
Coordinate
Geometry (Analytic Geometry) in Threedimensional Space 

Line
and Plane

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

The angle between line and plane






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

Example:
Through a line which is written as the intersection of two planes, P_{1}
::
x 
2y
+ 3z 
4 = 0 and
P_{2}
::
3x + y

z + 1 = 0, lay a plane which passes through the point
A(
1,
2,
1).

Determine the equation of the plane. 
Solution: Write the equation of a sheaf,
P_{1}
+ lP_{2} = 0
or (x  2y
+
3z 
4)
+ l(3x +
y 
z + 1) = 0. 
Plug
the point A
into the equation to determine l
such that the plane passes through the given point, so 
A(1,
2,
1)
=>
P_{1}
+ lP_{2} = 0
gives (1
 2 · 2 +
3 · 1 
4)
+ l[3 · (1) +
2 
1 + 1] = 0, l =

6. 
Therefore,
(x  2y
+
3z 
4) 
6 · (3x + y 
z + 1) = 0
or 
17x 
8y
+ 9z 
10 = 0 
is the plane passing through
the given line and the point A. 

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.


The angle between line and plane

The angle j
between a line and a plane is the angle subtended by the line and its orthogonal projection onto
the plane.

Since the normal vector
N =
Ai + Bj + Ck
of the plane forms with the direction vector s =
ai + bj + ck
of the line the angle y
= 90° 
j,
the angle j
between a line and a plane we calculate indirectly, that is


If a line is perpendicular to a
plane, its projection onto the plane is a point and
therefore its direction vector and the normal vector of the plane are collinear, i.e.,


If
a line is parallel to a plane, then

s
^ N
=>
s
· N = 0,
j =
0°.




Example:
Given is a line 

and
a plane x

3y + 2z 
8 = 0, find the angle between


the line
and the plane.

Solution: From the equations of the line and the plane,
s = 3i

j 
2k and N =
i 
3j + 2k
so that











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