Sec 4 E-Math · Vectors

Position vectors questions with answers

Practice questions on position vectors, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

$A$ is the point $(-5, -4)$ and $O$ is the origin. Write down the position vector of $A$ , $\overrightarrow{OA}$ .

Step 1 of 1

The position vector of a point is its coordinates as a column vector

x\text{-coordinate} = -5

y\text{-coordinate} = -4

\overrightarrow{OA} = \begin{pmatrix} -5 \\ -4 \end{pmatrix}

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Questions

Question 1The position vector of a point
$A$ is the point $(-2, 2)$ and $O$ is the origin. Write down the position vector of $A$ , $\overrightarrow{OA}$ .
Show the worked solution
1. Step 1: The position vector of a point is its coordinates as a column vector $x\text{-coordinate} = -2$ $y\text{-coordinate} = 2$ $\overrightarrow{OA} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$
Question 2A vector between two points
$A$ is the point $(2, 6)$ and $B$ is the point $(6, -1)$ . Express $\overrightarrow{AB}$ as a column vector.
Show the worked solution
1. Step 1: AB = OB − OA $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $x: 6 - 2 = 4$ $y: -1 - 6 = -7$
2. Step 2: Write the column vector $\overrightarrow{AB} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}$
Question 3Coordinates from vectors
$\overrightarrow{OA} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . Find the coordinates of $B$ .
Show the worked solution
1. Step 1: OB = OA + AB $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}$ $x: 1 + 1 = 2$ $y: -4 + (-2) = -6$
2. Step 2: The coordinates are the components of OB $B = (2, -6)$
Question 4Length between two points
$A$ is the point $(-4, -3)$ and $B$ is the point $(-5, 1)$ . Find $|\overrightarrow{AB}|$ , correct to 3 significant figures.
Show the worked solution
1. Step 1: AB = OB − OA $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $x: -5 - (-4) = -1$ $y: 1 - (-3) = 4$
2. Step 2: Magnitude: square, add, square root $|\overrightarrow{AB}| = \sqrt{(-1)^2 + 4^2}$ $|\overrightarrow{AB}| = \sqrt{17}$ $|\overrightarrow{AB}| = 4.12$ (3 s.f.)
Question 5The midpoint
$\overrightarrow{OA} = \begin{pmatrix} -4 \\ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$ . $M$ is the midpoint of $AB$ . Find the position vector of $M$ .
Show the worked solution
1. Step 1: OM is the average of OA and OB $\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$ $x: \dfrac{1}{2}(-4 + (-2)) = -3$ $y: \dfrac{1}{2}(-3 + 2) = -0.5$
2. Step 2: Write the column vector $\overrightarrow{OM} = \begin{pmatrix} -3 \\ -0.5 \end{pmatrix}$

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Worked example

A

is the point

(-5, -4)

and

O

is the origin. Write down the position vector of

A

\overrightarrow{OA}

Step 1 of 1

The position vector of a point is its coordinates as a column vector

x\text{-coordinate} = -5

y\text{-coordinate} = -4

\overrightarrow{OA} = \begin{pmatrix} -5 \\ -4 \end{pmatrix}

Questions

Question 1The position vector of a point

$A$ is the point $(-2, 2)$ and $O$ is the origin. Write down the position vector of $A$ , $\overrightarrow{OA}$ .

Show the worked solution

Step 1: The position vector of a point is its coordinates as a column vector $x\text{-coordinate} = -2$ $y\text{-coordinate} = 2$ $\overrightarrow{OA} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$

Question 2A vector between two points

$A$ is the point $(2, 6)$ and $B$ is the point $(6, -1)$ . Express $\overrightarrow{AB}$ as a column vector.

Show the worked solution

Step 1: AB = OB − OA $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $x: 6 - 2 = 4$ $y: -1 - 6 = -7$
Step 2: Write the column vector $\overrightarrow{AB} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}$

Question 3Coordinates from vectors

$\overrightarrow{OA} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . Find the coordinates of $B$ .

Show the worked solution

Step 1: OB = OA + AB $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}$ $x: 1 + 1 = 2$ $y: -4 + (-2) = -6$
Step 2: The coordinates are the components of OB $B = (2, -6)$

Question 4Length between two points

$A$ is the point $(-4, -3)$ and $B$ is the point $(-5, 1)$ . Find $|\overrightarrow{AB}|$ , correct to 3 significant figures.

Show the worked solution

Step 1: AB = OB − OA $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $x: -5 - (-4) = -1$ $y: 1 - (-3) = 4$
Step 2: Magnitude: square, add, square root $|\overrightarrow{AB}| = \sqrt{(-1)^2 + 4^2}$ $|\overrightarrow{AB}| = \sqrt{17}$ $|\overrightarrow{AB}| = 4.12$ (3 s.f.)

Question 5The midpoint

$\overrightarrow{OA} = \begin{pmatrix} -4 \\ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$ . $M$ is the midpoint of $AB$ . Find the position vector of $M$ .

Show the worked solution

Step 1: OM is the average of OA and OB $\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$ $x: \dfrac{1}{2}(-4 + (-2)) = -3$ $y: \dfrac{1}{2}(-3 + 2) = -0.5$
Step 2: Write the column vector $\overrightarrow{OM} = \begin{pmatrix} -3 \\ -0.5 \end{pmatrix}$