Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 7 / ii / Solution

No, as printed. Each displayed vector has only two components, so neither is an element of the stated ${\mathbb{R}}^3$ without an additional embedding convention. Even if one inserts a zero third component, two vectors span at most a two-dimensional linear subspace and cannot form a basis of ${\mathbb{R}}^3$. This conclusion does not require guessing a missing third entry.