Аналитическая геометрия. Часть II. Аналитическая геометрия пространства. Шурыгин В.В. - 45 стр.

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x
1
x
2
x
3
=
2
3
0
+ t
2
1
2
.
` a = {2; 1; 2}
M
0
(2; 3; 0)
` x
1
= 1+2t
x
2
= 2+t x
3
= 3+t `
` a = {2; 1; 1}
Ann (V
1
(`)) 2A
1
+A
2
+A
3
= 0
{A
1
; A
2
; A
3
}
e
A Ann (V
1
(`))
e
A = {−1; 2; 0}
e
B = {−1; 0; 2}
` M
0
(1; 2; 3)
(
(x
1
1) + 2(x
2
2) + 0(x
3
+ 3) = 0
(x
1
1) + 0(x
2
2) + 2(x
3
+ 3) = 0
(
x
1
+ 2x
2
3 = 0
x
1
+ 2x
3
+ 7 = 0.
π
M
0
(1; 1; 2) `
(
2x
1
3x
2
+ x
3
+ 2 = 0
x
1
4x
2
+ x
3
+ 3 = 0
π
λ(2x
1
3x
2
+ x
3
+ 2) + µ(x
1
4x
2
+ x
3
+ 3) = 0.
M
0
3λ + 2µ = 0 λ = 2 µ = 3 λ µ
λ = 2 µ = 3 π
x
1
+ 6x
2
x
3
5 = 0
π
x
1
= 2 + 3t
1
4t
2
, x
2
= 1 3t
1
+ 2t
2
, x
3
= 1 + t
1
2t
2
.
  %fžfMLf' )@0)4 --2)4 09)D ›ˆ 4)2+C+4 0--0 ?+-1) )-1+sB
E 9(-1)D ,+1(0)4~
                                      
                        x1       2         2
                       2              
                       x  =  −3  + t  1  .
                        x3       0         2
?3403 ` 4))2 0,0913A D 9).2+ a = {2; 1; 2}  ,+E+C2 ()): 2+(.
M0 (2; −3; 0)
              
   ‚ninƒn dÿ' ?3403 ` :0C00 ,004)2()-.4 09)34~ x1 = 1+2t O
                      
x2 = 2+t O x3 = −3+t 0C027 ,34A ` --2)4+D C9E 1)DE 09)D
                                                                          
   %fžfMLf' ?3403 ` 4))2 0,0913A D 9).2+ a = {2; 1; 1}  13 0B
E+sC)3 Ann (V1(`)) )@0)4 4)2+C+4 0--0 09)) 2A1 + A2 + A3 = 0
+2+-2)17+ .++C02 {A1; A2; A3} 1)D+D F+4 Ae ∈ Ann (V1(`)) 
„2+ 09)) 4))2 C90 1)D+ ):09-4E )@)3 Ae = {−1; 2; 0}
 Be = {−1; 0; 2}  ?+-.+17. ` ,+E+C2 ()): 2+(. M0(1; 2; −3) O 2+ +0
:0C0)2-3 --2)4+D 09)D
   (                                                 (
       −(x1 − 1) + 2(x2 − 2) + 0(x3 + 3) = 0             −x1 + 2x2 − 3 = 0
                                             ⇐⇒
       −(x1 − 1) + 0(x2 − 2) + 2(x3 + 3) = 0             −x1 + 2x3 + 7 = 0.
  ‚ninƒn dd' +-20927 09)) ,1+-.+-2 π O ,+E+C3 )D ()): 2+(.
M0 (1; 1; 2)  ,34A ` ~
                          (
                              2x1 − 3x2 + x3 + 2 = 0
                               x1 − 4x2 + x3 + 3 = 0
                                                     
  %fžfMLf' ?1+-.+-27 π ,0C1)s2 ,(. ,1+-.+-2)D
                                                                           
             λ(2x1 − 3x2 + x3 + 2) + µ(x1 − 4x2 + x3 + 3) = 0.         ›Š
?+C-209133 9 09)) ›Š .++C02 2+(. M O ,+1(0)4 09))
3λ + 2µ = 0 O +2.C0 0E+C4      O
                             λ = 2 µ = −3    :0()3
                                                     0
                                                       ,004)2+9 λ  µ -B
  )-29) - 2+(+-27A C+ ,+,+‡+017+-2 – ?+C-209133 0DC))
:0()3 λ = 2 O µ = −3 9 09)) ›ŠO 0E+C4 09)) ,1+-.+-2 π ~
x1 + 6x2 − x3 − 5 = 0
                      
   ‚ninƒn d™' ?1+-.+-27 π :0C00 ,004)2()-.4 09)34
                                                                           
     x1 = 2 + 3t1 − 4t2 , x2 = 1 − 3t1 + 2t2 , x3 = −1 + t1 − 2t2 .    ›—
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