
Analytic Geometry (Coordinate Geometry) in Threedimensional Space 

Line
and Plane

How determine two planes of which, a given line is their
intersection line

Intersection point of a line and a plane

Points, Lines and planes relations in 3D space, examples






How determine two planes of which a given line is their
intersection line

Through a given line 

lay two planes each of which is orthogonal to one of the 

coordinate planes. 
That way, given line will be determined by any of
the following pairs of
equations, 

as the intersection line of the corresponding planes
(each of which is perpendicular to one of the three coordinate
planes). 

Example:
Given is a line 

find two planes of which given line is the intersection line. 

Solution: By eliminating the part of equation which contains
xcoordinates determined is a plane which
passes through the given line orthogonal to the yz
coordinate plane that is 

Therefore,
P_{1 }::
y 
18z + 4 = 0
with N_{1}
= j 
18k and
P_{2}
:: x 
11z + 3 = 0 with N_{2}
= i 
11k. 
The obtained result is correct if the vector product of the corresponding normal vectors,
N_{1}
and N_{2}
of the planes
equals the direction vector s
of the given line, so 

By calculating the coordinates of the intersection point, of the line of intersection represented by obtained
two planes, and the coordinate planes, by plugging successively, z = 0,
y = 0 and
x = 0 into their equations, we can prove that the result coincides with that in the
first example of this section, because it is the same line. 

Intersection point of a line and a plane

The point of intersection is a common point of a line and a plane. Therefore, coordinates of intersection must
satisfy both equations, of the line and the plane. 
The parametric equation of a line, 
x =
x_{0} + at,
y = y_{0}
+ bt and
z
= z_{0} + ct 
represents coordinates of any point of
the line expressed as the function of a variable parameter t
which makes
possible to determine any point of the line according to given condition. 
Therefore, by plugging these variable coordinates of a point of the line into
the given plane determine what value
must have the parameter t,
this point to be the common point of the line and the plane. 

Example:
Given is a line 

and a plane
4x 
13y + 23z 
45 = 0, find the


intersection
point of the line and the plane. 
Solution: Transition from the symmetric to the parametric form
of the line


by plugging these variable coordinates
into the given plane we will find the value of the parameter t
such that these
coordinates represent common point of the line and the plane, thus 
x =
t +
4, y
= 4t 
3 and z
= 4t 
2 => 4x 
13y + 23z 
45 = 0 
which
gives, 4 · (t +
4)  13 · (4t 
3) + 23 · (4t 
2) 
45 = 0 Þ
t = 1. 
Thus,
for t = 1
the point belongs to the line and the plane, so 
x =
t +
4 = 
1+ 4 = 3, y =
4t 
3 = 4 · 1 
3 = 1 and z =
4t 
2 = 4 · 1 
2 = 2. 
Therefore, the intersection point
A(3,
1,
2) is the point which is at the same time on the line and the plane. 

Points, Lines and planes relations
in 3D space, examples

Example:
Through the line 

lay
a plane to be parallel to the
yaxis, determine


the
equation of the plane. 
Solution: Write the line as the intersection of two planes to be able to form the sheaf of planes, therefore


P_{1} and
P_{2}
are the planes of which the given line is intersection and which are perpendicular to the coordinate
planes,
xy
and yz
respectively.

Form the equation of a sheaf to determine the parameter
l according to
the given condition 
P_{1}
+ lP_{2} = 0
or (5x

3y 
1)
+ l(4y

5z + 3) = 0 
then,
5x
+ (
3
+ 4l) ·
y 
5lz +
(3l

1) = 0
and N_{s}
= 5i + (
3
+ 4l)
j 
5lk 
is the equation of planes that all intersect at the given line
and the normal vectors of which are perpendicular to the direction vector
s
of the line.

As the plane to be found should be parallel with the
yaxis, then the normal vector
of the sheaf N_{s}
must be perpendicular to the unit vector
j of the
yaxis that is, the scalar product of the normal vector of the plane
and the unit vector
j
must be zero, thus

N_{s}
^
j
=>
N_{s}
·
j = 0.

Since a scalar product of two vectors expressed by their coordinates is

a · b = a_{x}b_{x} +
a_{y}b_{y} + a_{z}b_{z}
and since j
= 0 · i + 1 ·
j + 0 · k

then,
N_{s}
·
j = 5 · 0 + (
3
+ 4l) · 1 + (5l)
· 0 = 0
=>
l
= 3/4.

By plugging l
= 3/4 into (5x

3y 
1)
+ l(4y

5z + 3) = 0 obtained
is 4x

3z + 1 = 0

the plane through the given line that
is parallel to the
yaxis.


Example:
Through the line 

(the
same as in the previous example) lay a plane


orthogonal to the
plane
4x

2y + z 
4 = 0 and determine its equation.

Solution: Write given line as the intersection of two planes (like in
the above example) to form the equation

of a
sheaf,
5x
+ (
3
+ 4l) ·
y 
5lz +
(3l

1) = 0
and N_{s}
= 5i + (
3
+ 4l)
j 
5lk

that is the plane parameter l
of which should be determined so the plane to be perpendicular to the given plane.

In this case, the normal vector of the sheaf
N_{s}
and the normal vector of the given plane N
= 4i 
2 j + k, are
orthogonal to each other, therefore

N_{s}
^ N
=>
N_{s}
· N = 0 that
is, 5 · 4 + (
3
+ 4l) · (2)
+ (5l) · 1 = 0
=> l =
2.

Then, by plugging l
= 2 into 5x
+ (
3
+ 4l) ·
y 
5lz +
(3l

1) = 0

obtained is 5x
+ 5y 
10z + 5 = 0
or x
+ y 
2z + 1 = 0

the equation of the plane which passes through the given line perpendicular to the given plane.









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