Introduction to Vectors (Cambridge International IGCSE Further Maths)

Basic Vectors

What is a vector?

• Vectors represent a movement of a certain magnitude (size) in a given direction

  • For example: two objects with velocities of 7 m/s and ‑7 m/s are travelling at the same speed but in opposite directions

• You should have already come across vectors when translating functions of graphs

• They appear in many contexts of maths including mechanics for modelling forces

• A vector in two directions has components in the direction of the x- and y- axes

  • Vector quantities can have positive or negative components

• Vectors can be represented in different ways such as a column vector or as an i and j unit vector

• Some examples of vector quantities you may come across are displacement, velocity or acceleration

Exam Hint

• Think of vectors like a journey from one place to another
• Diagrams can help, if there isn’t one, draw one
• In your exam you can’t write in bold so should underline your vector notation

Worked Example

Magnitude of a Vector

How do you find the magnitude of a vector?

• The magnitude of a vector tells us its size or length

• The magnitude of the vector $\overset{\rightarrow}{\mathrm{AB}}$ is denoted $\left| A \vec{B} \right|$

  • The magnitude of the vector a is denoted |a|

• The magnitude of a vector can be found using Pythagoras’ theorem

• The magnitude of a vector $v=v_{1}i+v_{2}j$ is found using

  • $\left| v \right| = \sqrt{v_{1}^{ 2} + v_{2}^{ 2} }$
  • where $v = ( \begin{gathered} v_{1} \\ v_{2} \end{gathered} )$

What is a unit vector?

• A unit vector has a magnitude of 1

• The vectors $i$ and $j$ are unit vectors

  • the direction of $i$ is in the positive $x$-direction
  • the direction of $j$ is in the positive $y$-direction

• To find a unit vector in the direction of a given vector divide the vector by its magnitude

Exam Hint

• Finding the magnitude of a vector is the same as finding the distance between two coordinates

  • Commit the formula to memory and be prepared to use it in the exam

Worked Example A vector $\vec{XY} = ( \begin{gathered} - 12 \\ 5 \end{gathered} )$.

a) Find $\vert \vec{XY} \vert$.

$\vert \vec{XY} \vert$ is the magnitude of the vector, so use Pythagoras' theorem.

$\vert \vec{XY} \vert = \sqrt{12^{2} + 5^{2}} = \sqrt{169}$

$\vert \vec{XY} \vert = 13$

b) Find the unit vector in the direction of $\overset{\rightarrow}{XY}$.

For a unit vector, divide the vector $\overset{\rightarrow}{XY}$ by its magnitude, which was found in part (a).

$\frac{\vec{XY}}{\vert \vec{XY} \vert} = \frac{- 12 i + 5 j}{13} = - \frac{12}{13} i + \frac{5}{13} j$

The unit vector in the direction of$\overset{\rightarrow}{XY}$is$( \begin{gathered} \text{bevelled} \frac{- 12}{13} \\ \text{bevelled} \frac{5}{13} \end{gathered} )$.

Position Vectors

What is a position vector?

• Position vectors describe the position of a point in relation to the origin

• They are different to displacement vectors which describe the direction and distance between any two points

• The position vector of point A is written with the notation a = $\overset{\rightarrow}{\mathrm{OA}}$

  • The origin is always denoted O

• The individual components of a position vector are the coordinates of its end point

  • For example the point with coordinates (3, -2) has position vector 3i – 2j

How do I find the distance between two points using vectors?

• The distance between two points is the magnitude of the vector between them

How do I find the magnitude of a displacement vector?

• You can use coordinate geometry to find magnitudes of displacement vectors from A to B

  • From the position vectors of A and B you know their coordinates

    • If $a = \vec{OA} = x_{1} i + y_{1} j = ( \begin{gathered} x_{1} \\ y_{1} \end{gathered} )$, then point A has coordinates $( x_{1} , y_{1} )$
    • If $b = \vec{OB} = x_{2} i + y_{2} j = ( \begin{gathered} x_{2} \\ y_{2} \end{gathered} )$, then point B has coordinates $( x_{2} , y_{2} )$

  • The distance between two points is given by $d = \sqrt{( x_{1} - x_{2} )^{2} + ( y_{1} - y_{2} )^{2}}$

    • So $\left| \vec{AB} \right| = \sqrt{( x_{1} - x_{2} )^{2} + ( y_{1} - y_{2} )^{2}}$

  • For example, if points A and B have position vectors $5i+3j$ and $3i-6j$ respectively

    • then $\left| \vec{AB} \right| = \sqrt{( 5 - 3 )^{2} + ( 3 - ( - 6 ) )^{2}} = \sqrt{85} = 9.22 ( 3 \mathrm{s} . \mathrm{f} . )$

• Alternatively, you could find $\left| \vec{AB} \right|$ by

  • first using $\overset{\rightarrow}{AB}=-\overset{\rightarrow}{OA}+\overset{\rightarrow}{OB}$ to find $\overset{\rightarrow}{AB}$ in vector form

    • and then calculating its magnitude directly

  • See the Worked Example below

Exam Hint

• Remember if asked for a position vector, you must find the vector all the way from the origin
• Diagrams can help, if there isn’t one, draw one

Worked Example