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| Coordinate
Geometry (Analytic Geometry) in Three-dimensional Space |
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Plane
in a three-dimensional (3D) coordinate system |
Equations of
a plane in a coordinate space
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The equation of a
plane in a 3D coordinate system |
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The distance between a point and a
plane - plane given in Hessian normal form
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Comparison of
the general form and the
Hessian normal form of equations of a plane
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The distance of a point to a plane -
plane given in general form
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Angle (dihedral angle) between two
planes
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| Equations of a plane in a coordinate space
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| 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
P1P
= r -
r1, drawn from the
given point P1(x1,
y1, z1) 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 |
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which is called the vector equation of a
plane. |
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Or,
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| Therefore, |
| (x
-
x1)
· A
+ (y -
y1)
· B
+ (z -
z1)
· C
= 0 |
| giving |
A
· x
+ B · y
+ C · z
+ D = 0 |
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where, D
=
-
(Ax1+
By1+
Cz1) |
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the general equation of a plane in 3D
space. |
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| 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. |
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| The distance between a point and a
plane, plane given in Hessian normal form
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| Distance from a point
A0(x0,
y0, z0)
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 |
| x0
· cosa
+ y0 · cosb
+ z0 · cosg
= p +
d. |
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Thus, the
distance of the point to the given
plane is
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d
= x0
· cosa
+ y0 · cosb
+ z0 · cosg -
p
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| 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. |
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| Comparison of general form and the
Hessian normal form of equations of a plane
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| Coefficients,
A,
B
and C,
beside coordinates x,
y
and z, of the general form of equation |
| 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. |
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| Thus, to convert from general form to the Hessian normal form, divide the general form of equation by |
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taking the sign of the square root opposite to the sign of
D, where
D
is not 0 |
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| The distance of a point to a plane -
plane given in general form
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| Let
replace in the Hessian formula for the distance d
= x0
· cosa
+ y0 · cosb
+ z0 · cosg -
p, the direction
cosines and the length of the normal p, by coefficients of the general form,
to obtain |
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| the formula for the
distance of a point to a plane given in the general
form. |
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| Angle (dihedral angle) between two
planes
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| 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 |
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| the
angle between two planes. |
| If planes are parallel normal vectors are collinear that is, |
| N1
= lN2
<=> j
= 0°, |
| and if planes are orthogonal, then their normal vectors are orthogonal too, so their scalar product is zero, |
| N1 · N2
= 0 <=> j
= 90°. |
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| Example:
Find the angle between planes,
P1
::
2x -
y + 4z -
3 = 0 and
P2
::
-x -
3y + 5z + 6 = 0.
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| Solution:
Thus, |
| N1 =
2i
-
j + 4k and
N2
=
-i
-
3j + 5k,
then |
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| Coordinate
geometry contents |
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