Find a set of vectors that spans the subspace

[mateomy]

Find a set of vectors in $\mathbb{R}^3$ that spans the subspace
$$S\,=\,\{\,u\,\in\,\mathbb{R}^3\,|\,u\cdot v\,=\,0\,\}$$
where v=<1,2,3>


Maybe 12 hours of studying is too much and I'm fried or, maybe I'm looking for excuses. Either way...

To solve this I'm trying to set up a matrix multiplication and augment it at zero. But, I just get a single linear equation which tells me that the only way I can have a span of this subspace is if my other vector is the zero vector <0,0,0>. I don't think that's right.

$$\begin{bmatrix} a & b & c \end{bmatrix} * \begin{bmatrix} 1\\2\\3 \end{bmatrix} = \mathbf{0}$$

Getting $a+2b+3c=0$

Where's my issue?


Thanks.

 

[clamtrox]

Perhaps you should stop and think what it is you're calculating here. You are looking for a subspace which is perpendicular to a vector, so it's a plane. Maybe that will help you interpret your result.

 

[HallsofIvy]

From a+ 2b+ 3c= 0, you have a= -2b- 3c so <a, b, c>= <-2b- 3c, b, c>.

 

[mateomy]

From a+ 2b+ 3c= 0, you have a= -2b- 3c so <a, b, c>= <-2b- 3c, b, c>.

Thanks. I was looking at that and didn't take the next step of actually putting it into a vector form. Again, thanks.