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

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

    • $y=mx+c$

    • 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 $\mathbf{i},\mathbf{j},\mathbf{k}$ reference unit vectors or as column vectors 

• It follows that in a vector equation of a line either form can be employed – for example,

 $\mathbf{r}=3\mathbf{i}+\mathbf{j}-7\mathbf{k}+t\left(\mathbf{i}-2\mathbf{j}\right)$  and  $\mathbf{r}=\left(\begin{array}{c}3\\ 1\\ -7\end{array}\right)+t\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$   

                             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

  • $x-2y=5$ and $3x-6y=15$

• For vector equations this is not true – two equations might look different but still represent the same line:

  • $\mathbf{r}=\left(\begin{array}{c}5\\ 0\end{array}\right)+t\left(\begin{array}{c}2\\ 1\end{array}\right)$ and $\mathbf{r}=\left(\begin{array}{c}1\\ -2\end{array}\right)+t\left(\begin{array}{c}-2\\ -1\end{array}\right)$

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 $\mathit{r}= \left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)$ then $\mathit{r}=\mathit{a}+\lambda \mathit{b}$ becomes

  • $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{a_1}{\begin{array}{c}{a}_2\\ a_3\end{array}}\right)+\lambda \left(\genfrac{}{}{0pt}{}{b_1}{\begin{array}{c}{b}_2\\ b_3\end{array}}\right)$

    • Where $\left(\genfrac{}{}{0pt}{}{a_1}{\begin{array}{c}{a}_2\\ a_3\end{array}}\right)$ is a position vector and $\left(\genfrac{}{}{0pt}{}{b_1}{\begin{array}{c}{b}_2\\ b_3\end{array}}\right)$ is a direction vector

  • This vector equation can then be split into its three separate component forms:

    • $x= a_1+ \lambda b_1$

    • $y= a_2+ \lambda b_2$

    • $z= a_3+ \lambda b_3$

Equation of a line in Cartesian form

What is the Cartesian equation of a line in 3D?

• 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 $\lambda$ from the parametric equations

    • $\lambda$ can be eliminated by making it the subject of each of the parametric equations

    • For example: $x= x_0+ \lambda l$gives $\lambda = \frac{ x- x_0}{l}$

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

  •  $y=mx+c$

  • $ax+by+d=0$

  • $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$ by rearranging $y-y_1=m\left(x-x_1\right)$

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

  • $\frac{x- a_1}{b_1}= \frac{y- a_2}{b_2}= \frac{z- a_3}{b_3}(=\lambda )$

  • where $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{a_1}{\begin{array}{c}{a}_2\\ a_3\end{array}}\right)+\lambda \left(\genfrac{}{}{0pt}{}{b_1}{\begin{array}{c}{b}_2\\ b_3\end{array}}\right)$

  • This is given in the formula booklet

• If one of your variables does not depend on $\lambda$ 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

  • $\frac{x- a_1}{b_1}= \frac{y- a_2}{b_2}= \frac{z- a_3}{b_3}(=\lambda )$

• A vector equation of the line can be found by

  • STEP 1: Set each part of the equation equal to $\lambda$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

    • $x= a_1+ \lambda b_1$

    • $y= a_2+ \lambda b_2$

    • $z=a_3+ \lambda b_3$

  • STEP 3: Write this in the vector form $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{a_1}{\begin{array}{c}{a}_2\\ a_3\end{array}}\right)+\lambda \left(\genfrac{}{}{0pt}{}{b_1}{\begin{array}{c}{b}_2\\ b_3\end{array}}\right)$

  • STEP 4: Set r  to equal $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)$

• If one part of the cartesian equation is given separately and is not in terms of $\lambda$ then the corresponding component in the direction vector is equal to zero