Finding points on a plane or line in $\mathbb{R}^3$

Question

I'm just getting familiar with analytical geometry and basically I have two probably very simple questions.

Question 1:

Let's assume I have some plane, i.e: $\pi: x - 2y + 4z - 8 = 0$

What should I do to find points on that plane?

My take on this: to find any point $P(x_{0}, y_{0}, z_{0}) \in \pi $ I can assume that two out of three $x_{0}, y_{0}, z_{0}$ are any constant values $\in \mathbb{R}$ and then find the value of third coordinate. For example let's assume: $x_{0} = 0, z_{0} = 8$, then:

$$0 - 2y_{0} + 32 - 8 = 0$$

$$y_{0} = 12$$

$$\Rightarrow P(0,12,8) \in \pi$$

Is that how should it be done?

Question 2:

Let's assume I have some line i.e.: $\displaystyle \frac{x}{2} = \frac{y - 1}{-2} = z$

I can directly see that point $P(0, 1, 0)$ will surely belong to that line. But let's say I want to find any other point(s) - how should I do it?

I suspect that I can switch the line equation into the parametric form, and then plug in for example value $5$ for parameter $t \in \mathbb{R}$.

Is this the way to go?

Answer 1

Your first idea is precisely what I would do (and have done before): on a plane, one has $2$ degrees of freedom of movement, corresponding to the fact that one has freedom in choosing values for any two of the variables$^\dagger$, and this will determine the third. Your second idea is also what I would do, wherein the line can be parametrized as:

$$\alpha(t) = \begin{bmatrix} 2t \\ -2t + 1 \\ t \end{bmatrix}$$

The trick, for anyone feeling like I pulled this out of a hat, is to simply let any one of the variables be the parametrizing variable $t$. Here, of course, we let $z=t$, and then, using the given relations, solved for the other variables so that we obtained expressions $x = f(t)$ and $y = g(t)$; these functions $f$ and $g$ serve as the $x$ and $y$ components of $\alpha(t)$ respectively. Finally, note that parametrizations of curves are never unique, so don't panic if a purported parametrization and yours don't perfectly visually compare.

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P.S. if you have two points $P_1$ and $P_2$ on any line in $\mathbb{R}^n$, one can parametrize this line as $\alpha(t) = (1-t)P_1 + tP_2$. Indeed, check that $\alpha(0) = P_1$ and $\alpha(1) = P_2$, and you can see $\alpha$ is linear because, for instance, it has constant derivative. Parametrizing in this way has added utility: one can tell where points are relative to $P_1$ and $P_2$. Note that points corresponding to $t \leq 0$ lie "before" $P_1$; points corresponding to $0 < t < 1$ lie between $P_1$ and $P_2$, and so forth.

I only mention this last bit as this discussion reminded me of a coding project I'm working on where this idea saved me so much time (needed to implement a subroutine that checks where a line intersects another relative to two points on the line being intersected). Hopefully in my sharing, it serves someone equally well.

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$^\dagger$This "...any two of the variables..." bit isn't quite true in every case. For example, if the plane was horizontal with $z=1$ fixed (or, in general, when the plane is parallel to one of the planes spanned by any two of the coordinate axes), clearly one can't choose an arbitrary value for $z$ (or whichever variable is fixed). But one nevertheless still has two degrees of choice freedom: $x$ and $y$ can be anything.

Answer 2

I am not sure if this can be considered a complete answer, or if this is what you want, but I hope this is at least helpful:

Question 1: So, a point in that plane is a point satisfying the equation, and the way you found it is the right idea. For instance, say you fix $x,y$. Then, solve for $z$. From $x-2y+4z-8=0$. You get $z=\frac{8-x+2y}{4}$. So, any point in the plane in general will have the form $\left(x,y,\frac{8-x+2y}{4}\right)$. In particular if you choose $x=0,y=12$ you get the point you found $(0,12,8)$.

Question 2: You have $\frac{x}{2}=\frac{y-1}{-2}=z$. Let's write $\frac{x}{2}=\frac{y-1}{-2}=z=t$ And find each variable in terms of the parameter $t$. We have $x=2t$, $y=-2t+1$, $z=t$. So, any point of this line is of the form $(2t,-2t+1,t)$. In particular for $t=0$, you get the point you found $(0,1,0)$.

Answer 3

Another way is find just one point say v=(0,0,2) putting x and y equal to 0. Then solve x-2y+4z=0 to get {w=(a,b,(2b-a)/4)}. Your required soln set is {v+w}