IB Mathematics Analysis and Approaches Topic AHL…

Why A Line Needs A Point And A Direction

In two dimensions you describe a line with $y = mx + c$. In three dimensions that fails: a single equation like $z = 2x + 1$ describes a whole plane, not a line.
A better description works in any dimension: name one point the line passes through, and one direction it travels in. "Start here, head that way."
Vectors capture this exactly.
A position vector fixes the starting point, a direction vector fixes the heading, and a parameter slides you along.
This article builds the vector equation, rewrites it in parametric and Cartesian form, finds the angle between two lines, and reads a line as the motion of an object in time.

The Vector Equation r = a + λb

Let $\mathbf{a}$ be the position vector of a known point $A$ on the line, and $\mathbf{b}$ a direction vector parallel to the line.
To reach any point $R$ on the line, travel out to $A$, then move some multiple of $\mathbf{b}$:
1. $\vec{OR} = \vec{OA} + \vec{AR}$, and $\vec{AR}$ is a multiple of the direction, $\lambda\mathbf{b}$.
2. So $\vec{OR} = \mathbf{a} + \lambda\mathbf{b}$. $$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$
As the scalar $\lambda$ runs through all real values, $\mathbf{r}$ traces out every point on the line.

Note

The vector, parametric, and Cartesian forms of a line are all in the formula booklet under AHL 3.14.
The equation is not unique: any point on the line works as $\mathbf{a}$, and any scalar multiple of $\mathbf{b}$ works as the direction.
Two different-looking equations can describe the same line.

Example

A line through $(1, 2, 3)$ parallel to $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ has equation: $$\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$$

Parametric And Cartesian Forms

Writing the vector equation out component by component gives the parametric form, with $(x_0, y_0, z_0)$ the point and $(l, m, n)$ the direction: $$x = x_0 + \lambda l, \qquad y = y_0 + \lambda m, \qquad z = z_0 + \lambda n$$
To get the Cartesian form, make $\lambda$ the subject of each equation:
1. $\lambda = \dfrac{x - x_0}{l}$, and likewise for $y$ and $z$.
2. Since all three equal the same $\lambda$, set them equal: $$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$$
This form has no parameter, which makes it handy for testing whether a point lies on the line or for finding intersections.

Example

For the line above:

parametric $x = 1 + 2\lambda,\ y = 2 - \lambda,\ z = 3 + 3\lambda$;
Cartesian $\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{3}$.

The Angle Between Two Lines

The angle between two lines depends only on their directions, so use the dot product of the direction vectors $\mathbf{b}_1$ and $\mathbf{b}_2$. $$\cos\theta = \frac{\mathbf{b}_1 \cdot \mathbf{b}_2}{|\mathbf{b}_1|\,|\mathbf{b}_2|}$$
By convention the angle between two lines is taken as the acute one, so use the absolute value of the dot product, $|\mathbf{b}_1 \cdot \mathbf{b}_2|$, on top.

Example question

Find the acute angle between the lines with direction vectors $\mathbf{b}_1 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

Solution

Dot product: $$\mathbf{b}_1\cdot\mathbf{b}_2 = 1(2) + 2(-1) + 2(2) = 4$$
Magnitudes: $$|\mathbf{b}_1| = 3 \text{ and } |\mathbf{b}_2| = 3$$

$$\cos\theta = \dfrac{|4|}{3 \times 3} = \dfrac{4}{9}$$

$$\theta = \arccos\tfrac49 \approx 63.6^\circ$$

Kinematics: A Line That Moves In Time

If the parameter is time, the vector equation becomes the path of a moving object.
Replace $\lambda$ with $t$:
1. $\mathbf{a}$ is the initial position (at $t = 0$).
2. $\mathbf{b}$ is the velocity, giving both the direction of motion and the distance covered per unit time.
3. $|\mathbf{b}|$ is the speed, the magnitude of the velocity.

$$\mathbf{r} = \mathbf{a} + t\mathbf{b}$$

Example question

An object starts at $\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$ and moves with velocity $\begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$ (units per second). Find its position after $t$ seconds and its speed.

Solution

Position: $$\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + t\begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$$
Speed is the magnitude of the velocity: $$|\mathbf{b}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{21}$$
So the speed is $\sqrt{21} \approx 4.58$ units per second.

Exam technique

Keep the time unit in $t$ and the velocity unit consistent, so the position and speed come out in matching units.
If two objects are described by separate equations, equal position vectors means they are at the same place; the same $t$ as well means they actually collide.

Active recall

Explain why a line in 3D needs a point and a direction, not a single equation like $y = mx + c$.
Write the vector equation of a line and say what $\mathbf{a}$, $\mathbf{b}$, and $\lambda$ each represent.
Convert a parametric line to Cartesian form by eliminating the parameter.
How do you find the acute angle between two lines?
In a kinematics context, what do $\mathbf{a}$, $\mathbf{b}$, and $|\mathbf{b}|$ mean?

Why A Line Needs A Point And A Direction

In two dimensions you describe a line with $y = mx + c$. In three dimensions that fails: a single equation like $z = 2x + 1$ describes a whole plane, not a line.
A better description works in any dimension: name one point the line passes through, and one direction it travels in. "Start here, head that way."
Vectors capture this exactly.
A position vector fixes the starting point, a direction vector fixes the heading, and a parameter slides you along.
This article builds the vector equation, rewrites it in parametric and Cartesian form, finds the angle between two lines, and reads a line as the motion of an object in time.

The Vector Equation r = a + λb

Let $\mathbf{a}$ be the position vector of a known point $A$ on the line, and $\mathbf{b}$ a direction vector parallel to the line.
To reach any point $R$ on the line, travel out to $A$, then move some multiple of $\mathbf{b}$:
1. $\vec{OR} = \vec{OA} + \vec{AR}$, and $\vec{AR}$ is a multiple of the direction, $\lambda\mathbf{b}$.
2. So $\vec{OR} = \mathbf{a} + \lambda\mathbf{b}$. $$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$
As the scalar $\lambda$ runs through all real values, $\mathbf{r}$ traces out every point on the line.

Note

The vector, parametric, and Cartesian forms of a line are all in the formula booklet under AHL 3.14.
The equation is not unique: any point on the line works as $\mathbf{a}$, and any scalar multiple of $\mathbf{b}$ works as the direction.
Two different-looking equations can describe the same line.

Example

Parametric And Cartesian Forms

Writing the vector equation out component by component gives the parametric form, with $(x_0, y_0, z_0)$ the point and $(l, m, n)$ the direction: $$x = x_0 + \lambda l, \qquad y = y_0 + \lambda m, \qquad z = z_0 + \lambda n$$
To get the Cartesian form, make $\lambda$ the subject of each equation:
1. $\lambda = \dfrac{x - x_0}{l}$, and likewise for $y$ and $z$.
2. Since all three equal the same $\lambda$, set them equal: $$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$$
This form has no parameter, which makes it handy for testing whether a point lies on the line or for finding intersections.

Example

For the line above:

parametric $x = 1 + 2\lambda,\ y = 2 - \lambda,\ z = 3 + 3\lambda$;
Cartesian $\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{3}$.

The Angle Between Two Lines

The angle between two lines depends only on their directions, so use the dot product of the direction vectors $\mathbf{b}_1$ and $\mathbf{b}_2$. $$\cos\theta = \frac{\mathbf{b}_1 \cdot \mathbf{b}_2}{|\mathbf{b}_1|\,|\mathbf{b}_2|}$$
By convention the angle between two lines is taken as the acute one, so use the absolute value of the dot product, $|\mathbf{b}_1 \cdot \mathbf{b}_2|$, on top.

Example question

Find the acute angle between the lines with direction vectors $\mathbf{b}_1 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

Solution

Dot product: $$\mathbf{b}_1\cdot\mathbf{b}_2 = 1(2) + 2(-1) + 2(2) = 4$$
Magnitudes: $$|\mathbf{b}_1| = 3 \text{ and } |\mathbf{b}_2| = 3$$

$$\cos\theta = \dfrac{|4|}{3 \times 3} = \dfrac{4}{9}$$

$$\theta = \arccos\tfrac49 \approx 63.6^\circ$$

Kinematics: A Line That Moves In Time

If the parameter is time, the vector equation becomes the path of a moving object.
Replace $\lambda$ with $t$:
1. $\mathbf{a}$ is the initial position (at $t = 0$).
2. $\mathbf{b}$ is the velocity, giving both the direction of motion and the distance covered per unit time.
3. $|\mathbf{b}|$ is the speed, the magnitude of the velocity.

$$\mathbf{r} = \mathbf{a} + t\mathbf{b}$$

Example question

Solution

Position: $$\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + t\begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$$
Speed is the magnitude of the velocity: $$|\mathbf{b}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{21}$$
So the speed is $\sqrt{21} \approx 4.58$ units per second.

Exam technique

Keep the time unit in $t$ and the velocity unit consistent, so the position and speed come out in matching units.
If two objects are described by separate equations, equal position vectors means they are at the same place; the same $t$ as well means they actually collide.

Active recall

Explain why a line in 3D needs a point and a direction, not a single equation like $y = mx + c$.
Write the vector equation of a line and say what $\mathbf{a}$, $\mathbf{b}$, and $\lambda$ each represent.
Convert a parametric line to Cartesian form by eliminating the parameter.
How do you find the acute angle between two lines?
In a kinematics context, what do $\mathbf{a}$, $\mathbf{b}$, and $|\mathbf{b}|$ mean?

IB Mathematics Analysis and Approaches AHL 3.14—vector Equation of Line Notes

IB Mathematics Analysis and Approaches AHL 3.14—vector Equation of Line Notes

Why A Line Needs A Point And A Direction

The Vector Equation r = a + λb

Parametric And Cartesian Forms

The Angle Between Two Lines

Kinematics: A Line That Moves In Time

Why A Line Needs A Point And A Direction

The Vector Equation r = a + λb

Parametric And Cartesian Forms

The Angle Between Two Lines

Kinematics: A Line That Moves In Time

IB Mathematics Analysis and Approaches AHL 3.14—vector Equation of Line Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches AHL 3.14—vector Equation of Line Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Why A Line Needs A Point And A Direction

The Vector Equation r = a + λb

Parametric And Cartesian Forms

The Angle Between Two Lines

Kinematics: A Line That Moves In Time

Why A Line Needs A Point And A Direction

The Vector Equation r = a + λb

Parametric And Cartesian Forms

The Angle Between Two Lines

Kinematics: A Line That Moves In Time