What? What basis is to be extended? S1= {(x, y, z)| x- y= 3z} is a plane in . It is NOT a set of vectors in R and so cannot be even a part of a basis for R. In any case, R (as a vector space over the real numbers) isone-dimensional. Any single number constitutes a "basis" and cannot be "extended".

Perhaps you have written the problem wrongly. Are you, perhaps, given that u and v are independent vectors in S1 and to find a third vector that extends the set to a basis for ? In that case, any vector that is NOT in S1 will do but the simplest to get would be the vectorperpendicularto S1, i- j- 3k.

(Well, perhaps not the simplest.Noneof i= 1i+0j+ 0k so x= 1, y= 0, z=0, j= 0i+ 1j+ 0k so x= 0, y= 1, z= 0, or k= 0i+ 0j+0k so x= 0, y= 0, z= 1, satisfy x- y= 3z so any of those will do.)