keisti
Tegu duota plokštuma
A
x
+
B
y
+
C
z
+
D
=
0
{\displaystyle Ax+By+Cz+D=0}
ir tiesė
x
−
x
0
m
=
y
−
y
0
n
=
z
−
z
0
p
,
{\displaystyle {\frac {x-x_{0}}{m}}={\frac {y-y_{0}}{n}}={\frac {z-z_{0}}{p}},}
M
(
x
0
;
y
0
;
z
0
)
{\displaystyle M(x_{0};y_{0};z_{0})}
s
→
=
(
m
;
n
;
p
)
{\displaystyle {\vec {s}}=(m;n;p)}
Tada:
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
.
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}.}
Po to, kad rasti tiesės ir plokštumos susikirtimo taško
M
1
(
x
1
;
y
1
;
z
1
)
{\displaystyle M_{1}(x_{1};y_{1};z_{1})}
t
1
{\displaystyle t_{1}\;}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
Duotos lygtis
(
N
→
,
r
→
)
+
D
=
0
{\displaystyle ({\vec {N}},\;{\vec {r}})+D=0}
P , kur
N
→
=
(
A
;
B
;
C
)
,
r
→
=
(
x
;
y
;
z
)
,
{\displaystyle {\vec {N}}=(A;B;C),\;{\vec {r}}=(x;y;z),}
r
→
=
r
0
→
+
s
→
t
{\displaystyle {\vec {r}}={\vec {r_{0}}}+{\vec {s}}t}
l , kur
s
→
=
(
m
;
n
;
p
)
{\displaystyle {\vec {s}}=(m;n;p)}
r
0
→
=
(
x
0
;
y
0
;
z
0
)
.
{\displaystyle {\vec {r_{0}}}=(x_{0};y_{0};z_{0}).}
Tegu
M
1
{\displaystyle M_{1}}
l su plokštuma P . Pažymėsime per
t
1
{\displaystyle t_{1}\;}
t taškui
M
1
{\displaystyle M_{1}}
Kadangi prie šito reikšmė parametro vektorius
r
→
=
r
0
→
+
s
→
t
1
{\displaystyle {\vec {r}}={\vec {r_{0}}}+{\vec {s}}t_{1}}
(
N
→
r
→
)
+
D
=
0
,
{\displaystyle ({\vec {N}}{\vec {r}})+D=0,}
(
N
→
,
(
r
0
→
+
s
→
t
1
)
)
+
D
=
0
,
{\displaystyle ({\vec {N}},\;({\vec {r_{0}}}+{\vec {s}}t_{1}))+D=0,}
t
1
(
N
→
,
s
→
)
+
(
N
→
,
r
0
→
)
+
D
=
0
,
{\displaystyle t_{1}({\vec {N}},\;{\vec {s}})+({\vec {N}},{\vec {r_{0}}})+D=0,}
iš kur, jeigu
(
N
→
,
s
→
)
≠
0
,
{\displaystyle ({\vec {N}},\;{\vec {s}})\neq 0,}
t
1
=
−
(
N
→
,
r
0
→
)
+
D
(
N
→
,
s
→
)
.
{\displaystyle t_{1}=-{\frac {({\vec {N}},{\vec {r_{0}}})+D}{({\vec {N}},\;{\vec {s}})}}.}
Įstatę reikšmę
t
1
{\displaystyle t_{1}\;}
l , rasime spindulį-vektorių
r
1
→
{\displaystyle {\vec {r_{1}}}}
l ir plokštumos P :
r
1
→
=
r
0
→
+
s
→
t
1
.
{\displaystyle {\vec {r_{1}}}={\vec {r_{0}}}+{\vec {s}}t_{1}.}
Tegu dabar lygtys plokštumos ir tiesės parašytos koordinatinėje formoje:
(
P
)
.
.
.
A
x
+
B
y
+
C
z
+
D
=
0
,
{\displaystyle (P)...\;Ax+By+Cz+D=0,}
(
l
)
.
.
.
x
−
x
0
m
=
y
−
y
0
n
=
z
−
z
0
p
.
{\displaystyle (l)...\;{\frac {x-x_{0}}{m}}={\frac {y-y_{0}}{n}}={\frac {z-z_{0}}{p}}.}
Tada panaudoję lygybę
t
1
=
−
(
N
→
,
r
0
→
)
+
D
(
N
→
,
s
→
)
,
{\displaystyle t_{1}=-{\frac {({\vec {N}},{\vec {r_{0}}})+D}{({\vec {N}},\;{\vec {s}})}},}
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
.
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}.}
Po to, kad rasti tiesės ir plokštumos susikirtimo taško
M
1
(
x
1
;
y
1
;
z
1
)
{\displaystyle M_{1}(x_{1};y_{1};z_{1})}
t
1
{\displaystyle t_{1}\;}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
Duota lygtis
A
x
+
B
y
+
C
z
+
D
=
0
{\displaystyle Ax+By+Cz+D=0}
P , kur plokštumos normalė stati plokštumai yra
N
→
=
(
A
;
B
;
C
)
{\displaystyle {\vec {N}}=(A;B;C)}
r
→
=
(
x
;
y
;
z
)
,
{\displaystyle {\vec {r}}=(x;y;z),}
tiesės
l
,
{\displaystyle l,\;}
s
→
=
(
m
;
n
;
p
)
{\displaystyle {\vec {s}}=(m;n;p)}
M
0
(
x
0
;
y
0
;
z
0
)
.
{\displaystyle M_{0}(x_{0};y_{0};z_{0}).}
Tegu
M
1
(
x
1
;
y
1
;
z
1
)
{\displaystyle M_{1}(x_{1};y_{1};z_{1})}
l su plokštuma P . Pažymėsime per
t
1
{\displaystyle t_{1}\;}
t taškui
M
1
{\displaystyle M_{1}}
Kadangi prie šito reikšmė parametro vektorius
r
→
=
{\displaystyle {\vec {r}}=}
{
x
=
x
0
+
m
t
1
,
y
=
y
0
+
n
t
1
,
z
=
z
0
+
p
t
1
{\displaystyle {\begin{cases}x=x_{0}+mt_{1},&\\y=y_{0}+nt_{1},&\\z=z_{0}+pt_{1}&\end{cases}}}
yra spindulys-vektorius taško, gulinčio ant plokštumos, tai, įstačius jį į lygtį plokštumos
(
N
→
r
→
)
+
D
=
A
x
+
B
y
+
C
z
+
D
=
0
,
{\displaystyle ({\vec {N}}{\vec {r}})+D=Ax+By+Cz+D=0,}
(
N
→
,
(
r
0
→
+
s
→
t
1
)
)
+
D
=
A
(
x
0
+
m
t
1
)
+
B
(
y
0
+
n
t
1
)
+
C
(
z
0
+
p
t
1
)
+
D
=
0
,
{\displaystyle ({\vec {N}},\;({\vec {r_{0}}}+{\vec {s}}t_{1}))+D=A(x_{0}+mt_{1})+B(y_{0}+nt_{1})+C(z_{0}+pt_{1})+D=0,}
t
1
(
N
→
,
s
→
)
+
(
N
→
,
r
0
→
)
+
D
=
t
1
(
A
m
+
B
n
+
C
p
)
+
A
x
0
+
B
y
0
+
C
z
0
+
D
=
0
,
{\displaystyle t_{1}({\vec {N}},\;{\vec {s}})+({\vec {N}},\;{\vec {r_{0}}})+D=t_{1}(Am+Bn+Cp)+Ax_{0}+By_{0}+Cz_{0}+D=0,}
iš kur, jeigu
(
N
→
,
s
→
)
=
A
x
0
+
B
y
0
+
C
z
0
≠
0
,
{\displaystyle ({\vec {N}},\;{\vec {s}})=Ax_{0}+By_{0}+Cz_{0}\neq 0,}
t
1
=
−
(
N
→
,
r
0
→
)
+
D
(
N
→
,
s
→
)
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
.
{\displaystyle t_{1}=-{\frac {({\vec {N}},{\vec {r_{0}}})+D}{({\vec {N}},\;{\vec {s}})}}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}.}
Įstatę reikšmę
t
1
{\displaystyle t_{1}\;}
l
{\displaystyle l\;}
r
1
→
{\displaystyle {\vec {r_{1}}}}
l
{\displaystyle l\;}
P :
r
1
→
=
r
0
→
+
s
→
t
1
=
{\displaystyle {\vec {r_{1}}}={\vec {r_{0}}}+{\vec {s}}t_{1}=}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
Tegu dabar lygtys plokštumos ir tiesės parašytos koordinatinėje formoje:
(
P
)
.
.
.
A
x
+
B
y
+
C
z
+
D
=
0
,
{\displaystyle (P)...\;Ax+By+Cz+D=0,}
(
l
)
.
.
.
x
−
x
0
m
=
y
−
y
0
n
=
z
−
z
0
p
.
{\displaystyle (l)...\;{\frac {x-x_{0}}{m}}={\frac {y-y_{0}}{n}}={\frac {z-z_{0}}{p}}.}
Tada panaudoję lygybę
t
1
=
−
(
N
→
,
r
0
→
)
+
D
(
N
→
,
s
)
,
{\displaystyle t_{1}=-{\frac {({\vec {N}},{\vec {r_{0}}})+D}{({\vec {N}},\;s)}},}
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
.
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}.}
Po to, kad rasti tiesės ir plokštumos susikirtimo taško
M
1
(
x
1
;
y
1
;
z
1
)
{\displaystyle M_{1}(x_{1};y_{1};z_{1})}
t
1
{\displaystyle t_{1}\;}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
Trumpai tariant, esmė buvo surasti tiesės l nusakytos parametrinėmis lygtimis
{
x
=
x
0
+
m
t
1
,
y
=
y
0
+
n
t
1
,
z
=
z
0
+
p
t
1
,
{\displaystyle {\begin{cases}x=x_{0}+mt_{1},&\\y=y_{0}+nt_{1},&\\z=z_{0}+pt_{1},&\end{cases}}}
t
1
,
{\displaystyle t_{1},\;}
x
0
+
m
t
1
{\displaystyle x_{0}+mt_{1}}
y
0
+
n
t
1
{\displaystyle y_{0}+nt_{1}}
z
0
+
p
t
1
{\displaystyle z_{0}+pt_{1}}
A
x
+
B
y
+
C
z
+
D
=
0
{\displaystyle Ax+By+Cz+D=0}
x , y ir z . Tokiu budu randamas tiesės ir plokštumos susikirtimo taškas
M
1
(
x
1
;
y
1
;
z
1
)
,
{\displaystyle M_{1}(x_{1};y_{1};z_{1}),}
t
1
{\displaystyle t_{1}\;}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
Kažkas panašaus kaip sprendžiama dviejų tiesių plokštumoje lygčių sistema (kad surasti tų tiesių susikirtimo tašką) keitimo budu.
Rasti projekciją taško M (9; -13; -18) į plokštumą P
9
x
−
17
y
−
15
z
+
23
=
0.
{\displaystyle 9x-17y-15z+23=0.}
Užduoties išsprendimui sudarysime lygtį tiesės, praeinančios per tašką M statmenai plokštumai P .
Lygtys betkurios tiesės, praeinančios per tašką M , turės pavidalą:
x
−
9
m
=
y
+
13
n
=
z
+
18
p
.
{\displaystyle {\frac {x-9}{m}}={\frac {y+13}{n}}={\frac {z+18}{p}}.}
Pagal statmenumo sąlygą tiesės ir plokštumos
(
A
m
=
B
n
=
C
p
)
{\displaystyle \left({\frac {A}{m}}={\frac {B}{n}}={\frac {C}{p}}\right)}
m , n , p įstatyti proporcingus jiems dydžius A , B , C :
x
−
9
9
=
y
+
13
−
17
=
z
+
18
−
15
.
{\displaystyle {\frac {x-9}{9}}={\frac {y+13}{-17}}={\frac {z+18}{-15}}.}
Kad surasti projekciją N taško M (9; -13; -18) į plokštumą reikia rasti susikirtimo tašką plokštumos
9
x
−
17
y
−
15
z
+
23
=
0
{\displaystyle 9x-17y-15z+23=0}
x
−
9
m
=
y
+
13
n
=
z
+
18
p
.
{\displaystyle {\frac {x-9}{m}}={\frac {y+13}{n}}={\frac {z+18}{p}}.}
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
=
−
9
⋅
9
+
(
−
17
)
⋅
(
−
13
)
+
(
−
15
)
⋅
(
−
18
)
+
23
9
⋅
9
−
17
⋅
(
−
17
)
−
15
⋅
(
−
15
)
=
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}=-{\frac {9\cdot 9+(-17)\cdot (-13)+(-15)\cdot (-18)+23}{9\cdot 9-17\cdot (-17)-15\cdot (-15)}}=}
=
−
81
+
221
+
270
+
23
81
+
289
+
225
=
−
1.
{\displaystyle =-{\frac {81+221+270+23}{81+289+225}}=-1.}
Iš čia koordinatės taško
N
(
x
N
;
y
N
;
z
N
)
{\displaystyle N(x_{N};y_{N};z_{N})}
{
x
1
=
x
0
+
m
t
1
,
y
1
=
y
0
+
n
t
1
,
z
1
=
z
0
+
p
t
1
.
{\displaystyle {\begin{cases}x_{1}=x_{0}+mt_{1},&\\y_{1}=y_{0}+nt_{1},&\\z_{1}=z_{0}+pt_{1}.&\end{cases}}}
{
x
N
=
x
0
+
m
t
1
=
9
+
9
⋅
(
−
1
)
=
0
,
y
N
=
y
0
+
n
t
1
=
−
13
−
17
⋅
(
−
1
)
=
−
13
+
17
=
4
,
z
N
=
z
0
+
p
t
1
=
−
18
−
15
⋅
(
−
1
)
=
−
18
+
15
=
−
3
,
{\displaystyle {\begin{cases}x_{N}=x_{0}+mt_{1}=9+9\cdot (-1)=0,&\\y_{N}=y_{0}+nt_{1}=-13-17\cdot (-1)=-13+17=4,&\\z_{N}=z_{0}+pt_{1}=-18-15\cdot (-1)=-18+15=-3,&\end{cases}}}
t. y. N (0; 4; -3). Duoti du tiesės taškai
M
0
(
3
;
5
;
2
)
,
{\displaystyle M_{0}(3;5;2),}
M
1
(
7
;
4
;
9
)
{\displaystyle M_{1}(7;4;9)}
6
x
+
7
y
+
8
z
+
12
=
0
{\displaystyle 6x+7y+8z+12=0}
M
1
M
2
{\displaystyle M_{1}M_{2}}
6
x
+
7
y
+
8
z
+
12
=
0
{\displaystyle 6x+7y+8z+12=0}
Sprendimas . Žinodami du tiesės T taškus
M
1
{\displaystyle M_{1}}
M
2
{\displaystyle M_{2}}
T kanoninę lygtį pasinaudodami formule:
x
−
x
1
x
2
−
x
1
=
y
−
y
1
y
2
−
y
1
=
z
−
z
1
z
2
−
z
1
;
{\displaystyle {\frac {x-x_{1}}{x_{2}-x_{1}}}={\frac {y-y_{1}}{y_{2}-y_{1}}}={\frac {z-z_{1}}{z_{2}-z_{1}}};}
x
−
3
7
−
3
=
y
−
5
4
−
5
=
z
−
2
9
−
2
;
{\displaystyle {\frac {x-3}{7-3}}={\frac {y-5}{4-5}}={\frac {z-2}{9-2}};}
x
−
3
4
=
y
−
5
−
1
=
z
−
2
7
.
{\displaystyle {\frac {x-3}{4}}={\frac {y-5}{-1}}={\frac {z-2}{7}}.}
Tiesės T krypties vektorius yra
s
→
=
(
4
;
−
1
;
7
)
.
{\displaystyle {\vec {s}}=(4;-1;7).}
Tiesės ir plokštumos susikirtimo tašką pažymėkime
N
(
x
2
;
y
2
;
z
2
)
{\displaystyle N(x_{2};y_{2};z_{2})}
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
=
−
6
⋅
3
+
7
⋅
5
+
8
⋅
2
+
12
6
⋅
4
+
7
⋅
(
−
1
)
+
8
⋅
7
=
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}=-{\frac {6\cdot 3+7\cdot 5+8\cdot 2+12}{6\cdot 4+7\cdot (-1)+8\cdot 7}}=}
=
−
18
+
35
+
16
+
12
24
−
7
+
56
=
81
73
=
−
1.109589041.
{\displaystyle =-{\frac {18+35+16+12}{24-7+56}}={\frac {81}{73}}=-1.109589041.}
Paverskime tiesės T kanonines lygtys
x
−
3
4
=
y
−
5
−
1
=
z
−
2
7
=
t
1
{\displaystyle {\frac {x-3}{4}}={\frac {y-5}{-1}}={\frac {z-2}{7}}=t_{1}}
{
x
−
3
4
=
t
1
,
y
−
5
−
1
=
t
1
,
z
−
2
7
=
t
1
;
{\displaystyle {\begin{cases}{\frac {x-3}{4}}=t_{1},&\\{\frac {y-5}{-1}}=t_{1},&\\{\frac {z-2}{7}}=t_{1};&\end{cases}}}
{
x
−
3
=
4
t
1
,
y
−
5
=
−
t
1
,
z
−
2
=
7
t
1
;
{\displaystyle {\begin{cases}x-3=4t_{1},&\\y-5=-t_{1},&\\z-2=7t_{1};&\end{cases}}}
{
x
=
3
+
4
t
1
,
y
=
5
−
t
1
,
z
=
2
+
7
t
1
.
{\displaystyle {\begin{cases}x=3+4t_{1},&\\y=5-t_{1},&\\z=2+7t_{1}.&\end{cases}}}
Toliau vietoje šios
6
x
+
7
y
+
8
z
+
12
=
0
{\displaystyle 6x+7y+8z+12=0}
x , y , z reikšmių įstatome tiesės reikšmes
3
+
4
t
1
{\displaystyle 3+4t_{1}}
5
−
t
1
{\displaystyle 5-t_{1}}
2
+
7
t
1
{\displaystyle 2+7t_{1}}
6
(
3
+
4
t
1
)
+
7
(
5
−
t
1
)
+
8
(
2
+
7
t
1
)
+
12
=
0
,
{\displaystyle 6(3+4t_{1})+7(5-t_{1})+8(2+7t_{1})+12=0,}
18
+
24
t
1
+
35
−
7
t
1
+
16
+
56
t
1
+
12
=
0
,
{\displaystyle 18+24t_{1}+35-7t_{1}+16+56t_{1}+12=0,}
24
t
1
−
7
t
1
+
56
t
1
=
−
18
−
35
−
16
−
12
,
{\displaystyle 24t_{1}-7t_{1}+56t_{1}=-18-35-16-12,}
73
t
1
=
−
81
,
{\displaystyle 73t_{1}=-81,}
t
1
=
−
81
73
=
−
1.109589041.
{\displaystyle t_{1}=-{\frac {81}{73}}=-1.109589041.}
Dabar galime surasti tiesės T ir plokštumos
6
x
+
7
y
+
8
z
+
12
=
0
{\displaystyle 6x+7y+8z+12=0}
N
(
x
2
;
y
2
;
z
2
)
{\displaystyle N(x_{2};y_{2};z_{2})}
N koordinatės yra:
{
x
2
=
x
0
+
m
t
1
=
3
+
4
⋅
(
−
81
73
)
=
3
−
324
73
=
−
1.438356164
,
y
2
=
y
0
+
n
t
1
=
5
−
(
−
81
73
)
=
5
+
81
73
=
6.109589041
,
z
2
=
z
0
+
p
t
1
=
2
+
7
⋅
(
−
81
73
)
=
2
−
567
73
=
−
5.767123288
,
{\displaystyle {\begin{cases}x_{2}=x_{0}+mt_{1}=3+4\cdot \left(-{\frac {81}{73}}\right)=3-{\frac {324}{73}}=-1.438356164,&\\y_{2}=y_{0}+nt_{1}=5-\left(-{\frac {81}{73}}\right)=5+{\frac {81}{73}}=6.109589041,&\\z_{2}=z_{0}+pt_{1}=2+7\cdot \left(-{\frac {81}{73}}\right)=2-{\frac {567}{73}}=-5.767123288,&\end{cases}}}
vadinasi, N (-1,438356164; 6,109589041; -5,767123288).
Rasti atstumą nuo taško M (-2; 3; 4) iki tiesės
(
l
)
.
.
.
x
−
3
1
=
y
+
1
2
=
z
−
1
1
.
{\displaystyle (l)...{\frac {x-3}{1}}={\frac {y+1}{2}}={\frac {z-1}{1}}.}
Pavyzdžio sprendimo planas sekantis:
1. Sudarysime lygtį plokštumos P , praeinančios per tašką M statmenai tiesei l .
2. Rasime tašką K susikirtimo tiesės l su plokštuma P .
3. Nustatysime atstumą nuo taško M iki tiesės l , kaip atstumą tarp taškų M ir K .
Sprendimas . Sudarysime lygtį įvairių plokštumų su centru taške M (plokštumos kurios eina per tašką M ):
A
(
x
+
2
)
+
B
(
y
−
3
)
+
C
(
z
−
4
)
=
0.
{\displaystyle A(x+2)+B(y-3)+C(z-4)=0.}
Plokštumos P normalės vektorius (statmenas plokštumai P ) yra
n
→
=
(
A
;
B
;
C
)
.
{\displaystyle {\vec {n}}=(A;B;C).}
l krypties vektorius yra
s
→
=
(
m
;
n
;
p
)
=
(
1
;
2
;
1
)
.
{\displaystyle {\vec {s}}=(m;n;p)=(1;2;1).}
n
→
=
(
A
;
B
;
C
)
=
(
1
;
2
;
1
)
.
{\displaystyle {\vec {n}}=(A;B;C)=(1;2;1).}
P ir tiesės l galima užrašyti lygtį plokštumos P :
(
x
+
2
)
+
2
(
y
−
3
)
+
(
z
−
4
)
=
0
,
{\displaystyle (x+2)+2(y-3)+(z-4)=0,}
x
+
2
+
2
y
−
6
+
z
−
4
=
0
,
{\displaystyle x+2+2y-6+z-4=0,}
x
+
2
y
+
z
−
8
=
0.
{\displaystyle x+2y+z-8=0.}
Taškui K
t
1
=
−
A
x
0
+
B
y
0
+
C
z
0
+
D
A
m
+
B
n
+
C
p
=
−
1
⋅
3
+
2
⋅
(
−
1
)
+
1
⋅
1
−
8
1
⋅
1
+
2
⋅
2
+
1
⋅
1
=
{\displaystyle t_{1}=-{\frac {Ax_{0}+By_{0}+Cz_{0}+D}{Am+Bn+Cp}}=-{\frac {1\cdot 3+2\cdot (-1)+1\cdot 1-8}{1\cdot 1+2\cdot 2+1\cdot 1}}=}
=
−
3
−
2
+
1
−
8
1
+
4
+
1
=
−
−
6
6
=
1
,
{\displaystyle =-{\frac {3-2+1-8}{1+4+1}}=-{\frac {-6}{6}}=1,}
iš kur
{
x
K
=
x
0
+
m
t
1
=
3
+
1
⋅
1
=
4
,
y
K
=
y
0
+
n
t
1
=
−
1
+
2
⋅
1
=
1
,
z
K
=
z
0
+
p
t
1
=
1
+
1
⋅
1
=
2.
{\displaystyle {\begin{cases}x_{K}=x_{0}+mt_{1}=3+1\cdot 1=4,&\\y_{K}=y_{0}+nt_{1}=-1+2\cdot 1=1,&\\z_{K}=z_{0}+pt_{1}=1+1\cdot 1=2.&\end{cases}}}
Gavome tiesės l ir plokštumos P susikirtimo tašką K (4; 1; 2). Atstumas nuo taško M (-2; 3; 4) iki taško K (4; 1; 2) yra:
M
K
=
(
4
+
2
)
2
+
(
1
−
3
)
2
+
(
2
−
4
)
2
=
36
+
4
+
4
=
44
=
2
11
.
{\displaystyle MK={\sqrt {(4+2)^{2}+(1-3)^{2}+(2-4)^{2}}}={\sqrt {36+4+4}}={\sqrt {44}}=2{\sqrt {11}}.}
Pastebėsime, kad taškas K yra projekcija taško M į tiesę l .
keisti
Tarkime, tiesė T nusakoma kanoninėmis lygtimis
x
−
x
0
l
=
y
−
y
0
m
=
z
−
z
0
n
,
{\displaystyle {\frac {x-x_{0}}{l}}={\frac {y-y_{0}}{m}}={\frac {z-z_{0}}{n}},}
o plokštuma
π
{\displaystyle \pi }
A
x
+
B
y
+
C
z
+
D
=
0.
{\displaystyle Ax+By+Cz+D=0.}
ϕ
{\displaystyle \phi }
T ir plokštumos
π
{\displaystyle \pi }
π
{\displaystyle \pi }
ϕ
+
α
=
π
2
,
{\displaystyle \phi +\alpha ={\frac {\pi }{2}},}
cos
α
=
cos
(
π
2
−
ϕ
)
=
sin
ϕ
;
{\displaystyle \cos \alpha =\cos \left({\frac {\pi }{2}}-\phi \right)=\sin \phi ;}
čia
α
{\displaystyle \alpha }
T krypties vektoriaus
s
→
=
(
l
;
m
;
n
)
{\displaystyle {\vec {s}}=(l;m;n)}
π
{\displaystyle \pi }
n
→
=
(
A
;
B
;
C
)
.
{\displaystyle {\vec {n}}=(A;B;C).}
α
{\displaystyle \alpha }
T ir plokštumos normalės
n
→
=
(
A
;
B
;
C
)
.
{\displaystyle {\vec {n}}=(A;B;C).}
n
→
{\displaystyle {\vec {n}}}
s
→
{\displaystyle {\vec {s}}}
cos
α
=
n
→
⋅
s
→
‖
n
→
‖
⋅
‖
s
→
‖
=
A
l
+
B
m
+
C
n
A
2
+
B
2
+
C
2
l
2
+
m
2
+
n
2
.
{\displaystyle \cos \alpha ={\frac {{\vec {n}}\cdot {\vec {s}}}{\|{\vec {n}}\|\cdot \|{\vec {s}}\|}}={\frac {Al+Bm+Cn}{{\sqrt {A^{2}+B^{2}+C^{2}}}{\sqrt {l^{2}+m^{2}+n^{2}}}}}.}
Tada
sin
ϕ
=
A
l
+
B
m
+
C
n
A
2
+
B
2
+
C
2
l
2
+
m
2
+
n
2
,
{\displaystyle \sin \phi ={\frac {Al+Bm+Cn}{{\sqrt {A^{2}+B^{2}+C^{2}}}{\sqrt {l^{2}+m^{2}+n^{2}}}}},}
čia
ϕ
{\displaystyle \phi }
T ir plokštumos
A
x
+
B
y
+
C
z
+
D
=
0.
{\displaystyle Ax+By+Cz+D=0.}
T lygiagreti plokštumai
π
{\displaystyle \pi }
s
→
=
(
l
;
m
;
n
)
{\displaystyle {\vec {s}}=(l;m;n)}
n
→
=
(
A
;
B
;
C
)
,
{\displaystyle {\vec {n}}=(A;B;C),}
n
→
⋅
s
→
=
0.
{\displaystyle {\vec {n}}\cdot {\vec {s}}=0.}
tiesės ir plokštumos lygiagretumo sąlygą :
A
⋅
l
+
B
⋅
m
+
C
⋅
n
=
0.
{\displaystyle A\cdot l+B\cdot m+C\cdot n=0.}
T statmena plokštumai
π
{\displaystyle \pi }
s
→
=
(
l
;
m
;
n
)
{\displaystyle {\vec {s}}=(l;m;n)}
n
→
=
(
A
;
B
;
C
)
,
{\displaystyle {\vec {n}}=(A;B;C),}
tiesės ir plokštumos statmenumo sąlyga :
A
l
=
B
m
=
C
n
.
{\displaystyle {\frac {A}{l}}={\frac {B}{m}}={\frac {C}{n}}.}
Pavyzdis . Su kuria B reikšme tiesė T , nusakoma lygtimis
{
4
x
−
5
y
+
z
−
3
=
0
,
x
+
2
y
−
3
z
+
9
=
0
,
{\displaystyle {\begin{cases}4x-5y+z-3=0,&\\x+2y-3z+9=0,&\end{cases}}}
bus lygiagreti su plokštuma
π
{\displaystyle \pi }
2
x
−
B
y
−
2
z
−
3
=
0
{\displaystyle 2x-By-2z-3=0}
Sprendimas . Kai tiesė T lygiagreti su plokštuma
π
{\displaystyle \pi }
s
→
=
(
l
;
m
;
n
)
{\displaystyle {\vec {s}}=(l;m;n)}
n
→
=
(
2
;
−
B
;
−
2
)
{\displaystyle {\vec {n}}=(2;-B;-2)}
s
→
⋅
n
→
=
0.
{\displaystyle {\vec {s}}\cdot {\vec {n}}=0.}
Pažymėkime:
n
1
→
=
(
1
;
−
2
;
3
)
,
n
2
→
=
(
4
;
−
3
;
4
)
.
{\displaystyle {\vec {n_{1}}}=(1;-2;3),\;\;{\vec {n_{2}}}=(4;-3;4).}
s
→
=
n
1
→
×
n
2
→
,
{\displaystyle {\vec {s}}={\vec {n_{1}}}\times {\vec {n_{2}}},}
s
→
⋅
n
→
=
0
{\displaystyle {\vec {s}}\cdot {\vec {n}}=0}
(
n
1
→
×
n
2
→
)
⋅
n
→
=
0.
{\displaystyle ({\vec {n_{1}}}\times {\vec {n_{2}}})\cdot {\vec {n}}=0.}
|
1
−
2
3
4
−
3
4
2
−
B
−
2
|
=
0
;
{\displaystyle {\begin{vmatrix}1&-2&3\\4&-3&4\\2&-B&-2\end{vmatrix}}=0;}
|
1
−
2
3
4
−
3
4
2
−
B
−
2
|
=
1
(
−
3
⋅
(
−
2
)
−
4
⋅
(
−
B
)
)
−
(
−
2
)
(
4
⋅
(
−
2
)
−
4
⋅
2
)
+
3
(
4
⋅
(
−
B
)
−
(
−
3
)
⋅
2
)
=
0
;
{\displaystyle {\begin{vmatrix}1&-2&3\\4&-3&4\\2&-B&-2\end{vmatrix}}=1(-3\cdot (-2)-4\cdot (-B))-(-2)(4\cdot (-2)-4\cdot 2)+3(4\cdot (-B)-(-3)\cdot 2)=0;}
|
1
−
2
3
4
−
3
4
2
−
B
−
2
|
=
(
6
+
4
B
)
+
2
(
−
8
−
8
)
+
3
(
−
4
B
+
6
)
=
6
+
4
B
−
32
−
12
B
+
18
=
−
8
B
−
8
=
0
;
{\displaystyle {\begin{vmatrix}1&-2&3\\4&-3&4\\2&-B&-2\end{vmatrix}}=(6+4B)+2(-8-8)+3(-4B+6)=6+4B-32-12B+18=-8B-8=0;}
−
8
B
−
8
=
0
;
{\displaystyle -8B-8=0;}
−
8
B
=
8
;
{\displaystyle -8B=8;}
B
=
−
1.
{\displaystyle B=-1.}
Pavyzdis . Sudaryti lygtį plokštumos P , praeinančios pro tašką M (2; -1; 3) lygiagrečiai dviems tiesiems:
T
1
.
.
.
x
−
2
3
=
y
+
17
4
=
z
−
8
−
5
{\displaystyle T_{1}...{\frac {x-2}{3}}={\frac {y+17}{4}}={\frac {z-8}{-5}}}
ir
T
2
.
.
.
x
+
8
2
=
y
−
14
−
3
=
z
1
.
{\displaystyle T_{2}...{\frac {x+8}{2}}={\frac {y-14}{-3}}={\frac {z}{1}}.}
Parašysime lygtį visų plokštumų su centru taške M :
A
(
x
−
2
)
+
B
(
y
+
1
)
+
C
(
z
−
3
)
=
0.
{\displaystyle A(x-2)+B(y+1)+C(z-3)=0.}
Panaudojame lygiagretumo sąlyga (
A
⋅
l
+
B
⋅
m
+
C
⋅
n
=
0
{\displaystyle A\cdot l+B\cdot m+C\cdot n=0}
P tiesei
T
1
{\displaystyle T_{1}}
T
2
{\displaystyle T_{2}}
3
A
+
4
B
−
5
C
=
0
,
{\displaystyle 3A+4B-5C=0,}
2
A
−
3
B
+
C
=
0.
{\displaystyle 2A-3B+C=0.}
Iš šitos sistemos giminingų lygčių nustatysime santykį koeficientų A , B , C ir paskui į lygtį
A
(
x
−
2
)
+
B
(
y
+
1
)
+
C
(
z
−
3
)
=
0
{\displaystyle A(x-2)+B(y+1)+C(z-3)=0}
A , B , C įstatysime proporcionalius jiems dydžius:
A
:
B
:
C
=
|
4
−
5
−
3
1
|
:
|
−
5
3
1
2
|
:
|
3
4
2
−
3
|
=
−
11
:
−
13
:
−
17
,
{\displaystyle A:B:C={\begin{vmatrix}4&-5\\-3&1\end{vmatrix}}:{\begin{vmatrix}-5&3\\1&2\end{vmatrix}}:{\begin{vmatrix}3&4\\2&-3\end{vmatrix}}=-11:-13:-17,}
11
(
x
−
2
)
+
13
(
y
+
1
)
+
17
(
z
−
3
)
=
0
,
{\displaystyle 11(x-2)+13(y+1)+17(z-3)=0,}
11
x
+
13
y
+
17
z
−
60
=
0.
{\displaystyle 11x+13y+17z-60=0.}
