Equations of Lines in 3D (Edexcel A Level Further Maths: Core Pure)

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Amber

Last updated

12 December 2023

Equation of a Line in Vector Form

How do I find the vector equation of a line?

You need to know:
- The position vector of one point on the line
- A direction vector of the line (or the position vector of another point)

There are two formulas for getting a vector equation of a line:
- r = a + t (b - a)
  - use this formula when you know the position vectors a and b of two points on the line
- r = a + t d
  - use this formula when you know the position vector a of a point on the line and a direction vector d
- Both forms could be compared to the Cartesian equation of a 2D line
  - $y = m x + c$
  - The point on the line a is similar to the “+c” part
  - The direction vector d or b – a is similar to the “m” part
The vector equation of a line shown above can be applied equally well to vectors in 2 dimensions and to vectors in 3 dimensions
Recall that vectors may be written using $i, j, k$ reference unit vectors or as column vectors
It follows that in a vector equation of a line either form can be employed – for example,

$r = 3 i + j - 7 k + t (i - 2 j)$ and $r = (\begin{matrix} 3 \\ 1 \\ - 7 \end{matrix}) + t (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix})$

show the same equation written using the two different forms

How do I determine if a point is on a line?

Each different point on the line corresponds to a different value of t
- For example: if an equation for a line is r = 3i + 2j - k + t (i + 2j)
  - the point with coordinates (2, 0, -1) is on the line and corresponds to t = -1
- However we know that the point with coordinates (-7, 5, 0) is not on this line
  - No value of t could make the k component 0

Can two different equations represent the same line?

Why do we say a direction vector and not the direction vector? Because the magnitude of the vector doesn’t matter; only the direction is important
- we can multiply any direction vector by a (non-zero) constant and this wouldn’t change the direction
Therefore there are an infinite number of options for a (a point on the line) and an infinite number of options for the direction vector
For Cartesian equations – two equations will represent the same line only if they are multiples of each other
- $x - 2 y = 5$ and $3 x - 6 y = 15$
For vector equations this is not true – two equations might look different but still represent the same line:
- $r = (\begin{matrix} 5 \\ 0 \end{matrix}) + t (\begin{matrix} 2 \\ 1 \end{matrix})$ and $r = (\begin{matrix} 1 \\ - 2 \end{matrix}) + t (\begin{matrix} - 2 \\ - 1 \end{matrix})$

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 reference unit vectors or as column vectors – use the form that you prefer!
If, for example, an exam question uses column vectors, then it is usual to leave the answer in column vectors, but it isn’t essential to do so - you’ll still get the marks!

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.

al-fm-6-1-1-vector-equation-of-line-we-solution-b

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
- $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}}) + λ (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$
  - Where $(\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}})$ is a position vector and $(\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$ is a direction vector
- This vector equation can then be split into its three separate component forms:
  - $x = a_{1} + λ b_{1}$
  - $y = a_{2} + λ b_{2}$
  - $z = a_{3} + λ b_{3}$

Worked example

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

al-fm-6-1-1-parametric-equation-of-line-we-solution

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

$\frac{x - a_{1}}{b_{1}} = \frac{y - a_{2}}{b_{2}} = \frac{z - a_{3}}{b_{3}} (= λ)$
where $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}}) + λ (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$
This 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: $b_{2} = 0 \Rightarrow y = a_{2}$ gives $\frac{x - a_{1}}{b_{1}} = \frac{z - a_{3}}{b_{3}}, y = a_{2}$

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
- $\frac{x - a_{1}}{b_{1}} = \frac{y - a_{2}}{b_{2}} = \frac{z - a_{3}}{b_{3}} (= λ)$
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
  - $x = a_{1} + λ b_{1}$
  - $y = a_{2} + λ b_{2}$
  - $z = a_{3} + λ b_{3}$
- STEP 3: Write this in the vector form $(\binom{x}{\begin{matrix} y \\ z \end{matrix}}) = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}}) + λ (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$
- 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.

al-fm-6-1-1-cartesian-equation-of-line-we-solution-a