# VECTOR MULTIPLICATION BY A SCALAR/UNIT VECTOR

**FURTHER MATHEMATICS SS 1 SECOND TERM**

**WEEK 5 DATE(S)……………………………………….**

**SUBJECT: FURTHER MATHEMATICS**

**CLASS: SS1**

** TOPIC: VECTOR 2**

**CONTENT**:

- Scalar multiplication of vectors
- Unit vector
- Direction cosines
- Scalar (dot) product: Application of scalar (dot) product.
- Projections of vectors
- Application of scalar product

**Sub-Topic: VECTOR MULTIPLICATION BY A SCALAR/UNIT VECTOR**

Let say and m are scalars then the following laws are true for any vectors **a** and **b:**

**a** is also a vector

(** a**+

**) =**

*b***+**

*a***(distributive)**

*b*( + m)a = ** a** + m

**.**

*a*If k is a scalar, *then k**a** is a vector which is parallel to **a** but k times the magnitude of **a**.* If k>0 then ka is in the same direction of a.

However, if k<0, then k** a** is in a direction opposite to a

**THE UNIT VECTOR**

The unit vector is an important concept in the study of vectors.

The definition of unit vector was given earlier as a vector which has an absolute value of unity

We now amplify on this concept.

If ** OP = ai + bj,** then we represent a unit vector in the direction of

**by . since a unit vector has a magnitude equal to unity**

*p*i.e and ll = 1

ll = .

Thus a unit vector in the direction of is given by

Example 1: Given that **, , . **Find a unit vector in the direction of and write down a vector parallel to

Solution:

** = **

Unit vector in the direction of ** = **

Example 2: Given that **, , , **where and are scalars, find an. If ** + . **

Solution: ** + **

2(

**CLASS ACTIVITY**

- Given that the vectors
**and**are parallel, find the constant - Given that and
**,**find the unit vector in the direction of

**Sub-Topic: PROJECTION OR RESOLUTION OF VECTORS**

b

O

⍬

N

A

B

If the position vector of the point **A** relative to a reference point ** o** is a and that of the point

**B**relative to

**O**is

**b**, we call the length ON the projection of the vector b on the vector

**a**. If

**a**denotes the unit vector in the direction of the vector a then

a =/ậ/ and /ậ/ = 1

Now

/on/ = /OB/ cos Ө

= /b/ cos Ө

=/ȃ/ /b/ cos Ө

Then the projection of the vector b on a is ȃ •b where ȃ is the unit vector in the direction of the vector a.

Also the projection of the vector **a** on **b **is **a.**Where is the unit vector in the direction of the vector** b**.

As the resolved part of **a** in the direction of** b** the projection of the vector **a** on **b** can also be viewed.

**Examples: **Find the projection of the vector **p** on the vector **q**if:

**SOLUTION:**

- Let the projection of q on p be u, then

U_{1}=

/ P/ =

=

**p.q** = (2)(-1) + (3)(4) = -3 + 12 = 9

Therefore,

U_{1} = =

- Let the projection
**q**on**p**be,**U**, then_{2}

U_{2} =

/p/ = =

**p.q** = (4)(-1) + (-5)(1) = -9

Therefore,

U_{2} = X =

- Let the projection q on p be U3, then

U_{3} =

/P/ = =

**p.q** = (6)(2) + (2)(3) = 12 + 6 = 18

Therefore,

U_{3} = X =

- Let the projection of
**q**and**p**be U_{4}, then

U_{4} =

/p/ = =

**p.q** = (3)(1) + (2)(3) = 3+6 = 9

Therefore,

U_{4} = X =

**CLASS ACTIVITY**

For each pair of the following, find the projection of the second vector on the first vector:

**a**= 3 – 4 and**b**= 5 + 8**m**= 7+ 3 and**n**= 2–5**x**= 4- 5 and**y**= 3 – 7

**SUB TOPIC:- SCALAR (DOT) PRODUCT/DIRECTION COSINES**

**Scalar Product:**

The scalar product of two vectors **a** and **b** is written **a.b** and is defined as the product of the lengths of the two vectors and the cosine of the angle between them.

Thus if Ө is the angle between the vectors a and b then a.b =/a/

Let and be two unit vectors which are perpendicular to each other.

Let

a= a_{x} + a_{y}

b=b_{x} + b_{y}

Then

a.b = (a_{x} + a_{y}) **.** (b_{x} + b_{y})

= a_{x}b_{x} . + a_{x}b_{y}. + a_{y}b_{x}. + a_{y}b_{y} .

As and are unit vectors which are mutually perpendicular to each other we have

- i. i = 1X1 cos0
^{0}= 1 - j.j = 1 X 1 cos0
^{0}= 1 - i.j =1 x 1 cos 90
^{0}=0 - j.i = 1 x 1 cos 90
^{0}= 0

So a.b =a_{x}b_{x} + a_{y}b_{y}

The scalar product can be sometimes be called the dot product.

The angle between two vectors Ө can be defined from the definition of scalar product.

Thus from a.b = /a/ /b/ cosӨ , we have

Cos Ө =

If **a** and **b** are parallel then Ө = 0 and a.b =/a/ /b/

If a and are perpendicular then Ө = and **a.b** =0

Example:

- Find the dot product of the following pairs of vectors:
- a = + 4 and b = 5 -3
- p = 6 – and q = 2 – 8

SOLUTION:

- a.b = ( + 4j) . (5 – 3

=(1)(5) + (4)(-3) = 7

- p.q = (6i – j ) . (2 – 8 )

= (6)(2) + (-1)(-8) = 20

- Find the cosine of the angle between the following pairs of vectors;
- m = 4 + 3 and n= 2 + 5
- S = 7 – 4 and t = 3 – 2

**SOLUTION:**

- Let the angle between the vectors m and n be 𝝰
_{1}then

/m/ =

/m/ =

=5

/n/ =

=

mn = (4)(2) + (3)(5) = 23

Cos𝝰_{1}= = X = =

(b). Let the angle between the vectors **s** and t be then

/s/ = =

/t/ = =

s.t = (7)(3) + (-4)(-2) = 29

Cos𝝰_{2} = = = X = =

**PROPERTIES OF A SCALAR (DOT) PRODUCT **

- Commutative property

Let a = a_{x} + a_{y} , b = b_{x} + b_{y}

Then

**a**.**b** =**b**.**a**

- Distribution property
- if //. In particular
^{2}//^{2} - if then
- Multiplication by a scalar (𝛌
- If ⍬ is the angle between and then

cos ⍬=

Condition for parallelism If the vector x+ ay is parallel to the vector

b =b_{x} + b_{y} ,

i.e =𝛌is a scalar) then** = = **𝛌

Condition for perpendicularity

If b, i.e then _{xx} + _{yy} = 0

Example:

- Given that /p / =3 ,/q/ =4 and =-6, find the angle between p and q (WAEC)

**SOLUTION: **

Let ⍬

⍬ = cos^{-1} -0.5 = 120^{0}.

- If = 3- 4 and b = 6 – 8 find the scalar product of and

SOLUTION**:**

**DIRECTION COSINES**

The direction cosines of vector are:

Where α, β and λ are angles which **OR **makes with **OX, OY **and **OZ **axes respectively, where **O **is the origin and X,Y and Z are mutually perpendicular directions in a 3 – dimensional plane.

**CLASS ACTIVITY**

- Evaluate:
- Given that:
- Find the direction cosines of vectors: (i) (ii)

**SUB – TOPIC: **APPLICATION OF SCALAR PRODUCT

Example 1: The vertices A,B and C of a triangle have position vectors and respectively , relative to the origin .prove that .

Solution:

a

B

A

C

b

c

From the triangle above, taking A is the origin and applying dot product we get

By triangle law, **a** = **b** – **c**,

and **a**.**a** = (**b** – **c**).(**b** – **c**)

Substituting for ** , **we get

Example 2: A particle moves from a point with position vector to a point with position vector. A constant force, among other forces acting on the particle is responsible for the movement. Find the work done by the force.

**Solution**: work done is defined as the product of force and displacement, and it is a scalar quantity.

work done by ** , **where refers to the displacement of the particle.

Work done by

** = **10 + 24 = 34joules

**PRACTICE EXERCISE**

- Find the unit vector direction of vectors: (a) (b)
- Find modulus of each of the following vectors :(i) (ii) (iii)
- Find the dot product of the following pairs of vectors:(i)
- If
**U,V**are any two vectors, prove that**.**Give a geometrical interpretation of this result when: (i) (ii) - For what values of λ are the vectors and perpendicular

**ASSIGNMENT**

**F**ind the values of µ and λ so that the forces may be in equilibrium.- For the following pair of vectors, find the projection of the second vector on the first vector:
**a**= 3 – 4 and**b**= 5 + 8 - What is the unit vector which is parallel to the vector ?
- If
**m**= 3 – 4 and**n**= 5 + 8 what is the cosine of the angle between**m**and**n ?** - Given that
**a**= -3 + 4 and**b**= 2 -, evaluate

**KEYWORDS**

**PROJECTION****SCALAR/DOT PRODUCT****MODULUS/MAGNITUDE****UNIT VECTOR**