vector journeys!

Block 3

Vector Journeys

What is to be learned?

• How to get components of a vector using other vectors

Useful (Indeed Vital!) To KnowParallel vectors with the same magnitude will

have the same…………………

If vector AB has components ai + bj + ck, then BA will have components……………...

components

-ai – bj – ck

DiversionsSometimes needalternative route!

AB

C

DE

AE = AB + BC + CD + DE

Using ComponentsAB = 2i + 4j + 5k and BC = 5i + 8j + 3k

AC = 7i + 12j + 8k

DE = 6i + 5j + k and FE = -3i + 4j + 5k

DF = 9i + j – 4k

= AB + BC

= DE + EF

= 3i – 4j – 5kEF

Wee Diagrams and Components

A

B

C

D

ABCD is a parallelogramTT is mid point of DC

AB = 846( )

BC = 351

( )AT

Info Given

= DC

= AD

= AD + DT= AD + ½ DC

351( ) 4

23

( )+774( )=

Find AT

Wee Diagrams and Letters

A B

CDAB = 4DC

uv

Find AD in terms of u and v

?

4u

-v-u

AD = 4u – v – u

= 3u – v

Vector JourneysFind alternative routes using diversionsWith Letters A

B C

D

u

v

AD = 3BC

Find CD in terms of u and v

CD = CB + BA + AD

= -v + u + 3v= 2v + u

with components

A B

CD

E

EABCD is a rectangular based pyramid

TT divides AB in ratio 1:2

EC = 2i + 4j + 5kBC = -3i + 2j – 3kCD = 3i + 6j + 9k

Find ET

ET = EC + CB + BT

2/3 of CD

1 2

245( )=

3 -2 3( )+ +

negative 246( )

= 7i + 6j + 14k

Key Question

A

B

C

D

ABCD is a parallelogram

T

T is mid point of BC

AB = 846( )

BC = 462( )

DT

= DC

= DC + CT= DC + ½ CB

846

( ) -2 -3 -1 ( )+

615( )=

Find DT

Questions

Cunningly Acquired!

Hint

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VABCD is a pyramid with rectangular base ABCD.

The vectors are given by

Express in component form.

, andAB AD AV

8 2 2AB

i j k 2 10 2AD

i j k

7 7AV

i j k

CV

AC CV AV

CV AV AC

BC AD

AB BC AC

Ttriangle rule ACV Re-arrange

Triangle rule ABC also

CV AV AB AD 1 8 2

7 2 107 2 2

CV

55

7CV

9 5 7CV

i j k

VECTORS: Question 3

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Reveal answer only

EXIT

P

QR

S

T

U V

W

A

BPQRSTUVW is a cuboid in whichPQ , PS & PW are represented by the vectors

[ ], 4 2 0 [ ]and

-2 4 0 [ ]resp.

0 0 9

A is 1/3 of the way up ST & B is the midpoint of UV.ie SA:AT = 1:2 & VB:BU = 1:1.

Find the components of PA & PB and hence the size of angle APB.

VECTORS: Question 3

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Reveal answer only

EXIT

P

QR

S

T

U V

W

A

BPQRSTUVW is a cuboid in whichPQ , PS & PW are represented by the vectors

[ ], 4 2 0 [ ]and

-2 4 0 [ ]resp.

0 0 9


Find the components of PA & PB and hence the size of angle APB.

|PA|

= 29

|PB|

= 106

= 48.1°APB

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Question 3

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PA =

PS + SA =

PS + 1/3ST

PS + 1/3PW

=

[ ]-2 4 0 [ ] =

0 0 3

+ [ ] -2 4 3

=

PB =

PQ + QV + VB

PQ + PW + 1/2PS

=

[ ] 4 2 0 [ ]+

0 0 9

+ [ ]= -1 2 0 [ ] 3

4 9

=

PQRSTUVW is a cuboid in whichPQ , PS & PW are represented by vectors

[ ], 4 2 0 [ ]and

-2 4 0 [ ]resp.

0 0 9


Find the components of PA & PB

and hence the size of angle APB.

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Question 3

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PA = [ ] -2

4 3

PB = [ ] 3

4 9PQRSTUVW is a cuboid in which

PQ , PS & PW are represented by vectors

[ ], 4 2 0 [ ]and

-2 4 0 [ ]resp.

0 0 9




(b) Let angle APB =

A

P

B

ie

PA .

PB =

[ ] -2 4 3 [ ] 3

4 9

.

= (-2 X 3) + (4 X 4) + (3 X 9)

= -6 + 16 + 27

= 37

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Question 3

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PQRSTUVW is a cuboid in whichPQ , PS & PW are represented by vectors

[ ], 4 2 0 [ ]and

-2 4 0 [ ]resp.

0 0 9




PA .

PB =

37

|PA| = ((-2)2 + 42 + 32)

= 29

|PB| = (32 + 42 + 92)

= 106

Given that PA.PB = |PA||PB|cos

then cos = PA.PB

|PA||PB| = 3729 106

so = cos-1(37 29 106)

= 48.1°

Ex

Suppose that AB = ( ) 3-1 and BC = ( ) .5

8

Find the components of AC .

AC = AB + BC =********

( ) + ( ) 3-1

58 = ( ) .8

7

Ex

Simplify PQ - RQ

********

PQ - RQ = PQ + QR = PR

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