#
fp= => f
f:E → E pNfp=Ef
1
Ef:E → E f
1
f
P: 5x2y11z+3 = 0 D:R(2,1,1)
A=R2R2, f(x, y)=(ax, bx +cy +d)a6= 0 c6= 0
A
(A, )
E3
x0= 3x+ 4y+ 2z4
y0=2x3y2z+ 4
z0= 4x+ 8y+ 5z8
fm:R2R2
M(x, y)7−M0(x0, y0)
x0= (m+ 1)x+ 4my m
y0= (m2)x+ (m3)y+ 2(m1)
fm
m fm
f1f1=
f0
D
x+ 2y= 1 D03yx= 0 2
3
f
x0=λ+1
3x2
3y+2
3z
y0= 2(1 λ)2
3x+1
3y+2
3z
z0=λ+2
3x+2
3y+1
3z
f
X E RX
X
R= (O,
e1,
e2,
e3)
x+ 2y+z= 1 (
e1+
e2+
e3)
(x+y+ 1 = 0
2y+z+ 2 = 0,
3x+ 3y2z= 0
R= (O,
e1,
e2,
e3)
x0= 3x+ 4y+ 2z4
y0=2x3y2z+ 4
z0= 4x+ 8y+ 5z8.
f
MM0
EA, B, C, D
f f(A) = B f(B) = C f(C) = D f(D) = A
f3=
Pf:P → P f3=f6=
A6=f(A)A, f(A), f2(A)
f
P D P f, g D
0
λ µ
fg
π, π0E
F
λRπλM7−(π(M) : λ, π0(M):1λ)
πλ
F
f:E → E f s
t f =st=ts
sB
Ft
u
st=ts
u
B
f(s, t)f=st=ts
f f f f
fR= (O,
e1,
e2,
e3)
x0= (x2y2z+ 1)/3
y0= (2x+y2z+ 2)/3
z0= (2x2y+z1)/3.
EP, Q, P 0, Q0P6=Q
f f(P) = P0f(Q) = Q0
P P0A, B, C ∈ P O /∈ P
A0, B0, C0P0(OA) (OB) (OC)
α, β, γ [B, C] [C, A] [A, B]
(A0α) (B0β) (C0γ)
A1, . . . , AnnE
B1, . . . , BnAi= (Bi, Bi+1)An= (Bn, B1))
OPO
f M 7−M0M0P(OM)P
O
f
f
Ef:E → E
fdet(
f) = ±1
s(M)=2P(M)M
Mp(M) = α
u p(M)=(x1, y1, z1)M
Px1=x+ 2α y1=yα z1=z+α α =5x+ 2y+ 11z3
(x1=9x+ 4y+ 22z6
y1= 5xy11z+ 3
z1=5x+ 2y+ 12z3
p(M)s(M) =
Mp(M)
x0=19x+ 8y+ 44z12
y0= 10x3y22z+ 6
z0=10x+ 4y+ 23z6
fa b
0cf
f(0, y) = (0, cy +d)f
(0, e) (0,ed
c)
g: (x, y)7−(ax+a0y+a00 , bx +b0y+b00
g(0, y) = (0, b0y+b00)a0y+a00 = 0 a0=a00 =
0
g(0, e) (0,b00 +e
b0)b06= 0
a6= 0 a6= 0 g(1, y)
x+ 2y+z2=0
MM0= (2x+ 4y+ 2z4)(
i
j+ 2
k)
MM0
(1,1,2)
λ f(
v) = λ
v
x0=λx = 3x+ 4y+ 2z4
y0=λy =2x3y2z+ 4
z0=λz = 4x+ 8y+ 5z8
(3 λ)x+ 4y+ 2z4 = 0
2x(3 + λ)y2z+ 4 = 0
4x+ 8y+ (5 λ)z8 = 0
3λ4 2
23λ2
4 8 5 λ
= 0 (3 λ)(λ1)2= 0
k= 3 k= 1
M f(M) = M mx + 4my m= 0 (D1) (m2)x+
(m4)y+ 2m1 = 0 (D2)m= 0 2x+ 4y+ 1 = 0
m6= 0 m=4
3
m6=4
3A(3,1)
fmAm=m+ 1 4m
m2m33(m1)2
m6= 1 fm
A1=2 4
12A2
1= 0
f1
f1= 0Ef1f1(M) =
f1f1(A) +
f1
f1(
AM)
| {z }
=OE
=A
Mf0(M) = (2x4y1)
j
j= (0,1) ∆
2x4y1 = 0 p
jp(M)
p(M)M+
j p(M) = M+λ
j2x4λ4y1=0 λ=2x+ 4y+ 1
4
p(M)f0(M) = 5
p(M)M f0
j
5
x0=1
5(4x2y+ 1)
y0=1
15 (x+ 13y+ 1)
A=
1
3
2
3
2
3
2
3
1
3
2
3
2
3
2
3
1
3
(ϕ(
i), ϕ(
j), ϕ(
k))
det(A) = 1ϕ
x+yz= 0 f
x=1
3x2
3y+2
3z+λ
y=2
3x+1
3y+2
3z+ 2(1 λ)
z=2
3x+2
3y+1
3zλ
x+yz3
2λ= 0
x+yz+ 3(1 λ) = 0
t
u f t f=ft A
M
MA =
u
f(M)f(A) =
MA f
sAsBX(A, B)X2X sA
sBX sBsA=t2
AB X(sBsA)n=t2n
AB
f(M) = t2n
AB sA(M) = A
AM + 2n
AB =A+n
AB (M(A+n
AB)) Cn
f(M) = C
CM =sCn(M)A6=B Ck6=Cnn6=k
2x0=x2yz+ 1
2y0=xz+ 1
2z0=x2y+z+ 1
2x0=5x3y+ 2z3
2y0= 3x+y2z1
z0=3x3y+z3
P:x+ 2y+z= 2 (
e1
e2+ 2
e3)
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