# 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:

1. Scalar multiplication of vectors
2. Unit vector
3. Direction cosines
4. Scalar (dot) product: Application of scalar (dot) product.
5. Projections of vectors
6. 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 + b (distributive)

( + m)a = a + ma.

If k is a scalar, then ka 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 ka 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 p by . since a unit vector has a magnitude equal to unity

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

1. Given that the vectors and are parallel, find the constant
2. 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 qif:

SOLUTION:

1. Let the projection of q on p be u, then

U1=

/ P/ =

=

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

Therefore,

U1 = =

1. Let the projection q on p be, U2, then

U2 =

/p/ = =

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

Therefore,

U2 = X =

1. Let the projection q on p be U3, then

U3 =

/P/ = =

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

Therefore,

U3 = X =

1. Let the projection of q and p be U4, then

U4 =

/p/ = =

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

Therefore,

U4 = X =

CLASS ACTIVITY

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

1. a = 3 – 4 and b = 5 + 8
2. m = 7+ 3 and n = 2–5
3. x = 4- 5 andy = 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= ax + ay

b=bx + by

Then

a.b = (ax + ay) . (bx + by)

= axbx . + axby. + aybx. + ayby .

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

1. i. i = 1X1 cos00 = 1
2. j.j = 1 X 1 cos00 = 1
3. i.j =1 x 1 cos 900 =0
4. j.i = 1 x 1 cos 900 = 0

So a.b =axbx + ayby

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:

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

SOLUTION:

1. a.b = ( + 4j) . (5 – 3

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

1. p.q = (6i – j ) . (2 – 8 )

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

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

SOLUTION:

1. Let the angle between the vectors m and n be 𝝰1 then

/m/ =

/m/ =

=5

/n/ =

=

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

Cos𝝰1= = X = =

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

/s/ = =

/t/ = =

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

Cos𝝰2 = = = X = =

PROPERTIES OF A SCALAR (DOT) PRODUCT

1. Commutative property

Let a = ax + ay , b = bx + by

Then

a.b =b.a

1. Distribution property
2. if //. In particular 2//2
3. if then
4. Multiplication by a scalar (𝛌
5. If ⍬ is the angle between and then

cos ⍬=

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

b =bx + by ,

i.e =𝛌is a scalar) then = = 𝛌

Condition for perpendicularity

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

Example:

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

SOLUTION:

Let ⍬

⍬ = cos-1 -0.5 = 1200.

1. 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

1. Evaluate:
2. Given that:
3. 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 = bc,

and a.a = (bc).(bc)

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

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

ASSIGNMENT

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

KEYWORDS

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