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Given the following vectors $\vec u$ and $\vec v$, find a vector $\vec w$ in $\mathbb R^3$ so that $\{\vec u, \vec v, \vec w\}$ spans $\mathbb R^3$ and a non-zero vector $\vec z$ in $\mathbb R^3$ so that $\{\vec u, \vec v, \vec z\}$ does not span $\mathbb R^3$.

I know that $\vec u=[-9\ -10\ -1]$ and $\vec v=[8\ -4\ \ 1]$. I was thinking of making a matrix and trying to solve it to get the identity matrix. I am not really sure how to approach this.

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    For two different vectors $u,v \in \mathbb R^3$, the cross product $u \times v$ is always orthogonal to both, so for if, $u \neq 0, v \neq 0$ and $u \neq v$, the set $u,v, u\times v$ always spans $\mathbb R^3$. – 2017-01-13

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For $w$ you can take $w=[1,0,0]$. Then the determinant of the matrix $A$ with rows $u,v$ and $w$ is $-14$, so $A$ has full rank.

For $z$ you may take $z=u$, so that $u,v,z$ are linearly dependent. Or $z=u+v$

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In $\mathbb R^3$ you can use the cross product. As long as the two given vectors are linearly independent (i.e., not parallel), you will get a non-zero vector orthogonal to both of them, and thus linearly independent of them. Any set of three linearly independent vectors spans $\mathbb R^3$, so you’re done.