Answer
Since the two vectors are linearly independent, they form the basis for the set and,
the dimension is 2.
Work Step by Step
We are required to find the basis and state the dimension
Let the given subspace be represented by a vector $\mathbf{\vec{z}}$ such that:
$\mathbf{\vec{z}}=\begin{bmatrix}4s\\-3s\\-t\end{bmatrix}$
Writing the given set in parametric form;
$\mathbf{\vec{z}}=\begin{bmatrix}4s\\-3s\\-t\end{bmatrix}=s\begin{bmatrix}4\\-3\\0\end{bmatrix}+t\begin{bmatrix}0\\0\\-1\end{bmatrix}$
The given subspace is a linear combination of two vectors
$\begin{bmatrix}4\\-3\\0\end{bmatrix}$ and $\begin{bmatrix}0\\0\\-1\end{bmatrix}$
Since the two vectors are linearly independent, they form the basis for the set and the dimension is 2
