Magnitude of a Vector (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

• If there is no diagram, sketch one!

  • You can sketch a vector and use it to form a right-angled triangle

Consider two points $A\left(-3,5\right)$ and $B\left(7,1\right)$.

(a) Write down the column vector $\overrightarrow{AB}$.

Answer:

Find the horizontal and vertical distances between the two points
Subtract the x and y components of A from B

$\overrightarrow{AB}=\left(\begin{array}{c}7--3\\ 1-5\end{array}\right)$

$\overrightarrow{\mathit{A}\mathit{B}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{10}\\ \mathbf{-}\mathbf{4}\end{array}\mathbf{\right)}$

(b) Find the modulus of vector $\overrightarrow{AB}$.

Answer:

Sketching a diagram of the vector $\overrightarrow{AB}$ can help

Apply Pythagoras' theorem to the x and y components of $\overrightarrow{AB}$

$\begin{array}{rcl}\left|\overrightarrow{AB}\right|& =& \sqrt{{10}^2+{\left(-4\right)}^2}\\ & =& \sqrt{100+16}\\ & =& \sqrt{116}\end{array}$

$\begin{array}{rcl}\mathit{\left|}\overrightarrow{\mathit{A}\mathit{B}}\mathit{\right|}& \mathbf{=}& \mathbf{2}\sqrt{\mathbf{29}}\end{array}$

(c) Briefly explain why $\left|\overrightarrow{BA}\right|=\left|\overrightarrow{AB}\right|$.

Answer:

The magnitude of a vector is it's 'size'

Direction of the vector is ignored

$\left|\overrightarrow{BA}\right|=\left|\overrightarrow{AB}\right|$ since both vectors have the same distance

Another vector, $\overrightarrow{CD}$, has three times the magnitude of vector $\overrightarrow{AB}$.

(d) Write down a possible column vector for $\overrightarrow{CD}$.

Answer:

Being three times $\left|\overrightarrow{AB}\right|$ means the vector $\overrightarrow{AB}$ is three times longer

One way to find a vector is to multiply each component of the vector $\overrightarrow{AB}$ by 3 or -3

$\begin{array}{rcl}\overrightarrow{CD}& =& 3\overrightarrow{AB}=3\left(\begin{array}{c}10\\ -4\end{array}\right)\end{array}$

$\overrightarrow{\mathit{C}\mathit{D}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{30}\\ \mathbf{-}\mathbf{12}\end{array}\mathbf{\right)}$ or $\overrightarrow{CD}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{30}\\ \mathbf{12}\end{array}\mathbf{\right)}$