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Analytic geometry in space

8.6.1. Canonical equation of a line

z (l)
M

−→
s
M 0

y
. . . . . . . . .
0

x

Figure 8.9 – Canonical equation of a line

To solve the task, some special equations of the line in space can be used. Let’s suppose
− →
−→
line l and non-zero vector s = (m, n, p), which is parallel to this line. Vector s is called the
directed vector of the line. Point M 0 (x 0 , y 0 , z 0 ) belongs to the line (fig. 8.9). In order to obtain
−→
the equation of the line with directed vector s , passing through point M 0 , let’s take any point
−−−→
M(x, y, z) and consider vector M 0 M = (x − x 0 , y − y 0 , z − z 0 ). It is obvious, that regardless
−→ −→
of the position of point M on the line, vectors M 0 M and s are colinear. Therefore, according
to the condition of colinearity:

x − x 0 y − y 0 z − z 0
= = . (8.16)
m n p

Equation (8.16) is called the canonical equation of the line.

−→
Remark 8.4. In case one of coordinates of vector s , for example, m = 0, i.e.
this vector is perpendicular to axis Ox, the canonical equation of the line looks like:
x−x 0 = y−y 0 = z−z 0 . So, the line, which is perpendicular to axis Ox, can be presented
0 n p
{
x − x 0 = 0,
−→
by means of two equations: y−y 0 z−z 0 In case two coordinates of vector s ,
n = p .
for example, m = 0 and n = 0, i.e. this vector is perpendicular to axes Ox and Oy (or
is parallel to axis Oz), the canonical equation of the line looks like: x−x 0 = y−y 0 = z−z 0 .
0 0 p
So, the line, which is perpendicular to axis Oz, can be presented by means of two
{
x = x 0 ,
equations: For example, equation x = y = z is the canonical equation of
y = y 0 . 0 0 1
axis Oz.

Remark 8.5. In order to pass from the general equation of the line to its canon-
ical one, it is necessary:
— to find any point M 0 (x 0 , y 0 , z 0 ), which belongs to the line. In order to do that,
it is sufficient to assume any variable and obtain another coordinates from
equation (8.15);
−→
— to find normal vector s = (m, n, p). Since line l is the line of crossing two planes
− →
(p 1 ) and (p 2 ), then this line is perpendicular to each vector n 1 (A 1 , B 1 , C 1 ) and

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