 /

)

1 −1

A

√ ( / +(− ) ) 2

lengt @

2

)

/ −

@

)

@0

⃗  O T  T

'osition vector o% T @

@

−2

@

/

B(3,4)

)

⃗ O  R

 A 6

A

2

)

⃗   R T  T 8

@

)

0

so coordinates @ T *8# 0,

M A(6,2

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

(ctivit$ /

O

B(-5,7)

C(12, -2)

C(1,4)

1

⃗  @ i% ; @ mid ( ten  B M  @

1

/

2

−2

)

@

2

⃗  B  A

A(-8,4)

)

1.0 −1

M O

 ?ow te coordinates coordinates o% ; can be %ound %rom te te ⃗  position vector O M  were

⃗ O  M 

 @ @

⇒

⃗ + B⃗ M    O  B / 

A

1.0

−1

@

.0 /

The objective is to prove that 

*1,

 ?ow

coordinates @ ;* .0# /,

 = M  C      and *2, A  B C ⃗ =  A O ⃗ + O⃗ B    A B @

(lso

⇒ so 

)

−  − 6 

A

⃗ =  M  O ⃗ + O⃗ C   M  C 

  @ N O  M   @ O  M 

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 = C    M   A  B

O   A

−2 1

@ N

)

−0 7

)

@

)

/ /

 but ;( @ 1

    −6 

@

−2 1

−  − 2 1

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    1 − −2  1

  =  M  C  C

so

/ /

=

 = M  C     A  B C

⇒

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⇒

1 1 a   b− a  / 2

 @  P  M 

E.... *1,

1

@

8

  /b − a 

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

 = C   We now need to prove that : M   A  B

'oints '# ; and ? are collinear i%

 = k P  M    M   ,  /

⃗ @  M  A

now

C  ⃗  B

also



⃗ O  A

@

/

−6





−8

@

/

)

 so we need to %ind  M  , 

= C ⃗ O +O⃗  B @ −

1

now

−0

A



7

−8

@

E

⃗  @  M  B ⃗ + B ,  ⃗  M  , 

  @ were  M  B

/

  M   , 

so

 = C   M   A  B

⇒

)

*  is a constant,

*2,

  b − a   and I  b −a 

@ @

Ab

  /b − a 

⃗  @ now note tat  M   , 

4rom *1, and *2,# (C; is a parallelogram parallelogram.

  @ b  B , 

⃗ / P  M 

⇒ @/

ter%ore ' # ; and ? are on a strigt line * collinear,

Activit, 0 N

  consider triangle (?  A , 

To %ind te lengt o%

b

B

⇒

O    ,  =O   A  A    , 

⇒

 =O ,   - O  A  @  A , 

M A

⇒

A    , 

@2

b 1/3 (a)

P

@ 2/3 (a)

2b − a

      1 2

2 

-

-

8 2

8 2

@

− 2

O

Lengt @

indin!:

⃗ + O  B ⃗  A O

 @  A B

−a + b

@

indin!

now

 P    A =

1 /

2

@

.7

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

b−a

 a

⃗ + A M  ⃗  P  A

 @ now A M  = MB =I  A  B

⃗ P  M 

 @

⃗ + O⃗  B − A O

2

⃗  P  A

⃗  @ indin!  P  M 

so

@

−  2  −

 @

1 /

a

+

1 2

I  b −a 

( a− b)

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