Position Vectors (AQA A-Level Mathematics): Revision Notes

In two dimensions (2D), if the coordinates of point $P$ are $(x, y)$ , the position vector $\mathbf{r}$ is:

$\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}$

In three dimensions (3D), if the coordinates of point $P$ are $(x, y, z)$ , the position vector $\mathbf{r}$ is:

$\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

This vector points from the origin $O(0, 0)$ in 2D or $O(0, 0, 0)$ in 3D to the point $P(x, y)$ or $P(x, y, z)$ .

📑Example:

If point $A$ is located at $(4, 3)$ in 2D space, the position vector $\mathbf{OA}$ is:

$\mathbf{OA} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$

In 3D space, if point $B$ is located at $(2, -1, 5)$ , the position vector $\mathbf{OB}$ is:

$\mathbf{OB} = \begin{pmatrix} 2 \\ -1 \\ 5 \end{pmatrix}$

1. Addition: If you have two points $A$ and $B$ with position vectors $\mathbf{a}$ and $\mathbf{b}$ , the vector $\mathbf{AB}$ from $A$ to $B$ is:

$\mathbf{AB} = \mathbf{b} - \mathbf{a}$

This represents the displacement from $A$ to $B$ . 2. Magnitude: The magnitude (length) of a position vector $\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}$ in 2D is given by:

$|\mathbf{r}| = \sqrt{x^2 + y^2}$

In 3D, for $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ :

$|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$

3. Direction: The direction of a position vector is given by its direction cosines or by normalising the vector (dividing by its magnitude).

📝Consider the points$A(2, 4), B(4, 8)$, and $C(20, 40)$. Determine whether these points are collinear.

• A position vector locates a point in space relative to the origin.
• It is represented by coordinates in vector form, either in 2D or 3D.
• Position vectors can be added, subtracted, and their magnitudes can be calculated to understand the spatial relationships between points.