Vector Equations of Lines (DP IB Maths: AA HL)

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Amber

Last updated

2 January 2024

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Equation of a Line in Vector Form

How do I find the vector equation of a line?

The formula for finding the vector equation of a line is
- $r = a + λ b$
  - Where r is the position vector of any point on the line
  - a is the position vector of a known point on the line
  - b is a direction (displacement) vector
  - $λ$ is a scalar
- This is given in the formula booklet
- This equation can be used for vectors in both 2- and 3- dimensions
This formula is similar to a regular equation of a straight line in the form $y = m x + c$ but with a vector to show both a point on the line and the direction (or gradient) of the line
- In 2D the gradient can be found from the direction vector
- In 3D a numerical value for the direction cannot be found, it is given as a vector
As a could be the position vector of any point on the line and b could be any scalar multiple of the direction vector there are infinite vector equations for a single line
Given any two points on a line with position vectors a and b the displacement vector can be written as b - a
- So the formula r = a + λ(b - a) can be used to find the vector equation of the line
- This is not given in the formula booklet

How do I determine whether a point lies on a line?

Given the equation of a line $r = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}}) + λ (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$ the point c with position vector $(\binom{c_{1}}{\begin{matrix} c_{2} \\ c_{3} \end{matrix}})$ is on the line if there exists a value of $λ$ such that

- $(\binom{c_{1}}{\begin{matrix} c_{2} \\ c_{3} \end{matrix}}) = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}}) + λ (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$
- This means that there exists a single value of $λ$ that satisfies the three equations:
  - $c_{1} = a_{1} + λ b_{1}$
  - $c_{2} = a_{2} + λ b_{2}$
  - $c_{3} = a_{3} + λ b_{3}$
A GDC can be used to solve this system of linear equations for
- The point only lies on the line if a single value of $λ$ exists for all three equations
Solve one of the equations first to find a value of $λ$ that satisfies the first equation and then check that this value also satisfies the other two equations
If the value of $λ$ does not satisfy all three equations, then the point c does not lie on the line

Examiner Tip

Remember that the vector equation of a line can take many different forms
- This means that the answer you derive might look different from the answer in a mark scheme
You can choose whether to write your vector equations of lines using unit vectors or as column vectors
- Use the form that you prefer, however column vectors is generally easier to work with

Worked example

Find a vector equation of a straight line through the points with position vectors a = 4i – 5k and b = 3i - 3k

Determine whether the point C with coordinate (2, 0, -1) lies on this line.

3-10-1-ib-aa-hl-vector-equation-of-a-line-we-b

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Equation of a Line in Parametric Form

How do I find the vector equation of a line in parametric form?

By considering the three separate components of a vector in the x, y and z directions it is possible to write the vector equation of a line as three separate equations
- Letting $r = (\binom{x}{\begin{matrix} y \\ z \end{matrix}})$ then $r = a + λ b$ becomes
- $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{x_{0}}{\begin{matrix} y_{0} \\ z_{0} \end{matrix}}) + λ (\binom{l}{\begin{matrix} m \\ n \end{matrix}})$
  - Where $(\binom{x_{0}}{\begin{matrix} y_{0} \\ z_{0} \end{matrix}})$ is a position vector and $(\binom{l}{\begin{matrix} m \\ n \end{matrix}})$ is a direction vector
- This vector equation can then be split into its three separate component forms:
  - $x = x_{0} + λ l$
  - $y = y_{0} + λ m$
  - $z = z_{0} + λ n$
- These are given in the formula booklet

Worked example

Write the parametric form of the equation of the line which passes through the point (-2, 1, 0) with direction vector $(\binom{3}{\begin{matrix} 1 \\ - 4 \end{matrix}})$ .

3-10-1-ib-aa-hl-parametric-we

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Equation of a Line in Cartesian Form

The Cartesian equation of a line can be found from the vector equation of a line by

Finding the vector equation of the line in parametric form
Eliminating $λ$ from the parametric equations

$λ$ can be eliminated by making it the subject of each of the parametric equations
For example: $x = x_{0} + λ l$ gives $λ = \frac{x - x_{0}}{l}$

In 2D the cartesian equation of a line is a regular equation of a straight line simply given in the form

$y = m x + c$
$a x + b y + d = 0$
$\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}$ by rearranging $y - y_{1} = m (x - x_{1})$

In 3D the cartesian equation of a line also includes z and is given in the form

$\frac{x - x_{0}}{l} = \frac{y - y_{0}}{m} = \frac{z - z_{0}}{n}$
where $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{x_{0}}{\begin{matrix} y_{0} \\ z_{0} \end{matrix}}) + λ (\binom{l}{\begin{matrix} m \\ n \end{matrix}})$
This is given in the formula booklet

If one of your variables does not depend on $λ$ then this part can be written as a separate equation

For example: $m = 0 \Rightarrow y = y_{0}$ gives $\frac{x - x_{0}}{l} = \frac{z - z_{0}}{n}, y = y_{0}$

How do I find the vector equation of a line given the Cartesian form?

If you are given the Cartesian equation of a line in the form
- $\frac{x - x_{0}}{l} = \frac{y - y_{0}}{m} = \frac{z - z_{0}}{n}$
A vector equation of the line can be found by
- STEP 1: Set each part of the equation equal to $λ$ individually
- STEP 2: Rearrange each of these three equations (or two if working in 2D) to make x, y, and z the subjects
  - This will give you the three parametric equations
  - $x = x_{0} + λ l$
  - $y = y_{0} + λ m$
  - $z = z_{0} + λ n$
- STEP 3: Write this in the vector form $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{x_{0}}{\begin{matrix} y_{0} \\ z_{0} \end{matrix}}) + λ (\binom{l}{\begin{matrix} m \\ n \end{matrix}})$
- STEP 4: Set r to equal $(\binom{x}{\begin{matrix} y \\ z \end{matrix}})$
If one part of the cartesian equation is given separately and is not in terms of $λ$ then the corresponding component in the direction vector is equal to zero

Worked example

A line has the vector equation $r = (\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}) + λ (\begin{matrix} 4 \\ - 2 \\ 1 \end{matrix})$ . Find the Cartesian equation of the line.

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