## 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 $latex \vec{AB}$.

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) $latex \bf{RP} = \bf{RO}+\bf{OP}$

• $latex \bf{RP} = \begin{pmatrix} -7\\ -10 \end{pmatrix} + \begin{pmatrix} 1\\ 5 \end{pmatrix}$
• $latex \bf{RP} = \begin{pmatrix} -6\\ -5 \end{pmatrix}$