Parametric to non parametric conversion of a line in 3d I can't for the life of me figure out how to convert this parametric equation to a non parametric equation for a line in 3D. Our lecture notes didn't cover it and I feel like it should be simple but whenever I try to figure it out I end up with nothing. Can someone try to explain how this can be done? The question is:
Convert the parametric equation:
(x, y, z) = (2, −1, −2) + t(3, 1, 2)
to a nonparametric equation.
I know that you're probably meant to come up with the equations for x, y and z:
x= 2 + 3t
y = -1 + t
z = -2 + 2t
but I'm not sure what to do with these
Thanks in advance for any help.
 A: A non-parametric representation of a line in 3D won't be a single equation but rather a system of (two) equations in the Cartesian coordinates $x$, $y$ and $z$.
As the name suggests, you should try to eliminate the parameter by first solving for $t$:
$$\left\{ \begin{array}{rcl}
x &=& 2 + 3t \\
y &=& -1 + t \\
z &=& -2 + 2t
\end{array} \right. \; (t \in \mathbb{R}) \quad\Rightarrow\quad
\left\{ \begin{array}{rcl}
t &=& \frac{x-2}{3} \\
t &=& y+1 \\
t &=& \frac{z+2}{2}
\end{array} \right. \; (t \in \mathbb{R})$$
and now equating these three expressions for $t$.

In general (if $v_i \ne 0$ for $i=1,2,3$), you'd have:
$$\left\{ \begin{array}{rcl}
x &=& a_1 + v_1t \\
y &=& a_2 + v_2t \\
z &=& a_3 + v_3t
\end{array} \right. \; (t \in \mathbb{R}) \quad\Rightarrow\quad
\left\{ \begin{array}{rcl}
t &=& \frac{x-a_1}{v_1} \\
t &=& \frac{y-a_2}{v_2} \\
t &=& \frac{z-a_3}{v_3}
\end{array} \right. \; (t \in \mathbb{R})$$
This leads to the following standard form:
$$\color{blue}{\frac{x-a_1}{v_1} = \frac{y-a_2}{v_2} = \frac{z-a_3}{v_3}}$$
Note: this will not always be possible (one or more of the $v_i$'s might be $0$), but then you can still obtain a system of two equations by elimination of the parameter through substitution.
