Position vectors, distance and dividing a line segment

A position vector locates a point relative to a fixed origin. This turns geometric questions about points, distances and line segments into vector calculations.

The central fact is simple but easy to reverse:

$\boxed{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.}$

In words, the vector from $A$ to $B$ is end minus start.

Before you begin

You should be able to:

• add, subtract and multiply vectors by scalars
• find a vector’s magnitude using Pythagoras’ theorem
• work confidently with fractions and ratios
• use coordinates in two and three dimensions

Review vector arithmetic or magnitude and direction if needed.

What is a position vector?

Let $O$ be the origin and let $A$ have coordinates $(3,-2)$. The position vector of $A$ is the vector from $O$ to $A$:

$\overrightarrow{OA}=\begin{pmatrix}3\\-2\end{pmatrix}.$

Position vectors are often represented by lower case bold letters:

$\mathbf a=\overrightarrow{OA},\qquad \mathbf b=\overrightarrow{OB}.$

So if $A(3,-2)$ and $B(-1,5)$, then

$$\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-1\\5\end{pmatrix}.$$

Coordinates describe a point; a position vector describes the displacement from the origin to that point. Their components have the same numbers, but the objects are conceptually different.

In three dimensions, $P(2,-4,7)$ has position vector

$\overrightarrow{OP}=\begin{pmatrix}2\\-4\\7\end{pmatrix}.$

The vector between two points

Travelling from $O$ to $B$ directly has the same overall effect as travelling first from $O$ to $A$, then from $A$ to $B$:

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

Therefore

$\boxed{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf b-\mathbf a.}$

This derivation is safer than memorising an unexplained subtraction rule.

Example 1: find a displacement vector

Let $A(-2,4)$ and $B(5,1)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.

For $A$ to $B$, subtract the position vector of the start from that of the end:

$$\overrightarrow{AB} =\begin{pmatrix}5\\1\end{pmatrix} -\begin{pmatrix}-2\\4\end{pmatrix} =\begin{pmatrix}7\\-3\end{pmatrix}.$$

Reversing the journey reverses every component:

$$\overrightarrow{BA} =\begin{pmatrix}-2\\4\end{pmatrix} -\begin{pmatrix}5\\1\end{pmatrix} =\begin{pmatrix}-7\\3\end{pmatrix} =-\overrightarrow{AB}.$$

Example 2: an unknown endpoint

Point $A$ has position vector $\begin{pmatrix}4\\-1\\2\end{pmatrix}$ and

$\overrightarrow{AB}=\begin{pmatrix}-3\\5\\4\end{pmatrix}.$

Find the coordinates of $B$.

Since $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$,

$$\overrightarrow{OB} =\begin{pmatrix}4\\-1\\2\end{pmatrix} +\begin{pmatrix}-3\\5\\4\end{pmatrix} =\begin{pmatrix}1\\4\\6\end{pmatrix}.$$

Hence $\boxed{B(1,4,6)}$.

Direction matters

$\overrightarrow{AB}$ means from $A$ to $B$, so calculate $\mathbf b-\mathbf a$. A quick check is that $\mathbf a+\overrightarrow{AB}$ must equal $\mathbf b$.

Distance between two points

Distance is the magnitude of the displacement vector:

$\boxed{AB=\left|\overrightarrow{AB}\right|=|\mathbf b-\mathbf a|.}$

For $A(x_1,y_1)$ and $B(x_2,y_2)$,

$\boxed{AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.}$

For $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$,

$\boxed{AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.}$

The formula is Pythagoras’ theorem applied to the perpendicular component changes. Unlike $\overrightarrow{AB}$, the distance $AB$ is a non-negative scalar and has no direction.

Example 3: distance in three dimensions

Find the distance between $P(-1,2,3)$ and $Q(3,-1,9)$.

First find the displacement:

$$\overrightarrow{PQ} =\begin{pmatrix}3-(-1)\\-1-2\\9-3\end{pmatrix} =\begin{pmatrix}4\\-3\\6\end{pmatrix}.$$

Then find its magnitude:

$$PQ=\sqrt{4^2+(-3)^2+6^2} =\sqrt{16+9+36} =\boxed{\sqrt{61}}.$$

Do not simplify $\sqrt{61}$ to a decimal unless the question asks for an approximation.

Midpoints

If $M$ is the midpoint of $AB$, then it lies halfway from $A$ to $B$:

$$\overrightarrow{OM} =\overrightarrow{OA}+\frac12\overrightarrow{AB} =\mathbf a+\frac12(\mathbf b-\mathbf a) =\boxed{\frac{\mathbf a+\mathbf b}{2}}.$$

Thus the midpoint is found by averaging corresponding coordinates.

Example 4: midpoint and a missing endpoint

$A$ has position vector $\mathbf a=\begin{pmatrix}-2\\5\end{pmatrix}$ and the midpoint $M$ of $AB$ has position vector $\mathbf m=\begin{pmatrix}4\\1\end{pmatrix}$. Find $\mathbf b$.

Use

$\mathbf m=\frac{\mathbf a+\mathbf b}{2}.$

Rearranging gives $\mathbf b=2\mathbf m-\mathbf a$, so

$$\mathbf b =2\begin{pmatrix}4\\1\end{pmatrix} -\begin{pmatrix}-2\\5\end{pmatrix} =\begin{pmatrix}10\\-3\end{pmatrix}.$$

Check by averaging $(-2,5)$ and $(10,-3)$: the result is $(4,1)$.

Dividing a line segment in a ratio

Suppose $P$ divides $AB$ internally in the ratio

$AP:PB=m:n.$

The whole segment contains $m+n$ equal parts, and $P$ is $m/(m+n)$ of the way from $A$ to $B$. Therefore

$$\begin{aligned} \mathbf p &=\mathbf a+\frac{m}{m+n}(\mathbf b-\mathbf a)\\ &=\boxed{\frac{n\mathbf a+m\mathbf b}{m+n}}. \end{aligned}$$

Notice the apparently crossed coefficients: $\mathbf a$ is multiplied by $n$, while $\mathbf b$ is multiplied by $m$. The journey form

$\boxed{\mathbf p=\mathbf a+\frac{m}{m+n}(\mathbf b-\mathbf a)}$

is often easier to reconstruct and less prone to errors.

Example 5: internal division

$A$ and $B$ have position vectors

$$\mathbf a=\begin{pmatrix}1\\-2\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}11\\8\end{pmatrix}.$$

Point $P$ lies on $AB$ such that $AP:PB=2:3$. Find $\mathbf p$.

$P$ is $2/(2+3)=2/5$ of the way from $A$ to $B$:

$$\begin{aligned} \mathbf p &=\mathbf a+\frac25(\mathbf b-\mathbf a)\\ &=\begin{pmatrix}1\\-2\end{pmatrix} +\frac25\begin{pmatrix}10\\10\end{pmatrix}\\ &=\boxed{\begin{pmatrix}5\\2\end{pmatrix}}. \end{aligned}$$

Check the ratio:

$$\overrightarrow{AP}=\begin{pmatrix}4\\4\end{pmatrix}, \qquad \overrightarrow{PB}=\begin{pmatrix}6\\6\end{pmatrix}.$$

Since $\overrightarrow{AP}=\frac23\overrightarrow{PB}$, their lengths are in the ratio $2:3$ and the directions agree.

Example 6: find an endpoint from a ratio

$P(4,7)$ divides $AB$ in the ratio $AP:PB=1:2$, and $A(-2,3)$. Find $B$.

Here

$\mathbf p=\frac{2\mathbf a+\mathbf b}{3}.$

Therefore

$$\mathbf b=3\mathbf p-2\mathbf a =3\begin{pmatrix}4\\7\end{pmatrix} -2\begin{pmatrix}-2\\3\end{pmatrix} =\boxed{\begin{pmatrix}16\\15\end{pmatrix}}.$$

Indeed, $\overrightarrow{AP}=(6,4)$ and $\overrightarrow{PB}=(12,8)=2\overrightarrow{AP}$.

Proving that points are collinear

Points are collinear if they lie on one straight line. To prove that $A$, $B$ and $C$ are collinear, show that two displacement vectors from a common point are scalar multiples:

$\boxed{\overrightarrow{AC}=\lambda\overrightarrow{AB}}$

for some real scalar $\lambda$.

This establishes parallel direction. Because both vectors begin at $A$, it also establishes that $B$ and $C$ lie on the same line through $A$.

Example 7: prove collinearity and determine order

Let

$A(1,-1,2),\qquad B(3,2,0),\qquad C(7,8,-4).$

Then

$$\overrightarrow{AB} =\begin{pmatrix}2\\3\\-2\end{pmatrix}$$

and

$$\overrightarrow{AC} =\begin{pmatrix}6\\9\\-6\end{pmatrix} =3\begin{pmatrix}2\\3\\-2\end{pmatrix} =3\overrightarrow{AB}.$$

Therefore $A$, $B$ and $C$ are collinear. Since the multiplier $3$ is greater than $1$, $B$ lies between $A$ and $C$. Also $AB:BC=1:2$ because

$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=2\overrightarrow{AB}.$

A negative multiplier would place the two endpoints on opposite sides of the common point. Merely obtaining equal vector magnitudes is not enough: equal lengths do not prove equal or parallel directions.

Example 8: determine an unknown coordinate

Points $A(2,-1)$, $B(6,5)$ and $C(k,11)$ are collinear. Find $k$.

Calculate vectors from $A$:

$$\overrightarrow{AB}=\begin{pmatrix}4\\6\end{pmatrix}, \qquad \overrightarrow{AC}=\begin{pmatrix}k-2\\12\end{pmatrix}.$$

For collinearity, $\overrightarrow{AC}=\lambda\overrightarrow{AB}$. The second components give

$12=6\lambda\quad\Rightarrow\quad\lambda=2.$

The first components then give

$k-2=2(4)=8,$

so $\boxed{k=10}$.

Common misconceptions

• Reversing the subtraction: $\overrightarrow{AB}=\mathbf b-\mathbf a$, not $\mathbf a-\mathbf b$.
• Confusing a point with a vector: write $A(2,3)$ but $\overrightarrow{OA}=\begin{pmatrix}2\\3\end{pmatrix}$.
• Confusing distance with displacement: $AB$ is a scalar; $\overrightarrow{AB}$ is a vector.
• Using the wrong fraction in a ratio: if $AP:PB=2:3$, then $P$ is $2/5$ of the way from $A$ to $B$, not $2/3$.
• Claiming collinearity from proportional lengths: the vectors themselves must be scalar multiples, with all corresponding components in the same ratio.
• Rounding too early: keep exact roots and fractions until the final line unless an approximation is required.

Self-check

1. $A(-3,2)$ and $B(4,-6)$. Find $\overrightarrow{AB}$ and $AB$.
2. $P(1,-2,4)$ and $Q(5,1,-2)$. Find the midpoint of $PQ$.
3. $R$ divides $AB$ internally in the ratio $AR:RB=3:1$, where $\mathbf a=\begin{pmatrix}-2\\4\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}6\\-8\end{pmatrix}$. Find $\mathbf r$.
4. Show that $A(-1,2)$, $B(2,8)$ and $C(4,12)$ are collinear. Which point lies between the other two?
5. Point $M(3,5)$ is the midpoint of $A(-4,1)$ and $B$. Find $B$.

Answers

1. $\overrightarrow{AB}=\begin{pmatrix}7\\-8\end{pmatrix}$ and $AB=\sqrt{7^2+(-8)^2}=\sqrt{113}$.
2. The midpoint is $\left(\frac{1+5}{2},\frac{-2+1}{2},\frac{4+(-2)}{2}\right)=\left(3,-\frac12,1\right)$.
3. $\mathbf r=\mathbf a+\frac34(\mathbf b-\mathbf a)=\begin{pmatrix}4\\-5\end{pmatrix}$.
4. $\overrightarrow{AB}=\begin{pmatrix}3\\6\end{pmatrix}$ and $\overrightarrow{AC}=\begin{pmatrix}5\\10\end{pmatrix}=\frac53\overrightarrow{AB}$, so the points are collinear. Since $1<5/3$, $B$ lies between $A$ and $C$.
5. $\mathbf b=2\mathbf m-\mathbf a=\begin{pmatrix}10\\9\end{pmatrix}$, so $B(10,9)$.

What to learn next

You can now translate between points, position vectors and displacements, calculate distances, divide segments in a given ratio, and prove collinearity.

Next, study vectors in modelling to use these ideas in geometric and physical contexts. For equations and intersections of coordinate lines, revisit straight lines in coordinate geometry.