Vectors (O level Maths)
This tutorial is about O level vectors for Zimsec and Cambridge students. First we are going to define a vector, then represent it on a Cartesian plane and finally calculate problems involving vectors.
Vector representation
A vector is a quantity that has both magnitude (size, length or modulus) an direction. On a Cartesian plane, a vector is represented as a arrow joining two points.
The diagram above shows a vector $latex \vec{AB}$ which represents the displacement from point A(1; 2) to point B(5; 3).
A vector can be written in many ways:
- either $latex \bar{AB}$, $latex \vec{AB}$, $latex \bf{AB}$
- or $latex \bar{a}$, $latex \vec{a}$, $latex \bf{a}$
The values of a vector are usually given as a column matrix. On the diagram above, there are dashed lines showing the horizontal displacement (in the x-axis) and vertical displacement (in the y-axis) of the vector. From the diagram we can see from the dashed lines that x = 4 units and y = 1 unit. In column form, the vector is represented as:
$latex \vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}$
Negative vectors
When dealing with vectors, direction is very important. For example, $latex \bar{AB}$ and $latex \bar{BA}$ are not the same because they are going in opposite directions. $latex \bar{BA}$ is negative of $latex \bar{AB}$.
This means that if $latex \vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}$
Then $latex \vec{BA} = \begin{pmatrix} -4\\ -1 \end{pmatrix}$
Which therefore means $latex \vec{AB}$ = – $latex \vec{AB}$
Magnitude
The magnitude of a vector (also known as the modulus) is the length or size of the vector.
Magnitude of $latex \vec{AB}$ is usually shortened to |$latex \vec{AB}$|.
Magnitude = $latex \sqrt{x^2+y^2}$
Using our earlier example, |$latex \vec{AB}$| = $latex \sqrt{4^2+1^2}$
$latex = \sqrt{17}$ (in surd form)
The modulus of a vector is always positive.
Now let us find the modulus of $latex \vec{BA}$.
|$latex \vec{BA}$| = $latex \sqrt{(-4)^2+(-1)^2}$
$latex = \sqrt{17}$
Scalar multiplication
If vector $latex \vec{AB}$ is multiplied a scalar k, whereby k is any number, the resulting vector is a vector k times as big as $latex \vec{AB}$ and parallel to $latex \vec{AB}$.
For example, vector $latex 5\vec{AB}$ is 5 times as big as vector $latex \vec{AB}$ and also parallel to vector $latex \vec{AB}$.
Addition and Subtraction
Vectors can be added and subtracted as follows:
If $latex \vec{a} = \begin{pmatrix} 5\\ 2 \end{pmatrix}$ and $latex \vec{b} = \begin{pmatrix} 1\\ 6 \end{pmatrix}$
then $latex \vec{a} + \vec{b}= \begin{pmatrix} 5\\ 2 \end{pmatrix} + \begin{pmatrix} 1\\ 6 \end{pmatrix}$
$latex = \begin{pmatrix} 6\\ 8 \end{pmatrix}$
and $latex \vec{a} – \vec{b}= \begin{pmatrix} 5\\ 2 \end{pmatrix} – \begin{pmatrix} 1\\ 6 \end{pmatrix}$
$latex = \begin{pmatrix} 4\\ -4 \end{pmatrix}$
Position vectors
A position vector is a vector that is tied to the origin. Let us illustrate it using a diagram.
If we have a point A, the position vector of point A is $latex \vec{OA}$. By the same token, the position vector of point B is $latex \vec{OB}$.
Now if a point has the coordinates (x; y), its position vector is $latex \begin{pmatrix} x\\ y \end{pmatrix}$.
Therefore, using the diagram above we can see that:
- the position vector of A = $latex OA = \begin{pmatrix} 1\\ 2 \end{pmatrix}$ since A has the coordinates (1; 2)
- the position vector of B = $latex OB = \begin{pmatrix} 5\\ 3 \end{pmatrix}$ since B has the coordinates (5; 3)
- $latex \vec{b} = \vec{OA}$
- $latex \vec{c} = \vec{OB}$
- $latex \vec{a} = \vec{AB}$
Now here is the thing about vectors, a displacement of $latex \vec{AO}$ followed by $latex \vec{OB}$ is equivalent to a resultant displacement of $latex \vec{AB}$.
Therefore $latex \vec{AB} = \vec{AO} + \vec{OB}$
$latex \vec{AB} = \vec{AO} + \vec{OB}$
$latex \vec{AB} = \begin{pmatrix} -1\\ -2 \end{pmatrix} + \begin{pmatrix} 5\\ 3 \end{pmatrix}$
$latex \vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}$
Examples
Question 1
Given points A(7; 8) and B(2; 1), find:
a) $latex \bf{AB}$
b) $latex \bf{BA}$
Solution 1
To make it easy when solving vectors you first convert all the given points to position vectors: $latex \bf{OA} = \begin{pmatrix} 7\\ 8 \end{pmatrix}$ and $latex \bf{OB} = \begin{pmatrix} 2\\ -1 \end{pmatrix}$
a) $latex \bf{AB} = \bf{AO}+\bf{OB}$
- $latex \bf{AB} = \begin{pmatrix} -7\\ -8 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}$
- $latex \bf{AB} = \begin{pmatrix} -5\\ -9 \end{pmatrix}$
b) $latex \bf{BA} = \bf{-AB}$
- $latex \bf{AB} = -\begin{pmatrix} -5\\ -9 \end{pmatrix}$
- $latex \bf{AB} = \begin{pmatrix} 5\\ 9 \end{pmatrix}$
Question 2
The points O, P, Q, R, S have coordinates (0; 0), (1; 5), (3; 8), (7; 10), (10; 3) respectively. Express each of the following as a column vector.
a) $latex \bf{OQ}$
b) $latex \bf{OS}$
c) $latex \bf{PQ}$
d) $latex \bf{QR}$
e) $latex \bf{QS}$
f) $latex \bf{RP}$
Solution 2
First convert the points to position vectors
- $latex \bf{OP} = \begin{pmatrix} 1\\ 5 \end{pmatrix}$
- $latex \bf{OQ} = \begin{pmatrix} 3\\ 8 \end{pmatrix}$
- $latex \bf{OR} = \begin{pmatrix} 7\\ 10 \end{pmatrix}$
- $latex \bf{OS} = \begin{pmatrix} 10\\ 3 \end{pmatrix}$
a) $latex \bf{OQ} = \begin{pmatrix} 3\\ 8 \end{pmatrix}$
b) $latex \bf{OS} = \begin{pmatrix} 10\\ 3 \end{pmatrix}$
c) $latex \bf{PQ} = \bf{PO}+\bf{OQ}$
- $latex \bf{PQ} = \begin{pmatrix} -1\\ -5 \end{pmatrix} + \begin{pmatrix} 3\\ 8 \end{pmatrix}$
- $latex \bf{PQ} = \begin{pmatrix} 2\\ 3 \end{pmatrix}$
d) $latex \bf{QR} = \bf{QO}+\bf{OR}$
- $latex \bf{QR} = \begin{pmatrix} -3\\ -8 \end{pmatrix} + \begin{pmatrix} 7\\ 10 \end{pmatrix}$
- $latex \bf{QR} = \begin{pmatrix} 4\\ 2 \end{pmatrix}$
e) $latex \bf{QS} = \bf{QO}+\bf{OS}$
- $latex \bf{QS} = \begin{pmatrix} -3\\ -8 \end{pmatrix} + \begin{pmatrix} 10\\ 3 \end{pmatrix}$
- $latex \bf{QS} = \begin{pmatrix} 7\\ -5 \end{pmatrix}$
f) $latex \bf{RP} = \bf{RO}+\bf{OP}$
- $latex \bf{RP} = \begin{pmatrix} -7\\ -10 \end{pmatrix} + \begin{pmatrix} 1\\ 5 \end{pmatrix}$
- $latex \bf{RP} = \begin{pmatrix} -6\\ -5 \end{pmatrix}$
