Week 6 – SS1 Third Term Further Mathematics Notes

WEEK 6
MAGNITUDE OF A VECTOR
The magnitude of a vector a, sometimes called the modulus of the vector is represented by |a|.
Zero Vector:    The zero vector is a vector with zero magnitude.
Unit Vector:    The unit vector is the vector represented by a and is such that a = |a| a
Negative Vector:    The negative vector of a is written as – a
Equality of vector:    Two vectors are equal when they have same magnitude and direction.
Example:    Find the modulus of each of the following vectors

1. 3i + 4j
2. -2i – 5j
3. Solution
   

4. Let r = 3i + 4j ; then |r| =
5. Let r = -2i – 5j ; then |r| =
6. Let r = ; then |r| =

    

Example:    If ; find the modulus and direction cosines of:

2.  Solution
   

1. |r₁ + r₂| =

Let cos be the direction cosines of
            cos
                

 

1. || =
   Let cos be the direction cosines of
               cos

                
Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun
UNIT VECTOR
Example:    Find the unit vectors in the directions of the following vectors

1. r = 21 + 3j
2. q = 4i – 5j
3. p = 7i + 2j – 3k
4. t = 3i -5j -3k

   Solution
   

5. Let be the unit vector in the direction of r; then
6. Let be the unit vector in the direction of q; then
7. Let be the unit vector in the direction of p; then
8. Let be the unit vector in the direction of t; then

Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 12
ARITHMETIC OPERATIONS ON VECTORS
Example:    If p = 2i – 3j; q = 3i + 5j and r = i + j; Find the values of

1. 2p + q + 3r
2. 3p – 2q

   Solution
   

3. 2p = 2(2i – 3j ) = 4i – 6j

   3r = 3( i + j ) = 3i + 3j
   Therefore; 2p + q + 3r = (4i – 6j) + (3i + 5j) + (3i + 3j)
                   = 10i + 2j

4. 3p = 3(3i – 3j) = 9i – 9j

   2q = 2(3i + 5j) = 6i + 10j
   Therefore 3p – 2q = (9i – 9j) – (6i + 10j) =3i – 19j
   
    Example:    Given that = a – b and = 2a + 3b, where a = 2i + 3j and b = 3i – 2j, find     

                           = (2a + 3b) – (a – b)
                       = 2a + 3b – a + b = a + 4b
                       = (2i + 3j) + 4(3i – 2j) = 14i – 5j            

Evaluation:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 5
Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.
Assignment:
New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 6