Position Vectors (Edexcel A Level Maths: Pure): Revision Note

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

Distance between two points

• The distance between two points is the magnitude of the vector between them (see Magnitude Direction)

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