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Equations of the straight line in space

−→ −→ −→
n 2 (A 2 , B 2 , C 2 ). Therefore, the vector product of vectors n 1 and n 2 can be
−→ −→
considered as the directed vector s : s = N 1 ×N 2 or the view of coordinates:
−→ −→ −→
i j k ( )

−→ B 1 C 1 C 1 A 1 A 1 C 1
;
;
= .

s = A 1 B 1 C 1
B 2 C 2 C 2 A 2 A 2 C 2

A 2 B 2 C 2
8.6.2. Equation of a line passing through two points

z (l)
M 2

− →
s
M 1

y
. . . . . . . . . .
−→ 0
s
x

Figure 8.10 – Line passing through two points

Let’s suppose in the system of coordinates Oxyz line l is given by means of two points
− →
M 1 (x 1 , y 1 , z 1 ) and M 2 (x 2 , y 2 , z 2 ) (fig. 8.10). In this case the directed vector s equals to vector
−−−−→
M 1 M 2 = (x 2 − x 1 , y 2 − y 1 , z 2 − z 1 ). Then the equation of the line, which passes through point
−−−−→
M with the directed vector M 1 M 2 , looks like:

x − x 1 y − y 1 z − z 1
= = . (8.17)
x 2 − x 1 y 2 − y 1 z 2 − z 1

This equation is called the equation of the line passing through two points.

8.6.3. Vector equation of a line. Parametric equation of a line
−→
Let’s suppose the line is given by point M 0 (x 0 , y 0 , z 0 ) and directed vector s = (m, n, p). Let’s
consider any point M(x, y, z) on the line. From (fig. 8.11), such equality is obvious:

−−→ −−−→ −−−→
OM = OM 0 + M 0 M. (8.18)

−→
Vector M 0 M on the line, regardless of the position of point M on the line, is colinear to vector
−−−→
−→ − →
s . That’s why M 0 M = t · s , where t is a scalar multiplier, which is called the parameter, it
can take any value, depending on the position of point M. Having marked points’ M 0 and M
− →
−→
radius-vector as r 0 and r , we’ll rewrite the equality (8.18) in such a way:
− → −→ − →
r = r 0 + t · s . (8.19)

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