- #1

loli12

U = {(a1, a2, a3) in R^3: a1 = 3(a2) and a3 = -a2}

V = {(a1, a2, a3) in R^3: a1 - 4(a2) - a3 = 0}

I used the information in U and substituted it into the equation in V and I got 0 = 0. So, does it mean that the intersection of U and V is the whole R^3 which has no restrictions on a1, a2 and a3 (they are free)? Or do the original restrictions on both the original subspaces still being applied to the intersection?