
Coordinate
Geometry (Analytic Geometry) in Threedimensional Space 

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

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 
The distance between a point and a
plane  plane given in Hessian normal form






Equations of a plane in a coordinate space


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 


The distance between a point and a
plane  plane given in Hessian normal form

Distance from a point
A_{0}(x_{0},
y_{0}, z_{0})
to a plane is taken to be positive if the given point is on the one side while the origin is on the other side regarding to the
plane, as is in
the right figure. 
Through the point
A
lay a plane parallel to the given plane. 
The length of a normal segment from the origin to the plane through
A,
written in the normal form, is p +
d. 
As the point A
lies in that plane its coordinates must satisfy the equation 
x_{0}
· cosa
+ y_{0} · cosb
+ z_{0} · cosg_{ }
= p +
d. 
Thus, the
distance of the point to the given
plane is


d
= x_{0}
· cosa
+ y_{0} · cosb
+ z_{0} · cosg 
p






To express the distance
d
in terms of the coefficients of the equation of a plane given in the general or
standard form, we should examine the relations of
the two forms of equations. 










Precalculus contents
J 



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