Introduction to Vectors (Cambridge (CIE) IGCSE Additional Maths): Revision Note

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

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 $\overrightarrow{\mathrm{AB}}$ is denoted $\left|\overrightarrow{\mathrm{AB}}\right|$

  • 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 $\mathit{v}=v_1\mathbf{i}+\mathbf{ }{v}_2\mathbf{j}$ is found using

  • $\left|\mathit{v}\right|= \sqrt{v_1^2+\mathbf{ }{v}_2^{ 2}}$

  • where $v= \left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)$

What is a unit vector?

• A unit vector has a magnitude of 1

• The vectors $\mathbf{i}$ and $\mathbf{j}$ are unit vectors

  • the direction of $\mathbf{i}$ is in the positive $x$-direction

  • the direction of $\mathbf{j}$ is in the positive $y$-direction

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

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 = $\overrightarrow{\mathrm{OA}}$ 

  • 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  $\mathbf{a}=\overrightarrow{OA}=x_1\mathbf{i}+y_1\mathbf{j}=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)$,  then point A has coordinates $\left(x_1,y_1\right)$

    • If  $\mathbf{b}=\overrightarrow{OB}=x_2\mathbf{i}+y_2\mathbf{j}=\left(\begin{array}{c}{x}_2\\ y_2\end{array}\right)$,  then point B has coordinates $\left(x_2,y_2\right)$

  • The distance between two points is given by $d= \sqrt{{\left(x_1- x_2\right)}^2+{\left(y_1- y_2\right)}^2}$

    • So  $\left|\overrightarrow{AB}\right|= \sqrt{{\left(x_1- x_2\right)}^2+{\left(y_1- y_2\right)}^2}$

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

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

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

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

    • and then calculating its magnitude directly

  • See the Worked Example below