1.8.3 Vectors - CCEA Maths Revision Guide

What you'll learn

How a position vector describes the location of a point relative to the origin.
How to find the vector from one point to another using $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ .
How to calculate the distance between two points from their position vectors.
How to use vector equations to find unknown points and check geometric relationships.

Vector notation

A vector is a quantity with both magnitude and direction. It can be represented by a directed line segment or by its components.

In two dimensions, a vector is usually written as a column vector:

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

The first component gives the horizontal movement and the second gives the vertical movement. For example,

(4−3)\begin{pmatrix} 4\\ -3 \end{pmatrix}

means a movement of 4 units to the right and 3 units down.

Vectors are often shown using bold letters such as $a\mathbf a$ , $b\mathbf b$ and $v\mathbf v$ . A vector from point $A$ to point $B$ is written $AB→\overrightarrow{AB}$ .

In three dimensions, a vector has three components:

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

Definition

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v\mathbf v$ is written $∣v∣|\mathbf v|$ .

For a two-dimensional vector,

∣(xy)∣=x2+y2.\left| \begin{pmatrix} x\\ y \end{pmatrix} \right| =\sqrt{x^2+y^2}.

This is an application of Pythagoras' theorem.

Example

Finding the magnitude of a vector

Find the magnitude of

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

Square the horizontal and vertical components:
$(−5)2=25,122=144.(-5)^2=25,\qquad 12^2=144.$
Add the squared components and take the square root:
$∣v∣=25+144=169.|\mathbf v|=\sqrt{25+144}=\sqrt{169}.$
Therefore,
$∣v∣=13.|\mathbf v|=13.$

Tip

Signs disappear when finding magnitude

A negative component still contributes positively to the length because it is squared.

Position vectors

Choose a fixed point called the origin, labelled $O$ . The position vector of a point $A$ is the vector from the origin to $A$ .

Definition

Position vector

If the position vector of $A$ is $a\mathbf a$ , then

OA→=a.\overrightarrow{OA}=\mathbf a.

The components of $a\mathbf a$ give the coordinates of $A$ relative to the origin.

For example, if

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

then $A$ has coordinates $(3, - 2)$ .

A position vector is different from a general displacement vector because its starting point is fixed at the origin.

Example

Interpreting a position vector

The point $P$ has position vector

p=(−47).\mathbf p= \begin{pmatrix} -4\\ 7 \end{pmatrix}.

The first component gives the horizontal coordinate, so the point lies 4 units to the left of the origin.
The second component gives the vertical coordinate, so the point lies 7 units above the origin.
Therefore, the coordinates of $P$ are $(- 4, 7)$ .

Common Mistake

Confusing a point with a vector

A point may be written as $A (3, - 2)$ , while its position vector is written as $a=(3−2)\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$ . They contain the same components but represent different mathematical objects.

The vector between two points

Suppose that points $A$ and $B$ have position vectors

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

Travelling from $O$ to $B$ can be split into a journey from $O$ to $A$ , followed by a journey from $A$ to $B$ :

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

Substituting the position vectors gives

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

Therefore,

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

Key Idea

End minus start

To find the 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.

Vector subtraction is performed component by component:

(b1b2)−(a1a2)=(b1−a1b2−a2).\begin{pmatrix} b_1\\ b_2 \end{pmatrix} - \begin{pmatrix} a_1\\ a_2 \end{pmatrix} = \begin{pmatrix} b_1-a_1\\ b_2-a_2 \end{pmatrix}.

Example

Finding the vector from one point to another

Points $A$ and $B$ have position vectors

a=(2−1),b=(73).\mathbf a= \begin{pmatrix} 2\\ -1 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} 7\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ .

Since the direction is from $A$ to $B$ , use ending position minus starting position:
$AB→=b−a.\overrightarrow{AB}=\mathbf b-\mathbf a.$
Subtract corresponding components:
$AB→=(73)−(2−1)=(7−23−(−1)).\overrightarrow{AB} = \begin{pmatrix} 7\\ 3 \end{pmatrix} - \begin{pmatrix} 2\\ -1 \end{pmatrix} = \begin{pmatrix} 7-2\\ 3-(-1) \end{pmatrix}.$
Simplify:
$AB→=(54).\overrightarrow{AB}= \begin{pmatrix} 5\\ 4 \end{pmatrix}.$

Reversing the direction reverses the vector:

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

The vectors $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ have the same magnitude but opposite directions.

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate $b−a\mathbf b-\mathbf a$ , not $a−b\mathbf a-\mathbf b$ . Read the letters as “from $A$ to $B$ ”, then use end minus start.

Distance between two points

The vector $AB→\overrightarrow{AB}$ gives both the direction and displacement from $A$ to $B$ . The distance $A B$ is the magnitude of this vector.

Definition

Distance from position vectors

If $A$ and $B$ have position vectors $a\mathbf a$ and $b\mathbf b$ , then

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

In two dimensions, if

a=(a1a2),b=(b1b2),\mathbf a= \begin{pmatrix} a_1\\ a_2 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} b_1\\ b_2 \end{pmatrix},

then

AB=(b1−a1)2+(b2−a2)2.AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}.

Example

Calculating the distance between two points

Points $P$ and $Q$ have position vectors

p=(−25),q=(41).\mathbf p= \begin{pmatrix} -2\\ 5 \end{pmatrix}, \qquad \mathbf q= \begin{pmatrix} 4\\ 1 \end{pmatrix}.

Find the exact distance $P Q$ .

Find the displacement from $P$ to $Q$ :
$PQ→=q−p=(41)−(−25)=(6−4).\overrightarrow{PQ} =\mathbf q-\mathbf p = \begin{pmatrix} 4\\ 1 \end{pmatrix} - \begin{pmatrix} -2\\ 5 \end{pmatrix} = \begin{pmatrix} 6\\ -4 \end{pmatrix}.$
Take the magnitude of the displacement vector:
$PQ=62+(−4)2=36+16=52.PQ=\sqrt{6^2+(-4)^2} =\sqrt{36+16} =\sqrt{52}.$
Simplify the surd:
$PQ=4×13=213.PQ=\sqrt{4\times13}=2\sqrt{13}.$

Tip

Distance cannot be negative

Although a displacement vector can have negative components, its magnitude is always non-negative. Your final distance should be a scalar, not a column vector.

Distance in three dimensions

For three-dimensional position vectors,

a=(a1a2a3),b=(b1b2b3),\mathbf a= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix},

the same method gives

AB=(b1−a1)2+(b2−a2)2+(b3−a3)2.AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}.

Example

Calculating a three-dimensional distance

The points $A$ and $B$ have position vectors

a=(1−24),b=(51−4).\mathbf a= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} 5\\ 1\\ -4 \end{pmatrix}.

Calculate the displacement:
$AB→=b−a=(43−8).\overrightarrow{AB} =\mathbf b-\mathbf a = \begin{pmatrix} 4\\ 3\\ -8 \end{pmatrix}.$
Find its magnitude:
$AB=42+32+(−8)2=16+9+64.AB=\sqrt{4^2+3^2+(-8)^2} =\sqrt{16+9+64}.$
Hence the exact distance is
$AB=89.AB=\sqrt{89}.$

Finding an unknown point

The relationship $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ can be rearranged. If you know the position of $A$ and the displacement from $A$ to $B$ , then

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

Example

Finding a position vector from a displacement

Point $A$ has position vector

a=(−32),\mathbf a= \begin{pmatrix} -3\\ 2 \end{pmatrix},

and

AB→=(7−5).\overrightarrow{AB}= \begin{pmatrix} 7\\ -5 \end{pmatrix}.

Find the position vector of $B$ .

Use $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ and rearrange for $b\mathbf b$ :
$b=a+AB→.\mathbf b=\mathbf a+\overrightarrow{AB}.$
Add the vectors component by component:
$b=(−32)+(7−5)=(−3+72−5).\mathbf b= \begin{pmatrix} -3\\ 2 \end{pmatrix} + \begin{pmatrix} 7\\ -5 \end{pmatrix} = \begin{pmatrix} -3+7\\ 2-5 \end{pmatrix}.$
Therefore,
$b=(4−3).\mathbf b= \begin{pmatrix} 4\\ -3 \end{pmatrix}.$

Exam technique

In the exam

Write $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ before substituting, so the direction and subtraction order are clear.
Subtract vectors component by component, taking particular care when subtracting negative components.
If a distance is required, find the displacement vector first and then take its magnitude; simplify surds unless a decimal is requested.
Check the form of your answer: a displacement is a vector, while a distance is a non-negative scalar.

Self review

Check yourself

If $a=(3−4)\mathbf a=\begin{pmatrix}3\\-4\end{pmatrix}$ and $b=(−16)\mathbf b=\begin{pmatrix}-1\\6\end{pmatrix}$ , what are $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ ?
How would you calculate the exact distance between points with position vectors $(25)\begin{pmatrix}2\\5\end{pmatrix}$ and $(8−3)\begin{pmatrix}8\\-3\end{pmatrix}$ ?
Point $P$ has position vector $p\mathbf p$ , and $PQ→=v\overrightarrow{PQ}=\mathbf v$ . What is the position vector of $Q$ ?

What you'll learn

How a position vector describes the location of a point relative to the origin.
How to find the vector from one point to another using $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ .
How to calculate the distance between two points from their position vectors.
How to use vector equations to find unknown points and check geometric relationships.

Vector notation

A vector is a quantity with both magnitude and direction. It can be represented by a directed line segment or by its components.

In two dimensions, a vector is usually written as a column vector:

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

The first component gives the horizontal movement and the second gives the vertical movement. For example,

(4−3)\begin{pmatrix} 4\\ -3 \end{pmatrix}

means a movement of 4 units to the right and 3 units down.

Vectors are often shown using bold letters such as $a\mathbf a$ , $b\mathbf b$ and $v\mathbf v$ . A vector from point $A$ to point $B$ is written $AB→\overrightarrow{AB}$ .

In three dimensions, a vector has three components:

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

Definition

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v\mathbf v$ is written $∣v∣|\mathbf v|$ .

For a two-dimensional vector,

∣(xy)∣=x2+y2.\left| \begin{pmatrix} x\\ y \end{pmatrix} \right| =\sqrt{x^2+y^2}.

This is an application of Pythagoras' theorem.

Example

Finding the magnitude of a vector

Find the magnitude of

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

Square the horizontal and vertical components:
$(−5)2=25,122=144.(-5)^2=25,\qquad 12^2=144.$
Add the squared components and take the square root:
$∣v∣=25+144=169.|\mathbf v|=\sqrt{25+144}=\sqrt{169}.$
Therefore,
$∣v∣=13.|\mathbf v|=13.$

Tip

Signs disappear when finding magnitude

A negative component still contributes positively to the length because it is squared.

Position vectors

Choose a fixed point called the origin, labelled $O$ . The position vector of a point $A$ is the vector from the origin to $A$ .

Definition

Position vector

If the position vector of $A$ is $a\mathbf a$ , then

OA→=a.\overrightarrow{OA}=\mathbf a.

The components of $a\mathbf a$ give the coordinates of $A$ relative to the origin.

For example, if

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

then $A$ has coordinates $(3, - 2)$ .

A position vector is different from a general displacement vector because its starting point is fixed at the origin.

Example

Interpreting a position vector

The point $P$ has position vector

p=(−47).\mathbf p= \begin{pmatrix} -4\\ 7 \end{pmatrix}.

The first component gives the horizontal coordinate, so the point lies 4 units to the left of the origin.
The second component gives the vertical coordinate, so the point lies 7 units above the origin.
Therefore, the coordinates of $P$ are $(- 4, 7)$ .

Common Mistake

Confusing a point with a vector

The vector between two points

Suppose that points $A$ and $B$ have position vectors

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

Travelling from $O$ to $B$ can be split into a journey from $O$ to $A$ , followed by a journey from $A$ to $B$ :

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

Substituting the position vectors gives

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

Therefore,

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

Key Idea

End minus start

To find the 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.

Vector subtraction is performed component by component:

(b1b2)−(a1a2)=(b1−a1b2−a2).\begin{pmatrix} b_1\\ b_2 \end{pmatrix} - \begin{pmatrix} a_1\\ a_2 \end{pmatrix} = \begin{pmatrix} b_1-a_1\\ b_2-a_2 \end{pmatrix}.

Example

Finding the vector from one point to another

Points $A$ and $B$ have position vectors

a=(2−1),b=(73).\mathbf a= \begin{pmatrix} 2\\ -1 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} 7\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ .

Since the direction is from $A$ to $B$ , use ending position minus starting position:
$AB→=b−a.\overrightarrow{AB}=\mathbf b-\mathbf a.$
Subtract corresponding components:
$AB→=(73)−(2−1)=(7−23−(−1)).\overrightarrow{AB} = \begin{pmatrix} 7\\ 3 \end{pmatrix} - \begin{pmatrix} 2\\ -1 \end{pmatrix} = \begin{pmatrix} 7-2\\ 3-(-1) \end{pmatrix}.$
Simplify:
$AB→=(54).\overrightarrow{AB}= \begin{pmatrix} 5\\ 4 \end{pmatrix}.$

Reversing the direction reverses the vector:

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

The vectors $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ have the same magnitude but opposite directions.

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate $b−a\mathbf b-\mathbf a$ , not $a−b\mathbf a-\mathbf b$ . Read the letters as “from $A$ to $B$ ”, then use end minus start.

Distance between two points

The vector $AB→\overrightarrow{AB}$ gives both the direction and displacement from $A$ to $B$ . The distance $A B$ is the magnitude of this vector.

Definition

Distance from position vectors

If $A$ and $B$ have position vectors $a\mathbf a$ and $b\mathbf b$ , then

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

In two dimensions, if

a=(a1a2),b=(b1b2),\mathbf a= \begin{pmatrix} a_1\\ a_2 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} b_1\\ b_2 \end{pmatrix},

then

AB=(b1−a1)2+(b2−a2)2.AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}.

Example

Calculating the distance between two points

Points $P$ and $Q$ have position vectors

p=(−25),q=(41).\mathbf p= \begin{pmatrix} -2\\ 5 \end{pmatrix}, \qquad \mathbf q= \begin{pmatrix} 4\\ 1 \end{pmatrix}.

Find the exact distance $P Q$ .

Find the displacement from $P$ to $Q$ :
$PQ→=q−p=(41)−(−25)=(6−4).\overrightarrow{PQ} =\mathbf q-\mathbf p = \begin{pmatrix} 4\\ 1 \end{pmatrix} - \begin{pmatrix} -2\\ 5 \end{pmatrix} = \begin{pmatrix} 6\\ -4 \end{pmatrix}.$
Take the magnitude of the displacement vector:
$PQ=62+(−4)2=36+16=52.PQ=\sqrt{6^2+(-4)^2} =\sqrt{36+16} =\sqrt{52}.$
Simplify the surd:
$PQ=4×13=213.PQ=\sqrt{4\times13}=2\sqrt{13}.$

Tip

Distance cannot be negative

Although a displacement vector can have negative components, its magnitude is always non-negative. Your final distance should be a scalar, not a column vector.

Distance in three dimensions

For three-dimensional position vectors,

a=(a1a2a3),b=(b1b2b3),\mathbf a= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix},

the same method gives

AB=(b1−a1)2+(b2−a2)2+(b3−a3)2.AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}.

Example

Calculating a three-dimensional distance

The points $A$ and $B$ have position vectors

a=(1−24),b=(51−4).\mathbf a= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix}, \qquad \mathbf b= \begin{pmatrix} 5\\ 1\\ -4 \end{pmatrix}.

Calculate the displacement:
$AB→=b−a=(43−8).\overrightarrow{AB} =\mathbf b-\mathbf a = \begin{pmatrix} 4\\ 3\\ -8 \end{pmatrix}.$
Find its magnitude:
$AB=42+32+(−8)2=16+9+64.AB=\sqrt{4^2+3^2+(-8)^2} =\sqrt{16+9+64}.$
Hence the exact distance is
$AB=89.AB=\sqrt{89}.$

Finding an unknown point

The relationship $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ can be rearranged. If you know the position of $A$ and the displacement from $A$ to $B$ , then

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

Example

Finding a position vector from a displacement

Point $A$ has position vector

a=(−32),\mathbf a= \begin{pmatrix} -3\\ 2 \end{pmatrix},

and

AB→=(7−5).\overrightarrow{AB}= \begin{pmatrix} 7\\ -5 \end{pmatrix}.

Find the position vector of $B$ .

Use $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ and rearrange for $b\mathbf b$ :
$b=a+AB→.\mathbf b=\mathbf a+\overrightarrow{AB}.$
Add the vectors component by component:
$b=(−32)+(7−5)=(−3+72−5).\mathbf b= \begin{pmatrix} -3\\ 2 \end{pmatrix} + \begin{pmatrix} 7\\ -5 \end{pmatrix} = \begin{pmatrix} -3+7\\ 2-5 \end{pmatrix}.$
Therefore,
$b=(4−3).\mathbf b= \begin{pmatrix} 4\\ -3 \end{pmatrix}.$

Exam technique

In the exam

Write $AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a$ before substituting, so the direction and subtraction order are clear.
Subtract vectors component by component, taking particular care when subtracting negative components.
If a distance is required, find the displacement vector first and then take its magnitude; simplify surds unless a decimal is requested.
Check the form of your answer: a displacement is a vector, while a distance is a non-negative scalar.

Self review

Check yourself

If $a=(3−4)\mathbf a=\begin{pmatrix}3\\-4\end{pmatrix}$ and $b=(−16)\mathbf b=\begin{pmatrix}-1\\6\end{pmatrix}$ , what are $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ ?
How would you calculate the exact distance between points with position vectors $(25)\begin{pmatrix}2\\5\end{pmatrix}$ and $(8−3)\begin{pmatrix}8\\-3\end{pmatrix}$ ?
Point $P$ has position vector $p\mathbf p$ , and $PQ→=v\overrightarrow{PQ}=\mathbf v$ . What is the position vector of $Q$ ?

1.9.3 Vectors

What you'll learn

Vector notation

Position vectors

The vector between two points

Distance between two points

Distance in three dimensions

Finding an unknown point

Review 1.8.3 Vectors by teaching Genie

1.9.3 Vectors

What you'll learn

Vector notation

Position vectors

The vector between two points

Distance between two points

Distance in three dimensions

Finding an unknown point

Review 1.8.3 Vectors by teaching Genie

1.9.3 Vectors

What you'll learn

Vector notation

Position vectors

The vector between two points

Distance between two points

Distance in three dimensions

Finding an unknown point

Review 1.8.3 Vectors by teaching Genie

1.8.3 Vectors Revision Guide

1.9.3 Vectors

What you'll learn

Vector notation

Position vectors

The vector between two points

Distance between two points

Distance in three dimensions

Finding an unknown point

Review 1.8.3 Vectors by teaching Genie

1.8.3 Vectors Revision Guide