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Point of crossing point of a line with a plane

−→
s
z
(l)
−→
n

(p)

y
. . . . . . . . .
0

x

Figure 8.15 – Perpendicularity of a line and a plane

8.10. Point of crossing point of a line with a plane

Suppose we have plane (p), given by general equation and line l, given by parametric equation:


 x = x 0 + mt,
y = y 0 + nt,

z = z 0 + pt.
Let’s formulate the task to find a point of their crossing. This point’s coordinates have to satisfy
both equations. That’s why let’s consider the system with four equations and four variables:


 x = x 0 + mt,


y = y 0 + nt,
 z = z 0 + pt,


Ax + By + Cz + D = 0.
Having substituted x, y, z from the first three equations into the fourth and regrouped terms,
we will get: (Am + Bn + Cp) · t = −(Ax 0 + By 0 + Cz 0 + D).When solving, following three
cases are possible:

a) Am + Bn + Cp ̸= 0, that is line (l) and plane (p) are not parallel. Then the equation has
Ax 0 +By 0 +Cz 0 +D
the only solution: t = − . Having substituted the found value of t into the
Am+Bn+Cp
first three equations, we will obtain x, y, z — coordinates of the crossing point between
the line and the plane.

b) Am + Bn + Cp = 0, but Ax 0 + By 0 + Cz 0 + D ̸= 0, i.e. l is parallel to (p), but point
M 0 (x 0 , y 0 , z 0 ) doesn’t belong to plane. Then the equation doesn’t have any solution, that
is the line doesn’t cross the plane.

c) Am + Bn + Cp = 0 and Ax 0 + By 0 + Cz 0 + D = 0, i.e. l is parallel to (p) and one
of line’s points M 0 (x 0 , y 0 , z 0 )belongs to the plane. It means, that the line itself belongs
to the plane. In this case the equation has many solutions, i.e. every point is the point of
crossing.

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