and a vertical horizontal profile projection plane r 1 r 2 p p The horizontal projection of t he intersection point N of the line ID: 652832
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Slide1
Intersection of a line and a vertical/horizontal/profile projection plane
r
1
r
2
p’’
p’
The
horizontal
projection
of the intersection point N of the line p and the plane P lies on the 1st trace of the plane. Why?
N’
N’’
xSlide2
1. p 2. = q 3. q p = N
1x2
r
1
r2
p’’
p’
= d
1
d
2
Q
1’
Q
1’’
Q
2’
Q
2
’’
q’’
=
q’
N’’
N’
Intersection
of
a line
and
a plane
determined
with
its
traces
Outline
of
the
solution
:
Becouse
of
the
simplicity
of
the
construction
the
plane
is a
vertical
projection
plane
.Slide3
1. Determine the intersection of the line p and the plane .s1s2
x
p’= p’’
z
y
y
Remark
. A line
parallel with
3 is
not uniquely determined by its
horizontal and vertical projection
. It has to be determined with the projections of two of its points.
B’
B’’
A’
A’’
s
3
A’’’
B’’’
p’’’
N’’’
N’’
N’
Exercises
.Slide4
2. Construct a line segment of the length d lying on the line p
starting from
the
intersection
with the
plane P
.
r
2
r
1
xp’p”
d
2
d
1
Q
1
’
Q
1
”
Q
2
”
Q
2
’
q’
q”
S’
S”
S
0
K’
K”
K
0
p
0
T
0
T’
T”
dSlide5
Two basic constructions (perpendicularity)1. Construct a line perpendicular to a given plane through a given point.
r
2
r
1
x
T’’
T’
2
.
Construct
a plane
through a given point perpendicular to a given line.x
p’’
p’
T’’
T’
Instruction
.
The
2nd
principle
line (or 1st
princile
line) is
used
to
construct
the
traces
of
a plane
passing
through
the
point
T
.
r
1
r
2
n’
n’’
.
.
.
s”
s’
S
1
’
S
1
”Slide6
Exercise 3. Construct the symmetry plane of the line segment AB. Definition. The set of point in space which have the same distance from the end-points of a line segment lie in the
symmetry plane of that
line segment.
A’
B’
A’’
B’’
Symmetry
plane
passes
through
the midpoint of the line segment and is perpendicular to the line segment.
P’
P’’
m”
m’
M
1
”
M
1
’
s
1
.
s
2
.Slide7
Exercise 4. Construct a plane perpendicular to the plane P and passes through line p.x
r
1
r
2
p’
p’’
S’
S’’
s
2
Remark
!
A plane is
perpendicular
to a plane is one of it containes
at least one line perpendicular to the
other plane
.
s
1
n’
.
n”
.Slide8
Metrical exercise1. Determine the distance between the point T and the plane P.r2
r
1
x
T’’
T’
n’
n’’
Outline
of
the solution: 1) T n, n P2) n P = N
d
1
d
2
q’
q’’
N’’
N’
3)
d
(
T
,
N
)
T
0
N
0
dSlide9
2. Construct line segment of lenght d from point S on a line perpendicular to the plane P.xr2r1
S’
s’
s’’
S’’
n’
n’’
d
S
0
p
0
n
0
.
V
0
V’
V’’
dSlide10
xyyz3. Determine the distance between point A and the plane .
s
2
s
1
A’’
A’
s
3
A’’’
N’’’
N’’
N’
Remark
1.
The
plane
is
the
profile
projecting
plane.
Remark
2.
This
construction
is
the
same for
the
following
task
:
Construct
a line segment
of
given
lenght
from
a
point
in
a
plane on a
perpendicular
line to
the
plane
.
d
.Slide11
1. Intersection of a trinagle and a line
p’’
p’
x
A’
B’
C’
A’’
B’’
C’’
Instruction
.
Because
of the
simplicity of the
construction a vertical
projection plane through
the line is used, and afterwards
its
intersection
with
the
triangle plane is
constructed
.
d
2
d
1
q’’
M’’
N’’
M’
N’
q’
P’
P’’
Remark
.
The
inetrsection
of
a
parallelogram
and
a
line is
constructed
in
the
same
way
.
Solved
exercisesSlide12
2. Intersection of the line p and a plane determined with
the
intersecting
lines
(a, b)
1)
p
1
Pravcem je postavljena prva projicirajuća ravnina .
2) P
= qPravci a i b probadaju ravninu u točkama 1 i 2, a njihova je spojnica presječnica q ravnina P i .
a’
b’
p’
1’
2’
2’’
1’’
M’
M’’
q’’
a’’
b’’
p”
N’’
N’
=q’
=d
1
Napomena
. Na isti se način konstruira probodište pravca i ravnine zadane dvama paralelnim pravcima.
3)
q
p
=
NSlide13
p’p’’xs1
s
2
d
2
q’’
The intersection of the
vertical projection plane through
the line p
and the plane
is the principal line
q of the plane
.q’
S’
S’’
3.
Construct
the
intersection of
the
line
p
(
parallel
to
the
x
axis
)
and
the
plane
.Slide14
s1 s2 k1 k24. Intersection of a line and the
symmetry
plane/coincidence
plane
with
the use of
the
profile peojection
.
x
z
ys3k3
p”
p’
P
1
’
P
1
”
P
2
’
P
1
’’’
P
2
”
P
2
’’’
p’’’
N’’’
N”
N’
N
=
p
R’’’
R
=
p
K
R’= R”Slide15
r1d1d2
Metrical
exercises.
1. Determine the distance between
the point T
and the line p
.
x
T’
T’’
p’’
p’
Outline of the
solution:
1) T P
, P p
r
2
2)
n
P
=
N
N’
N’’
3)
d
(
T
,
N
)
T
0
dSlide16
2. Construct a line segment of lenght d from the point A on a line perpendicular to the plane P.xy
z
r2
r
1
A’
r
3
A’’’
A”
n’’’
n’= n”
d
B’’’
B”
B’
Remark
.
There
exist
two
solutions
.
d