Answer
$\mathbf{F}(x,y) = 2\mathbf{i} - \mathbf{j}$.
All vectors have the same direction (rightward and slightly downward)
and the same magnitude, indicating that the field is uniform.
Work Step by Step
\draw[->] (-2,0) -- (4,0) node[right] {$x$};
\draw[->] (0,-3) -- (0,2) node[above] {$y$};
% Define and draw vector field F(x,y) = 2i - j
\foreach \x in {-2,-1,0,1,2,3}
\foreach \y in {-2,-1,0,1}
{
% Compute vector components scaled for visibility
\pgfmathsetmacro{\vx}{2*0.5} % scaled x-component (2)
\pgfmathsetmacro{\vy}{-1*0.5} % scaled y-component (-1)
% Draw the arrow (vector)
\draw[->, thick, blue] (\x,\y) -- ++(\vx,\vy);
}
