Аналитическая геометрия. Часть II. Аналитическая геометрия пространства. Шурыгин В.В. - 15 стр.

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3
[λb + µc, a] = λ[b, a] + µ[c, a]
1
a||b [a, b] = 0 = [b, a] a b
c = [a, b] d = [b, a] 2
3
d = ±c 4
{a, b, c}
{a, b, c} {b, a, c} {b, a, c}
0 1 0
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
.
{b, a, c} 1
3
1
2
2
Ψ
a
= [a, ·] : E
3
3 b 7→ [a, b] E
3
a E
3
a = 0 2
a 6= 0
λ 6= 0
[λa, b] = λ[a, b].
λ = 0
λ 6= 0 a||b
a b [a, b] = c [λa, b] = d 2
3
d||c |d| = |λ||c| d = ±λc
{a, b, c} {λa, b, λc}
λ 0 0
0 1 0
0 0 λ
= λ
2
> 0.
e
a
a a = |a|e
a
[a, b] = |a|[e
a
, b].
     
  3◦ [λb + µc, a] = λ[b, a] + µ[c, a]
                                      
   ½hjnonqfŸ¾pqkh'
                                         ?                      (
   1◦ -1 a||b O 2+ [a, b] = 0 = −[b, a] )C,+1+s4 2),)7O 2+ a  b
) .+11)0O  c = [a, b] O 0 d = [b, a]  : -9+D-29 2∗  3∗ -1)C)2O (2+
                       (                            ‹
d = ±c : 4∗ -1)C)2O 2+ {a, b, c} ; ,09D *0:- 02‡ ,))E+C0 +2
{a, b, c} . {b, a, c}  {b, a, −c} 4)A2O -++29)2-29)+O 9C
                                                   
                       0 1 0                   0 1 0
                     
                     1 0 0
                                                   
                                              1 0 0 .
                       0 0 1                   0 0 −1
,)C)12)17 ,)9+D 402‡ +2‡02)1)O 0 92++D ; ,+1+s2)1)O ,+B
†2+4 ,094 3913)2-3 *0:- {b, a, −c} O (2+  C+.0:90)2 -9+D-29+ 1◦ 
   9+D-29+ 3◦ 92).0)2 : -9+D-29 1◦  2◦  +.0s)4 2◦  „2+ -9+D-29+
+:0(0)2O (2+ +2+*0s))
                     Ψa = [a, ·] : E3 3 b 7→ [a, b] ∈ E3
1)D+ , 1A*+4 a ∈ E3  ? a = 0 -9+D-29+ 2◦ 9,+13)2-3 +()9CB
4 +*0:+4 01)) ,)C,+100)4O (2+ a 6= 0  ?+9)C)4 C+.0:02)17-29+ 9
)-.+17.+ @0+9 
   € +.0s)4 -0(010O (2+ , λ 6= 0 9,+13)2-3
                                                                            
                               [λa, b] = λ[a, b].                       €Š
++2+@)) €Š +()9C+ 9,+13)2-3  , λ = 0 O + + 0- 2))-)2
-1(0D λ 6= 0  ? a||b -++2+@)) €Š 9,+13)2-3 +()9C4 +*0:+4
?-27 9).2+ a  b ) .+11)0  [a, b] = c O [λa, b] = d  : 2∗  3∗
-1)C)2O (2+ d||c  |d| = |λ||c| O ,+†2+4 d = ±λc  -20)2-3 ,+9)27O (2+
*0:- {a, b, c}  {λa, b, λc} +C0.+9+ +)2+90  + †2+ -1)C)2 :
2++O (2+
                             λ 0 0
                             0 1 0 = λ2 > 0.
                             0 0 λ
   ?-27 e ; )C(D 9).2+O 4)A D 2+ s) 0,091))O (2+  9).B
2+ a O 2+ )-27O a = |a|ea O 2+C0 : €Š -1)C)2O (2+
            a



                             [a, b] = |a|[ea , b].
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