Vector equation of a line | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Separate anchor, direction and parameter

r = a + td

t = −1t = 0: at = 1d

Rescaling d changes which t reaches a point, not the line.

The anchor fixes one point; the direction vector fixes the line; the parameter moves along it.

\mathbf r=\mathbf a+t\mathbf d

$\mathbf r$ is the position vector of a general point on the line.
$\mathbf a$ is the position vector of one fixed point.
$\mathbf d$ is any non-zero direction vector.
$t$ is a real parameter; changing it moves through all points on the line.

Key idea

Reversing or rescaling the direction gives a different-looking equation, but the set of points on the line is unchanged.

Two points give the direction by subtraction

Worked example

Current-paper line through B and C

Let $B=(5,2,0)$ and $C=(8,5,-3)$ .

\overrightarrow{BC}=(3,3,-3)

\overrightarrow{BC}=3(1,1,-1)

Answer

\begin{aligned}\mathbf r&=(5,2,0)\\&\quad+\lambda(1,1,-1)\end{aligned}

(9709/32/M/J/24 Q8(a))

Examiner note

You may anchor the line at B or C, and you may reverse or rescale the direction. These equations can describe the same line.

Read the three parametric equations

The vector equation $\mathbf r=(a_1,a_2,a_3)+t(d_1,d_2,d_3)$ means:

\begin{aligned}x&=a_1+td_1\\y&=a_2+td_2\\z&=a_3+td_3\end{aligned}

These are the parametric equations. The same value of $t$ must be used in all three because the three coordinates belong to one point.

Your turnIndependent check [4 marks]

Find a vector equation of the line through $A=(1,-2,3)$ and $B=(4,0,-1)$ . Hence write its parametric equations.

Show worked answer

\overrightarrow{AB}=(3,2,-4)

Answer

\begin{aligned}\mathbf r&=(1,-2,3)\\&\quad+t(3,2,-4)\end{aligned}

Answer

x=1+3t

y=-2+2t

z=3-4t

At $t=0$ the line gives A; at $t=1$ it gives B.

Separate anchor, direction and parameter

r = a + td

t = −1t = 0: at = 1d

Rescaling d changes which t reaches a point, not the line.

The anchor fixes one point; the direction vector fixes the line; the parameter moves along it.

\mathbf r=\mathbf a+t\mathbf d

$\mathbf r$ is the position vector of a general point on the line.
$\mathbf a$ is the position vector of one fixed point.
$\mathbf d$ is any non-zero direction vector.
$t$ is a real parameter; changing it moves through all points on the line.

Key idea

Reversing or rescaling the direction gives a different-looking equation, but the set of points on the line is unchanged.

Two points give the direction by subtraction

Worked example

Current-paper line through B and C

Let $B=(5,2,0)$ and $C=(8,5,-3)$ .

\overrightarrow{BC}=(3,3,-3)

\overrightarrow{BC}=3(1,1,-1)

Answer

\begin{aligned}\mathbf r&=(5,2,0)\\&\quad+\lambda(1,1,-1)\end{aligned}

(9709/32/M/J/24 Q8(a))

Examiner note

You may anchor the line at B or C, and you may reverse or rescale the direction. These equations can describe the same line.

Read the three parametric equations

The vector equation $\mathbf r=(a_1,a_2,a_3)+t(d_1,d_2,d_3)$ means:

\begin{aligned}x&=a_1+td_1\\y&=a_2+td_2\\z&=a_3+td_3\end{aligned}

These are the parametric equations. The same value of $t$ must be used in all three because the three coordinates belong to one point.

Your turnIndependent check [4 marks]

Find a vector equation of the line through $A=(1,-2,3)$ and $B=(4,0,-1)$ . Hence write its parametric equations.

Show worked answer

\overrightarrow{AB}=(3,2,-4)

Answer

\begin{aligned}\mathbf r&=(1,-2,3)\\&\quad+t(3,2,-4)\end{aligned}

Answer

x=1+3t

y=-2+2t

z=3-4t

At $t=0$ the line gives A; at $t=1$ it gives B.