I want to know if my reasoning can hold in general, of if there are caveats or if it's simply a load of nonsenses.
Say I have to determine the position of a line with respect to a plane. Say I have the equation of a plane in cartesian form, for example:
$$\sigma: x + y – z + 1 = 0$$
and a parametric equation of a line, like
$$\begin{cases}
x = 2t\\
y = t+1 \\
z = -t
\end{cases}
$$
Now the question: instead of using matrices representation and Rouché-Capelli's Theorem, could I just say that
-
If by substituting the parametric equations into the equation of a plane I get $0 = 0$ or $2 = 2$ and so on, then the line belongs to the plane (always true system/identity);
-
If by substituting the parametric equations into the equation of a plane I get something like $1 = 0$ or $3 = -1$ and so on, then the line and the plane are parallel (impossible system);
-
If by substituting the parametric equations into the equation of a plane I get $3t = 0$ or $t = -5$ and so on, then the line and the plane are incident (the line passes through the plane, unique solution system).
So, can I, or is this method a failure?
I actually do not see any reason for this to be a bad way to solve such kind of problems, though. The only question is: is there a way to understand the difference between a line that passes through a plane perpendicularly or not perpendicularly?
Best Answer
In short: Yes, you're correct.
In more detail:
$\newcommand{\R}{\mathbb{R}}$ Suppose your plane is given in Hesse normal form by $\sigma=\{\vec{x}\in\R^3 \, \colon\, \vec{x}\cdot \vec{n} = d \}$ where $d$ is the distance to the origin and $n$ is the (outward-pointing) unit normal vector of $\sigma$. By "outward-pointing" I mean that $\vec{x}\cdot\vec{n} = d \geq 0$ for any point $\vec{x}\in \sigma$. The line is given by $L = \{\vec{x}\in\R^3 \, \colon\, \vec{x} = \vec{p} + t\vec{v}, t\in\R\}$ where $p$ is any point on $L$ and $v$ is the direction of $L$.
We want to find out if there is an intersection of $\sigma$ and $L$. So we plug the equation for the line into the equation for the plane, since any point in the intersection has to satisfy both equations and we thus get $$ \vec{p}\cdot \vec{n} + t (\vec{v}\cdot \vec{n}) = d. $$ First, note that the vector $(\vec{p}\cdot \vec{n}) \vec{n}$ is the projection of the point $\vec{p}$ onto the line through the origin perpendicular to $\sigma$. So the number $\vec{p}\cdot \vec{n}$ gives you the (signed) distance of this projection to the origin. Second, the number $\vec{v}\cdot \vec{n} = |\vec{v}| \cos\angle(\vec{v},\vec{n})$ tells you if plane and line are parallel (which includes the case of plane containing line) or transversal.
Case 1: $\vec{v}$ and $\vec{n}$ are not perpendicular, i.e. $\vec{v}\cdot \vec{n}\neq 0$. In other words, the plane and line are transversal (so there is an intersection point) Then we find for the parameter $$ t = \tfrac{d - \vec{p}\cdot \vec{n}}{\vec{v}\cdot \vec{n}} $$ and we also find the intersection point $\vec{z}$ by plugging this parameter back into the line equation $$ \vec{z} = \vec{p} + \tfrac{d - \vec{p}\cdot \vec{n}}{\vec{v}\cdot \vec{n}}\vec{v}. $$
Case 2: $\vec{v}$ and $\vec{n}$ are perpendicular, i.e. $\vec{v}\cdot \vec{n}=0$. In other words, the plane and line are parallel and we need to decide if $L$ is contained in $\sigma$ or not. Then the equation above reduces to $\vec{p}\cdot \vec{n} = d$. If this equation holds that means any point $\vec{x}$ on the line fulfils the equation of the plane and thus $L$ is contained in $\sigma$. Conversely, if this equation does not hold then no point $\vec{x}$ on the line lies in the plane and they are parallel with positive distance. The distance is given by $|d-\vec{p}\cdot \vec{n}|$ if $\vec{p}\cdot \vec{n}\geq0$ or $d+\vec{p}\cdot \vec{n}$ if $\vec{p}\cdot \vec{n}<0$.
For your example, you may check this with:
$d=\tfrac{1}{\sqrt{3}}$, $\vec{n} = \tfrac{1}{\sqrt{3}} \begin{pmatrix}-1\\-1\\1\end{pmatrix}$, $\vec{p} = \begin{pmatrix}0\\1\\0\end{pmatrix}$ and $\vec{v} = \begin{pmatrix}2\\1\\-1\end{pmatrix}$.