Parametric form of a line in 3D

Question

I always get confused when i see an equation of line Given in a shape like r :2x=y=2-2z How can i write it in a parametric form Like r:

$$ \left\{ \begin{array}{c} x=a+lt \\ Y=a’+mt \\ Z=a”+nt \end{array} t\in R \right. $$ For example to define the direction vector...

Answer 1

As these are always linear equations, you want to find a particular solution and a solution to the homogenous equation. What does this mean?

To find a particular solution, you want to solve for a single point, that fulfills your equations. E.g. $2x=y=2-2z$ has solution $(x,y,z)=(0, 0, 1)$.

Next you want to find a solution to the homogenous equation. This means removing all constants from your equation and solving that. E.g. $2x=y=2-2z$ becomes $2x=y=-2z$. Here you want to find all solution. For a line there will only be one solution $(x, y, z)$ and its multiples. E.g. $2x=y=-2z$ has solution $k\cdot (1, 2, -1), k\in \mathbb R$.

Adding a homogenous solution to your particular solution will keep the equation valid. This should help you convince yourself, that all solution to you equation are of the form: a particular solution + all homogenous solutions. E.g. $2x=y=2-2z$ has solutions: $\{(0, 0, 1)+k\cdot (1, 2, -1), k\in \mathbb R\}$

Answer 2

$2X = Y = 2 - 2Z$ means also that $Y = 2X$ and $Z = 1 - X$, just solve for $Y$ and for $Z$ separately. One possible parametrization of the line could be given like this:

$$\begin{bmatrix}X \\ Y\\Z\end{bmatrix} = \begin{bmatrix}X\\2X\\1-X\end{bmatrix} = \begin{bmatrix}t\\2t\\1-t\end{bmatrix}= \begin{bmatrix}0+1\cdot t\\0+2\cdot t\\1+(-1)\cdot t\end{bmatrix}= \begin{bmatrix}a+l\cdot t\\a'+m\cdot t\\a''+n\cdot t\end{bmatrix}, \ \ \ t\in\mathbb{R}$$

You take $t=X$. The last or third component $Z$ of this 3D vector is equal to $1-X$, because from the given equation $2X = Y = 2 - 2Z$ we can see that $2X = 2 - 2Z$. Solving this last equation gives $X = 1-Z$ or $Z = 1-X$.

The equation $2X = Y = 2 - 2Z$ is just a compressed form of expressing three equations at once, which are: $Y = 2X$ and $Y = 2 - 2Z$ and $2X = 2 - 2Z$.

Answer 3

Since you have three equal values, set all of them equal to $t$, in your case, $$2x=t \\ y=t \\ 2-2z=t$$ and then solve these three equations for $x$, $y$ and $z$, respectively: $$\begin{align} x &= \frac12t \\ y &= t \\ z &= -\frac12t + 1. \end{align}$$ Incidentally, to convert from parametric to symmetric form, you basically reverse this process by solving each individual equation for $t$.