I am given two vectors, $\vec u$ and $\vec v$. I'm not sure how to find the third vector. I know for linearly independent, the coefficients for the linear combination must all equal zero in order to get the zero vector. For linearly dependent, I know that the not all of the coefficients can equate to zero.
Original Problem:
Given the following vectors $\vec u$ and $\vec v$, find a vector $\vec w$ in $\Bbb R^4$ so that $\{\vec u, \vec v, \vec w\}$ is linearly independent and a non-zero vector $\vec z$ in $\Bbb R^4$ so that $\{\vec u, \vec v, \vec z\}$ is linearly dependent:
$$\vec u=\begin{bmatrix}-9\\-1\\-1\\4\end{bmatrix} \quad\vec v=\begin{bmatrix}5\\10\\-8\\-1\end{bmatrix}$$ $$\vec w=\begin{bmatrix}0\\0\\1\\1\end{bmatrix} \quad\vec z=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$
The second question is easier. The answer exploits the fact that every set containing at least a vector and the zero vector is linearly dependent. Alternatively, you can solve it simply choosing two scalars $\alpha, \beta \in \Bbb R$ so that $z=\alpha u+ \beta v$.
As to the first one, the formal way would be use the definition of linear independence and go through a linear system in four equations. Nonetheless with a bit of logic you can easily solve it without using any kind of computation. For example, if you call $w=(3,3,0,0)$ the set will still be linearly independent. Why? Try to think about the linear combinations of the first two components of the vectors $v,u$. Can you find two scalars such that their linear combination with the first two components of the vectors $v,u$ gives you back the first two components of the vector $w$? In this way you can construct as many example as you want.