
Coordinate
Geometry (Analytic Geometry) in Threedimensional Space 

Plane
in a threedimensional (3D) coordinate system 
Equations of
a plane in a coordinate space

The equation of a
plane in a 3D coordinate system 
The Hessian normal form of the
equation of a plane

The intercept
form of the equation of a plane 
A plane in a 3D
coordinate system, examples 
Equation of the
plane through three points example 






Equations of a plane in a coordinate space

The equation of a
plane in
a
3D coordinate system 
A plane in space is defined by three points (which don’t all lie on the same line) or by a point and a normal
vector to the plane. 
Then, the scalar product of the vector
P_{1}P
= r 
r_{1}, drawn from the
given point P_{1}(x_{1},
y_{1}, z_{1}) of the plane to any point
P(x,
y, z)
of the plane, and the normal vector N
= Ai + Bj
+ Ck, is zero, that is 


which is called the vector equation of a
plane. 

Or,




Therefore, 
(x_{
} 
x_{1})
· A
+ (y_{ }
y_{1})
· B
+ (z_{ }
z_{1})
· C
= 0 
giving 
A
· x
+ B · y
+ C · z
+ D = 0 


where, D
= _{
}
(Ax_{1}+
By_{1}+
Cz_{1}) 
the general equation of a plane in 3D
space. 
If plane passes through the origin
O
of a coordinate system then its coordinates, x =
0, y
= 0, and z
= 0 plugged into the equation of the plane, give 
A
· 0
+ B · 0
+ C · 0
+ D = 0 => D = 0. 
Thus, the condition that plane passes through the 





The Hessian normal form of the
equation of a plane

Position of a plane in space can also be defined by the length of the normal, drawn from the origin to the plane
and by angles,
a, b and
g, that the normal forms with the
coordinate axes.

If
N°
is the unit vector of the normal then its components are, cosa,
cosb
and
cosg.

Then the projection of the position vector
r, of any point
P(x,
y, z) of the plane, onto the normal has the length
p. 
Since the projection is determined by expression 

that is,
r
· N°
= p or
written in the coordinates 

x
· cosa
+ y · cosb
+ z · cosg

p = 0 


represents the
equation of a plane in the Hessian
normal
form. 




The intercept form of the equation of a
plane

If,
l,
m
and n
are the intercepts of
x,
y
and z
axes and a plane respectively, then projections of these segments in
direction of the normal drawn from the origin to the
plane are all equal to the length of the normal, that is


By plugging these values of cosines into Hessian normal
form of the equation of plane, obtained is


the intercept form of the equation of plane. 




A
plane in
a
3D coordinate system
examples 
Example:
Find the equation of
the plane through three points, A(1,
1, 4), B(2,
4, 3)
and C(3,
4, 1).

Solution: 
Equation of a plane
Ax
+ By
+ Cz
+ D = 0 is determined by
the components (the
direction cosines)
of the normal vector N
= Ai + Bj
+ Ck and
the coordinates of any point of the plane. 
The normal vector is perpendicular to the plane determined by given
points, and as the vector 
AB
= r_{B }
r_{A}
= (2i + 4j + 3k)

(i
+ j + 4k)

r_{B
}
r_{A}
= 3i + 3j 
k,
and the vector

AC =
r_{C}_{ }
r_{A}
= 2i
+ 3j 
3k

then,





By plugging the coordinates, of any given point, into equation
of the plane determines the value of the coefficient
D.
Thus, plugging 
A(1,
1, 4) into

6x
+ 11y + 15z + D = 0
gives 
6
· (1)
+ 11
· 1
+ 15
· 1
+ D = 0, D = 77. 
Therefore,

6x + 11y + 15z 
77 = 0 is the equation of the plane
the given points lie on. 
The coordinate of the remaining two points
B
and C
must also satisfy the obtained equation, prove. 

Example:
Given are points,
A(1,
1, 1)
and B(3,
2,
6), find the equation of a plane which is normal to
the
vector
AB
and passes through the point A.

Solution:
According to the given conditions
the vector
AB
= N
so,

N
= r_{B }
r_{A}
= (3i 
2j + 6k)

(i
+ j + k) = 4i

3j + 5k

As the plane should pass through the point
A 
plug
A(1,
1, 1) into
4x 
3y + 5z + D = 0
=>
4 · (1)

3 · 1
+ 5
· 1
+ D = 0,
D = 2 
Therefore,
the equation of the plane P
::
4x 
3y + 5z + 2 = 0. 

Example:
Segments that a plane cuts on the axes,
x
and y, are
l = 1 and
m = 2
respectively, find the standard or general equation of the plane if it passes through the point
A(3, 4, 6).

Solution:
As the given point A(3, 4,
6) and the segments, l =
1 and
m = 2
must satisfy the intercept form of the equation of plane,
then











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