CxcDirect Institute -

Vectors - Lesson 1 & 2

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Vectors Introduction- Lesson1 To solve vector problems at the CSEC level, you will need to understand the following terms: • Position vector, • Displacement vector, • Collinear vectors, • Equal vectors, • Parallel vectors and resultant vectors.

The three representations of the Position vector of point A 2 are: a , or O ⃗ A 3

()

Displacement Vector

An example will be used to illustrate each point Vector Definition:

A vector is a quantity that has a magnitude (size) and a direction. (shown by an arrow) In the diagram above O ⃗ A and O ⃗ B are called position vectors because their starting points are taken relative to the origin (O). The starting point of vector A ⃗ B is not the origin, so the term displacement vector is used to differentiate between this vector and the position vector.

Vector Representation Vector AB is the vector going from A to B. Three different representations of the vector AB are: 1. 2. 3.

A⃗ B

() 4 3

- as two letters with an overhead arrow - as a column matrix (column vector)

Example 1 O P

Points P(3, 2) and Q(-1, -3) have position vectors  relative to the origin O. and O Q

 as column vectors OQ

1.

Express

O P

and

2.

Express

 PQ

as a column vector

3.

Find the length of

 PQ

m - as a lowercase letter

The size ( modulus, magnitude, length ) of vector AB is found using Pythagoras:  4 2 3 2 = 5

Solution: The position vectors can be found directly from the coordinates of P and Q:

1)

Position vector

()

O⃗ P=

3 2

 

 = −1 OQ −3

a

Point A(2,3) can be viewed as being displaced from the origin O, by a vector called the position vector where: © cxcDirect Institute - 876 469-2775 mail: [email protected] website: www.cxcdirect.org

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CxcDirect Institute -

Vectors - Lesson 1 & 2

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⃗ Finding P Q The displacement vector ⃗ PQ is the vector going from P to Q where : so how do we find this vector?

Find

Vector equation

⃗ = AC

b+c

⃗ AD = ⃗ OD = ⃗ BO = ⃗ BM =

⃗ MA =

Imagine that your starting position is point P and you wish to get to point Q. Note that the only know path or course is to travel first from P to O, and then from O to Q. This means that we can get from P to Q using the two vectors that we already know. i.e

⃗ PQ

⃗ PO

=

Parallel Vectors and Equal Vectors

⃗ OQ

+

Note carefully that we have the vector O ⃗ P , but what ⃗ . we need is the vector P O This is however easily found however, since the vector ⃗ is simply the reverse of the vector O ⃗ PO P i.e

⃗ PO

2) so :

⃗ PQ =

=

⃗ = −OP

()

−

3 2

+

() ( )

−

Vector c is parallel to vector b if : c = kb, and k is a constant (scalar ).

3 2

−1 −3

so if a vector is a constant (scalar) multiple of another vector, then they are parallel.

( ) −4 −5

=

Example: 3. Length (magnitude) of

  −42 −5 2 P Q=

= 6.4

()

Given: a b=

c=

() 12 8

Solution:

D f

Now

M d e

A

;

Prove that the vectors are parallel.

Vector Equation Exercise:

a

3 2

so

C

O

⇒

c

b

() 12 8

can be written as

c=4×b

4

() 3 2

(take 4 as a factor)

( k = 4 a constant)

c = 4b ( so b and c are parallel)

B

In the diagram above, the points are A, B, C , M , and O, and the vectors are a,b,c,d,e, and f. ⃗ So for example: M C=d Use this diagram to complete the table. (Note the direction of the arrows) © cxcDirect Institute - 876 469-2775 mail: [email protected] website: www.cxcdirect.org

Equal Vectors

Vector e is equal to vector h if they both have the same magnitude and direction. ( so e = h) It follows that equal vectors are also parallel vectors

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CxcDirect Institute -

Vectors - Lesson 1 & 2

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Collinear vectors ( on a straight line) 6

7

Lesson 2 Example1 -Vectors

8 C

Two points A and B have position vectors −2 4 ; O B = , where O is the O A = 5 2 origin ( 0,0). The point G lies on the line AB such that 1 x AG = AB . Express in the form , 3 y  ; position vector O G  A B ; AG

 

B





A collinear vectors

Two vectors are collinear if one vector is a scalar ⃗ are multiple of the other vector. If A ⃗B and B C ⃗ ⃗ collinear (on a straight line), then B C = k.A B … where k is a scalar (constant) (Note that this is the same condition for parallel vectors)

A(- 2, 5)

G

B(4,2)

Example: Given Points A(-2,1); B(2,3) and C(8,6) . Use a vector method to prove that the points are collinear

0

**************************************************

A⃗ B

1) Finding

Solution:

Now A ⃗ B is the vector going from A to B That is: First go from A to O and then from O to B

C B

so:

A⃗ B

=

but

⃗ AO

=

so

A B =

⃗ AO

O⃗ B

+

−O ⃗A

=

A

( )

−

o

2) Finding

We need to prove that : ⃗ = k.A ⃗ BC B …. ( the condition for collinearity)

now

−2 5

⃗ +O ⃗ AO B

=

⃗ BC

⃗ ⃗ = B O+O C

( )   () () 

−2 2 = − + 1 3 2 8 = − + 3 6

=

=

4 2 6 3

1 ⃗ AB 3

⃗ = AG

3) Finding Now

−2 5

4 2

6 −3

⃗ AG

so we need to find the two displacement vectors A  B  and then establish the relationship between and B C them. A⃗ B

+

( )  =   −

= 1/3

  6 −3

=

  2 −1

 OG

 OG

= =

 O A + AG −2 2 + 5 −1

    =  0 4

**************************************************

but

 6 3

=

()

4 1.5 2

so

⃗ = k.A ⃗ BC B

k = 1.5

so the three points are on a straight line (collinear.)

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CxcDirect Institute -

Vectors - Lesson 1 & 2

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Class Activity 1 The position vectors of points A, B and C are: O ⃗A =

() 6 2

,

O⃗ B=

Express in the form

() a b

() 3 4

( )

, and O C ⃗ = 12 −2

, vectors

B⃗ A ,

3D ⃗ A ⃗ DA

giving so:

⃗ BC

⃗ = DC

a + ½ ( b – 3a) = a + ½ b - 1.5 a = ½b-½a = ½ ( b – a)  = DO ⃗ +O X ⃗ DX ⃗ +O X ⃗ = −O D = −2a+2b = 2 (b−a )

⇒

Finding

2) State one geometrical relationship between BA and BC 3) If Point M is the mid point of AB; Find the coordinates of M. Example 2: A

3a a

= =

Two Geometrical Relationships: ii) DX and DC are parallel, (2) DX = 4 DC iii) D, C and X are collinear ( on a straight line)

Class Activity 2 The Position vectors of R and J are :

D C

O

 

O R = −2 3 B

and

O J=

  1 −1

X

1. Express

In the diagram above: C is the mid point of AB and B is the mid point of OX, and D is such that OD = 2DA. The vectors a and b are such that: O ⃗A=3a and O⃗ B =b

R J in the form

2. Find the length

 a b

R J

3. Given that another point is such that



R T = 8 2

Find the coordinates of T

Express the following in terms of  ; DC  and D X  A B ; AC

a

and

b *********************************************************

State two geometrical relationship between DX and DC State one geometrical relationship between the points D, C and X **************************************************

Hint: The coordinates of T comes from the position vector O T *********************************************************

Class Activity 3 ABCD is a quadrilateral such that: Finding A B now

=

⃗ +O ⃗ AO B −O ⃗ A+O ⃗ B =

=

b−3a

A B =

Finding

O⃗ A=

−8 4

;

O⃗ B=

( ) −5 7

;

()

⃗= OC

1 4

−3a +b

OM is a point on OA such that the ratio OM:OA = 1:4 Prove that ABCM is a parallelogram

 AC

now: C is mid point AB  = ½ A  B ⇒ AC

( )

= ½ ( b – 3a)

Solution: *********************************************************

Finding now

 DC  =D   , DC A A C

where OD = 2DA

Hint: The opposite sides of a parallelogram are equal., so this question is testing that you know how to prove that two vectors are equal. *********************************************************

and so

⃗ ⃗ = 3a O D+D A 2D⃗ A+D ⃗ A = 3a

( given)

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CxcDirect Institute -

Vectors - Lesson 1 & 2

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Class Avtivity 4 The position vectors A and B relative to the origin are a and b respectively. The point P is on OA such that OP = 2PA The point M is on BA such that BM = MA A

B

b a

O

OB is produced to N such that OB = ON Express in terms of a and b, the vectors: 1.

A B;

P A;

 PM

2.

use a vector method to Prove that Points P, M and N are collinear

3.

Calculate the length AN given that : a=

 6 2

and b =

 1 2

***************************************************

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Vectors - Lesson 1 & 2

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Activity 2

Vector lesson - Solutions to activity questions

T

Activity 1

⃗ RT

R(-2,3)

B(3,4)

⃗ OT

A(6,2

O

O

J(1, -1)

C(12, -2)

R⃗ J

()  ( ) () ( ) ( )

⃗ ⃗A = B O+O

3 − 4

⃗ = B O+O ⃗ ⃗ = BC C

3 − 4

B⃗ A =

so:

⃗ BC

3B⃗ A

=

+

6 2

+

12 −2

3 −2

=

=

( geometrical relationship)

9 −6

=

⃗ +O ⃗ RO J

=

−

( ) −2 3

( )=( ) 1 −1

+

√( 3 +(−4 ) ) 2

length =

2

=5 O T⃗

Position vector of T =

−2 3

B(3,4)

O⃗ R

+

() 8 2

+

R T⃗

=

() 6 5

so coordinates = T (6, 5)

M A(6,2

*********************************************

Activity 3

O

B(-5,7)

C(12, -2)

if M = mid AB; then =

⃗ BM

C(1,4)

1 ⃗ BA 2

=

A(-8,4)

( )=( )

1 3 2 −2

1.5 −1

M O

Now the coordinates of M can be found from the ⃗ position vector O M ⃗ OM

= =

⇒

=

( )

=

where

3 −4

⃗ O⃗ B+ B M

() + ( ) = ( ) 3 4

1.5 −1

4.5 3

The objective is to prove that: (1) Now:

 ; and (2) A B =M C

⃗ ⃗ A⃗ B = A O+O B

coordinates = M( 4.5, 3)

= Also: ⇒

so : Giving © cxcDirect Institute - 876 469-2775 mail: [email protected] website: www.cxcdirect.org

M  A= C  B

( )

−

−8 4

+

⃗ = MO ⃗ +O C ⃗ MC  OM

=¼ O A = ¼

 = OM ⃗ MO

=

( ) () −5 7

=

3 3

but OM:OA = 1:4 −8 −2 = 4 1

   

( ) ( ) −2 1

−

−2 1

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CxcDirect Institute -

Vectors - Lesson 1 & 2

    1 − −2 4 1

 = MC

so:

3 3

=

 A B =M C

⇒

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⇒

1 1 a b− a 3 2

 = PM

….... (1)

1 3b− a 6

=

********************************************** Points P, M and N are collinear if:

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

 =k P M  M N

( )

3 ⃗ OA = 4

now:

M⃗ A =

also

⃗ +O ⃗ C⃗ B = CO B

()

= −

1 4

+

3 −8 4 4

−6 3

( ) ( ) −5 7

…

−6 3

=

 M N

so we need to find

⃗ M N

=

⃗ M⃗ B+B N

where:

M  B

=

½ b−a ; and

so

 = M N

now

M  A= C  B

⇒

( )

=

( k is a constant)

(2)

now note that

=b

½ b −a  + b

=

From (1) and (2), ABCM is a parallelogram.

 BN

½ 3b− a 

⃗ M N

⃗ 3PM

=

⇒ k=3

therfore P , M and N are on a stright line ( collinear)

Activity 4 N

 : consider triangle OAN: AN

To find the length of

b

B

M

A

b

⇒

 =O   ON A A N

⇒

 =O N  – O A = AN

⇒

 AN

=2

1/3 (a)

P

=

2/3 (a)

2b− a

      1 2

2 4

-

-

6 2

6 2

=

−4 2

O

Length =

Finding:

A B =

= Finding now:

⃗ +O ⃗ AO B

= =

−a+b

⃗ ⃗ −A O+O B

2

 −4  2

2

=

4.47

************************************************

b−a

P⃗ A

1 P A= a 3

⃗ Finding P M

=

⃗ P⃗ A+ A M

now A M = MB =½ A B =

⃗ so P M

=

½ b −a 

1 1 a + ( a−b) 3 2

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