Mathematics 9709 · Pure Mathematics 3

The vector equation of a line

A line is a point plus a direction: start at $\mathbf{a}$ , then slide along $\mathbf{b}$ by any amount $t$ .
The one equation $\mathbf{r}=\mathbf{a}+t\mathbf{b}$ draws the whole line as $t$ takes every value.

$\mathbf{r}=\mathbf{a}+t\mathbf{b}$ : $\mathbf{a}$ is any point on the line and $\mathbf{b}$ is its direction. If you are given two points, the direction is the step between them, $\mathbf{c}-\mathbf{a}$ .

Read the equation row by row to get the parametric form $x=a_1+tb_1,\ y=a_2+tb_2,\ z=a_3+tb_3$ . A point is on the line only if a single $t$ works for all three rows.

r = a + t b: a point and a direction

$\mathbf{r}=\mathbf{a}+t\mathbf{b}\quad(\text{point }\mathbf{a},\ \text{direction }\mathbf{b})$

Given two points $A$ and $C$ , the direction is $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$ . You can use either point as the start $\mathbf{a}$ .
A parallel line keeps the same direction $\mathbf{b}$ and changes only the point.

Each t gives one point on r = a + t b: t = 0 is A, and b steps you along.

The same line has many correct equations

Two right answers can look different. You might start from a different point, or use a direction multiplied by any non-zero number ( $(2,-3,4)$ and $(-2,3,-4)$ point along the same line). The mark scheme accepts any valid pair, so don't panic if your $\mathbf{a}$ or $\mathbf{b}$ looks unlike the worked solution. Just check it is a point on the line and a multiple of the direction.

Worked examples

Worked example

Line through $(3,-2,4)$ parallel to $(1,-1,1)$

Parallel keeps the same direction, so the start is $(3,-2,4)$ and the direction is $(1,-1,1)$ :

$\mathbf{r}=\begin{pmatrix}3\\-2\\4\end{pmatrix}+t\begin{pmatrix}1\\-1\\1\end{pmatrix}$ i.e. $x=3+t,\ y=-2-t,\ z=4+t$ .

Answer

\mathbf{r}=(3\mathbf{i}-2\mathbf{j}+4\mathbf{k})+t(\mathbf{i}-\mathbf{j}+\mathbf{k})

Worked example

Real question: line $m$ through $A(-2,2,-1)$ and $B(3,-1,1)$

9709/33 O/N 2023 Q11(b): turn two points into a line equation.

Direction = step between the points: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}3\\-1\\1\end{pmatrix}-\begin{pmatrix}-2\\2\\-1\end{pmatrix}=\begin{pmatrix}5\\-3\\2\end{pmatrix}$ .
Use A as the start point:

Answer

\mathbf{r}=(-2\mathbf{i}+2\mathbf{j}-\mathbf{k})+\mu(5\mathbf{i}-3\mathbf{j}+2\mathbf{k})

Worked example

Real question: line through $A(2,1,1)$ and $B(1,-2,2)$

9709/32 F/M 2022 Q10(a). Same two-point recipe: find the direction first, then pick a point.

Direction = step between the points: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}1\\-2\\2\end{pmatrix}-\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-3\\1\end{pmatrix}$ .
Use $A$ as the start point (either point is fine):

Answer

\mathbf{r}=(2\mathbf{i}+\mathbf{j}+\mathbf{k})+\lambda(-\mathbf{i}-3\mathbf{j}+\mathbf{k})

This direction $(-1,-3,1)$ comes back in part (b) to find the acute angle with another line: see §9.4½.

Where the marks go

Common mistake

From two points, the direction is

\mathbf{c}-\mathbf{a}

, not

\mathbf{a}-\mathbf{c}

. A parallel line keeps the same direction and changes only the point. And

\mathbf{a}

must be the start point, never the direction.

Common mistake

To test whether a point lies on a line, you need one

t

that fits every row. If the

x

and

y

rows give different

t

values, the point is off the line. One row agreeing is not enough.

Now you try

Does $P(5,0,1)$ lie on $\mathbf{r}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}$ ? (9709-style)

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

You need one $t$ that fits all three rows: $1+2t=5\Rightarrow t=2$ ; $-2+t=0\Rightarrow t=2$ ; $3-t=1\Rightarrow t=2$ . All agree:

Answer

\text{yes, at } t=2

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The vector equation of a line

A line is a point plus a direction: start at $\mathbf{a}$ , then slide along $\mathbf{b}$ by any amount $t$ .
The one equation $\mathbf{r}=\mathbf{a}+t\mathbf{b}$ draws the whole line as $t$ takes every value.

Read the equation row by row to get the parametric form $x=a_1+tb_1,\ y=a_2+tb_2,\ z=a_3+tb_3$ . A point is on the line only if a single $t$ works for all three rows.

r = a + t b: a point and a direction

$\mathbf{r}=\mathbf{a}+t\mathbf{b}\quad(\text{point }\mathbf{a},\ \text{direction }\mathbf{b})$

Given two points $A$ and $C$ , the direction is $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$ . You can use either point as the start $\mathbf{a}$ .
A parallel line keeps the same direction $\mathbf{b}$ and changes only the point.

Each t gives one point on r = a + t b: t = 0 is A, and b steps you along.

The same line has many correct equations

Worked examples

Worked example

Line through $(3,-2,4)$ parallel to $(1,-1,1)$

Parallel keeps the same direction, so the start is $(3,-2,4)$ and the direction is $(1,-1,1)$ :

$\mathbf{r}=\begin{pmatrix}3\\-2\\4\end{pmatrix}+t\begin{pmatrix}1\\-1\\1\end{pmatrix}$ i.e. $x=3+t,\ y=-2-t,\ z=4+t$ .

Answer

\mathbf{r}=(3\mathbf{i}-2\mathbf{j}+4\mathbf{k})+t(\mathbf{i}-\mathbf{j}+\mathbf{k})

Worked example

Real question: line $m$ through $A(-2,2,-1)$ and $B(3,-1,1)$

9709/33 O/N 2023 Q11(b): turn two points into a line equation.

Direction = step between the points: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}3\\-1\\1\end{pmatrix}-\begin{pmatrix}-2\\2\\-1\end{pmatrix}=\begin{pmatrix}5\\-3\\2\end{pmatrix}$ .
Use A as the start point:

Answer

\mathbf{r}=(-2\mathbf{i}+2\mathbf{j}-\mathbf{k})+\mu(5\mathbf{i}-3\mathbf{j}+2\mathbf{k})

Worked example

Real question: line through $A(2,1,1)$ and $B(1,-2,2)$

9709/32 F/M 2022 Q10(a). Same two-point recipe: find the direction first, then pick a point.

Direction = step between the points: $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}1\\-2\\2\end{pmatrix}-\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-3\\1\end{pmatrix}$ .
Use $A$ as the start point (either point is fine):

Answer

\mathbf{r}=(2\mathbf{i}+\mathbf{j}+\mathbf{k})+\lambda(-\mathbf{i}-3\mathbf{j}+\mathbf{k})

This direction $(-1,-3,1)$ comes back in part (b) to find the acute angle with another line: see §9.4½.

Where the marks go

Common mistake

From two points, the direction is

\mathbf{c}-\mathbf{a}

, not

\mathbf{a}-\mathbf{c}

. A parallel line keeps the same direction and changes only the point. And

\mathbf{a}

must be the start point, never the direction.

Common mistake

To test whether a point lies on a line, you need one

t

that fits every row. If the

x

and

y

rows give different

t

values, the point is off the line. One row agreeing is not enough.

Now you try

Does $P(5,0,1)$ lie on $\mathbf{r}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}$ ? (9709-style)

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

You need one $t$ that fits all three rows: $1+2t=5\Rightarrow t=2$ ; $-2+t=0\Rightarrow t=2$ ; $3-t=1\Rightarrow t=2$ . All agree:

Answer

\text{yes, at } t=2