Hello,
I need help solving this question. thanks.
In each case,Extend the basis u,v to form a basis for R
S_{1}={(x y z)|x-y=3z}.
What? What basis is to be extended? S1= {(x, y, z)| x- y= 3z} is a plane in $\displaystyle R^2$ . 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) is one-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 $\displaystyle R^3$? In that case, any vector that is NOT in S1 will do but the simplest to get would be the vector perpendicular to S1, i- j- 3k.
(Well, perhaps not the simplest. None of 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.)