1.11.4 Position vectors - Revision Guide

What you'll learn

How a position vector describes the location of a point.
How to find the vector joining two points by subtracting position vectors.
How to calculate the distance between two points in two or three dimensions.
How to interpret and check your answers geometrically.

Prerequisites: vectors and components

A vector is a quantity with both magnitude and direction. The magnitude is the length or size of the vector.

In two dimensions, a vector can be written as a column vector:

(xy)\begin{pmatrix} x\\ y \end{pmatrix}

The number $x$ is the horizontal component and $y$ is the vertical component. A positive horizontal component points right, while a positive vertical component points upwards.

In three dimensions, a vector has three components:

(xyz)\begin{pmatrix} x\\ y\\ z \end{pmatrix}

Vectors may also be written using the perpendicular unit vectors $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ :

xi+yj+zkx\mathbf{i}+y\mathbf{j}+z\mathbf{k}

Here, $i\mathbf{i}$ points in the positive $x$ -direction, $j\mathbf{j}$ in the positive $y$ -direction and $k\mathbf{k}$ in the positive $z$ -direction.

Adding and subtracting vectors

To add or subtract vectors, work with corresponding components.

For example:

(ab)−(cd)=(a−cb−d)\begin{pmatrix} a\\ b \end{pmatrix} - \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} a-c\\ b-d \end{pmatrix}

The order matters when subtracting because changing the order reverses the direction of the resulting vector.

Example

Subtracting two vectors

Given

p=(7−2)andq=(34),\mathbf{p}= \begin{pmatrix} 7\\ -2 \end{pmatrix} \qquad\text{and}\qquad \mathbf{q}= \begin{pmatrix} 3\\ 4 \end{pmatrix},

find $p−q\mathbf{p}-\mathbf{q}$ .

Subtract the horizontal components: $7 - 3 = 4$ .
Subtract the vertical components: $- 2 - 4 = - 6$ .
Therefore,

p−q=(4−6).\mathbf{p}-\mathbf{q} = \begin{pmatrix} 4\\ -6 \end{pmatrix}.

What is a position vector?

Choose a fixed point called the origin, labelled $O$ . The position vector of a point $A$ is the vector that starts at $O$ and ends at $A$ .

Definition

Position vector

The position vector of $A$ is $OA→\overrightarrow{OA}$ . It describes the position of $A$ relative to the origin $O$ .

If $A$ has coordinates $(3, - 2)$ , then its position vector is

OA→=(3−2).\overrightarrow{OA} = \begin{pmatrix} 3\\ -2 \end{pmatrix}.

This tells you to travel 3 units to the right and 2 units down from the origin to reach $A$ .

A position vector is often represented by a lowercase bold letter. For example:

OA→=a.\overrightarrow{OA}=\mathbf{a}.

Similarly, if $OB→=b\overrightarrow{OB}=\mathbf{b}$ , then $a\mathbf{a}$ and $b\mathbf{b}$ give the positions of $A$ and $B$ .

Key Idea

The fixed starting point

A vector is a position vector only when its starting point is the origin. The vector $AB→\overrightarrow{AB}$ describes a displacement from $A$ to $B$ , so it is not usually a position vector.

Finding the vector between two points

Suppose the position vectors of $A$ and $B$ are

OA→=aandOB→=b.\overrightarrow{OA}=\mathbf{a} \qquad\text{and}\qquad \overrightarrow{OB}=\mathbf{b}.

To travel from $A$ to $B$ , you can first travel from $A$ back to $O$ , then from $O$ to $B$ :

AB→=AO→+OB→.\overrightarrow{AB} = \overrightarrow{AO}+\overrightarrow{OB}.

Since $AO→=−OA→=−a\overrightarrow{AO}=-\overrightarrow{OA}=-\mathbf{a}$ , this gives

AB→=−a+b=b−a.\overrightarrow{AB} = -\mathbf{a}+\mathbf{b} = \mathbf{b}-\mathbf{a}.

Position vectors from the origin and the vector from A to B

Key Idea

End minus start

To find a vector from one point to another, subtract the position vector of the starting point from the position vector of the ending point:

AB→=b−a.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

Example

Finding a vector between two points

Points $A$ and $B$ have position vectors

a=(2−1)andb=(83).\mathbf{a}= \begin{pmatrix} 2\\ -1 \end{pmatrix} \qquad\text{and}\qquad \mathbf{b}= \begin{pmatrix} 8\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ .

For $AB→\overrightarrow{AB}$ , use end minus start:

AB→=b−a=(83)−(2−1)=(64).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 8\\ 3 \end{pmatrix} - \begin{pmatrix} 2\\ -1 \end{pmatrix} = \begin{pmatrix} 6\\ 4 \end{pmatrix}.

Reverse the subtraction to find $BA→\overrightarrow{BA}$ :

BA→=a−b=(−6−4).\overrightarrow{BA} = \mathbf{a}-\mathbf{b} = \begin{pmatrix} -6\\ -4 \end{pmatrix}.

The answers satisfy $BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}$ , as expected because they have equal lengths but opposite directions.

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate $b−a\mathbf{b}-\mathbf{a}$ , not $a−b\mathbf{a}-\mathbf{b}$ . The arrow tells you that $A$ is the start and $B$ is the end.

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v\mathbf{v}$ is written as $∣v∣\lvert\mathbf{v}\rvert$ .

For a two-dimensional vector

v=(xy),\mathbf{v}= \begin{pmatrix} x\\ y \end{pmatrix},

Pythagoras' theorem gives

∣v∣=x2+y2.\lvert\mathbf{v}\rvert=\sqrt{x^2+y^2}.

In three dimensions, if

v=(xyz),\mathbf{v}= \begin{pmatrix} x\\ y\\ z \end{pmatrix},

then

∣v∣=x2+y2+z2.\lvert\mathbf{v}\rvert=\sqrt{x^2+y^2+z^2}.

Example

Finding the magnitude of a vector

Find the magnitude of

v=(−512).\mathbf{v}= \begin{pmatrix} -5\\ 12 \end{pmatrix}.

Square each component and add:

5)^2+12^2=25+144=169.

Take the positive square root because a length cannot be negative:

∣v∣=169=13.\lvert\mathbf{v}\rvert=\sqrt{169}=13.

Common Mistake

Losing negative components

A negative component becomes positive when squared. Write brackets, such as $5)^2$ , to avoid incorrectly treating it as $5^2$ .

Distance between two points

The distance from $A$ to $B$ is the magnitude of the vector $AB→\overrightarrow{AB}$ .

If the points have position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , then

AB=∣AB→∣=∣b−a∣.AB=\lvert\overrightarrow{AB}\rvert =\lvert\mathbf{b}-\mathbf{a}\rvert.

Notice that $AB→\overrightarrow{AB}$ is a vector, but $A B$ is a distance and therefore a non-negative scalar.

Example

Calculating a distance in two dimensions

Points $A$ and $B$ have position vectors

a=(−25)andb=(4−3).\mathbf{a}= \begin{pmatrix} -2\\ 5 \end{pmatrix} \qquad\text{and}\qquad \mathbf{b}= \begin{pmatrix} 4\\ -3 \end{pmatrix}.

Find the exact distance $A B$ .

Find the displacement from $A$ to $B$ :

AB→=b−a=(4−(−2)−3−5)=(6−8).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 4-(-2)\\ -3-5 \end{pmatrix} = \begin{pmatrix} 6\\ -8 \end{pmatrix}.

Find its magnitude:

\sqrt{6^2+(-8)^2} = \sqrt{36+64}.

Simplify the exact value:

AB=100=10.AB=\sqrt{100}=10.

Tip

A useful check

Distance does not depend on direction: $∣AB→∣=∣BA→∣\lvert\overrightarrow{AB}\rvert=\lvert\overrightarrow{BA}\rvert$ . Reversing the vector changes every sign but not its magnitude.

Position vectors in three dimensions

The same method works in three dimensions. First subtract the position vectors, then use all three components when finding the magnitude.

Example

Calculating a distance in three dimensions

Points $P$ and $Q$ have position vectors

p=(1−24)andq=(51−2).\mathbf{p}= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix} \qquad\text{and}\qquad \mathbf{q}= \begin{pmatrix} 5\\ 1\\ -2 \end{pmatrix}.

Find the exact distance $P Q$ .

Calculate the displacement from $P$ to $Q$ :

PQ→=q−p=(5−11−(−2)−2−4)=(43−6).\overrightarrow{PQ} = \mathbf{q}-\mathbf{p} = \begin{pmatrix} 5-1\\ 1-(-2)\\ -2-4 \end{pmatrix} = \begin{pmatrix} 4\\ 3\\ -6 \end{pmatrix}.

Apply the three-dimensional magnitude formula:

\sqrt{4^2+3^2+(-6)^2} = \sqrt{16+9+36}.

Hence,

PQ=61.PQ=\sqrt{61}.

Since 61 has no square factors greater than 1, this is already in exact form.

Exam technique

In the exam

Label the position vectors clearly, then use end minus start to form the required displacement vector.
When finding a distance, square every component, add the results and take the positive square root.
Keep distances exact unless a decimal is requested, and check that reversing the direction would leave your distance unchanged.

Self review

Check yourself

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , what is $BA→\overrightarrow{BA}$ ?
How would you find the distance between points with position vectors $(23)\begin{pmatrix}2\\3\end{pmatrix}$ and $(−48)\begin{pmatrix}-4\\8\end{pmatrix}$ ?
Which extra term is needed in the magnitude formula when you move from two dimensions to three dimensions?

What you'll learn

How a position vector describes the location of a point.
How to find the vector joining two points by subtracting position vectors.
How to calculate the distance between two points in two or three dimensions.
How to interpret and check your answers geometrically.

Prerequisites: vectors and components

A vector is a quantity with both magnitude and direction. The magnitude is the length or size of the vector.

In two dimensions, a vector can be written as a column vector:

(xy)\begin{pmatrix} x\\ y \end{pmatrix}

The number $x$ is the horizontal component and $y$ is the vertical component. A positive horizontal component points right, while a positive vertical component points upwards.

In three dimensions, a vector has three components:

(xyz)\begin{pmatrix} x\\ y\\ z \end{pmatrix}

Vectors may also be written using the perpendicular unit vectors $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ :

xi+yj+zkx\mathbf{i}+y\mathbf{j}+z\mathbf{k}

Here, $i\mathbf{i}$ points in the positive $x$ -direction, $j\mathbf{j}$ in the positive $y$ -direction and $k\mathbf{k}$ in the positive $z$ -direction.

Adding and subtracting vectors

To add or subtract vectors, work with corresponding components.

For example:

(ab)−(cd)=(a−cb−d)\begin{pmatrix} a\\ b \end{pmatrix} - \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} a-c\\ b-d \end{pmatrix}

The order matters when subtracting because changing the order reverses the direction of the resulting vector.

Example

Subtracting two vectors

Given

p=(7−2)andq=(34),\mathbf{p}= \begin{pmatrix} 7\\ -2 \end{pmatrix} \qquad\text{and}\qquad \mathbf{q}= \begin{pmatrix} 3\\ 4 \end{pmatrix},

find $p−q\mathbf{p}-\mathbf{q}$ .

Subtract the horizontal components: $7 - 3 = 4$ .
Subtract the vertical components: $- 2 - 4 = - 6$ .
Therefore,

p−q=(4−6).\mathbf{p}-\mathbf{q} = \begin{pmatrix} 4\\ -6 \end{pmatrix}.

What is a position vector?

Choose a fixed point called the origin, labelled $O$ . The position vector of a point $A$ is the vector that starts at $O$ and ends at $A$ .

Definition

Position vector

The position vector of $A$ is $OA→\overrightarrow{OA}$ . It describes the position of $A$ relative to the origin $O$ .

If $A$ has coordinates $(3, - 2)$ , then its position vector is

OA→=(3−2).\overrightarrow{OA} = \begin{pmatrix} 3\\ -2 \end{pmatrix}.

This tells you to travel 3 units to the right and 2 units down from the origin to reach $A$ .

A position vector is often represented by a lowercase bold letter. For example:

OA→=a.\overrightarrow{OA}=\mathbf{a}.

Similarly, if $OB→=b\overrightarrow{OB}=\mathbf{b}$ , then $a\mathbf{a}$ and $b\mathbf{b}$ give the positions of $A$ and $B$ .

Key Idea

The fixed starting point

A vector is a position vector only when its starting point is the origin. The vector $AB→\overrightarrow{AB}$ describes a displacement from $A$ to $B$ , so it is not usually a position vector.

Finding the vector between two points

Suppose the position vectors of $A$ and $B$ are

OA→=aandOB→=b.\overrightarrow{OA}=\mathbf{a} \qquad\text{and}\qquad \overrightarrow{OB}=\mathbf{b}.

To travel from $A$ to $B$ , you can first travel from $A$ back to $O$ , then from $O$ to $B$ :

AB→=AO→+OB→.\overrightarrow{AB} = \overrightarrow{AO}+\overrightarrow{OB}.

Since $AO→=−OA→=−a\overrightarrow{AO}=-\overrightarrow{OA}=-\mathbf{a}$ , this gives

AB→=−a+b=b−a.\overrightarrow{AB} = -\mathbf{a}+\mathbf{b} = \mathbf{b}-\mathbf{a}.

Position vectors from the origin and the vector from A to B

Key Idea

End minus start

To find a vector from one point to another, subtract the position vector of the starting point from the position vector of the ending point:

AB→=b−a.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

Example

Finding a vector between two points

Points $A$ and $B$ have position vectors

a=(2−1)andb=(83).\mathbf{a}= \begin{pmatrix} 2\\ -1 \end{pmatrix} \qquad\text{and}\qquad \mathbf{b}= \begin{pmatrix} 8\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ .

For $AB→\overrightarrow{AB}$ , use end minus start:

AB→=b−a=(83)−(2−1)=(64).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 8\\ 3 \end{pmatrix} - \begin{pmatrix} 2\\ -1 \end{pmatrix} = \begin{pmatrix} 6\\ 4 \end{pmatrix}.

Reverse the subtraction to find $BA→\overrightarrow{BA}$ :

BA→=a−b=(−6−4).\overrightarrow{BA} = \mathbf{a}-\mathbf{b} = \begin{pmatrix} -6\\ -4 \end{pmatrix}.

The answers satisfy $BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}$ , as expected because they have equal lengths but opposite directions.

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate $b−a\mathbf{b}-\mathbf{a}$ , not $a−b\mathbf{a}-\mathbf{b}$ . The arrow tells you that $A$ is the start and $B$ is the end.

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v\mathbf{v}$ is written as $∣v∣\lvert\mathbf{v}\rvert$ .

For a two-dimensional vector

v=(xy),\mathbf{v}= \begin{pmatrix} x\\ y \end{pmatrix},

Pythagoras' theorem gives

∣v∣=x2+y2.\lvert\mathbf{v}\rvert=\sqrt{x^2+y^2}.

In three dimensions, if

v=(xyz),\mathbf{v}= \begin{pmatrix} x\\ y\\ z \end{pmatrix},

then

∣v∣=x2+y2+z2.\lvert\mathbf{v}\rvert=\sqrt{x^2+y^2+z^2}.

Example

Finding the magnitude of a vector

Find the magnitude of

v=(−512).\mathbf{v}= \begin{pmatrix} -5\\ 12 \end{pmatrix}.

Square each component and add:

5)^2+12^2=25+144=169.

Take the positive square root because a length cannot be negative:

∣v∣=169=13.\lvert\mathbf{v}\rvert=\sqrt{169}=13.

Common Mistake

Losing negative components

A negative component becomes positive when squared. Write brackets, such as $5)^2$ , to avoid incorrectly treating it as $5^2$ .

Distance between two points

The distance from $A$ to $B$ is the magnitude of the vector $AB→\overrightarrow{AB}$ .

If the points have position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , then

AB=∣AB→∣=∣b−a∣.AB=\lvert\overrightarrow{AB}\rvert =\lvert\mathbf{b}-\mathbf{a}\rvert.

Notice that $AB→\overrightarrow{AB}$ is a vector, but $A B$ is a distance and therefore a non-negative scalar.

Example

Calculating a distance in two dimensions

Points $A$ and $B$ have position vectors

a=(−25)andb=(4−3).\mathbf{a}= \begin{pmatrix} -2\\ 5 \end{pmatrix} \qquad\text{and}\qquad \mathbf{b}= \begin{pmatrix} 4\\ -3 \end{pmatrix}.

Find the exact distance $A B$ .

Find the displacement from $A$ to $B$ :

AB→=b−a=(4−(−2)−3−5)=(6−8).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 4-(-2)\\ -3-5 \end{pmatrix} = \begin{pmatrix} 6\\ -8 \end{pmatrix}.

Find its magnitude:

\sqrt{6^2+(-8)^2} = \sqrt{36+64}.

Simplify the exact value:

AB=100=10.AB=\sqrt{100}=10.

Tip

A useful check

Distance does not depend on direction: $∣AB→∣=∣BA→∣\lvert\overrightarrow{AB}\rvert=\lvert\overrightarrow{BA}\rvert$ . Reversing the vector changes every sign but not its magnitude.

Position vectors in three dimensions

The same method works in three dimensions. First subtract the position vectors, then use all three components when finding the magnitude.

Example

Calculating a distance in three dimensions

Points $P$ and $Q$ have position vectors

p=(1−24)andq=(51−2).\mathbf{p}= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix} \qquad\text{and}\qquad \mathbf{q}= \begin{pmatrix} 5\\ 1\\ -2 \end{pmatrix}.

Find the exact distance $P Q$ .

Calculate the displacement from $P$ to $Q$ :

PQ→=q−p=(5−11−(−2)−2−4)=(43−6).\overrightarrow{PQ} = \mathbf{q}-\mathbf{p} = \begin{pmatrix} 5-1\\ 1-(-2)\\ -2-4 \end{pmatrix} = \begin{pmatrix} 4\\ 3\\ -6 \end{pmatrix}.

Apply the three-dimensional magnitude formula:

\sqrt{4^2+3^2+(-6)^2} = \sqrt{16+9+36}.

Hence,

PQ=61.PQ=\sqrt{61}.

Since 61 has no square factors greater than 1, this is already in exact form.

Exam technique

In the exam

Label the position vectors clearly, then use end minus start to form the required displacement vector.
When finding a distance, square every component, add the results and take the positive square root.
Keep distances exact unless a decimal is requested, and check that reversing the direction would leave your distance unchanged.

Self review

Check yourself

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , what is $BA→\overrightarrow{BA}$ ?
How would you find the distance between points with position vectors $(23)\begin{pmatrix}2\\3\end{pmatrix}$ and $(−48)\begin{pmatrix}-4\\8\end{pmatrix}$ ?
Which extra term is needed in the magnitude formula when you move from two dimensions to three dimensions?

1.13.4 Position vectors

What you'll learn

Prerequisites: vectors and components

Adding and subtracting vectors

What is a position vector?

Finding the vector between two points

Magnitude of a vector

Distance between two points

Position vectors in three dimensions

Review 1.11.4 Position vectors by teaching Genie

1.13.4 Position vectors

What you'll learn

Prerequisites: vectors and components

Adding and subtracting vectors

What is a position vector?

Finding the vector between two points

Magnitude of a vector

Distance between two points

Position vectors in three dimensions

Review 1.11.4 Position vectors by teaching Genie

1.13.4 Position vectors

What you'll learn

Prerequisites: vectors and components

Adding and subtracting vectors

What is a position vector?

Finding the vector between two points

Magnitude of a vector

Distance between two points

Position vectors in three dimensions

Review 1.11.4 Position vectors by teaching Genie

1.11.4 Position vectors Revision Guide

1.13.4 Position vectors

What you'll learn

Prerequisites: vectors and components

Adding and subtracting vectors

What is a position vector?

Finding the vector between two points

Magnitude of a vector

Distance between two points

Position vectors in three dimensions

Review 1.11.4 Position vectors by teaching Genie

1.11.4 Position vectors Revision Guide