3.14 Vector equation of a line (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

The vector equation of a line

A line is fixed by one point on it and a direction. If $\mathbf{a}$ is the position vector of a point on the line and $\mathbf{b}$ is a direction vector, then every point on the line has position vector $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b},\qquad\lambda\in\mathbb{R}.$ Each value of the parameter $\lambda$ gives one point. The same form works in two and three dimensions. The equation is not unique: any point on the line and any non-zero multiple of $\mathbf{b}$ give the same line. For the line through $P$ and $Q$ , use $\mathbf{r}=\mathbf{p}+\lambda(\mathbf{q}-\mathbf{p})$ .

Key termsdirection vectorparameter

Common mistake

Using $\mathbf{q}$ (the position vector of $Q$ ) as the direction. The direction is $\mathbf{q}-\mathbf{p}$ .

Section 2

Parametric and Cartesian forms

Writing $\mathbf{a}=(x_0,y_0,z_0)$ and $\mathbf{b}=(l,m,n)$ , the components give the parametric form $x=x_0+\lambda l,\quad y=y_0+\lambda m,\quad z=z_0+\lambda n.$ Making $\lambda$ the subject of each and equating gives the Cartesian form $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}.$ Example: $\mathbf{r}=(1,-2,3)+\lambda(2,1,-2)$ becomes $\frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{-2}$ . If a direction component is 0, that coordinate is constant, e.g. $z=3$ , and you write it separately. In 2D the Cartesian form rearranges to $y=mx+c$ .

Key termsparametric formCartesian form

Exam tip

To convert Cartesian to vector form, read the point from the numerators (change the signs) and the direction from the denominators.

Section 3

Points on a line

To test whether a point lies on a line, find $\lambda$ from one component and check it gives the other components too. To find where a line meets a condition (e.g. $z=-5$ , or crossing the $x$ -axis where $y=0$ ), set the relevant component equal to the value, solve for $\lambda$ and substitute back.

Key termschecking a point

Section 4

The angle between two lines

The angle between two lines is the angle between their direction vectors $\mathbf{b}_1$ and $\mathbf{b}_2$ : $\cos\theta=\frac{|\mathbf{b}_1\cdot\mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}.$ The modulus gives the acute angle. The position vectors play no part. For the angle a line makes with the horizontal, compare its direction with the horizontal projection of that direction: for $(6,3,-2)$ use $(6,3,0)$ , or use $\sin\alpha=\frac{2}{7}$ directly.

Key termsacute angle between lines

Common mistake

Using the position vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ in the scalar product. Only the directions matter.

Section 5

Kinematics: $\lambda$ as time

If an object moves with constant velocity, its position at time $t$ is $\mathbf{r}=\mathbf{a}+t\mathbf{v}$ : $\mathbf{a}$ is the initial position, $\mathbf{v}$ is the velocity and $|\mathbf{v}|$ is the speed.

The closest approach to a fixed point $H$ happens when $\overrightarrow{HS}$ (from $H$ to the object) is perpendicular to $\mathbf{v}$ , so solve $\overrightarrow{HS}\cdot\mathbf{v}=0$ . For $\mathbf{r}=(2,5)+t(3,-4)$ and $H(12,0)$ : $\overrightarrow{HS}=(3t-10,\,5-4t)$ , giving $25t-50=0$ , $t=2$ , distance 5 km.

Key termsvelocityspeed

Must know

$\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$ : $\mathbf{a}$ a point, $\mathbf{b}$ a direction.
Parametric: $x=x_0+\lambda l$ , etc. Cartesian: $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$ .
Angle between lines: $\cos\theta=\frac{|\mathbf{b}_1\cdot\mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}$ .
Kinematics: $\mathbf{r}=\mathbf{a}+t\mathbf{v}$ , speed $=|\mathbf{v}|$ .

That's the notes covered.

Carry on to the next subtopic.