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SERIE DE EJERCICIOS PROPUESTOS

1. Si = (2,-3), = (5,4), = (3,1), P₀ = (1,-1) y P₁ = (4,3), encontrar:
a)
Del dato:
= (2,-3) + (5,4) = (2+5, -3+4) = (7,1)
b)Del dato:
= (2,-3) + (-5,-4) = (2-5, -3-4) = (-3,-7)
c)
Del dato:
= 3 (2,-3) + 4 (5,4) = (6,-9) + (20,16)
= (6+20, -9+16) = (26,7)
d) , si 4 + =4 + (2,-3) = 3 (5,4) 4 + (2,-3) = (15,12) 4 = (15,12) – (2,-3) 4 = (15,12) +(-2,3) = (13/4,15/4)
e) P₀ + 2(P₁ - P₀)
Del dato:
P₀ + 2P₁ - 2P₀ = 2P₁ - P₀ = 2 (4,3) – (1,-1) = (8,6) – (1,-1)= (8,6) + (-1,1) = (7,7)
f) (P₀ + P₁)/2
Del dato:
P₀ + P₁ = (1,-1) + (4,3) = (5,2) /2 = (5/2,1)


g) P₀ + t , para t=0,1, 2, 3
* Para t= 0, P₀ + 0 = (1,-1) + 0 (2,-3) = (1,-1) Para t =1, P₀ + 1 = (1,-1) + 1(2,-3) = (3,-4) Para t =2, P₀ + 2 = (1,-1) + 2 (2,-3) = (5,-7) Para t =3, P₀ + 3 = (1,-1) + 3 (2,-3) = (7,-10)
Para t = -1, P₀ + -1 = (1,-1) + -1(2,-3) = (-1,2) Para t = -2, P₀ +-2 = (1,-1) + -2(2,-3) = (-3,5) Para t = -3, P₀ + -3 = (1,-1) + -3(2,-3) = (-5,8)

2. Demuestre que si t 0 entonces s + t = tiene la única solución = 1/t ( - s )
De la ecuacióndada de vectores: * Despejamos: t = - s = 1/t ( - s ), para que este definido en los R, t 0

3. Resolver para el vector incógnita :
a) 2 (0,3) + 8 =(1,-6) (2,6) + 8 = (1,-6) 8 = (1,-6) - (2,6) 8 = (-1,-12) = (-1/8, -12/8)
b) -3(1,3) + 2 = 5(0,-2) + 4 (-3,-9) + 2 = (5,-10) + 4...