LECTURE NOTES VECTOR

CHAPTER 5:

SUBTOPICS

5.1 5.2 5.3 5.4

VECTORS IN SCALAR VECTOR APPLICATIONS
THREE DIMENSIONS PRODUCT PRODUCT OF VECTORS IN

GEOMETRY

LECTURE 1 OF 5

Subtopic:
5.1 Vectors in three dimensions
5.2 Scalar Product

LEARNING OUTCOMES

At the end of the lesson, students
should be able to

5.1 a) state the types of vectors.

b) find the magnitude of a vector and unit
vector.

c) perform addition, subtraction and scalar
multiplication of vectors.

d) find the direction cosines and directions
angles for a non-zero vector.

5.2 a) find the scalar product.
b) use the properties of scalar product.

c) find the angle between two vectors.
d) find the direction cosines and directions

angles for a non-zero vector.

Vectors In Three Dimensions

A vector is a quantity having both magnitude
and direction.

In three dimensions, the Cartesian Coordinate
systems uses x, y and z-axes to locate a point.

These axes are mutually perpendicular to each

other. z

y

O
x

Any point P in space can be specified by an
ordered triple of numbers (a, b, c) where a, b
and c are the steps in the direction of x, y
and z-axes respectively to P.

z
c

P(a, b, c)

y
aO b
x

The vector OP has initial point at the origin O
and terminate point at P(a, b, c).

We now take unit vectors i , j and k in the

direction of x, y and z-axes respectively.

Vector OP can be written as OP = ai + b j + ck ,

OP =  a, b, c  or  a
OP = b  .
 c 

5.1 a) Types of Vectors

Position Vector

A vector that start at the origin and terminate
at any arbitrary point is called position
vector.

The position vector of A is the vector that
starts from the origin and ends at A ,

OA = a .

Zero Vector

Zero vector or null vector is a vector with zero
magnitude and no direction and is denoted as

0.

Parallel Vectors

Two vectors are parallel if they have the same
direction or are in exactly opposite directions.

Two vectors u and v are parallel, if

u =v ,  R.

Perpendicular Vectors

Two vectors u and v are perpendicular if

angle between them is 900 .

Note : AB = OB − OA

5.1 b) Magnitude of a Vector and Unit Vector

The unit vector of a vector u is a vector

whose magnitude is 1 unit in the direction

of u .

If u = ai + b j + ck , then
a) the magnitude of u , u = a2 + b2 + c2 .

b) the unit vector of u ,

^ u = ai + b j + ck .

u= u a2 + b2 + c2

Example 1
Find the magnitude and unit vector of
a = 2i + 2 j + k .

Solution u = ai + b j + ck
a = 2i + 2 j + k

a = (2)2 + (2)2 + (1)2 = 3 u = a2 + b2 + c2

^ a = 1 (2i + 2 j + k) ^ u

a= u= u

a3

5.1 c) Addition, Subtraction and Scalar
Multiplication of Vectors

For any vectors a = a1 i + a2 j + a3 k and

b = b1 i + b2 j + b3 k and for any scalar   R

a) a + b = (a1 + b1)i + (a2 + b2 ) j + (a3 + b3)k
b) a − b = (a1 − b1)i + (a2 − b2 ) j + (a3 − b3)k

c)  a = a1 i +a2 j +a3 k

Example 2
If a = 2i + 3 j + k and b = i + 2 j − 2k , find

a) a + b
b) b − a
c) 3a

Solution

a) a + b = (2i + 3 j + k) + (i + 2 j − 2k)
= 3i + 5 j − k

b) b − a = (i + 2 j − 2k) − (2i + 3 j + k)
= −i − j − 3k

c) 3a = 3(2i + 3 j + k)
= 6i + 9 j + 3k

5.1 d) Direction Cosines and Direction Angles
for a Non-zero Vector

Consider the vector OP where P is the point
(a, b, c).

z P(a,b,c)

 y


o

x

If OP makes an angles of size  ,  and 

with the x-axis, y -axis and z -axis

respectively, then cos = a , cos  = b ,

OP OP

cos = c .

OP

cos , cos  and cos are known as the

direction cosines of OP and the angles
 ,  and  are known as the direction
angles of OP with cos2  + cos2  + cos2  = 1.

Example 3

Find the direction cosines and direction
angles of a = 2i + 3 j − k .

Solution

a = 2i + 3 j − k

a = (2)2 + (3)2 + (−1)2 = 14

The direction cosines of a are

cos = 2 , cos  = 3 , cos = − 1
14 14 14

The direction angles of a are
 = cos−1 2  = 57.70

 14 

 = cos−1 3  = 36.70

 14 

 = cos−1 − 1  =105.50

 14 

5.2 a) Scalar Product

The scalar product (dot product) of two

vectors a and b is defined as

a  b = a b cos

where  is the angle between a and b

b



a

The scalar product between a = a1 i + a2 j + a3 k

and b = b1 i + b2 j + b3 k is

a  b = a1b1 + a2b2 + a3b3

If a b = 0 then a and b are perpendicular.

Note:
a) i i = j  j = k  k =1
b) i  j = j i = j  k = k  j = k i = i  k = 0

Example 4
If a = 2i + 9 j + 6k and b = −3i + 4 j − 6k .
Find a b .

Solution

a b = (2i + 9 j + 6k)(−3i + 4 j − 6k)
= (2)(−3) + (9)(4) + (6)(−6)
= −6 + 36 − 36
= −6

Example 5
Given the vectors a = i + 2 j − k and

b = 2i + j +10k . Determine the value of 
such that a and b are perpendicular.

Solution
a and b are perpendicular

ab = 0

(i + 2 j − k) (2i + j +10k) = 0
2 + 2 −10 = 0
2 = 8
 =4

5.2 b) Properties of Scalar Product

a) a  a = a 2

b) a b = b a

c) a (b + c) = ab + a c

d) ( a) b = (ab) = a (b) ,  is scalar

e) 0 a = 0

Example 6
Simplify

a) (a − b)  (a + b)

b) (a + b)  c − (a + c)  b

Solution

a) (a − b)  (a + b) = a  a + a  b − b  a − b  b aa = a 2

= a 2 +ab−ab− b 2 ab = ba

= a 2− b 2

b) (a + b)  c − (a + c)  b bc = cb
= ac+bc−ab−cb
= ac+bc−ab−bc
= ac−ab
= a  (c − b)

Example 7

Given a = 3 , b = 5 and a b =1, find a + b .

Solution aa = a 2
ab = ba
a + b 2 = (a + b) (a + b)
= aa+ab+ba+bb
= a 2 +ab+ab+ b 2
= a 2 + 2a b + b 2
= (3)2 + 2(1) + (5)2

a + b 2 = 36

a+b =6

5.2 c) Angle Between Two Vectors

If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k

are two vectors and  is the angle between

them.

From the definition of a  b

a b = a b cos

cos = a b

ab

 = cos−1 ab 
ab

Example 8

Given a = 2i + 3 j + 6k and b =i − j + 2 k ,

find the angle between vectors a and b.

Solution  = cos−1  ab 
a b = (2i +3 j + 6k ) (i − j + 2k)  ab 
= 2 − 3 +12

= 11

a = (2)2 + (3)2 + (6)2 = 7

b = (1)2 + (−1)2 + (2)2 = 6

 = cos−1 ab 
ab

 = cos−1 11 6) 
(7)(

= 50.10

LECTURE 2 OF 5

Subtopic:
5.3 Vector Product

LEARNING OUTCOMES
At the end of the lesson, students
should be able to:

a) find the vector product.

b) use the properties of vector product.

c) find the area of parallelogram and
a triangle.

5.3 a) Vector Product

The vector product (cross product) of two

vectors a and b is defined as

^

ab = a b sin u

where  is the angle between a and b

^

and u is a unit vector in the direction of
ab .

Vector a b is perpendicular to both vectors
a and b.

If ab = 0 , then a and b are parallel.

The vector product of a = a1 i + a2 j + a3 k

and b = b1 i + b2 j + b3 k can be expressed as

ijk
a  b = a1 a2 a3

b1 b2 b3

= (a2b3 − a3b2 )i − (a1b3 − a3b1) j + (a1b2 − a2b1) k

Note:
a) i i = j  j = k  k = 0
b) i  j = k , j  k = i , k i = j
c) j i = −k , k  j = −i , i  k = − j

Example 1
Given a = 2i + 3 j − 2k and b = 4i − 2 j + 3 k ,
find a  b .

Solution

ijk
ab = 2 3 −2

4 −2 3
= (9 − 4)i − (6 + 8) j + (−4 −12)k
= 5i −14 j −16k

Example 2
Find a unit vector perpendicular to a = 4i − j + 3k
and b = 2i + 3 j − 2k .

Solution

ijk
ab = 4 −1 3

2 3 −2
= (2 − 9)i − (−8 − 6) j + (12 + 2)k
= −7i +14 j +14k

a b = (−7)2 + (14)2 + (14)2

= 21

^ ab

u= ab

= 1 (−7i +14 j +14 k)
21

= 1 (−i + 2 j + 2k)
3

5.3 b) Properties of Vector Product

a) a a = 0
b) ab = −b a

c) ( a)b = (ab) = a(b) ,  is scalar

d) a (b + c) = ab + a c
e) (a + b)c = ac + bc

f ) a (bc) = (ab) c

g) a(bc) = (ac)b − (a b)c

Example 3

If a and b are non-zero vectors, show that
a (a b) = 0 .

Solution a (bc) = (ab) c
a (a b) = (a  a) b

= 0b
=0

Example 4

The vectors a , b and c such that b  c = 3 i
and c  a = 2 j + k , where i , j and k are

unit vectors. Express ( a + b )  ( a + b + 4c )

in terms of i , j and k .

Solution

( a + b )( a + b + 4c ) = (a + b)(a + b) + (a + b)(4c)

= 0 + 4((a + b)c) aa =0

= 4(a c + b c) a b = −b a

= 4(−c a + bc) ca= 2 j +k
= 4(−(2 j + k) + 3i)
bc= 3i

= 4(−2 j − k + 3i)

= 12i − 8 j − 4k

5.3 c) Area of a Parallelogram and a Triangle
A parallelogram is a flat shape with opposite
sides parallel and equal in length.
Given a parallelogram with sides a and b .

a

b
Area of parallelogram = a b

Given a triangle with sides a and b .

a
b

Area of triangle = 1 a b
2

Example 5

Find the area of parallelogram with sides
a = 3i + j + 2k and b = i − 2 j − 2k .

Solution

a

b

ijk
ab = 3 1 2

1 −2 −2
= (−2 + 4)i − (−6 − 2) j + (−6 − 1)k
= 2i + 8 j − 7k

Area of parallelogram = a b

= (2)2 + (8)2 + (−7)2
= 10.8 unit 2

Example 6
Given the points A(1, 2, -2), B(2, 4, 6) and
C(-4, 3, -1). Find the area of the triangle ABC.

Solution
A(1, 2, -2), B(2, 4, 6) , C(-4, 3, -1)

AB =OB − OA

= ( 2i + 4 j + 6k) − (i + 2 j − 2k)
= i+2j+8k

AC =OC − OA

= ( − 4i + 3 j − k) − (i + 2 j − 2k)

= −5i + j + k

i jk
AB  AC = 1 2 8

−5 1 1

= (2 − 8)i − (1+ 40) j + (1+10)k
= −6i − 41j +11k

AB AC = (−6)2 + (−41)2 + (11)2 = 42.87

Area of triangle ABC = 1 AB AC

2
= 1 (42.87)

2

= 21.4 unit 2