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Specific Objectives
By the end of the topic the learner should be able to:
- Define vector and scalar
- Use vector notation
- Represent vectors both single and combined geometrically
- Identify equivalent vectors
- Add vectors
- Multiply vectors by scalars
- Define position vector and column vector
- Find magnitude of a vector
- Find mid-point of a vector
- Define translation as a transformation.
Content
- Vector and scalar quantities
- Vector notation
- Representation of vectors
- Equivalent vectors
- Addition of vectors
- Multiplication of a vector by a scalar
- Column vectors
- Position vectors
- Magnitude of a vector
- Midpoint of a vector
- Translation vector.
Introduction
A vector is a quantity with both magnitude and direction, e.g. acceleration velocity and force. A quantity with magnitude only is called scalar quantity e.g. mass temperature and time.
Representation of vectors
A vector can be presented by a directed line as shown below:
The direction of the vector is shown by the arrow.
Magnitude is the length of AB
Vector AB can be written as
Magnitude is denoted by |AB|
A is the initial point and B the terminal point
Equivalent vectors
Two or more vectors are said to be equivalent if they have:
- Equal magnitude
- The same direction.
Addition of vectors
A movement on a straight line from point A to B can be represented using a vector. This movement is called displacement
Consider the displacement from followed by
The resulting displacement is written as
Zero vector
Consider a diplacement from A to B and back to A .The total displacement is zero denoted by O
This vector is called a Zero or null vector.
AB + BA = O
If a + b = 0 , b = -a or a = – b
Multiplication of a vector by a scalar
Positive Scalar
If AB= BC =CD=a
A______B______C______D>
AD = a + a +a =3a
Negative scalar
Subtraction of one vector from another is performed by adding the corresponding negative
Vector. That is, if we seek a − b we form a + (−b).
DA = (- a) + (-a) + (-a)
= -3a
The zero Scalar
When vector a is multiplied by o, its magnitude is zero times that of a. The result is zero vector.
a.0 = 0.a = 0
Multiplying a Vector by a Scalar
If k is any positive scalar and a is a vector then ka is a vector in the same direction as a but k times longer. If k is negative, ka is a vector in the opposite direction to a andk times longer.
More illustrations……………………………………………
A vector is represented by a directed line segment, which is a segment with an arrow at one end indicating the direction of movement. Unlike a ray, a directed line segment has a specific length. The direction is indicated by an arrow pointing from thetail(the initial point) to the head (the terminal point). If the tail is at point A and the head is at point B, the vector from A to B is written as: |
|
The length (magnitude) of a vector v is written |v|. Length is always a non-negative real number. As you can see in the diagram at the right, the length of a vector can be found by forming a right triangle and utilizing the Pythagorean Theorem or by using the Distance Formula. The vector at the right translates 6 units to the right and 4 units upward. The magnitude of the vector is
|
|
from the Pythagorean Theorem, or from the Distance Formula:
|
The direction of a vector is determined by the angle it makes with a horizontal line. In the diagram at the right, to find the direction of the vector (in degrees) we will utilize trigonometry. The tangent of the angle formed by the vector and the horizontal line (the one drawn parallel to the x-axis) is 4/6 (opposite/adjacent). |
|
A free vector is an infinite set of parallel directed line segments and can be thought of as a translation. Notice that the vectors in this translation which connect the pre-image vertices to the image vertices are all parallel and are all the same length. You may also hear the terms “displacement” vector or “translation” vector when working with translations. | |
Position vector: Unlike a free vector, a position vector is “tied” or “fixed” to the origin. A position vector describes the spatial position of a point relative to the origin. TRANSLATION VECTOR Translation vector moves every point of an object by the same amount in the given vector direction. It can be simply be defined as the addition of a constant vector to every point.
Example The points A (-4 ,4 ) , B (-2 ,3) , C (-4 , 1 ) and D ( – 5 , 3) are vrtices of a quadrilateral. If the quadrilateral is given the translation T defined by the vector |
or 
Solution
Summary on vectors
Components of a Vector in 2 dimensions: To get from A to B you would move: 2 units in the x direction (x-component) 4 units in the y direction (y-component)
The components of the vector are these moves in the form of a column vector. thus |
Similarly: |
Magnitude of a Vector in 2 dimensions: We write the magnitude of u as | u |
The magnitude of a vector is the length of the directed line segment which represents it.
Use Pythagoras’ Theorem to calculate the length of the vector. |
The magnitude of vector u is |u| (the length of PQ)
and so |
Examples: 1. Draw a directed line segment representing 2.
3. P is (1, 3) and Q is (4, 1) find |
1.
2. Q is ( 2 + 4, 1 + 3) ® Q(6, 4)
3. |
Vector: A quantity which has magnitude and direction. Scalar: A quantity which has magnitude only. | Examples: Displacement, force, velocity, acceleration. Examples: Temperature, work, width, height, length, time of day. |
or 
or 
and P is (2, 1), find co-ordinates of Q
Did you understand everything? If not ask a teacher, friends or anybody and make sure you understand before going to sleep! |
Past KCSE Questions on the topic.
1. Given that 
and 
find
- (i)
(3 mks)- |
| (3 mks)
- |
- Show that A (1, -1), B (3, 5) and C (5, 11) are collinear (4 mks)
(3 mks)
| (3 mks)2. Given the column vectors 
and that 
- (i) Express p as a column vector (2mks)
- (ii) Determine the magnitude of p (1mk)
3. Given the points P(-6, -3), Q(-2, -1) and R(6, 3) express PQ and QR as column vectors. Hence show that the points P, Q and R are collinear. (3mks)
4. The position vectors of points x and y are 
and 
respectively. Find x y as a column vector (2 mks)
5. Given that 
(3mks)
6. The position vectors of A and B are 2 and 8 respectively. Find the coordinates of M
5 -7
which divides AB in the ratio 1:2. (3 marks)
7. The diagram shows the graph of vectors 
and 
.
Find the column vectors;
(a) 
(1mk)
(b) |
| (2mks)
8. 
. Find 
(2mks)
9. Find scalars m and n such that
m 4 + n -3 = 5
3 2 8
10. Given that p = 2i – j + k and q = i + j +2k, determine
(a.) │p + q│ (1 mk)
(b) │ ½ p – 2q │ (2 mks)
MATHEMATICS (121)
PAPER TWO
ALTERNATIVE A
INTRODUCTION
- Questions in this paper will mainly test topics from Form 3 and 4. However knowledge and skills acquired in form 1 and form 2 will be required
- The time allocated for this paper is 2 ½ hours
- The paper consist of a total of 100 marks
- The paper shall consist of two section: : Section 1 and II
Section I
This section will have 50 marks and sixteen (16) compulsory short- answer questions
Section II
This section will have 50 marks and a choice of eight (8) open ended question, for candidates to answer any five (5).The students should note that any attempted questions in this section will be marked if they are not cancelled
