Аналитическая геометрия. Часть II. Аналитическая геометрия пространства. Шурыгин В.В. - 13 стр.

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f
1
=
−−→
EA
1
|
−−→
EA
1
|
; f
2
=
−−−→
A
3
A
2
|
−−−→
A
3
A
2
|
.
−−→
EB
1
= |
−−→
EA
1
|(f
1
cos ϕ + f
2
sin ϕ)
−−→
EB
2
=
|
−−→
EA
1
|(f
1
cos(ϕ +
2π
3
) + f
2
sin(ϕ +
2π
3
))
−−→
EB
3
= |
−−→
EA
1
|(f
1
cos(ϕ −
2π
3
) + f
2
sin(ϕ −
2π
3
)) e
i
0
=
−−→
OE +
−−→
EB
i
ϕ =
π
3
−−→
EB
1
= −
−−→
EA
3
−−→
EB
2
= −
−−→
EA
1
−−→
EB
3
= −
−−→
EA
2
(A
3
EB
1
) = (A
1
EB
2
) = (A
2
EB
3
) = −2
B
1
B
2
B
3
e
1
0
=
−−→
OB
1
= {
2
3
;
2
3
; −
1
3
}
e
2
0
=
−−→
OB
2
= {−
1
3
;
2
3
;
2
3
} e
3
0
=
−−→
OB
3
= {
2
3
; −
1
3
;
2
3
}
x
1
x
2
x
3
=
2
3
−
1
3
2
3
2
3
2
3
−
1
3
−
1
3
2
3
2
3
x
1
0
x
2
0
x
3
0
.
{O; e
i
}
{O
0
; e
i
0
}
e
i
=
−−→
OA
i
e
i
0
=
−−→
O
0
A
i
i = 1, 2, 3 O 6= O
0
O
A
1
A
3
O
0
E
A
2
E
A
1
A
2
A
3
(
1
3
;
1
3
;
1
3
) O
0
OE
−−→
OO
0
= 2
−−→
OE
O
0
(
2
3
;
2
3
;
2
3
)
e
1
0
=
−−−→
O
0
A
1
= {
1
3
; −
2
3
; −
2
3
}
e
2
0
=
−−−→
O
0
A
2
= {−
2
3
;
1
3
; −
2
3
} e
3
0
=
−−−→
O
0
A
3
=
{−
2
3
; −
2
3
;
1
3
}
x
1
x
2
x
3
=
1
3
−
2
3
−
2
3
−
2
3
1
3
−
2
3
−
2
3
−
2
3
1
3
x
1
0
x
2
0
x
3
0
+
2
3
2
3
2
3
.
).+2+D +2++4+90D *0:-O 0,4)O
                                       −−→               −−−→
                                       EA1               A3 A2
                                f1 = −−→ ; f2 = −−−→ .
                                      |EA1 |            |A3 A2 |
  ),)7 1).+ 0E+C32-3 9).2+ −            EB1 = |EA1 |(f1 cos ϕ + f2 sin ϕ) O EB2 =
                                               −→      −−→                           −−→
 −−→                                     2π O −−→         −−→
|EA1 |(f1 cos(ϕ + 2π ) +   f  sin(ϕ  +      ))  EB    =  |EA1 |(f1 cos(ϕ − 2π
                                                                            3 ) + f2 sin(ϕ −
2π    O :02)4  9).2+
                   3        2
                                    −−
                                     →    3
                                             −− → +9++
                                                   3
                                                            *0:-0  
 3 ))                       ei0 = OE + EBi
      0--4+24 ,+C+*+ -1(0D ϕ = π  †2+4 -1(0) −                     EB1 = −EA3 O
                                                                           −→           −−→

EB2 = −EA1 O EB3 = −EA2 O +2-AC0 0E+C4 ,+-2) +2+@)3 -4
−−→          −−→ −−→               −−→              3
                                                                                                
ª€ˆ« O ®ˆ< (A3EB1) = (A1EB2) = (A2EB3) = −2 O 0 :02)4 .++C02
2+(). B1 O B2  B3  ?+-1) 9(-1)D ,+1(0)4 e10 = −                 OB1 = { 23 ; 32 ; − 31 } O
                                                                        −→

e20 = OB2 = {− 31 ; 23 ; 32 } O e30 = OB3 = { 32 ; − 13 ; 32 } O ,+†2+4 F+41 ,)+*B
         −−→                            −−→
0:+903 .++C02 4)A2 9C~
                                                                          
                                              2
                             x  1
                                               3   − 13         2
                                                                3       x 10
                             2              2     2                 20 
                            x =             3     3        − 13    x .
                                            − 31     2          2         0
                             x3                                         x3
   ‚ninƒn Î' +-20927 F+41 ,)+*0:+903 .++C02 , ,))E+B
                                                     3          3


C) +2 +2++4+90++ ),)0 {O; ei} . +2++4+90+4 ),)
                                (
{O0 ; ei } O )-1 :9)-2+O 2+ ei = OAi O ei = O0 Ai O i = 1, 2, 3 O 0 O 6= O0
                                         −−→             −−→                   
   %fžfMLf' 4 -+. Š 0--4+24O
          0                                               0


                                                               O0
.0.  9 ,)CC )D :0C0()O 2+(. E O
3913A A-3 ‡)2+4 ,0917++ 2)B
+17.0 A1A2A3  „20 2+(.0 4))2 .+B                       E
                         Û     (      (
+C02 ( 1 ; 1 ; 1 ) -+O 2+ 2+ .0 O0 1)B A1
                                                                               A3

s2 0 ,34+D OE  −        OO0 = 2OE O ,+†2+B
                3 3 3
                             −→      −−→
4 2+(.0 O0 4))2 .++C02 ( 2 ; 2 ; 2 )                            A2
1)C+902)17+O e = −        O A1 = { 3 ; − 23 ; − 23 } O
                             −0−→      1
                                          3 3 3
                                                                  O
                                                              -  Š
                       10


e2 = O0 A2 = {− 32 ; 13 ; − 23 } O e3 = O0 A3 =
       −−−→                              −−−→

{− 2 ; − 2 ; 1 } O 0 ,)+*0:+90) .++C02 4))2 9C
  0                                     0


      3       3 3
                                          0                                     
                                 1    2   2                                        2
                      x1         3 −  3 − 3    x1                                  3
                     2  2         1    2   20                               2   
                     x  =  −3     3  − 3  x 
                                                     +                            3   .
                              − 23 − 23   1                                        2
                                                 0
                      x3                  3    x3                                  3
                                                   ׌

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