CURS 6: Pozitii relative drepte si plane, distante, unghiuriusers.utcluj.ro/~todeacos/curs6.pdfCURS...

CURS 6: Pozitii relative drepte si plane,distante, unghiuri

Conf. dr. Constantin-Cosmin Todea

Cluj-Napoca

6.1. Pozitiile relative a 2 plane:

Doua plane ın spatiu pot fi

⎧⎪⎪⎪⎨⎪⎪⎪⎩

-paralele-secante (concurente)-confundate

(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)

Thm. 6.11 Planele (P1) si (P2) sunt secante

⇐⇒ rang⎛⎜⎝

A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2

2 Planele (P1) si (P2) sunt paralele ⇐⇒ A1A2

= B1B2

= C1C2

≠ D1D2

3 Planele (P1) si (P2) sunt confundate⇐⇒ A1

A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩

-paralele

-secante (concurente)-confundate

(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩

-paralele-secante (concurente)

-confundate

(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→

Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→

Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2

2 Planele (P1) si (P2)

sunt paralele ⇐⇒ A1A2

= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2

2 Planele (P1) si (P2) sunt paralele

⇐⇒ A1A2

= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


=

B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

=

C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2

3 Planele (P1) si (P2)

sunt confundate⇐⇒ A1

A2= B1

B2= C1

C2= D1

D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2

3 Planele (P1) si (P2) sunt confundate

⇐⇒ A1A2

= B1B2

= C1C2

= D1D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2=

B1B2

= C1C2

= D1D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2=

C1C2

= D1D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2=

D1D2



⎧⎪⎪⎪⎨⎪⎪⎪⎩


(P1) ∶ A1x +B1y + C1z +D1 = 0→ Ð→n1(A1,B1,C1)

(P2) ∶ A2x +B2y + C2z +D2 = 0→ Ð→n2(A2,B2,C2)



A1 B1 C1

A2 B2 C2

⎞⎟⎠= 2


= B1B2

= C1C2

≠ D1D2


A2= B1

B2= C1

C2= D1

D2

6.2. Fascicul de plane

Definitie:

Se numeste fascicul de plane multimea tuturor planelor caretrec printr-o dreapta D.

Se da dreapta

(D) ∶ { P1 = 0P2 = 0

ın ecuatie generala

Ecuatia fasciculului determinat de (D) este:

(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R

Daca α ≠ 0 notam λnot= β

α ⇒ fasciculul are ecuatia

(Pλ) ∶ P1 + λP2 = 0.


Definitie:

Se numeste

fascicul de plane multimea tuturor planelor caretrec printr-o dreapta D.

Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:

Se numeste fascicul de plane

multimea tuturor planelor caretrec printr-o dreapta D.

Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:

Se numeste fascicul de plane multimea tuturor planelor

caretrec printr-o dreapta D.

Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:


Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:


Se da dreapta

(D) ∶ { P1 = 0P2 = 0


Ecuatia fasciculului

determinat de (D) este:

(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:


Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.


Definitie:


Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R

Daca α ≠ 0

notam λnot= β


(Pλ) ∶ P1 + λP2 = 0.


Definitie:


Se da dreapta

(D) ∶ { P1 = 0P2 = 0



(Pα,β) ∶ αP1 + βP2 = 0 , α, β ∈ R



(Pλ) ∶ P1 + λP2 = 0.

Pb:

Sa se scrie ecuatia planului (P)

care trece prin dreapta deintersectie a planelor:

(P1) ∶ 2x + 3y + z − 2 = 0

(P2) ∶ x + 2y − 3z + 1 = 0

si este perpendicular pe planul (P3) ∶ 3x − y + 2z + 2 = 0.

Pb:

Sa se scrie ecuatia planului (P) care trece prin dreapta deintersectie a planelor:

(P1) ∶ 2x + 3y + z − 2 = 0

(P2) ∶ x + 2y − 3z + 1 = 0


Pb:


(P1) ∶ 2x + 3y + z − 2 = 0

(P2) ∶ x + 2y − 3z + 1 = 0

si este perpendicular

pe planul (P3) ∶ 3x − y + 2z + 2 = 0.

Pb:


(P1) ∶ 2x + 3y + z − 2 = 0

(P2) ∶ x + 2y − 3z + 1 = 0


6.3. Pozitiile relative a doua drepte:

Doua drepte ın spatiu pot fi

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩

-concurente-paralele-confundate

-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm

Dreptele (D1) si (D2) sunt coplanare

⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)


Doua drepte ın

spatiu pot fi

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩

-concurente

-paralele-confundate

-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩

-concurente-paralele

-confundate-necoplanare

Fie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanare

Fie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒

Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1)

M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒

Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2)

M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1

p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1

p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)



⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

-coplanare

⎧⎪⎪⎪⎨⎪⎪⎪⎩


-necoplanareFie

(D1) ∶x − x1p1

= y − y1q1

= z − z1r1

⇒ Ð→v1(p1,q1, r1) M1(x1, y1, z1) ∈ (D1)

(D2) ∶x − x2p2

= y − y2q2

= z − z2r2

⇒ Ð→v2(p2,q2, r2) M2(x2, y2, z2) ∈ (D2)

Thm


⇐⇒

RRRRRRRRRRRRRRRRRRR

x2 − x1 y2 − y1 z2 − z1p1 q1 r1p2 q2 r2

RRRRRRRRRRRRRRRRRRR

= 0 (∗)

Demonstratie.

(D1) si (D2) sunt coplanare

⇐⇒ ÐÐÐ→M1M2,

Ð→v1 ,Ð→v2 sunt coplanari

⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.

b) Daca ın (∗) ultimele doua linii sunt proportionale

⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);

c) Daca ın (∗) toate cele trei linii sunt proportionale⇒ (D1) = (D2).

Demonstratie.

(D1) si (D2) sunt coplanare ⇐⇒ ÐÐÐ→M1M2,

Ð→v1 ,Ð→v2

sunt coplanari

⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗)

ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii

nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M},

coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand

sistemul cu trei ecuatii alese din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese

din (D1) si (D2);De exemplu:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:

a) Daca ın (∗) ultimele doua linii nu sunt proportionale⇒ (D1)⋂(D2) = {M}, coordonatele lui M se determinarezolvand sistemul cu trei ecuatii alese din (D1) si (D2);

De exemplu:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.

b) Daca ın (∗)

ultimele doua linii sunt proportionale

⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2)

(⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);

c) Daca ın (∗)

toate cele trei linii sunt proportionale⇒ (D1) = (D2).

Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);

c) Daca ın (∗) toate cele trei linii sunt proportionale

⇒ (D1) = (D2).

Demonstratie.



⇐⇒ (ÐÐÐ→M1M2,Ð→v1 ,Ð→v2) = 0

Observatii:


⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x−x1p1

= y−y1q1

x−x1p1

= z−z1r1

x−x2p2

= y−y2q2

.


⇒ (D1) ∥ (D2) (⇐⇒ p1p2

= q1q2

= r1r2⇐⇒ Ð→v1 ∥ Ð→v2);


CURS 6: Pozitii relative drepte si plane, distante, unghiuriusers.utcluj.ro/~todeacos/curs6.pdfCURS...

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Transcript of CURS 6: Pozitii relative drepte si plane, distante, unghiuriusers.utcluj.ro/~todeacos/curs6.pdfCURS...