

Coordinate geometry
(or Analytic geometry) in threedimensional space


::
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
origin is D
= 0. 




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



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






Comparison of general form and the
Hessian normal form of equations of a plane

Coefficients,
A,
B
and C
of coordinates x,
y
and z, of the general form of equation
of a plane 
Ax
+ By
+ Cz
+ D = 0 
are the components of the normal vector
N
= Ai + Bj
+ Ck, while in the Hessian’s equation 
x
· cosa
+ y · cosb
+ z · cosg

p
= 0, 
on that place, there are components (the direction cosines) of the unit vector
N°
of the same vector. 

Thus, to convert from general form to the Hessian normal form, divide the general form of equation by 

taking the sign of the square root opposite to the sign of
D, where
D
is not 0 



The distance of a point to a plane 
plane given in general form

Let
replace in the Hessian formula for the distance d
= x_{0}
· cosa
+ y_{0} · cosb
+ z_{0} · cosg 
p, 
the direction
cosines and the length of the normal p, by coefficients of the general form,
to obtain 

the formula for the
distance of a point to a plane given in the general
form. 

Angle (dihedral angle) between two
planes

The angle subtended by normals drawn from the origin to the first and second plane, that is, the angle which
form
the normal vectors of the given planes is 

the
angle between two planes. If planes are parallel normal vectors are collinear that is, 
N_{1}
= lN_{2}
<=> j
= 0°, 
and if planes are orthogonal, then their normal vectors are orthogonal too, so their scalar product is zero, 
N_{1} · N_{2}
= 0 <=> j
= 90°. 


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:
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 


Example:
Find the angle between planes,
P_{1}
::
2x 
y + 4z 
3 = 0 and
P_{2}
::
x 
3y + 5z + 6 = 0.

Solution: Thus, 
N_{1} =
2i

j + 4k and
N_{2}
=
i

3j + 5k,
then 


















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