5 Linear inequalities. In plane. Bundle of lines.
By
analogy with that a linear equation represents a line on a plane, a
linear inequality
determines a half-plane (the set of points the coordinates of which
and
satisfy this inequality) restricted by the line
Show that this assertion is true for the case when a line
divide a plane
into two half-planes denoted by
and
:
We
say that a point
with radius-vector
belongs to the half-plane
(or respectively
)
if there exists
(respectively
)
such that
where
is
the orthogonal projection of
on
the line
.
Theorem
5.
iff
Proof:
Let
,
i.e. there is
such that
.
Estimate the quantity
Since
,
and
then
by
positivity of
.
Let
and
.
Then by
we obtain
.
Then
since
,
we have
and consequently
.
Exercise
1.
The system of coordinates
on a plane and a line
with equation
are given. Find the distance between this line and a point
the radius-vector of which is
.
Solution:
Let
.
Then
.
The point
belongs to this line, therefore the following holds:
Consequently,
.
Substituting
in the expression for
,
we obtain
Let
the system of coordinates be orthonormal. For an equation
as showed the vector
is perpendicular to the line. Therefore,
.
Taking
in account that the point
lies on the line
and consequently
,
we can write the final answer in the form:
.
A bundle of lines on a plane is called the set of all lines passing through some given point named the vertex of the bundle.
Theorem
6.
Let a point that is common for all lines of a bundle is the point of
intersection of non-parallel lines
and
.
Then
1.
For any line of the bundle there is a pair of non-simultaneously
equal to zero numbers
and
such that
is an equation of this line.
2.
For any non-simultaneously equal to zero numbers
and
the equation
is an equation of some line of this bundle.
Proof:
1. Take some point
non-coinciding with the vertex of the bundle and assume
Observe
that
since the point
doesn’t belong to these lines simultaneously. Moreover, the line
passes
through both the point
and the vertex of the bundle, and consequently it belongs to the
bundle.
2.
Let
and
be a pair of intersecting lines from the considered bundle. Then
obviously
.
And the equation
is an equation of a line because from
and
follows that
.
Indeed,
assume the contrary:
.
The
lines
and
by construction have at least one common point. Therefore they either
coincide or are intersected. By Theorem 4 they are coinciding iff
for which
and
.
The last two equalities are equivalent to
.
In
the considered case the lines are intersected, therefore
and consequently the system (*) can have only one solution. On other
hand, obviously that this system has a trivial solution
,
but it contradicts to the inequality
.
An
equation
with non-simultaneously equal to zero parameters
and
is called anequation
of a bundle of lines
on a plane.
6. Plane in the space. Plane in the space
Let
a system of coordinates
in the space and a plane
passing through a point
with lying on it non-collinear vectors
and
be given. Vectors
and
are calleddirecting
vectors
of the plane
.
Theorem
1.
The set of radius-vectors of points on plane
is represented in the form
where
and
are arbitrary real parameters.
Proof:
Let
be some point on the plane. Then vectors
and
will be coplanar.
Whence
we obtain
and consequently an equation of the plane will have the form:
where
and
.
Theorem 2. Every plane in any Cartesian system of coordinates can be represented by an equation:
Proof:
The condition of coplanarity of vectors
,
and
in the coordinate form has the form:
.
Whence
,
or finally
where
numbers
and
are equal to:
;
;
,
and
.
Thus, we obtained that an equation of plane is an equation of the
first degree.
The
condition of non-simultaneous equality of numbers
and
to zero follows by non-collinearity of vectors
and
.
Theorem
3.
Every equation of the form
in any Cartesian system of coordinates is an equation of some plane.
Proof:
By
direct checking we are convinced that
in case
can be written in the form:
,
and in case
in the form:
.
In both cases these equations determine a plane passing through some given point and that is parallel to two non-collinear vectors.
Theorem
4.
Planes
and
are
parallel iff their principal vectors are collinear.
Proof:
If principal vectors are collinear then there is such a number
that
,
,
and the system of equations
can be rewritten in the form
.
If
,
there is no common point of these planes; if
,
all points are common that means parallelism of the planes.
Let
the planes
and
be parallel. Then they must intersect the same coordinate planes by
parallel lines.Let
for definiteness these coordinate planes are planes for which
and
.
Lines of intersection corresponding to the first of the coordinate
planes will be determined by the following systems of equations:
and
.
Parallelism
of these lines means an existence of
such that
,
.
Considering
the case
,
we obtain the analogous system of equations:
and
.
But
the condition
and parallelism of this pair of lines imply that
.
Corollary
2.
The equations
and
are
equations of the same plane iff there exists
such that
,
,
,
.
A bundle of planes in the space is called the set of all planes passing through a given line.
An
equation
of a bundle of planes
passing through a line determined by intersection of a pair of
non-parallel planes
and
is called the equation of the form
.
A sheaf of planes in the space is called the set of all planes passing through a given point.
If
a point
belonging simultaneously to three planes
and
is unique then an equation of the form
is
called an equation
of a sheaf of planes
passing through the point
.