Is there a subspace of $R^3$ of dimension $1$ that contains the vectors $v=(1,1,2)$ and $w=(1,-1,2)$?

Question

Is there a subspace of $R^3$ of dimension $1$ that contains the vectors $v=(1,1,2)$ and $w=(1,-1,2)$?

I see that $v$ and $w$ linearly independent so I think that there isn't a subspace of dimension $1$ that contains both vectors. But I think that there is a subspace of dimension 2 that contains $v$ and $w$.

My question is whether my reasoning is correct and how to justify it.

Answer 1

It suffices to note that since $\vec v$ and $\vec w$ are linearly independent they span, as a basis, a subspace with dimension $2$. Therefore there isn't a subspace of dimension $1$ that contains both vectors.

Indeed if such subspace would exist with basis $\{\vec u\}$ we had

• $\vec v=a \vec u \quad \vec w=b \vec u\implies \vec v=c\vec w$

which is not true.

Answer 2

Please note that $$\det \begin{vmatrix} 1 & 2 \\ -1 & 2\\ \end{vmatrix}=4 $$ so the vectors mentioned are linearly independant. That means that they produce a subspace of $\mathbb{R}^3$ of dimension $2$.