Mathematics 9709 · Pure Mathematics 3

Position vectors

A position vector fixes a point. Its arrow starts at the origin O: point $A$ has position vector $\mathbf{a}=\overrightarrow{OA}$ .
One rule is used again and again in this chapter: the step from $A$ to $B$ is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .

End minus start. To turn two points into the step between them, subtract their position vectors: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ . Almost every later topic starts with this step.

To reach a point a fraction of the way along a step, start at one end and add that fraction of the step: $\overrightarrow{ON}=\mathbf{b}+\tfrac{k}{k+1}\,\overrightarrow{BC}$ cuts $BC$ in the ratio $k:1$ . The midpoint $\tfrac12(\mathbf{a}+\mathbf{b})$ is just the case $k=1$ .

Position vectors, and the step between two points

A point $(x,y,z)$ has position vector $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ . That is the same thing as the column $\begin{pmatrix}x\\y\\z\end{pmatrix}$ .
To go from $A$ to $B$ , go back along $\,\mathbf{a}$ to O, then out to B with $\mathbf{b}$ :

$\overrightarrow{AB}=\mathbf{b}-\mathbf{a},\qquad \text{midpoint }M=\tfrac12(\mathbf{a}+\mathbf{b})$

Position vectors a, b from O; the step AB that closes the triangle is b − a.

Points $A,B,C$ are collinear (all on one straight line) when one step is a multiple of another: $\overrightarrow{AC}=k\,\overrightarrow{AB}$ . They point the same way and share point A, so they all sit on one line.

Worked examples

Worked example

Real question: $\overrightarrow{OA}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}$ , $\overrightarrow{OC}=3\mathbf{i}+4\mathbf{j}-7\mathbf{k}$ , find $\overrightarrow{AC}$ and its length

The opening move of 9709/32 F/M 2024 Q9: turn the points into a step first.

Step = destination minus start: $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=\begin{pmatrix}3\\4\\-7\end{pmatrix}-\begin{pmatrix}5\\-2\\1\end{pmatrix}=\begin{pmatrix}-2\\6\\-8\end{pmatrix}$ .
Length by Pythagoras on all three components: $|\overrightarrow{AC}|=\sqrt{(-2)^2+6^2+(-8)^2}=\sqrt{4+36+64}=\sqrt{104}$ .

Answer

\overrightarrow{AC}=\begin{pmatrix}-2\\6\\-8\end{pmatrix},\quad |\overrightarrow{AC}|=\sqrt{104}=2\sqrt{26}

Worked example

$\mathbf{a}=(2,5,-3)$ , $\mathbf{c}=(-4,2,24)$ : are $A$ and $C$ collinear with $B=(0,4,6)$ ?

Two steps from A: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}-2\\-1\\9\end{pmatrix}$ , $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=\begin{pmatrix}-6\\-3\\27\end{pmatrix}$ .
Is one a multiple of the other? $\overrightarrow{AC}=3\,\overrightarrow{AB}$ , since each component triples. Same direction from a shared point A, so the three points line up.

Answer

\text{collinear (}\overrightarrow{AC}=3\,\overrightarrow{AB}\text{)}

Worked example

Real question: $A(0,5,2)$ , $B(1,0,1)$ , $C(4,-3,-2)$ ; $M$ = midpoint of $AC$ , $N$ on $BC$ with $BN=2NC$

9709/33 O/N 2022 Q9(a): the two most common “find the position vector” types, a midpoint and a point that splits a step in a ratio.

Midpoint of $AC$ averages the two ends: $\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{c})=\tfrac12\begin{pmatrix}4\\2\\0\end{pmatrix}=\begin{pmatrix}2\\1\\0\end{pmatrix}$ .
$BN=2NC$ means $N$ is $\tfrac23$ of the way from $B$ to $C$ . First the step: $\overrightarrow{BC}=\mathbf{c}-\mathbf{b}=\begin{pmatrix}3\\-3\\-3\end{pmatrix}$ .
Start at $B$ , add $\tfrac23$ of that step: $\overrightarrow{ON}=\mathbf{b}+\tfrac23\overrightarrow{BC}=\begin{pmatrix}1\\0\\1\end{pmatrix}+\begin{pmatrix}2\\-2\\-2\end{pmatrix}=\begin{pmatrix}3\\-2\\-1\end{pmatrix}$ .

Answer

\overrightarrow{OM}=\begin{pmatrix}2\\1\\0\end{pmatrix},\quad \overrightarrow{ON}=\begin{pmatrix}3\\-2\\-1\end{pmatrix}

Where the marks go

Common mistake

A ratio splits the step, not the whole position vector.

BN=2NC

puts

N

\tfrac23

of the way along

BC

, not

\tfrac12

. The longer piece

BN

sits next to

B

. Always start at the named point and add the fraction of the step. (9709/33 Oct/Nov 2022 Q9)

Common mistake

Don't treat

\overrightarrow{AB}

as a position vector. It is

\mathbf{b}-\mathbf{a}

, not

\mathbf{b}

. Almost every later method dots or compares steps, not raw position vectors, so subtract the start point first.

Common mistake

For collinearity, showing the steps are parallel (

\overrightarrow{AC}=k\,\overrightarrow{AB}

) only works because they share the point A. Two parallel steps that don't share a point prove nothing. Always build both from the same corner.

Now you try

P, Q, R have position vectors $\mathbf{p}=\mathbf{i}+2\mathbf{j}-\mathbf{k}$ , $\mathbf{q}=3\mathbf{i}+5\mathbf{j}+\mathbf{k}$ and $\mathbf{r}=7\mathbf{i}+11\mathbf{j}+5\mathbf{k}$ . Show that P, Q, R all lie on one line. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Build both steps from P: $\overrightarrow{PQ}=\begin{pmatrix}2\\3\\2\end{pmatrix}$ and $\overrightarrow{PR}=\begin{pmatrix}6\\9\\6\end{pmatrix}=3\overrightarrow{PQ}$ , so they line up:

Answer

\overrightarrow{PR}=3\,\overrightarrow{PQ}\ \Rightarrow\ \text{collinear}

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Position vectors

A position vector fixes a point. Its arrow starts at the origin O: point $A$ has position vector $\mathbf{a}=\overrightarrow{OA}$ .
One rule is used again and again in this chapter: the step from $A$ to $B$ is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .

End minus start. To turn two points into the step between them, subtract their position vectors: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ . Almost every later topic starts with this step.

Position vectors, and the step between two points

A point $(x,y,z)$ has position vector $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ . That is the same thing as the column $\begin{pmatrix}x\\y\\z\end{pmatrix}$ .
To go from $A$ to $B$ , go back along $\,\mathbf{a}$ to O, then out to B with $\mathbf{b}$ :

$\overrightarrow{AB}=\mathbf{b}-\mathbf{a},\qquad \text{midpoint }M=\tfrac12(\mathbf{a}+\mathbf{b})$

Position vectors a, b from O; the step AB that closes the triangle is b − a.

Points $A,B,C$ are collinear (all on one straight line) when one step is a multiple of another: $\overrightarrow{AC}=k\,\overrightarrow{AB}$ . They point the same way and share point A, so they all sit on one line.

Worked examples

Worked example

Real question: $\overrightarrow{OA}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}$ , $\overrightarrow{OC}=3\mathbf{i}+4\mathbf{j}-7\mathbf{k}$ , find $\overrightarrow{AC}$ and its length

The opening move of 9709/32 F/M 2024 Q9: turn the points into a step first.

Step = destination minus start: $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=\begin{pmatrix}3\\4\\-7\end{pmatrix}-\begin{pmatrix}5\\-2\\1\end{pmatrix}=\begin{pmatrix}-2\\6\\-8\end{pmatrix}$ .
Length by Pythagoras on all three components: $|\overrightarrow{AC}|=\sqrt{(-2)^2+6^2+(-8)^2}=\sqrt{4+36+64}=\sqrt{104}$ .

Answer

\overrightarrow{AC}=\begin{pmatrix}-2\\6\\-8\end{pmatrix},\quad |\overrightarrow{AC}|=\sqrt{104}=2\sqrt{26}

Worked example

$\mathbf{a}=(2,5,-3)$ , $\mathbf{c}=(-4,2,24)$ : are $A$ and $C$ collinear with $B=(0,4,6)$ ?

Two steps from A: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}-2\\-1\\9\end{pmatrix}$ , $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=\begin{pmatrix}-6\\-3\\27\end{pmatrix}$ .
Is one a multiple of the other? $\overrightarrow{AC}=3\,\overrightarrow{AB}$ , since each component triples. Same direction from a shared point A, so the three points line up.

Answer

\text{collinear (}\overrightarrow{AC}=3\,\overrightarrow{AB}\text{)}

Worked example

Real question: $A(0,5,2)$ , $B(1,0,1)$ , $C(4,-3,-2)$ ; $M$ = midpoint of $AC$ , $N$ on $BC$ with $BN=2NC$

9709/33 O/N 2022 Q9(a): the two most common “find the position vector” types, a midpoint and a point that splits a step in a ratio.

Midpoint of $AC$ averages the two ends: $\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{c})=\tfrac12\begin{pmatrix}4\\2\\0\end{pmatrix}=\begin{pmatrix}2\\1\\0\end{pmatrix}$ .
$BN=2NC$ means $N$ is $\tfrac23$ of the way from $B$ to $C$ . First the step: $\overrightarrow{BC}=\mathbf{c}-\mathbf{b}=\begin{pmatrix}3\\-3\\-3\end{pmatrix}$ .
Start at $B$ , add $\tfrac23$ of that step: $\overrightarrow{ON}=\mathbf{b}+\tfrac23\overrightarrow{BC}=\begin{pmatrix}1\\0\\1\end{pmatrix}+\begin{pmatrix}2\\-2\\-2\end{pmatrix}=\begin{pmatrix}3\\-2\\-1\end{pmatrix}$ .

Answer

\overrightarrow{OM}=\begin{pmatrix}2\\1\\0\end{pmatrix},\quad \overrightarrow{ON}=\begin{pmatrix}3\\-2\\-1\end{pmatrix}

Where the marks go

Common mistake

A ratio splits the step, not the whole position vector.

BN=2NC

puts

N

\tfrac23

of the way along

BC

, not

\tfrac12

. The longer piece

BN

sits next to

B

. Always start at the named point and add the fraction of the step. (9709/33 Oct/Nov 2022 Q9)

Common mistake

Don't treat

\overrightarrow{AB}

as a position vector. It is

\mathbf{b}-\mathbf{a}

, not

\mathbf{b}

. Almost every later method dots or compares steps, not raw position vectors, so subtract the start point first.

Common mistake

For collinearity, showing the steps are parallel (

\overrightarrow{AC}=k\,\overrightarrow{AB}

) only works because they share the point A. Two parallel steps that don't share a point prove nothing. Always build both from the same corner.

Now you try

Your turn— tap to reveal the worked answer (9709-style)

Build both steps from P: $\overrightarrow{PQ}=\begin{pmatrix}2\\3\\2\end{pmatrix}$ and $\overrightarrow{PR}=\begin{pmatrix}6\\9\\6\end{pmatrix}=3\overrightarrow{PQ}$ , so they line up:

Answer

\overrightarrow{PR}=3\,\overrightarrow{PQ}\ \Rightarrow\ \text{collinear}