When does a line in $\mathbb{R}^3$ intersect a space?

Question

Let $p$ be a line in $\mathbb{R}^3$ denoted as follows: $ p = { \begin{bmatrix} 3, 1, 2 \end{bmatrix} + t \begin{bmatrix} -1, 1,0 \end{bmatrix} : t \in \mathbb{R} } $

a) For which $a, b, c, d \in \mathbb{R}$ does the space with the general equation $ax + by + cz = d$ contain the line $p$? b) Find some system of 2 linear equations with 3 unknowns whose set of all solutions is equal to p.

Note: In part (a), all quads satisfying the condition must be described. In part (a), It is required that the equation $ax + by + cz = d$ to describe a space/plane, i.e. excluding the degenerative forms. In part b), on the other hand, it is sufficient to find only a one system of equations.

My idea: For it to work, we need to find the normal vector (a vector that is perpendicular to the direction vector), and then plug in the starting coordinates from the input. Is there a more practical way of solving this? How should I approach the (b)?

Answer 1

For part a), you want any point on the line $p$ to obey the equation of the plane. $$a(3-t)+b(1+t)+2c=d$$ We can rewrite this as $$3a+b+2c+t(-a+b)=d$$ The numbers $a,b,c,d$ can't depend on $t$, so the solution is to choose $b=a$ and $3a+b+2c=d$.

For part b), think what is the meaning of an equation with three variables: it's a plane. A system with two such equation means an intersection of two planes. So all you need is to choose two planes with the form in part a), such that the normals are not parallel/antiparallel. Any pair of such planes is a solution for your question

Answer 2

What you have is a ray-to-plane intersection algorithm.

1. General Case

   The ray is described by the parametric vector equation

   $${\bf p}(t) = {\bf o} + t\,{\bf e}$$

   and a point belongs to the plane when

   $$ {\bf n} \cdot {\bf p}(t) = d $$

   where ${\bf n}= \pmatrix{a\\b\\c}$ is the normal vector of the plane described by $a x +b y + c z = d$ and $\cdot$ is the vector dot product.

   The equation above expands to

   $$ ({\bf n}\cdot {\bf o}) + t\,({\bf n}\cdot{\bf e}) = d$$

   with solution

   $$ t = \frac{d - ({\bf n}\cdot {\bf o})}{({\bf n}\cdot{\bf e})} $$

   _{Note that anything inside a parenthesis is a scalar value, as well as italicized variables. Vectors are boldface variables.}

2. Infinite Solutions

   If you want the above to have infinite solutions, then you chose $d = {\bf n}\cdot{\bf o}$ to force the plane on the origin, and ${\bf n} \cdot {\bf e} = 0$ to force the line lie on the plane.

   So any arbitrary vector ${\bf u}$ can be used to construct the plane normal vector ${\bf n}={\bf e}\times{\bf u}$ and then $d$ is evaluated as shown above to move the plane on the ray origin. I think there is a restriction that ${\bf u}$ cannot be parallel to ${\bf o}$ or ${\bf e}$ or the above will fail.

   One way to construct ${\bf n}$ is to use $${\bf n}={\bf e}\times\left({\bf o}\times{\bf e}\right)$$ given that the ray origin and direction are not parallel to each other. The resulting plane must have $$d=\left\Vert {\bf o}\times{\bf e}\right\Vert ^{2}$$ where $\left\Vert \, \right\Vert$ is the Eucleadian norm.

   For example, with the numbers given above

   $$\small \begin{aligned}{\bf o} & =\begin{pmatrix}3\\ 2\\ 3 \end{pmatrix} & {\bf e} & =\begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix}\\ {\bf n} & =\begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix}\times\left(\begin{pmatrix}3\\ 2\\ 3 \end{pmatrix}\times\begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix}\right)=\begin{pmatrix}5\\ 5\\ 6 \end{pmatrix} & d & =\left\Vert \begin{pmatrix}3\\ 2\\ 3 \end{pmatrix}\times\begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix}\right\Vert ^{2}=43 \end{aligned}$$

   So the plane $5x + 5y + 6z = 43$ will always lie on the line.

3. No Solutions

   If you want no solutions then ${\bf n} \cdot {\bf e} = 0$ and $d \neq {\bf n}\cdot{\bf o}$ are your conditions. This is interpreted as the plane ray being parallel to the plane, but not coincidental.

4. One Solution

   As long as the denominator ${\bf n}\cdot {\bf e}$ is not zero, and the numerator $d - {\bf n}\cdot{\bf o}$ is not zero there will be one solution to the problem.