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aa

a

B

A

a

TOPIC 5: VECTOR AND SCALAR 5.1 Vector

Vector is a quantity with both magnitude and direction, represented by an arrow which

indicates the direction of the quantity and the length of which is proportional to the magnitude. In example, velocity, displacement, acceleration, weight and force.

Scalar is a quantity possessing only magnitude. In example, speed, length, distance, power and work. a) Vector Notation

It usually written in the form of OB , a

a~ or ~a .

b) Vector Representation c) Equality of Vectors

Two vectors are said to be equal if they have the same magnitude and

direction. In the figure below, three equal vectors, a

, b

and c

have

been represented. The equality of vectors is expressed in the following

way cba

.

The three vectors are also same in magnitude.

cba

d) Negative Vector

Negative of a vector means the vector having the same magnitude but

opposite direction. For example, the negative vector for AB is

AB or BA .

O

B

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cba ~~~

y-axis

x-axis

x

y

O

B (x, y)

a

A

5.2 Vector in Cartesian Coordinate a) Position Vector is a vector which expresses the position of a point with respect to the origin. Point B (x, y) has a position vector of:

ABOAOB (Triangle Law)

yjxiOB (engineering notation in terms of i and j)

y

xOB (in the form of column vector)

b) Unit Vector in OB direction is,

22 yx

yjxi

=

y

x

yx 22

1in terms of i and j or column vector.

c) The magnitude of OB is given by the formula,

OB = 22 yx or

a

= 22 yx .

5.3 Operations of Vectors Resultant Vectors (addition and subtraction of vector) can be solved using Polygon Methods

a) Triangle Construction Method

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cba ~~~

b) Parallelogram Method

Example 1:

Draw the diagram for the following vector if given aOA ~ .

a) aOB ~

2

1

b) aOC ~2

Answer:

a) aOB ~

2

1

O

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O

B

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b) aOC ~2

Example 2: The following diagram shows a polygon OABCD. Find the resultant vectors for each of the following.

a) CDBCAB

b) ODAOBACB

Answer:

a) ADCDBCAB

O

C

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b) CDODAOBACB

Resultant Vectors can also be solved using by algebraic operations. Example 3:

Given baOA~

3~ , baOB~~2 , baOC

~~ and baOD~

2~3 .

Express the following vectors in terms of a~ and b~

.

a) AB

b) BC2

c) CD

d) CA2

1

Answers:

a) AB OBAO

OBOA

)~~2()

~3~( baba

bbaa~~

3~2~

b) BC2 OCBO 2 OCOB 2 )~~()~~2(2 baba baba ~~~~22 ba ~2~32 ba

~4~6

c) CD ODCO ODOC )~2~3()~~( baba

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baba ~2~3~~ ba ~3~4 ba

~3~4

d) CA2

1 OACO

2

1

OAOC 2

1

)~3~()~~(2

1baba

baba ~3~~~2

1

ba ~2~22

1

ba~~

Example 4:

Given jia 2~ and jib 43~

, find the following.

a) ba~~

b) ba~

2~

c) )~~(2 ba

d) ba~

4~3

Answers:

a) ba~~ jiji 432

ji 35

b) ba~

2~ jiji 4322 jiji 862

ji 94

c) )~~

(2 ab jiji 2432 ji 52 ji 102

d) ba~

4~3 jiji 43423 jiji 161236

ji 196

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Example 5:

If kjip 22 and kjiq 43 , express the following in terms of i, j and k.

a) qp

b) pq 23

Answers:

a) qp kjikji 4322

kji 3

b) pq 23 kjikji 222433 kjikji 4243129

kji 71413

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1. Find the resultant vectors for each of the following.

a)

i. BDAB

ii. ODCO

iii. BCCA

iv. DOOB

b)

i. AB + BC + CD

ii. AC + DE + CD

iii. AB + CD + EF + BC + DE + FA

2. Find the following vector in terms of ~a ,

~b ,

~c and

~d .

a)

i. DA

ii. CE

iii. EB

cbabadcb ~~~,

~~,~~~

b)

i. EA

ii. CE

iii. AD

cbadcdcba ~~~,

~~,~~~~

A B

C

D E

F

A

B C

D

E F

EXERCISE

A B

C D

O

A

B

C

D E

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3. Given baOA~

3~ , baOB~~2 , baOC

~~ and baOD~

2~3 . Express the

following vectors in terms of a~ and b~

.

a) AB

b) BC

c) CD

d) DA2

e) CA3

f) BA2

g) ACAD 2

h) CBDB 2

i) DCBD 2

j) DBBO 22

4. Points A, B and C has the following position vectors qp 2 , qp 6 and qp 412

respectively. Prove that the three points are collinear.

5. The position vector for points P, Q and R are ba~~2 , ba

~2~4 and ba

~3~6

respectively. Show that the three points are on a straight line. Then, determine the ratio

of PRPQ : .

6. Show that point L, M and N with position vectors ba~

2~ , ba~~3 and ba

~4~5 are

on a straight line. Hence, determine the ratio of LM and MN .

7. Points S, T and U has the following position vectors qp 3 , qp 22 and qp 35

respectively. Prove that the three points are collinear. Hence, find the ratio of

SUST : .

8. Points U, V and W has the position vectors ji 2 , ji 114 and ji 83

respectively. Prove that the three points are collinear. Hence, find the ratio of

UWUV : .

9. Given jia 2~ and jib 43~

, find the following. Hence, determine the

magnitude.

a) ba~~

b) ba~

2~

c) )~~(2 ba

d) ba~

4~3

e) )~2~

(3 ab

f) ba~

2~2

10. Given jia 32~ , jib 4~

and jic 53~ , calculate the following and

determine the magnitude.

a) cba ~~~

b) cab ~3~~

2

c) abc ~2~

3~2

d) bca~

)~~(2

e) )~~3(~ bac

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5.4 Apply Scalar (Dot) Product of Two Vectors

The scalar or dot product is often defined either algebraically or geometrically.

Algebraically, if given vectors jyixOA11

and jyixOB22

, then the

scalar product for the vector is calculated as:

2121yyxxOBOA (for 2D vectors) or

212121zzyyxxOBOA (for 3D vectors)

Example 1: Given that A(2, -3, 5), B(-3, 1, 4), C(-2, 7, -6). Find: a) BA

b) BC

Answers: a) BA )4(5)1)(3()3(2

2036

11 b) BC )4)(6()1)(7()3(2

2476

11 Example 2:

If jip 2~ and jiq 3~ , find;

a) qp ~~

b) qp ~3~2

Answers:

a) qp ~~ )3)(1()1(2

32

1

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b) qp ~3~2 )3)(3()2(2 jiji )93()24( jiji

9234 1812

6

Example 3:

If kjiA 32 and kjiB 24 , find;

a) BA

b) AB3

1

Answers:

a) BA 13)2)(1()4(2 328

7

b) AB3

1 kjikji 32

3

124

kjikji

3

1

3

224

11)3

1)(2()

3

2(4

1)3

2()

3

8(

3

7

Geometrically, if given vectors a~OA and b~

OB , then the scalar product for the vector is written as:

cosOBOAOBOA or

cos~~~~ baba

Theta, is the angle between vectors where a~ and b~

converges at point O or

diverges out of point O and 1800 .

O

A

B

Diverge Converge

O

A

B

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If cosOBOAOBOA then,

2

2

2

2

2

1

2

1

2121cosyxyx

yyxx

OBOA

OBOA

for 2D vector and

2

2

2

2

2

2

2

1

2

1

2

1

212121

zyxzyx

zzyyxx

for 3D vector.

i. If 0 then, baba

~~~~

Then, 2~~ aaa and 2

~~bbb

1~~ 2 iii and 1

~~ 2 jjj

ii. If 90 then, 0

~~ ba

0~~ ji and 0

~~ ij

Example:

Based on figure below, kjiOA 432 and kjiOB 34 . Find:

a) OBOA

b) The value of .

Answers:

a) OBOA )1(4)3)(3()4(2

498

13

B

A O

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b) The value of .

OBOA

OBOAcos

222222 134432

13

2629

13cos 1

74.61

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1. If jia 2~ and jib 3~

, find;

a) ba~~

b) ba~~2

c) )~~(

~bab

d) )~~(2)~

~2( baab

e) abab ~2~~2~

2. Given kjiAB 32 and kjiBC 24 , find;

a) BCAB2

12

b) )(2

1ABBCAB

c) )( ABBCAB

d) ))(( ABBCBCAB

e) )2)(( ABBCBCAB

EXERCISE

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5.5 Apply Vector (Cross) Product of Two Vectors Vector Product just like scalar product, is another way of multiplying vectors which see the most applications in physics and astronomy. In Mathematical context, vector product is used to calculate the area of triangle and parallelogram.

a) If kzjyixOA111

and kzjyixOB 222 are given, then

222

111

zyx

zyx

kji

OBOA k

y

y

j

x

x

i

j

z

z

k

x

x

i

i

z

z

k

y

y

j

2

1

2

1

2

1

2

1

2

1

2

1

kyxyxjzxzxizyzy122112211221

b) To find the Area of Triangle OAB = OBOA2

1

c) The Area of Parallelogram OACB = sinOBOA = OBOA

Example 1:

Find the cross product for kjiP 32

and kjiR 42

.

Answers:

421

312

kji

RP

kji 112231423241 kji 5112 kji 5112

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A

B

C

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Example 2:

Find the vector product for

A and

B if given kji 2 and kji 22 respectively.

Answers:

122

211

kji

BA

kji 122122112211

kji 055

ji 55

Example 3:

Find the area of triangle OAB if given

OA kji 34 and

OB kji 33

respectively.

Answers:

Area of triangle,

313

1342

1

2

1

kji

OBOA

kji )94()312()19(2

1

kji 59102

1

222 59102

1

2062

1

18.7 units2

O

A

B

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Example 4:

Find the area of parallelogram spanned by vectors kjiOS 32 and

kjiOT 54 .

Answer:

Area of parallelogram,

541

132

kji

OTOS

kji )38()110()415(

kji 51119

222 51119

507

52.22 units2

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1. Find the cross product for kjiP 32

and kjiR 42

.

2. Find the vector product for

A and

B if given kji 2 and kji 22 respectively.

3. Given kjiS

32 and kjiT2

123

, find the cross product of the vectors.

4. Solve

NM if kjiM 223

and kjiN 23

1

.

5. Given kjOA 2 and jiOB 4 . Find the area of triangle OAB.

6. Find the area of parallelogram spanned by vectors kjiOS 32 and

kjiOT 54 .

7. Find the area of triangle defined by A(-1, 2, 3) and B(4, -3, 5).

8. Given the vertices P(-2, -3, 0), Q(4, 5, -3) and R(6, 2, 1). Find the area of triangle PQR.

9. Given vector kjiOA 32 and kjiOB 22 and kjiOC 43 . Determine:

a) AB

b) BC

c) CA

d) AC

e) BA

f) CB

g) BCAB

h) CAAC

i) CBBACA j) BABC

k) BAACBC l) CACBABBC

10. A triangle PQR has the corners P(1,-2,4), Q(3,1,-2) and R(-2,3,1). Find PQ , QR and RP .

Then find the length of all the sides.

11. If jiOA 34 , jiOB 26 and jiOC 2 . Find AB , BC and CA . Hence,

determine the length of each side.

12. Determine the angles between the following vectors.

a) kjiA 3 and kjiB 253

b) jia 2~ and jib 43~

c) jia 2~ and jib 3~

d) kjiPQ 32 and kjiPR 24

e) kjiOA 32

and kjiOB 42

EXERCISE

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13. Given kjiOA 32 , kjiOB 22 , kjiOC 43 and kjiOD 25 .

Find the angles of the following vectors.

a) AB and AC

b) BA and CA

c) BA and BC

d) DA and DB

e) CD and CB

f) BD and BD