A line $l$ in $\mathbb{R}^3$ can be thought as the intersection of two planes and therefore can be represented by two simultaneous equations of planes of the form

(1)

Eliminating one of the variables $x$ , $y$ , $z$ from the pair of equations, one obtains an equation of another plane. Thus, if one first eliminates $x$ and then $y$ , one may obtain simultaneous equations of the form

(2)

Note that an ordered triple $(x_0,\,y_0,\,z_0)$ satisfies (1) if and only if it also satisfies (2), and hence the system of equations (2) represents the same line as well. Separately, both of the equations in (2) are equations of planes, of which the former is parallel to the $x$ -axis ($x$ can change independently of $y$ and $z$ ) and the latter is parallel to the $y$ -axis ($y$ can change independently of $z$ and $x$ ). The line itself is the intersection of these two planes, and it can be projected along the planes indicated by the individual equations in (2) onto the $yz$ -plane and the $zx$ -plane, respectively.

In a narrower meaning, the equation $y = mz+p$ represents the projection line of $l$ onto the $yz$ -plane and the equation $x = nz+q$ the projection line of $l$ onto the $zx$ -plane.

A line in space can be represented also without using planes. We can derive another way by starting with two vectors: one vector $\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called direction vector) $\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\vec{r}$ of an arbitrary point of the line can be expressed as

(3)

$$\begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} x_0\\y_0\\z_0 \end{pmatrix}+t \begin{pmatrix} a\\b\\c \end{pmatrix}\!,$$ and we obtain three scalar equations

(4)

Bibliography

1

L. LINDELÖF: Analyyttisen geometrian oppikirja. Kolmas painos. Suomalaisen Kirjallisuuden Seura, Helsinki (1924).