1.10.6 Distance between points

What you'll learn

How a position vector locates a point relative to the origin.
How subtracting position vectors gives the displacement between two points.
How to calculate distance by finding the magnitude of a displacement vector.
How to apply the method in two and three dimensions.

Position vectors

A vector has both a magnitude, meaning its size or length, and a direction.

A position vector describes the position of a point relative to the origin, usually labelled $O$ . For example, if point $A$ has position vector

OA→=a=(a1a2),\overrightarrow{OA}=\mathbf{a} = \begin{pmatrix} a_1\\ a_2 \end{pmatrix},

then you travel $a_1$ units horizontally and $a_2$ units vertically to move from $O$ to $A$ .

Bold lower-case letters such as $a\mathbf{a}$ and $b\mathbf{b}$ are commonly used to represent vectors.

Definition

Position vector

The position vector of a point $A$ is the vector $OA→\overrightarrow{OA}$ from the origin $O$ to $A$ .

In three dimensions, a position vector has three components:

OA→=(a1a2a3).\overrightarrow{OA} = \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}.

The third component represents movement in the additional spatial direction, usually the $z$ -direction.

The vector between two points

Suppose points $A$ and $B$ have position vectors $a\mathbf{a}$ and $b\mathbf{b}$ respectively. To find the vector taking you from $A$ to $B$ , subtract the starting position vector from the finishing position vector:

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

This follows from the vector journey

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

Rearranging gives

AB→=OB→−OA→.\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.

Vector diagram showing position vectors from O to A and B, with the displacement from A to B equal to b minus a

Key Idea

Finish minus start

For a vector from $A$ to $B$ , subtract the position vector of $A$ from the position vector of $B$ : $AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .

Subtracting components

Vector subtraction is carried out component by component. In two dimensions,

(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)andb=(73).\mathbf{a}= \begin{pmatrix} 2\\ -1 \end{pmatrix} \quad\text{and}\quad \mathbf{b}= \begin{pmatrix} 7\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ .

Since the required direction is from $A$ to $B$ , calculate $b−a\mathbf{b}-\mathbf{a}$ .
Subtract the 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 to obtain
$AB→=(54).\overrightarrow{AB} = \begin{pmatrix} 5\\ 4 \end{pmatrix}.$
Therefore, moving from $A$ to $B$ means moving 5 units in the first coordinate direction and 4 units in the second.

Common Mistake

Subtracting in the wrong order

The vector $AB→\overrightarrow{AB}$ is $b−a\mathbf{b}-\mathbf{a}$ , not $a−b\mathbf{a}-\mathbf{b}$ . Reversing the subtraction gives $BA→\overrightarrow{BA}$ , which has the opposite direction.

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},

its magnitude is

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

This is Pythagoras' theorem: the components $x$ and $y$ form the perpendicular sides of a right-angled triangle, and the vector is its hypotenuse.

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=(−68).\mathbf{v}= \begin{pmatrix} -6\\ 8 \end{pmatrix}.

Substitute the components into the two-dimensional magnitude formula:
$∣v∣=(−6)2+82.\lvert\mathbf{v}\rvert=\sqrt{(-6)^2+8^2}.$
Square each component. The square of a negative number is positive:
$∣v∣=36+64.\lvert\mathbf{v}\rvert=\sqrt{36+64}.$
Simplify:
$∣v∣=100=10.\lvert\mathbf{v}\rvert=\sqrt{100}=10.$

Common Mistake

Losing a negative sign when squaring

Use brackets when squaring a negative component. For example, $6)^2=36$ , whereas writing $6^2$ is interpreted as $6^2)=-36$ .

Distance between two points

Distance is a scalar quantity: it has size but no direction. The distance from $A$ to $B$ is therefore the magnitude of the vector $AB→\overrightarrow{AB}$ .

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

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

Definition

Distance between position vectors

For points with position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , their distance apart is

distance=∣b−a∣.\text{distance}=\lvert\mathbf{b}-\mathbf{a}\rvert.

In two dimensions, if

a=(a1a2)andb=(b1b2),\mathbf{a}= \begin{pmatrix} a_1\\ a_2 \end{pmatrix} \quad\text{and}\quad \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}.

This is the coordinate distance formula, but thinking in terms of vectors makes its origin clear: subtract to find the displacement, then find its magnitude.

Example

Calculating a distance in two dimensions

Points $P$ and $Q$ have position vectors

p=(−25)andq=(4−3).\mathbf{p}= \begin{pmatrix} -2\\ 5 \end{pmatrix} \quad\text{and}\quad \mathbf{q}= \begin{pmatrix} 4\\ -3 \end{pmatrix}.

Find the exact distance $P Q$ .

Find the displacement from $P$ to $Q$ :
$PQ→=q−p=(4−(−2)−3−5)=(6−8).\overrightarrow{PQ} = \mathbf{q}-\mathbf{p} = \begin{pmatrix} 4-(-2)\\ -3-5 \end{pmatrix} = \begin{pmatrix} 6\\ -8 \end{pmatrix}.$
Find the magnitude of this displacement:
$\left\lvert \begin{pmatrix} 6\\ -8 \end{pmatrix} \right\rvert = \sqrt{6^2+(-8)^2}.$
Simplify:
$PQ=36+64=100=10.PQ=\sqrt{36+64}=\sqrt{100}=10.$

Tip

Distance is unchanged by direction

Although $AB→\overrightarrow{AB}$ and $BA→\overrightarrow{BA}$ have opposite directions, they have the same magnitude. Therefore, $∣b−a∣=∣a−b∣\lvert\mathbf{b}-\mathbf{a}\rvert=\lvert\mathbf{a}-\mathbf{b}\rvert$ .

Distance in three dimensions

The same method works in three dimensions. Subtract the three corresponding components and then find the magnitude.

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

then

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 distance in three dimensions

Points $A$ and $B$ have position vectors

a=(1−24)andb=(51−2).\mathbf{a}= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix} \quad\text{and}\quad \mathbf{b}= \begin{pmatrix} 5\\ 1\\ -2 \end{pmatrix}.

Find the exact distance $A B$ .

Subtract the starting position vector from the finishing position vector:
$AB→=b−a=(5−11−(−2)−2−4)=(43−6).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 5-1\\ 1-(-2)\\ -2-4 \end{pmatrix} = \begin{pmatrix} 4\\ 3\\ -6 \end{pmatrix}.$
Apply the three-dimensional magnitude formula:
$AB=42+32+(−6)2.AB=\sqrt{4^2+3^2+(-6)^2}.$
Simplify, keeping the answer exact:
$AB=16+9+36=61.AB=\sqrt{16+9+36}=\sqrt{61}.$

Checking your answer

A distance must always be non-negative because it is a magnitude. If the points are different, their distance is positive. If they have the same position vector, their distance is zero.

You can also estimate whether your answer is reasonable. The distance must be at least as large as the absolute value of any single component of the displacement. For example, a displacement with one component of size 8 cannot have a magnitude smaller than 8.

Common Mistake

Adding components instead of squares

The magnitude of $(xy)\begin{pmatrix}x\\y\end{pmatrix}$ is not $x + y$ . The components are perpendicular, so Pythagoras gives $x2+y2\sqrt{x^2+y^2}$ .

Exam technique

In the exam

Write the displacement as finish minus start, matching the direction requested.
Find its magnitude by squaring every component, adding the results and taking the square root.
Use brackets around negative components when squaring, and leave the answer as an exact surd unless a decimal is requested.
Check that your final distance is non-negative and large enough compared with the individual component changes.

Self review

Check yourself

How would you find $AB→\overrightarrow{AB}$ from the position vectors $a\mathbf{a}$ and $b\mathbf{b}$ ?
What is the exact distance between points with position vectors $(14)\begin{pmatrix}1\\4\end{pmatrix}$ and $(7−4)\begin{pmatrix}7\\-4\end{pmatrix}$ ?
How does the distance formula change when the position vectors have three components?

What you'll learn

How a position vector locates a point relative to the origin.
How subtracting position vectors gives the displacement between two points.
How to calculate distance by finding the magnitude of a displacement vector.
How to apply the method in two and three dimensions.

Position vectors

A vector has both a magnitude, meaning its size or length, and a direction.

A position vector describes the position of a point relative to the origin, usually labelled $O$ . For example, if point $A$ has position vector

OA→=a=(a1a2),\overrightarrow{OA}=\mathbf{a} = \begin{pmatrix} a_1\\ a_2 \end{pmatrix},

then you travel $a_1$ units horizontally and $a_2$ units vertically to move from $O$ to $A$ .

Bold lower-case letters such as $a\mathbf{a}$ and $b\mathbf{b}$ are commonly used to represent vectors.

Definition

Position vector

The position vector of a point $A$ is the vector $OA→\overrightarrow{OA}$ from the origin $O$ to $A$ .

In three dimensions, a position vector has three components:

OA→=(a1a2a3).\overrightarrow{OA} = \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}.

The third component represents movement in the additional spatial direction, usually the $z$ -direction.

The vector between two points

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

This follows from the vector journey

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

Rearranging gives

AB→=OB→−OA→.\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.

Vector diagram showing position vectors from O to A and B, with the displacement from A to B equal to b minus a

Key Idea

Finish minus start

For a vector from $A$ to $B$ , subtract the position vector of $A$ from the position vector of $B$ : $AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .

Subtracting components

Vector subtraction is carried out component by component. In two dimensions,

(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)andb=(73).\mathbf{a}= \begin{pmatrix} 2\\ -1 \end{pmatrix} \quad\text{and}\quad \mathbf{b}= \begin{pmatrix} 7\\ 3 \end{pmatrix}.

Find $AB→\overrightarrow{AB}$ .

Since the required direction is from $A$ to $B$ , calculate $b−a\mathbf{b}-\mathbf{a}$ .
Subtract the 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 to obtain
$AB→=(54).\overrightarrow{AB} = \begin{pmatrix} 5\\ 4 \end{pmatrix}.$
Therefore, moving from $A$ to $B$ means moving 5 units in the first coordinate direction and 4 units in the second.

Common Mistake

Subtracting in the wrong order

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},

its magnitude is

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

This is Pythagoras' theorem: the components $x$ and $y$ form the perpendicular sides of a right-angled triangle, and the vector is its hypotenuse.

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=(−68).\mathbf{v}= \begin{pmatrix} -6\\ 8 \end{pmatrix}.

Substitute the components into the two-dimensional magnitude formula:
$∣v∣=(−6)2+82.\lvert\mathbf{v}\rvert=\sqrt{(-6)^2+8^2}.$
Square each component. The square of a negative number is positive:
$∣v∣=36+64.\lvert\mathbf{v}\rvert=\sqrt{36+64}.$
Simplify:
$∣v∣=100=10.\lvert\mathbf{v}\rvert=\sqrt{100}=10.$

Common Mistake

Losing a negative sign when squaring

Use brackets when squaring a negative component. For example, $6)^2=36$ , whereas writing $6^2$ is interpreted as $6^2)=-36$ .

Distance between two points

Distance is a scalar quantity: it has size but no direction. The distance from $A$ to $B$ is therefore the magnitude of the vector $AB→\overrightarrow{AB}$ .

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

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

Definition

Distance between position vectors

For points with position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , their distance apart is

distance=∣b−a∣.\text{distance}=\lvert\mathbf{b}-\mathbf{a}\rvert.

In two dimensions, if

a=(a1a2)andb=(b1b2),\mathbf{a}= \begin{pmatrix} a_1\\ a_2 \end{pmatrix} \quad\text{and}\quad \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}.

This is the coordinate distance formula, but thinking in terms of vectors makes its origin clear: subtract to find the displacement, then find its magnitude.

Example

Calculating a distance in two dimensions

Points $P$ and $Q$ have position vectors

p=(−25)andq=(4−3).\mathbf{p}= \begin{pmatrix} -2\\ 5 \end{pmatrix} \quad\text{and}\quad \mathbf{q}= \begin{pmatrix} 4\\ -3 \end{pmatrix}.

Find the exact distance $P Q$ .

Find the displacement from $P$ to $Q$ :
$PQ→=q−p=(4−(−2)−3−5)=(6−8).\overrightarrow{PQ} = \mathbf{q}-\mathbf{p} = \begin{pmatrix} 4-(-2)\\ -3-5 \end{pmatrix} = \begin{pmatrix} 6\\ -8 \end{pmatrix}.$
Find the magnitude of this displacement:
$\left\lvert \begin{pmatrix} 6\\ -8 \end{pmatrix} \right\rvert = \sqrt{6^2+(-8)^2}.$
Simplify:
$PQ=36+64=100=10.PQ=\sqrt{36+64}=\sqrt{100}=10.$

Tip

Distance is unchanged by direction

Distance in three dimensions

The same method works in three dimensions. Subtract the three corresponding components and then find the magnitude.

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

then

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 distance in three dimensions

Points $A$ and $B$ have position vectors

a=(1−24)andb=(51−2).\mathbf{a}= \begin{pmatrix} 1\\ -2\\ 4 \end{pmatrix} \quad\text{and}\quad \mathbf{b}= \begin{pmatrix} 5\\ 1\\ -2 \end{pmatrix}.

Find the exact distance $A B$ .

Subtract the starting position vector from the finishing position vector:
$AB→=b−a=(5−11−(−2)−2−4)=(43−6).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix} 5-1\\ 1-(-2)\\ -2-4 \end{pmatrix} = \begin{pmatrix} 4\\ 3\\ -6 \end{pmatrix}.$
Apply the three-dimensional magnitude formula:
$AB=42+32+(−6)2.AB=\sqrt{4^2+3^2+(-6)^2}.$
Simplify, keeping the answer exact:
$AB=16+9+36=61.AB=\sqrt{16+9+36}=\sqrt{61}.$

Checking your answer

A distance must always be non-negative because it is a magnitude. If the points are different, their distance is positive. If they have the same position vector, their distance is zero.

Common Mistake

Adding components instead of squares

The magnitude of $(xy)\begin{pmatrix}x\\y\end{pmatrix}$ is not $x + y$ . The components are perpendicular, so Pythagoras gives $x2+y2\sqrt{x^2+y^2}$ .

Exam technique

In the exam

Write the displacement as finish minus start, matching the direction requested.
Find its magnitude by squaring every component, adding the results and taking the square root.
Use brackets around negative components when squaring, and leave the answer as an exact surd unless a decimal is requested.
Check that your final distance is non-negative and large enough compared with the individual component changes.

Self review

Check yourself

How would you find $AB→\overrightarrow{AB}$ from the position vectors $a\mathbf{a}$ and $b\mathbf{b}$ ?
What is the exact distance between points with position vectors $(14)\begin{pmatrix}1\\4\end{pmatrix}$ and $(7−4)\begin{pmatrix}7\\-4\end{pmatrix}$ ?
How does the distance formula change when the position vectors have three components?

What you'll learn

Position vectors

The vector between two points

Subtracting components

Magnitude of a vector

Distance between two points

Distance in three dimensions

Checking your answer

Review 1.10.6 Distance between points by teaching Genie

1.10.6 Distance between points

What you'll learn

Position vectors

The vector between two points

Subtracting components

Magnitude of a vector

Distance between two points

Distance in three dimensions

Checking your answer

Review 1.10.6 Distance between points by teaching Genie

1.10.6 Distance between points

What you'll learn

Position vectors

The vector between two points

Subtracting components

Magnitude of a vector

Distance between two points

Distance in three dimensions

Checking your answer

Review 1.10.6 Distance between points by teaching Genie

1.10.6 Distance between points Revision Guide

1.10.6 Distance between points

What you'll learn

Position vectors

The vector between two points

Subtracting components

Magnitude of a vector

Distance between two points

Distance in three dimensions

Checking your answer

Review 1.10.6 Distance between points by teaching Genie

1.10.6 Distance between points Revision Guide