Finding a Basis for Subspace a of R^3

[kwal0203]

Homework Statement

Find bases for the following subspace a of r^3

Y+z=0

The Attempt at a Solution

First I found a normal to this plane n=(0,1,1)

Then I found two vectors which are orthogonal to the normal u=(0,-1,1), v=(1,0,0)

Is this correct the answer in my book has v=(7,0,0)

Thanks

 

[HallsofIvy]

You are correct. If that is what your book really says it is wrong. Any vector in $$R^3$$ is of the form (x, y, z). Any vector for which z= -y is of the form (x, y, -y)= (x, 0, 0)+ (0, y, -y)= x(1, 0, 0)+ y(0, 1, -1). Of course, (7, 0, 0) would work as well as (1, 0, 0) but this subspace is definitely two dimensional.

 

[kwal0203]

Great thanks for that. I can see why x=7 works it just confused me why they put that as the only answer @.@