Física 1 (Exactas) Practica 0 – 1. 8. Vectores y trigonometría

 Dados los vectores A, B y C, demostrar:

 

 a)    Que el producto vectorial no es asociativo y se cumple: A x (B x C) = B (A . C)  - C (A . B)

A = (ax i + ay j + az k)

B = (bx i + by j + bz k)

C = (cx i + cy j + cz k)

Reemplazando 

B x C = (bx i + by j + bz k) x (cx i + cy j + cz k) =

 

Distribuyendo

B x C = bx i x cx i + bx i x cy j + bx i x cz k +

          + by j x cx i + by j x cy j + by j x cz k +

          + bz k x cx i + bz k x cy j + bz k x cz k) =

 

Producto de versores

B x C = bx cy k + bx cz (-j) + by cx (-k) + by cz i + bz cx j + bz cy (-i) =

 

Reordenando

B x C = (by cz - bz cy) i + (bz cx - bx cz) j + (bx cy - by cx) k

 

 

A x (B x C) = (ax i + ay j + az k) x (by cz - bz cy) i + (bz cx - bx cz) j + (bx cy - by cx) k =

Distribuyendo

A x (B x C) = ax i x (by cz - bz cy) i + ax i x (bz cx - bx cz) j + ax i x (bx cy - by cx) k +

                    + ay j x (by cz - bz cy) i + ay j x (bz cx - bx cz) j + ay j x (bx cy - by cx) k +

                    + az k x (by cz - bz cy) i + az k x (bz cx - bx cz) j + az k x (bx cy - by cx) k)

 

Producto de versores

A x (B x C) = ax (bz cx - bx cz) k + ax (bx cy - by cx) (-j) +

                   + ay (by cz - bz cy) (-k) + ay (bx cy - by cx) i +

                   + az (by cz - bz cy) j + az (bz cx - bx cz) (-i)

Reordenando

 A x (B x C) = (ay (bx cy - by cx) - az (bz cx – bx cz)) i +

                   + (az (by cz - bz cy) - ax (bx cy - by cx)) j +

                   + (ax (bz cx - bx cz) - ay (by cz - bz cy)) k  

 

A x (B x C) = (ay bx cy – ay by cx - az bz cx + az bx cz) i +

                   + (az by cz – az bz cy - ax bx cy + ax by cx) j +

                   + (ax bz cx – ax bx cz - ay by cz + ay bz cy) k (I)

 

 

A . C = ax cx + ay cy + az cz

 

Reemplazando 

B (A . C) = bx i (ax cx + ay cy + az cz) +

               +  by j (ax cx + ay cy + az cz) +

               +  bz k (ax cx + ay cy + az cz)

Distribuyendo 

B (A . C) = i (ax bx cx + ay bx cy + az bx cz) +

               +   j (ax by cx + ay by cy + az by cz) +

               +   k (ax bz cx + ay bz cy + az bz cz)

 

A . B = ax bx + ay by + az bz

Reemplazando 

C (A . B) = cx i (ax bx + ay by + az bz) +

               +  cy j (ax bx + ay by + az bz) +

               +  cz k (ax bx + ay by + az bz)

Distribuyendo 

C (A . B) = i (ax bx cx + ay by cx + az bz cx) +

               +   j (ax bx cy + ay by cy + az bz cy) +

               +   k (ax bx cz + ay by cz + az bz cz)

 

B (A . C) – C (A . B) =

              =  ((ax bx cx + ay bx cy + az bx cz) - (ax bx cx + ay by cx + az bz cx)) i +

              + ((ax by cx + ay by cy + az by cz) - (ax bx cy + ay by cy + az bz cy)) j +

              + ((ax bz cx + ay bz cy + az bz cz) - (ax bx cz + ay by cz + az bz cz)) k

 

B (A . C) – C (A . B) =

              = (ay bx cy + az bx cz -  ay by cx - az bz cx) i +

              +  (ax by cx + az by cz - ax bx cy -  az bz cy) j +

              + (ax bz cx + ay bz cy - ax bx cz - ay by cz) k (II)

 

 

(I) = (II) à A x (B x C) = B (A . C)  - C (A . B)

 

b)    Que cualesquiera sean los vectores, se cumple:

 

A x (B x C) + B x (C x A) + C x (A x B) = 0

 

A x (B x C) = B (A . C) – C (A . B)

B x (C x A) = C (B . A) – A (B . C)

C x (A x B) = A (C. B) – B (C . A)

 

Sumando

A x (B x C) + B x (C x A) + C x (A x B) =

                   = B (A . C) – C (A . B) + C (B . A) – A (B . C) + A (C. B) – B (C . A)

 

El producto escalar es conmutativo

A . C = C . A

A . B = B . A

B . C = C . B

 

A x (B x C) + B x (C x A) + C x (A x B) =

                   = B (A . C) – C (A . B) + C (A . B) – A (B . C) + A (B . C) – B (A . C) = 0

  

c)     Que el producto mixto es igual al volumen del paralelepípedo construido sobre los mismos una vez llevado a partir de su origen común.

 

 (A x B) .C = producto mixto

 

Volumen del paralelepípedo ABC = Área de la base * altura

Área de la base = | A x B |

Altura = | C | cos ϕ

 

Reemplazando

Volumen = | (A x B) | * | C | cos ϕ = (A x B) . C   

 

d)   Que la condición necesaria y suficiente para que los tres vectores sean paralelos a un mismo plano es que su producto mixto sea nulo.

 

(A x B) . C   = 0 à volumen = 0 à A, B y C sean  paralelos

 

Vectores paralelos = linealmente dependientes que equivale a coplanares (en el mismo plano) y misma dirección