Converting parametric line to intersection of planes line

Question

I know how to convert "plane" form of equation of a line to parametric/canonical form. Basically I know all combinations, but this one:

I have an equation of a line in parametric form:

$$\begin{cases} x = -7t + 5 \\ y = 2t - 7 \\ z = -5t + 2 \quad \quad \quad t \in \mathbb{R}\end{cases}$$

and I need to find the very same line, but in "plane" form.

By "plane" form I mean such form:

$$\begin{cases} A_{1}x+B_{1}y + C_{1}y + D_{1} = 0 \\ A_{2}x + B_{2}y + C_{2}y + D_{2} = 0\end{cases}$$

Where $A_{i}, B_{i}, C_{i}$ is corresponding plane's normal vector $\vec{n_{i}}=[A_{i},B_{i},C_{i}]$

Side information:

$$\vec{n_{1}} \times \vec{n_{2}} = [-7, 2, -5]$$

The intersection of both planes is my line.

Answer 1

Hint: With $$t=\frac{y+7}{2}$$ we get $$x=-\frac{7}{2}(y+7)+5$$ $$z=-\frac{5}{2}(y+7)+2$$

Answer 2

These two representations of a line in $\mathbb R^3$ are dual to each other, and this duality is quite transparent when you use homogeneous coordinates.

The parametric equations in your question can be generalized to the join of two points $\mathbf p$ and $\mathbf q$, that is, the set of nonzero linear combinations $\lambda\mathbf p + \mu\mathbf q$—the span of $\mathbf p$ and $\mathbf q$ in linear-algebraic terms. We can represent this linear combination in matrix form as $$\begin{bmatrix}\lambda&\mu\end{bmatrix} \begin{bmatrix}\mathbf p^T\\\mathbf q^T\end{bmatrix},$$ but as the line consists of all such linear combinations, we only need the right-hand matrix to represent the line. This is sometimes called its span representation. The parameterization in your question corresponds to fixing $\lambda = 1$ and choosing for $\mathbf q$ the line’s point at infinity (which is the counterpart of the inhomogeneous Cartesian direction vector of the line).

On the other hand, a line can also be represented as the meet of two planes—their intersection. The Cartesian plane equation $ax+by+cz+d=0$ can be written in vector form as $(a,b,c,d)(x,y,z,1)^T=0$, so the plane can be represented by the vector $\mathbf\pi=(a,b,c,d)^T$ of coefficients in its equation. The set of points on the plane is therefore the null space of the matrix $\mathbf\pi^T$. If we have two planes $\mathbf\pi_1$ and $\mathbf\pi_2$, their intersection is of course the solution set of the system of equations $\mathbf\pi_1^T\mathbf x=0$, $\mathbf\pi_2^T\mathbf x=0$, i.e., it is the null space of the matrix $$\begin{bmatrix}\mathbf\pi_1^T\\\mathbf\pi_2^T\end{bmatrix}.$$ This representation is sometimes called the null space representation of the line.

Since both of the points in the span representation above lie on both planes in the null space representation, we have the relation $$\begin{bmatrix}\mathbf\pi_1^T\\\mathbf\pi_2^T\end{bmatrix} \begin{bmatrix}\mathbf p & \mathbf q\end{bmatrix} = 0.$$ This identity gives us a way to convert between these representations: the rows of each one span the null space of the other.

Using your example, the parametric equations correspond to the matrix $$\begin{bmatrix}5&-7&2&1\\-7&2&-5&0\end{bmatrix}.$$ A basis for its null space can be found in the usual ways, such as row-reduction to $$\begin{bmatrix}1&0&\frac{31}{39}&-\frac2{39}\\0&1&\frac{11}{39}&-\frac7{39}\end{bmatrix}$$ from which we read the null space representation $$\begin{bmatrix}31&11&-39&0\\-2&-7&0&-39\end{bmatrix}.$$ This corresponds to the system $$31x+11y-39z=0 \\ -2x-7y-39=0.$$