Vectores

Abstract

1. Para a  = (1, −2, 3) y b  = (3, −1, 4), halla: a) b a   + b) b a   − 2 c) b a   3 + − d) b a c     +  = Solución: a) b a   + = (1, −2, 3) + (3, −1, 4) = (4, –3, 7). b) b a   − 2 = 2 · (1, −2, 3) – (3, −1, 4) = (2 – 3, –4 + 1, 6 – 4) = (–1, –3, 2). c) b a   3 + − = – (1, −2, 3) + 3 · (3, −1, 4) = (–1 + 9, 2 – 3, –3 + 12) = (8, –1, 9). d) b a c     +  = = ( ) ( ) ( )  +   −  −  +  = −  + −  4 3 , 2 , 3 4 , 1 , 3 3 , 2 , 1 .