How exactly one can show parametric equations in a graph I'm trying to understand parametric equations, it looks simple but it tricks me.
$x = 3-2t\\y=1+t\\z=2+3t$
How do I plot it ? I mean, where's $t$ in the graph?
 A: If you view $t$ as a parameter you don't plot it.  If you just had the first two equations, you would have a straight line in $\Bbb R^2$.  With the third equation you have a straight line in $\Bbb R^3$.  You can plot that, but need to do it in perspective and it may not show up well.  This would be a line through $(3,1,2)$ in the direction of $(-2,1,3)$
A: Parametric equations are usually used to plot curves in the plane (or, in this case, three-dimensional space) with some parameter. Here, $t$ is that parameter, and is used to trace out the curve (this is actually a line in which $(3, 1, 2)$ lies on the line and moves in the direction of the vector $(-2, 1, 3)$. You would still plot this in $\mathbb{R}^3$ noting that movement depends on $t$.
A: You can ask Wolfram Alpha to "plot parametric curve (3-2t, 1+t, 2-3t) for -10<=t<=10" [link]. (The graph is a straight line that goes from corner to corner of a box in the output, so it's not all that enlightening to look at for understanding parametric equations.)
A: Um here you would indeed have to plot it in $\Bbb R^4$ to fully plot it. But you could also eliminate the $t$ to get an equation involving $x,y,z$. Then you can eliminate another. Also as another answer suggests when plotting a parameterized curve you do not typically plot the parameters.
But on to parameterized curve. Another way to say is $$l(t)=\langle 3-2t, 1+t, 2+3t \rangle$$
Or $$=\langle -2,1,3 \rangle t + \langle 3,1,2 \rangle$$
$$=(3-2t)\mathbf x+(1+t)\mathbf y + (2+3t)\mathbf z$$
Or the image of the map $$\{(x,y,z) \in \Bbb R^3 \ : \ y=\frac{x-1}{2} \cap y=\frac{z+1}{3}\}$$ which is what you want to plot. To be technical if you said you want to plot the graph then that would be in $\Bbb R^4$.
Another way to think of it is that the curve starts at $(3,1,2)$ and has constant velocity $$\frac{d}{dt}l(t)=\mathbf v=\langle -2,1,3 \rangle$$ and the plot you want is the projection unto $\Bbb R^3$.
A: Normally when we make such a parameterization we do not plot $t$. We simply vary $t$ and determine the point $(X(t),~ Y(t),~ Z(t))$. A circle is a circle, but a rational parameterization of the circle gives us a way to find rational points directly, which is rather non-trivial given the full quadratic.
